Preface
This manual contains more or less complete solutions for every problem in the
book. Should you find errors in any of the solutions, please bring them to my attention.
Over the years, I have tried to enrich my lectures by including historical
information on the significant developments in thermodynamics, and biographical
sketches of the people involved. The multivolume Dictionary of Scientific Biography,
edited by Charles C. Gillispie and published by C. Scribners, New York, has been
especially useful for obtaining biographical and, to some extent, historical information.
[For example, the entry on Anders Celsius points out that he chose the zero of his
temperature scale to be the boiling point of water, and 100 to be the freezing point.
Also, the intense rivalry between the English and German scientific communities for
credit for developing thermodynamics is discussed in the biographies of J.R. Mayer, J. P.
Joule, R. Clausius (who introduced the word entropy) and others.] Other sources of
biographical information include various encyclopedias, Asimov’s Biographical
Encyclopedia of Science and Technology by I. Asimov, published by Doubleday & Co.,
(N.Y., 1972), and, to a lesser extent, Nobel Prize Winners in Physics 1901-1951, by
N.H. deV. Heathcote, published by H. Schuman, N.Y.
Historical information is usually best gotten from reading the original literature.
Many of the important papers have been reproduced, with some commentary, in a series
of books entitled “Benchmark Papers on Energy” distributed by Halsted Press, a division
of John Wiley and Sons, N.Y. Of particular interest are:
Volume 1, Energy: Historical Development of the Concept, by R. Bruce Lindsay.
Volume 2, Applications of Energy, 19th Century, by R. Bruce Lindsay.
Volume 5, The Second Law of Thermodynamics, by J. Kestin and
Volume 6, Irreversible Processes, also by J. Kestin.
The first volume was published in 1975, the remainder in 1976.
v
vi
Other useful sources of historical information are “The Early Development of the
Concepts of Temperature and Heat: The Rise and Decline of the Caloric Theory” by D.
Roller in Volume 1 of Harvard Case Histories in Experimental Science edited by J.B.
Conant and published by Harvard University Press in 1957; articles in Physics Today,
such as “A Sketch for a History of Early Thermodynamics” by E. Mendoza (February,
1961, p.32), “Carnot’s Contribution to Thermodynamics” by M.J. Klein (August, 1974,
p. 23); articles in Scientific American; and various books on the history of science. Of
special interest is the book The Second Law by P.W. Atkins published by Scientific
American Books, W.H. Freeman and Company (New York, 1984) which contains a very
extensive discussion of the entropy, the second law of thermodynamics, chaos and
symmetry.
I also use several simple classroom demonstrations in my thermodynamics courses.
For example, we have used a simple constant-volume ideal gas thermometer, and an
instrumented vapor compression refrigeration cycle (heat pump or air conditioner) that
can brought into the classroom. To demonstrate the pressure dependence of the melting
point of ice, I do a simple regelation experiment using a cylinder of ice (produced by
freezing water in a test tube), and a 0.005 inch diameter wire, both ends of which are
tied to the same 500 gram weight. (The wire, when placed across the supported cylinder
of ice, will cut through it in about 5 minutes, though by refreezing or regelation, the ice
cylinder remains intact.—This experiment also provides an opportunity to discuss the
movement of glaciers.) Scientific toys, such as “Love Meters” and drinking “Happy
Birds”, available at novelty shops, have been used to illustrate how one can make
practical use of the temperature dependence of the vapor pressure. I also use some
professionally prepared teaching aids, such as the three-dimensional phase diagrams for
carbon dioxide and water, that are available from laboratory equipment distributors.
Despite these diversions, the courses I teach are quite problem oriented. My
objective has been to provide a clear exposition of the principles of thermodynamics, and
then to reinforce these fundamentals by requiring the student to consider a great
diversity of the applications. My approach to teaching thermodynamics is, perhaps,
similar to the view of John Tyndall expressed in the quotation
“It is thus that I should like to teach you all things; showing you the way to
profitable exertion, but leaving the exertion to you—more anxious to bring out
your manliness in the presence of difficulty than to make your way smooth by
toning the difficulties down.”
Which appeared in The Forms of Water, published by D. Appleton (New York, 1872).
Solutions to Chemical and Engineering Thermodynamics, 3e
vii
Finally, I usually conclude a course in thermodynamics with the following quotation
by Albert Einstein:
“A theory is more impressive the greater the simplicity of its premises is, the
more different kinds of things it relates, and the more extended its area of
applicability. Therefore, the deep impression classical thermodynamics made
upon me. It is the only physical theory of universal content which, within the
framework of the applicability of its basic concepts, I am convinced will never
by overthrown.”
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3.11 (a) This is a Joule-Thomson expansion
⇒ H$ (70 bar, T = ?) = H$ (10133
.
bar, T = 400° C) ≈ H$ (1 bar, T = 400° C)
= 3278.2 kJ kg
and T = 447° C , S$ = 6.619 kJ kg K
(b) If turbine is adiabatic and reversible S&gen = 0 , then S$out = S$in = 6.619 kJ kg K and P = 1013
.
c
h
bar. This suggests that a two-phase mixture is leaving the turbine
S$ V = 7.3594 kJ kg K
S$ L = 13026
.
kJ kg K
Let x = fraction vapor
) = 6.619 kJ kg K or x = 08778
.
.
.
Then x(7.3594) + (1 − x)(13026
fluid leaving turbine is
.
H$ = 08788
×
.
26755
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)×
.
+ (1 − 08778
417.46
H$ L (sat’d, 1 bar )
= 2399.6
Therefore the enthalpy of
kJ
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dV 0
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dt
but M& in = − M& out
W&
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.
M
kg
in
(c) Saturated vapor at 1 bar
S$ = 7.3594 kJ kg K ; H$ = 26755
. kJ kg
W
− &s
M in
. − 26755
. = 602.7 kJ kg
= 32782
Actual
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= 68.6%
.
8786
S&gen
= 7.3594 − 6.619 = 0.740 kJ Kh
M&
in
(d)
& = −M
&
0 = M& 1 + M& 2 ⇒ M
2
1
W
Steam
70 bar
447° C
Water
1 bar
25° C
Q
dV
0 = M& 1 H$1 − H$ 2 + W&s + Q& − P
dt
&
Q
0 = M& 1 S$1 − S$2 + + S&gen
T
c
c
h
h
Simplifications to balance equations
dV
S&gen = 0 (for maximum work); P
= 0 (constant volume)
dt
Q& Q&
where T0 = 25° C (all heat transfer at ambient temperature)
=
T T0
kJ
kJ
H$ (sat'd liq, 25° C) = 104.89 ; S$ (sat'd liq, T = 25° C) = 0.3674
kg
kg K
−W&
Q&
= H$1 − H$ 2 + T0 S$2 − S$1 = H$ 1 − T0 S$1 − H$ 2 − T0 S$2
= T0 S$2 − S$1 ; & s
&
M
M
c
−W&s
&
M
h
c
max
h c
h c
h
= 3278.2 − 29815
. × 6.619 − 104.89 − 29815
. × 0.3674
max
= 1304.75 + 4.65 = 1309.4 kJ kg
3.12
Take that portion of the methane initially in the tank that is also in the tank finally to be in the
system. This system is isentropic S f = Si .
(a) The ideal gas solution
S f = S i ⇒ Tf = Ti
FG P IJ
H PK
f
R Cp
i
N=
F 35. I
H 70 K
= 300
8.314 36
. K
= 1502
Pf V
PV
PV
. mol
⇒ Ni = i = 1964.6 mol; N f =
= 1962
RT
RTi
RTf
. mol
∆N = N f − Ni = −17684
(b) Using Figure 2.4-2.
70 bar ≈ 7 MPa, T = 300 K
S$i = 505
. kJ kg K = S$ f
3
m
V$i = 0.0195
, so that mi =
kg
35.90 kg × 1000
Ni =
0.7m3
. kg.
= 3590
m3
0.0195
kg
g
kg
= 1282 mol
g
mol
. kJ kg K ⇒ T ≈ 138 K. Also,
At 3.5 bar = 0.35 MPa and S$ f = 505
28
m3
0.7m3
V$f = 0192
.
, so that m f =
.
= 3646
kg.
kg
m3
.
0192
kg
g
.
3646
kg × 1000
kg
Nf =
= 130.2 mol
g
28
mol
∆N = N f − Ni = 130.2 − 1282 = −11518
. mol
3.13
dT
dV
+R
eqn. (3.4-1)
T
V
dS = C
∆S =
z LNM
so that
(a − R) + bT + cT 2 + dT 3 +
OP
Q
e dT
dV
+R
V
T2 T
z
T2
c
+ b T2 − T1 + T22 − T12
2
T1
V
d 3
e
+ T2 − T13 − T2−2 − T1−2 + R ln 2
3
2
V1
S T2 , V 2 − S T1, V 1 = (a − R ) ln
a
f a
f
c
a
f c
h
h c
h
Now using
PV = RT ⇒
V 2 T2 P1
= ⋅ ⇒
V 1 T1 P2
T2
c
+ b T2 − T1 + T22 − T12
2
T1
d 3
e
P
+ T2 − T13 − T2−2 − T1−2 − R ln 2
3
2
P1
S T2 , P2 − S T1 , P1 = a ln
a
f a
f
a
c
f c
h
h c
h
Finally, eliminating T2 using T2 = T1 P2 V 2 PV
1 1 yields
FG P V IJ + b a P V − PV f
H PV K R
c
+
a P V f − a PV f
2R
d
+
a P V f − a PV f
3R
S P2 ,V 2 − S P1,V 1 = a ln
a
f a
f
2
2
2
2
2
2
2
3
2
2
2
1 1
3
−
3.14
1 1
2
1 1
3
1 1
eR 2
P
−2
P2V 2 −2 − PV
− R ln 2
1 1
2
P1
d
i d
i
System: contents of valve (steady-state, adiabatic, constant volume system)
Mass balance
0 = N& 1 + N& 2
Energy balance
0
dV 0
0 = N& 1 H 1 + N& 2 H 2 + Q& 0 + Ws − P
dt
⇒ H1 = H 2
Q& 0
Entropy balance 0 = N& 1 S 1 + N& 2 S 2 + S&gen +
T
S&gen
N&
(a) Using the Mollier Diagram for steam (Fig. 2.4-1a) or the Steam Tables
⇒ ∆S = S 2 − S 1 =
T1 = 600 K P2 = 7 bar
T ≈ 293° C
⇒ $2
P1 = 35 bar H$ 2 = 30453
. Jg
S2 = 7.277 J g K
H$1 = H$ 2 = 3045.3 J g . Thus S$1 = 65598
.
J g K ; Texit = 293° C
∆S$ = S$ − S$ = 0.717 J g K
2
1
(b) For the ideal gas, H1 = H 2 ⇒ T1 = T2 = 600 K
∆ S = S T2 , P2 − S T1 , P1 = Cp ln
a
= − R ln
f a f
T2
P
− R ln 2
P1
T1
P2
. J mol K ⇒
= 1338
P1
∆S$ = 0.743 J mol K
3.15
From the Steam Tables
P = 15538
MPa
.
V$ L = 0.001157 m3 / kg V$V = 012736
m3 / kg
.
U$ L = 85065
U$ V = 25953
. kJ / kg
. kJ / kg
At 200oC,
L
L
$
$
H = 852.45 kJ / kg
H = 27932
. kJ / kg
L
V
S$ = 2.3309 kJ / kg ⋅ K S$ = 6.4323 kJ / kg ⋅ K
∆H$ vap = 1940.7 kJ / kg ∆S$ vap = 41014
.
kJ / kg ⋅ K
(a) Now assuming that there will be a vapor-liquid mixture in the tank at the end, the properties
of the steam and water will be
P = 0.4578 MPa
V$ L = 0.001091 m3 / kg V$V = 0.3928 m3 / kg
U$ L = 63168
. kJ / kg
H$ L = 632.20 kJ / kg
o
At 150 C,
U$ V = 25595
. kJ / kg
H$ V = 27465
. kJ / kg
V
$
S = 68379
.
kJ / kg ⋅ K
S$ L = 18418
.
kJ / kg ⋅ K
vap
$
∆H = 2114.3 kJ / kg ∆S$ vap = 4.9960 kJ / kg ⋅ K
(b) For simplicity of calculations, assume 1 m3 volume of tank.
Then
0.8 m3
Mass steam initially =
= 6.2814 kg
0.12736 m3 / kg
0.2 m3
= 172.86 kg
0.001157 m3 / kg
6.2814
Weight fraction of steam initially =
= 0.03506
179.14
6.2814
Weight fraction of water initially =
= 0.96494
179.14
The mass, energy and entropy balances on the liquid in the tank (open system) at any time
yields
dM L
dM LU$ L
dM L S$ L
= M& L ;
= M& L H$ V ; and
= M& L S$V
dt
dt
dt
dU$ L $ L dM L
dM L
or M L
+U
= M& L H$ V = H$ V
dt
dt
dt
L
L
$
dU
dM $ V $ L
=
ML
H −U
dt
dt
Also, in a similar fashion, from the entropy balance be obtain
dS$ L dM L $V $ L
dM L $ vap
=
ML
S −S =
∆S
dt
dt
dt
Mass water initially =
c
c
h
h
There are now several ways to proceed. The most correct is to use the steam tables, and to use
either the energy balance or the entropy balance and do the integrals numerically (since the
internal energy, enthalpy, entropy, and the changes on vaporization depend on temperature.
This is the method we will use first. Then a simpler method will be considered.
Using the energy balance, we have
dM L
dU$ L
= V
, or replacing the derivatives by finite differences
L
M
H$ − U$ L
MiL+1 − MiL U$iL+1 − U$iL
U$iL+1 − U$iL
L
L
or
finally
1
=
=
+
M
M
+
i
1
i
Mi L
H$ iV − U$iL
H$ iV − U$iL
So we can start with the known initial mass of water, then using the Steam Tables and the data
at every 5oC do a finite difference calculation to obtain the results below.
IJ
K
FG
H
i
1
2
3
4
5
6
7
8
9
10
11
T (oC)
200
195
190
185
180
175
170
165
160
155
150
U$iL (kJ/kg K)
850.65
828.37
806.19
784.10
762.09
740.17
718.33
696.56
674.87
653.24
631.68
H$ iV (kJ/kg K)
2793.2
2790.0
2786.4
2782.4
2778.2
2773.6
2768.7
2763.5
2758.1
2752.4
2746.5
MiL (kg)
172.86
170.88
168.95
167.06
165.22
163.42
161.67
159.95
158.27
156.63
155.02
So the final total mass of water is 155.02 kg; using the specific volume of liquid water at
150oC listed at the beginning of the problem, we have that the water occupies 0.1691 m3
leaving 0.8309 m3 for the steam. Using its specific volume, the final mass of steam is found to
be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.09%, the
final volume fraction of water is 16.91%, and the fraction of the initial steam + water that has
been withdrawn is
(172.86+6.28-155.02-2.12)/(172.86+6.28) = 0.1228 or 12.28%. A total of 22.00 kg of steam
has withdrawn, and 87.7% of the original mass of steam and water remain in the tank.
For comparison, using the entropy balance, we have
dM L
dS$ L
=
, or replacing the derivatives by finite differences
ML
S$V − S$ L
MiL+1 − MiL S$iL+1 − S$iL
S$iL+ 1 − S$iL
L
L
1
=
=
+
M
M
of
finally
i +1
i
Mi L
∆S$ivap
∆S$ivap
So again we can start with the known initial mass of water, then using the Steam Tables and
the data at every 5oC do a finite difference calculation to obtain the results below.
FG
H
i
T (oC)
1
2
3
4
5
6
7
8
9
200
195
190
185
180
175
170
165
160
S$iL (kJ/kg K)
2.3309
2.2835
2.2359
2.1879
2.1396
2.0909
2.0419
1.9925
1.9427
IJ
K
S$iL (kJ/kg K)
6.4323
6.4698
6.5079
6.5465
6.5857
6.6256
6.6663
6.7078
6.7502
MiL (kg)
172.86
170.86
168.92
167.02
165.17
163.36
161.60
159.87
158.18
10
11
155
150
1.8925
1.8418
6.7935
6.8379
156.53
154.91
So the final total mass of water is 154.91 kg; using the specific volume of liquid water at
150oC listed at the beginning of the problem, we have that the water occupies 0.1690 m3
leaving 0.8310 m3 for the steam. Using its specific volume, the final mass of steam is found to
be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.10%, the
final volume fraction of water is 16.90%, and the fraction of the initial steam + water that has
been withdrawn is
(172.86+6.28-154.91-2.12)/(172.86+6.28) = 0.1234 or 12.34%. A total of 22.11 kg of steam
has withdrawn, and 87.7% of the original mass of steam and water remain in the tank.
These results are similar to that from the energy balance. The differences are the result of
round off errors in the simple finite difference calculation scheme used here (i.e., more
complicated predictor-corrector methods would yield more accurate results.).
A simpler method of doing the calculation, avoiding numerical integration, is to assume that
the heat capacity and change on vaporization of liquid water are independent of temperature.
Since liquid water is a condensed phase and the pressure change is small, we can make the
following assumptions
U$ L ≈ H$ L and H$ V − H$ L = ∆H$ vap
dU$ L dH$ L
dT L
dS$ L CPL dT L
≈
≈ CPL
≈
; and
dt
dt
dt
dt
T dt
With these substitutions and approximations, we obtain from the energy balance
dU$ L dM L $ V $ L
dH$ L dM L $ vap
=
→ ML
=
ML
∆H
H −U
dt
dt
dt
dt
dT dM L $ vap
=
∆H
M LCPL
dt
dt
Now using an average value of CPL and ∆H$ vap over the temperature range we obtain
c
h
1 dM L
CPL dT
=
or
M L dt
∆H$ vap dt
M fL
CPL
(
)
150
200
ln
−
=
M iL
∆H$ vap
FG IJ
H K
and from the entropy balance
dS$ L dM L $ vap
C L dT dM L $ vap
ML
∆S
∆S
=
→ ML P
=
dt
dt
T dt
dt
Now using an average value of CPL and ∆S$ vap over the temperature range we obtain
CPL dT
1 dM L
=
M L dt
T∆S$ vap dt
or
I FG IJ
K H K
M fL
CPL
150 + 27315
.
=
ln
ln
Mi L
200 + 27315
.
∆S$ vap
From the Steam Table data listed above, we obtain the following estimates:
F
H
U$ (T = 200o C) − U$ (T = 150o C) 852.45 − 632.20
kJ
=
= 4.405
kg ⋅ K
200o C - 150o C
50
or using the ln mean value (more appropriate for the entropy calculation) based on
CPL =
FG T IJ = S$aT f − S$aT f
HTK
CPL ln
2
2
1
1
kJ
.
S$(T = 200o C) − S$(T = 150o C) 2.3309 − 18418
=
= 4.3793
200 + 27315
.
47315
.
⋅K
kg
ln
ln
150 + 27315
.
42315
.
Also, obtaining average values of the property changes on vaporization, yields
1
1
kJ
. = 2027.5
∆H$ vap = × ∆H$ vap T = 150o C + ∆H$ vap T = 200o C = × 2114.3 + 19407
2
2
kg
1
1
kJ
.
= 4.5487
∆S$ vap = × ∆S$ vap T = 150o C + ∆S$ vap T = 200o C = × 4.9960 + 41014
2
2
kg ⋅ K
With this information, we can now use either the energy of the entropy balance to solve the
problem. To compare the results, we will use both (with the linear average Cp in the energy
balance and the log mean in the entropy balance. First using the energy balance
M fL
−4.405 × 50
CPL
(
)
=
= −010863
150
−
200
=
ln
.
L
vap
$
20275
.
M
∆H
CPL =
F
H
I
K
F
H
b
g
b
g
I
K
b
g
b
g
FG IJ
H K
i
M
L
f
L
i
) = 089706
= exp(−010863
.
.
M
Now using the entropy balance
M fL
CPL
.
.
.
150 + 27315
4.3793
42315
42315
ln
=
ln
=
ln
= 0.9628 ln
L
vap
$
Mi
.
.
.
200 + 27315
4.5487
47315
47315
∆S
FG IJ
F
I
H
K
H K
M
. I
42315
=F
= 089805
.
H 47315
M
. K
L
f
L
i
F
H
I
K
F
H
I
K
0.9628
Given the approximations, the two results are in quite good agreement. For what follows, the
energy balance result will be used. Therefore, the mass of water finally present (per m3) is
× M L (initial) = 15506
M L (final) = 0897
.
. kg
L
L
o
$
occupying V = M ( final) × V 150 C = 15506
m3
. × 0.001091 = 01692
.
b
g
Therefore, the steam occupies 0.8308 m3 , corresponding to
0.8308 m3 0.8308 m3
=
= 2115
.
kg
m3
V$ V 150o C
0.3928
kg
So the fraction of liquid in the tank by mass at the end is 155.06/(155.06+2.12) = 0.9865,
though the fraction by volume is 0.1692. Similarly the fraction of the tank volume that is
steam is 0.8308, though steam is only 2.12/(155.06+2.12) = 0.0135 of the mass in the tank.
M V (final) =
b
g
(c) Initially there was 6.28 + 172.86 = 179.14 kg of combined steam and water, and finally from
the simpler calculation above there is 155.06 + 2.12 = 157.18 kg. Therefore, 87.7% of the
total amount of steam + water initially in the tank are there finally, or 12.3% has been
withdrawn. This corresponds to 21.96 kg being withdrawn. This is in excellent agreement
with the more rigorous finite difference calculations done above.
dN
= 0 = N& 1 + N& 2 ;
dt
3.16 (a)
or N& 2 = − N& 1
dU
dV
W&
= 0 = N& 1 H 1 + N& 2 H 2 + W&S + Q& − P
= W&S + N& 1 H1 − N& 1 H 2 or S = H 2 - H1
dt
dt
N& 1
dS
Q&
= 0 = N& 1 S 1 − N& 1 S 2 + + S&gen
dt
T0
S&gen
= S 2 − S1
N&
1
Tf
W&S
= H 2 - H1 =
CPdT = CP ⋅ Tf − 29815
. K
N& 1
298.15K
temperature. First consider the reversible case,
z
Tf
S 2 − S 1 = 0 gives
z
Ti
CP = 37151
.
c
h
10
CP
dP
dT = R
T
P
1
z
if the heat capacity is independent of
The
J
W&Srev
= CP ⋅ (49914
.
. − 29815
. K) = 7467
&
mol
N1
J
= CP ⋅ Tf − 29815
. WSrev = 9334
. K
Wact = 125
mol
The solultion is Tf = 549.39K
c
J
mol ⋅ K
solution
The
actual
is
work
499.14K.
is
25%
Then
greater
h
(b)
Repeat the calculation with a temperature-dependent heat capacity
⋅ 10−2 T − 3499
⋅ 10−5 T 2 + 7.464 ⋅ 10−9 T 3
.
.
CP (T ) = 22.243 + 5977
Assuming reversibility Tf = 479.44K. Repeating the calculations above with the temperaturedependent heat capacity we find Wact = 9191 J, and Tf =520.92K.
So there is a significant difference between the results for the constant heat capacity and variable
heat capacity cases.
3.17
Ti = 300 K, Tf = 800 K, and Pi = 1.0 bar
CP (T ) = 29.088 - 0.192 × 10-2 T + 0.4 × 10-5 T 2 - 0.870 × 10-9 T 3
T f = 800 K
Pf
CP (T )
dP
dT = P
T
P
T = 300 K
P =1
i
z
z
i
Calculated final pressure Pf = 3.092 × 106 Pa.
T f = 800K
Wrev =
z
× 104
.
CP (T )dT = 1458
Ti = 300K
3.18
J
mol
Stage 1 is as in the previous problem.
Stage 2
Following the same calculation as above.
Stage 2 allowed pressure Pf ,2 = 9.563 × 107 Pa
Wrev = 1.458 × 10-4
J
= Stage 2 work
mol
J
mol ⋅ K
Stage 3
Following the same calculation method
Pf ,3 = 2.957 × 10-9 Pa = Stage 3 allowed pressure.
J
= Stage 3 work
mol
Question for the student: Why is the calculated work the same for each stage?
Wrev = 1.458 × 104
3.19 The mass, energy and entropy balances are
dM
& +M
& = 0, M& = − M&
=M
1
2
2
1
dt
dU
& H$ + M
& H$ + Q& + W& ; M
& H$ − H$ + W& = 0;
=0= M
1 1
2 2
s
1
1
2
s
dt
& H$ − H$
W&s = + M
1
2
1
c
c
h
h
dS
Q&
= 0 = M& 1S$1 + M& 2 S$2 + + S&gen = M& 1 S$1 − S$2 + S&gen = 0
dt
T
$
$
&
&
Sgen = M1 S2 − S1
c
c
h
h
300° C, 5 bar = 05
. MPa
H$1 = 3064.2 kJ kg
S$ = 7.4599 kJ kg K
100° C, 1 bar = 01
. MPa
H$ 2 = 26762
. kJ kg
$
S = 7.3614 kJ kg K
1
2
W&s
= 2676.2 − 3064.2 = 388 kJ kg satisfied the energy balance.
M& 1
S&gen $
= S2 − S$1 = 7.3614 − 7.4599 = −0.0985 kJ kg K can not be. Therefore the process is impossible.
M& 1
3.20 Steam 20 bar = 2 MPa and 300° C
H$ = 30235
. kJ kg
$
S = 6.7664 kJ kg
(from Steam Tables)
U$ = 2772.6 kJ kg
Final pressure = 1 bar. For reference saturation conditions are
P = 01
. MPa, T = 99.63
L
$
H$ L = 417.46
S$ L = 13026
.
U = 417.36
V
V
H$ = 26755
.
S$ V = 7.3594
U$ = 25061
.
(a) Adiabatic expansion valve W& = 0 and Q& = 0
dM
& +M
& = 0 ; M& 2 = − M& 1 ;
=M
1
2
dt
dU
& H$ + M
& H$ = 0 ; H$ = H$
E.B.:
=M
1 1
2 2
2
1
dt
From Steam Tables
T = 250° C
⇒ H2 = 30235
.
H$ = 2974.3 kJ kg S$ = 8.0333 kJ kg K
T = 300° C H$ = 3074.3 kJ/kg S$ = 8.2158 kJ/kg K
P = 01
. MPa
By interpolation T = 275° C gives H$ = 3023.5 kJ / kg ⇒ all vapor
M.B.:
S$ = 81245
.
kJ kg K
dS
& $ + M& S$ + S& = 0
= MS
1
2 2
gen
dt
S&gen
.
kJ kg K
= S$2 − S$1 = 8.1254 − 6.7664 = 1359
M& gen
(b) Well designed, adiabatic turbine
E.B.: M& 1H$ 1 + M& 2 H$ 2 + W& = 0 ; W& = H$ 2 − H$1
c
h
& S$ = 0 ; S$ = S$ ; S$ = 6.7664 kJ kg K
S.B.: M& 1S$1 + M
2 2
2
1
2
⇒ Two-phase mixture. Solve for fraction of liquid using entropy balance.
x ⋅ (7.3594) + (1 − x )⋅ 13026
.
= 6.7664
x = 0.902 (not good for turbine!)
. + 0.098 × 417.46 = 2454.2 kJ kg
H$ 2 = 0.902 × 26755
&
W
= (2454.2 − 30235
. ) = −569.3 kJ kg
M&
W&
−
= 569.3 kJ kg
M&
(c) Isorthermal turbine ⇒ superheated vapor
T = 300° C
H$ = 3074.3 kJ kg
final state
P = 01
. MPa
S$ = 8.2158 kJ kg K
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a
f
dU
dV
dV
dV
= W&S + Q& − P
= W&S + Q& − P0
− P − P0
dt
dt
dt
dt
3.37 (a)
and
dS Q&
=
+ S&gen
dt T0
Now let
a
f
dV
dV
W& = W& S − P0
− P − P0
dt
dt
and
a
f
dV
dV
dU
dV
W&u = W& + P0
= W&S − P − P0
⇒
= W&u + Q& − P0
dt
dt
dt
dt
or
a
U 2 − U1 = Wu + Q − P0 V2 − V1
S2 − S1 =
f
Q
+ S gen ⇒ Q = T0 S2 − T0 S1 − T0 Sgen
T0
and
a
U 2 − U 1 = Wu + T0 S 2 − T0 S1 − T0 S gen − P0 V2 − V1
a
f a
f
f
Wu = U 2 + PV
0 2 − T0 S 2 − U 1 + PV
0 1 − T0 S1 + T0 S gen
since T0 Sgen ≥ 0
Wumax = A2 − A 1, where A = U + PV
0 − T0 S
(b)
& H + Q& + W&
0 = M& H 1 − M
S
2
&
Q
0 = M& S 1 − M& S 2 +
+ Sgen
T0
Here W&S = W&u
& H − M& H − Q& = M& H − M
& H + MT
& S − MT
& S +TS
⇒ W&u = M
0 1
0 2
0 gen
2
1
2
1
a
f a
f
& H −T S − M
& H −T S +T S
⇒ W&u = M
0 2
0 1
0 gen
2
1
Since T0 S gen ≥ 0
b
g
& B − B where B = H − T S
W&umax = M
0
2
1
(c) Using the Steam Tables we find
i)
at 30 bar = 3 MPa and 600°C
U$ = 32850
. kJ kg , S$ = 7.5085 kJ kg K , V$ = 013243
.
m3 kg
$ 1 = U$ + P0V$ − T0 S$
A
= 32850
. + 1.013 bar + 0.13243 m3 kg × 102 kJ bar ⋅ m3 − 298.15 × 7.5085
= 105976
. kJ kg
ii)
at 5 bar = 05
. MPa and 300°C
$
U = 2802.9 kJ kg , S$ = 7.4599 kJ kg K, V$ = 05226
.
m3 kg
$ 2 = 2802.9 + 1013
A
. × 05226
.
× 102 − 29815
. × 7.4599 = 63167
. kJ kg
$ 2 − A$ 1 = (63167
W$u = A
. − 1059.76) kJ kg = −428.09 kJ kg
This is the maximum useful work that can be obtained in the transformation with the environment
at 25°C and 1.013 bar. It is now a problem of clever engineering design to develop a device which
will extract this work from the steam in a nonflow process.
(d) Since the inlet and exit streams are at 25°C and P = 1013
.
bar, any component which passes
through the power plant unchanged (i.e., the organic matter, nitrogen and excess oxygen in the air,
etc.) does not contribute to the change in availability, or produce any useful work. Therefore, for
each kilogram of coal the net change is:
0.7 kg of carbon = 58.33 mol of C
+58.33 mol of O 2
to produce 58.33 mol CO 2
also
0.15 kg of water = 8.33 mol of H2 O undergoes a phase change
from liquid to vapor
Therefore
b
&B
$ in = ∑ N& i B
M
i
i
g
in
= 58.33 × 0+ 58.33 × 0 + 8.33 × (− 68.317 + 298.15 × 0.039 )
(carbon)
(oxygen )
(liquid water )
= −1976 kJ kg coal
&
$ out = ∑ N& i B
MB
= 58.33 × (− 94.052 )+ 8.33 × ( −57.8 + 298.15 × 0.0106)
i out
i
b
g
(carbon dioxide)
( water vapo r )
= −24858 kJ kg coal
Wumax
= −24858 − (1976) kJ kg coal = 22882 kJ kg coal
= −2.2 kW - hr kg coal = 7920 kJ kg coal
7920 × 100
Efficiency in % =
= 34.6%
22882
Wuactual
Thus a coal-fired electrical power generation plants converts slightly more than 1/3 of the useful
work obtainable from the coal it consumes. This suggests that it would be useful to look for
another method of generating electrical power from coal . . . for example, using an electrochemical fuel cell. Considering the amount of coal consumed each year in power generation, and
the consequences (strip mining, acid rain, greenhouse effect, etc.) the potential economic savings
and environmental impact of using only 1/3 as much coal is enormous.
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u Carnot efficiency (1100°C and 15°C)
=
T1 − T2 (1100 + 27315
. ) − (15 + 273.15)
=
× 100 = 79.02%
T1
(1100 + 27315
. )
For comparison, Carnot efficiency (700°C and 60.1°C) which are the temperature levels of working fluid
(steam) in the closed-loop power cycle
=
( 700 + 273.15) − ( 601
. + 27315
. )
× 100 = 6576%
.
700 + 27315
.
which is almost twice as high as the actual efficiency.
3.39 Three subsystems: unknowns T1 f , P1 f , T2 f , P2 f , T3 f , P3 f (6 unknowns)
After process P1 f = P2 f = P3f (2 equations)
Subsystem 1 has undergone a reversible adiabatic expansion
⇒ S 1f = S i1 , or T1 f = T1i
FG P IJ
HPK
f
R CP
1
(1 equation)
i
1
(#1)
Subsystem 3 has undergone a reversible adiabatic compression
⇒ S 3f = S i3 , or T3 f = T3i
FG P IJ
HP K
f
3
i
3
R CP
= T3i
FG P IJ
HPK
f
R CP
(1 equation)
i
3
(#2)
Mass balance subsystems 1 + 2
N1f + N 2f = N1i + N 2i ⇒
or
Pf
P1 f V1
T1 f
+
P2 f V2
T2 f
=
P1iV1
T1i
+
P2iV2i
T2i
FG 0.5 + V IJ = 10 × 0.5 + 1× 0.25 = 0.017909 (1 equation)
H T T K 29315
.
29315
.
f
1
f
2
f
2
(#3)
Energy balance on subsystems 1 + 2 + 3
N1f U 1f + N2f U 2f + N 3f U 3f = N1i U i1 + N2i U i2 + N 3i U i3
P1 f V1
P fV
P fV
PiV
PiV
PiV
CV T1 f + 2 f2 CV T2f + 3 f3 CV T3 f = 1 1i CV T1i + 2 2i CV T2i + 3 3i CV T3i
f
RT1
RT2
RT3
RT1
RT2
RT3
c
h
P f V1 + V2 f + V3 f = P1iV1 + P2iV2 + P3iV3
but V1 + V2 f + V3 f = V1 +V2i +V3i = (05
. + 025
. + 025
. ) = 1 m3
Pf =
using this result
10 × 0.5 + 1 × 0.25 + 1 × 0.25
= 5.5 bar
1
in eqn. (1) → T1 f = 252.45 K
in eqn. (2) → T3 f = 448.93 K
V3 f = V3i
P3i T3f
1 448.93
⋅
= 0.25 ×
×
= 0.06961 m 3
P3 f T3i
55
.
293.15
V2 f = 0.25 × 2 − 0.06961 = 0.4304 m3
Now using Eqn. (#3)
FG 0.5 + V IJ = 55. FG 0.5 + 0.4304 IJ = 0.017909 ⇒ T
H T T K H 252.45 T K
Pf
f
2
f
2
f
1
f
f
2
= 337.41 K
2
Thus the state of the system is as follows
T1
Initial
293.15 K
Final
252.45 K
P1
10 bar
5.5 bar
T2
293.15 K
337.41 K
P2
1 bar
5.5 bar
V2
0.25 m
0.4304 m3
T3
293.15 K
448.93 K
P3
1
5.5 bar
3
V3
3
0.0696 m3
CV T3i = P3 f V3 f
CV
C
− P3iV3i V = W
R
R
0.25 m
Work done on subsystem 3
Energy balance
f
z
i
N 3f U 3 − N3i U 3 = W = − PdV
P3f V3f
RT3 f
CV T3 f −
P3iV3i
RT3i
CP − R f f
P3 V3 − P3iV3i = 3(55
. × 0.0696 − 1 × 0.25)
R
= 0.3984 bar ⋅ m3 = 39.84 kJ
W=
c
h
3.40 For the mass and energy balances, consider the composite system of can + tire as the system. Also gas is
ideal for this system Q = 0 and W = 0
mass balance: N1f + N2f = N1i + N 2i ⇒
P1 f V1
T1 f
+
P2 f V2
T2 f
=
P1iV1
T1i
+
P2iV2
T2i
(1)
energy balance
N1f U 1f + N2f U 2f = N1i U i1 + N 2 U i2 ⇒ P1 f V1 + P2f V2 = P1iV1 + P2iV2
(see derivation of eqn. (c) of Illustration 2.5-5)
Also P1 f = P2 f = 2.6 bar(← 3) ; using eqn. (3) in eqn. (2) yields
P1i =
c
h
−2
−2
P1 f V1 + P2 f V2 − P2iV2 2.6 × 4.06 × 10 − 1 × 4 × 10
=
= 109 .27 bar
V1
6 × 10− 4
(2)
To use eqn. (1) to get final temperatures, need another independent equation relating T1 f and T2 f . Could do
an energy balance around tank 1, as in derivation of eqn. (f) of Illustration 2.5-5. A more direct way is to do
an entropy balance around a small fluid element, as in Illustration 3.5-2 and immediately obtain Eqn. (e) of
that illustration
FG T IJ
HT K
f
CP R
=
1
i
1
FG P IJ
HPK
f
1
i
1
Thus
f
T1 =
T1i
FG P IJ
HPK
f
1
i
1
R CP
FH 2.6 IK
109.27
= 295
8.314 30
= 104.69 K (very cold!)
Using this result in eqn. (1) gives T2 f = 30326
. K.
3.41 (a) System: Gas in the tank — system boundary is just before exit to the tank. System is open, adiabatic, and
of constant volume.
dN
M.B.:
= N&
dt
d ( N S)
S. B.:
= N& S
dt
dN
dS
dN
dS
⇒S
+N
= N& S =
S⇒N
= 0 or S = constant (since N ≠ 0)
dt
dt
dt
dt
Note: Gas just leaving system has the same thermodynamic properties as gas in the system by the “wellmixed” assumption.
For the ideal gas this implies
T2 = T1
FG P IJ
HPK
R CP
2
1
F 1 bar I
H 17.5 bar K
(b) T2 = (22 + 27315
. )
8.314 30
= 133.52 K
(c) System: Gas in the tank + engine (open, constant-volume, adiabatic)
M.B.: N f − Ni = ∆N
∆N = amount of mass (moles) that left the system
E.B.: N f U f − Ni U i = ∆N H out + WS
Note: H out = constant, since gas leaving engine is of constant properties.
Thus
+WS = N f U f − Ni U i − ∆N H out
= N f U f − Ni U i − N f H out + Ni H out
= N f CV − Ni CV Ti − N f CPTout + Ni CPTout ; but Tout = Ti
= N f CV T f − CPTi + Ni RTi
B reference temperature
H = C aT − T f
UV see Eqn. (2.4 - 8)
U = C aT − T f − RT W
Now PV = NRT ⇒ N =
Ni =
P
0
V
0
0
PV
RT
17.5 bar × 0.5 m 3
= 0.3566 kmol
295.15 K × 8.314 × 10− 2 bar ⋅ m3 kmol K
1 bar × 0.5 m 3
= 0.04504 kmol
13352
. K × 8.314 × 10 −2
Ws = 0.04504 ( 30 − 8.314 ) × 133.52 − 30 × 295.15 + 0.3566 × 8.314 × 29515
.
Nf =
= 606.6 kJ
Since WS > 0 , work must be put into the engine if the outlet temperature is to be maintained at 22°C.
(Alternatively, heat could be added and work extracted.)
We should check to see if the process considered above is indeed possible. Can do this by using the
entropy balance and ascertaining whether Sgen ≥ 0 .
Entropy balance
c
h
N f S f − Ni S i = N f − Ni S out + S gen
⇒ N f CP ln
Tf
Tout
− N f R ln
Pf
Pout
− Ni CP ln
Ti
P
+ Ni R ln i = Sgen
Tout
Pout
But Tout = Ti = 295.15 K , and Pout = Pf = 1 bar so
N f CP ln
Tf
Ti
+ Ni R ln
Pi
= Sgen
Pf
or
Sgen = 0.04505 × 30 ln
133.52
17.5
+ 0.3566 × 8.314 ln
= 7.414 kJ K
29515
.
1
Thus, Sgen > 0 , and the process is possible!
(d) Similar process, but now isothermal: system = gas in tank and engine.
M.B.: N f − Ni = ∆N
c
= cN
h
− N hS
E.B.: N f U f − Ni U i = ∆N H out + Q + Ws = N f − Ni H out + Q + Ws
Q
Q
+ S gen
+ S gen
f
i
out +
T
T
= 0 , since we want maximum work (see Sec. 3.2). Thus
S.B.: N f S f − Ni S i = ∆N S out +
Set Sgen
d
i c
h
Q = T N f S f − N i S i − N f − Ni T S out
d
i
a
= TN f S f − S out − TN i S i − S out
= − N f TR ln
Pf
Pout
+ Ni RT ln
f
Pi
Pout
But Pf = Pout = 1 bar and
F 17.5I = 2504.6 kJ
H 1K
Q = 0.3566 × 8.314 × 295.15 ln
Ws = N f U f − N i U i − N f H out + Ni H out − Q
c
h
= − N f RT + Ni RT − Q = Ni − N f RT − Q
PV
1 × 0.5
=
= 2.038 × 10 −2 kmol and
RT
29515
. × 8.314 × 10−2
Ws = (0.3566 − 0.0204) × 8.314 × 29515
. − 2504.6 = −1679.6 kJ . In this case we obtain work!
but N f =
3.42
a) For each stage of the compressor, assuming steady-state operation and reversible adiabatic operation we
have from the mass, energy and entropy balances, respectively
&
&
0 = M& in + M
or M& out = − M& in = − M
out
0 = M& H$ + M& H$ + W&
or W& = M& ( H$ − H$ )
in
in
out
out
out
in
and
& S$
0 = M& inS$in + M
or S$out = S$in
out out
So through each compressor (but not intercooler) stage, one follows a line on constant entropy in Fig. 2.4-2
Therefore, for first compressor stage we have
H$ in (T = 200 K, P = 1 bar ) = 767 kJ / kg and S$in ( T = 200 K, P = 1 bar ) = 6.5 kJ / kg K
H$ ( S$ = 6.5 kJ / kg K, P = 5 bar ) = 963 kJ / kg and T = 295 K
out
out
Therefore the first stage work per kg. of methane flowing through the compressor is
W& ( first stage) = 963 − 767 kJ / kg = 196 kJ / kg
After cooling, the temperature of the methane stream is 200 K, so that for the second compressor stage we
have
H$ in (T = 200 K, P = 5 bar ) = 760 kJ / kg and S$in (T = 200 K, P = 5 bar ) = 5.65 kJ / kg K
H$ ( S$ = 5.65 kJ / kg K, P = 25 bar ) = 960 kJ / kg and T = 300 K
out
out
Therefore the second stage work per kg. of methane flowing through the compressor is
W& (sec ond stage) = 960 − 760 kJ / kg = 200 kJ / kg
Similarly, after intercooling, the third stage compressor work is found from
H$ in (T = 200 K, P = 25 bar ) = 718 kJ / kg and S$in ( T = 200 K, P = 25 bar ) = 4.65 kJ / kg K
H$ ( S$ = 4.65 kJ / kg K, P = 100 bar) = 855 kJ / kg and T = 288 K
out
out
Therefore the third stage work per kg. of methane flowing through the compressor is
W& ( third stage) = 855 − 718 kJ / kg = 137 kJ / kg
Consequently the total compressor work through all three stages is
W& = 196 + 200 + 137 = 533 kJ / kg
b) The liquefaction process is a Joule-Thomson expansion, and therefore occurs at constant enthalpy. The
enthalpy of the methane leaving the cooler at 100 bar and 200 K is 423 kJ/kg. At 1 bar the enthalpy of the
saturate vapor is 582 kJ/kg, and that of the liquid is 71 kJ/kg. Therefore from the energy balance on the
throttling valve and flash drum we have
H$ in = H$ out or
H$ (200 K, 100 bar) = (1 − x) H$ (sat' d. vapor, 1 bar) + xH$ ( sat'd. liquid, 1 bar)
423
kJ
kJ
kJ
= (1 − x ) ⋅ 71 + x ⋅ 582
kg
kg
kg
where x = 0.689 is the fraction of vapor leaving the flash drum, and (1-x) = 0.311 is the fraction of the
methane that has been liquefied. Therefore, for each kilogram of methane that enters the simple
liquefaction unit, 689 grams of methane are lost as vapor, and only 311 grams of LNG are produced.
Further, since 533 kJ of work are required in the compressor to produce 311 grams of LNG, approximately
1713 kJ of compressor work are required for each kg. of LNG produced.
c)
As in the illustration, we choose the system for writing balance equations to be the subsystem consisting of
the heat exchanger, throttle valve and flash drum (though other choices could be made). The mass and
energy balances for this subsystem (since there are no heat losses to the outside or any work flows) are
& = M
& +M
& or taking M
& = 1 and letting x be the fraction of vapor
M
3
5'
6
3
1 = (1 − x) + x
& H$ = M& H$ + M& H$
M
3 3
5 ' 5'
6 6
1 ⋅ H$ (T = 200 K, P = 100 bar) = x ⋅ H$ (T = 200 K, P = 1 bar) + (1 − x) ⋅ H$ ( sat'd.liquid, P = 1 bar)
423
kJ
kJ
kJ
= x ⋅ 718
+ (1 − x) ⋅ 71
kg
kg
kg
The solution to this equation is x = 0.544 as the fraction of vapor which is recycled, and 0.456 as the
fraction of liquid.
The mass and energy balances for the mixing of the streams immediately before the compressor are
& +M
& = M
& ; then basing the calculation of 1 kg of flow into the compressor
M
5'
1
1'
&
&
& = 0.456
M = 1, M = 0.544 and M
1'
5'
1
However, since both the recycle vapor and the inlet vapor are at 200 K and 1 bar, the gas leaving the
mixing tee must also be at these conditions, so that the inlet conditions to the first compressor are the
kJ
same as in the simple liquefaction process, and H$ 1' = 718
. Also, all other compressor stages
kg
operate as in the simple liquefaction process.
Therefore, the total compressor work per kg of methane passed through the compressor is
W& = 196 + 200 + 137 = 533 kJ / kg of methane through the compressor. However, each kg. of methane
through the compressor results in only 0.456 kg. of LNG (the remainder of the methane is recycled).
Consequently the compressor work required per kg. of LNG produced is (533 kJ/kg)/0.456 kg = 1168 kJ/kg
of LNG produced. This is to be compared to 1713 kJ/kg of LNG produced in the simple liquefaction
process.
3.43 (also available as a Mathcad worksheet)
Problem 3.43 with MATHCAD
bar
101300 . Pa
mol
Heat capacity
1
8.314 .
RE
joule
mol. K
RG
0.00008314 .
bar . m
mol. K
3
2.5 . RE
Cp
Initial Conditions (Vt=total volume, m^3):
Find initial molar volume and number of moles
Start with initial guess for volume, m^3/mol
Initial molar volume and
number of moles
298 . K
Ti
Vi = 6.194 10
5
Vi
Pi
RG. Ti
Vi = 6.194 10
Pi
3
m
N
Vt
Vi
400 . bar
Vt
5
0.045 . m
3
3
m
N = 726.518 mol
Final pressure is 1.013 bar, and final temperature is unknown; will be found by equating the initial and final
entropies. Guess final temperature is 200 K
Pf
1.013 . bar
T
50 . K
Vf
RG. T
Pf
Solve for final
temperature using
S(final) - S(initial) = 0
Given
0
Cp . ln
T
Ti
Pf
RG. ln
Pi
Tf
FIND( T )
Tf = 26.432 K
Final temperature
W
3.
RE. ( Tf
Ti) . N
6
W = 2.461 10
2
TNTeq
W
4600000 .
joule
kg
joule
TNTeq = 0.535 kg
3.44 (also available as a Mathcad worksheet)
3.44 N2
R
8.314
8.314 . 10
RR
5
Note that in the 1st and 2nd printings, carbon dioxide was used as the fluid. This gave
unreasonable answers when this problem was revisited with the Peng-Robinson eqn.
of state, as both the initial and final states were found to be in the liquid state. Therefore
from the 3rd printing on, the fluid has been changed to nitrogen.
Heat capacity constants for nitrogen
Cp 0
Ti
28.883
0.157 . 10
Cp 1
298.15
2
Tf
0.808 . 10
Cp 2
5
Cp 3
2.871 . 10
9
100
Given
Tf
Cp 0
Cp 1 . T
Cp 2 . T
2
Cp 3 . T
3
d T R. ln
1.013
T
140
Ti
Tf
Tf = 72.054
find( Tf )
Number of moles = N = PV/RT
N.
W
Tf
Cp 1 . T
Cp 0
2
140 . 3.1416 . ( .01 ) . .06
RR. 298.15
N
Cp 2 . T
N = 0.106
moles
Cp 3 . T d T
2
3
Ti
W = 695.114
4600 J = 1 gram TNT
joules
W
Grams of TNT = G
G = 0.151
4600
3.44CO2
R
8.314
RR
8.314 . 10
grams of TNT
5
Note that in the 1st and 2nd printings, carbon dioxide was used as the fluid. This gave
unreasonable answers when this problem was revisited with the Peng-Robinson eqn.
of state, as both the initial and final states were found to be in the liquid state. Therefore
from the 3rd printing on, the fluid has been changed to nitrogen.
Heat capacity constants
Cp0
Ti
22.243
Cp1
298.15
Tf
5.977 . 10
2
3.499 . 10
Cp2
5
Cp3
200
Given
Tf
Cp0
Cp1. T
Cp2. T
T
Ti
2
Cp3. T
3
d T R. ln
1.013
140
7.464 . 10
9
Tf
Tf = 79.836
find ( Tf )
Number of moles = N = PV/RT
2
140 . 3.1416 . ( .01 ) . .06
RR. 298.15
N
N = 0.106
moles
To calculate work done (energy released), we need the internal energy change. Therefore need
Cv = Cp - R
W
N.
Tf
Cp0
8.314
Cp1. T
Cp2. T
2
Cp3. T d T
3
Ti
W = 555.558
4600 J = 1 gram TNT
joules
Grams of TNT = G
W
G = 0.121
4600
3.45 25 bar = 2.5 MPa; 600°C
H$ = 36863
. kJ kg ; S$ = 7.5960 kJ kg K
1 bar
∆ = 027
.
S$ = 7.3614
S$ = 7.6314
100°C
150
⇒ T = 143.44° C
1311
H$ = 27764
. − 131
. = 27633
. kJ kg
W = −3686.3 + 27633
. = −923 kJ kg
− 16614 kJ mol
a
f
ideal gas = −16830 kJ mol
Actual work 784.55
dU
dV
= 0 = M& 1 H$ 1 − H$ 2 + Q& − P
+ W&s
dt
dt
W&
H$ 2 = H$ 1 + s = 36863
. − 784 .55 = 290175
.
M& 1
c
h
Final state P = 1 bar ; H$ = 290175
.
∆T
50
H$ (1 bar, 200° C) = 28753
.
=
99.6 ∆H$ 99.6
H$ (1 bar, 250° C) = 2974.9
UV
W
290175
. − H$ (1 bar, 200° C) = 26.45
50
26.45 = 213.28
99.6
S$ (1 bar, 250) = 8.0333
S$ (1 bar, 200) = 7.8343
T = 200 +
S$ (1 bar, 213.28 ) = 7.8872
grams of TNT
c
h
dS
= 0 = M& S$1 − S$2 + S& gen ;
dt
S&gen
= − S$2 − S$1 = 7.8872 − 7.5960 = 0.2912 kJ kg K
&
M
5.2416 kJ kg K
(ideal gas = 5.468 )
PR: T = 600° C ; P = 25 bar
H = 216064
.
× 104
S = 14.74377
Now P = 1 bar , S = 14.74377 . Guess T = 213° C .
T
213
150
S
19.67116
14.74399
H
727195
.
503486
.
W&
= 503486
. − 21606.4 = −1657154
. J mol
M&
Actual work ⇒ H f = 752059
.
T
213
230
220
219.9
219.92
S
19.67116
20.90787
20.18472
20.17742
20.17888
H
7271.95
7883.64
7523.41
7519.80
7520.53
S = 2017888
.
− 1474377
.
= 543511
.
3.46
From simple statics the change in atmospheric pressure dP accompanying a change in height dh is
dP = −ρgdh
where ρ is the local mass density and g is the gravitational constant. Assuming a packet of air undergoes an
altitude change relatively rapidly (compared to heat transfer), the entropy change for this process is
C
R
d S = P dT − dP = 0 since both Q& and S&gen equal zero.
T
P
Combining the two equations above we have
CP
R
R
R N
M
dT = dP = − ρgdh = −
Mgdh = −
gdh
T
P
P
PV
T
dT
Mg
or
=−
dh
CP
dT
K
dT
≅ − 9.7
. Note that
is referred to as the adiabat ic lapse rate.
dh
km
dh
Also, its value will be less than that above as the humidity increases.
In fact, if the humidity is 100%, so water will condense as the pressure decreases, the adiabatic lapse rate
will be almost zero.
For dry air
Solutions to Chemical and Engineering Thermodynamics, 3e
4
4.1
Using the Mollier diagram
µ=
(510 − 490)° C
F ∂T I = F ∆T I =
H ∂P K H ∆P K c1.241 × 10 − 7.929 × 10 hPa
H
7
H
6
= 4.463 × 10 −6 ° C Pa = 4.463 ° C MPa
κS =
(510 − 490)° C
F ∂T I ≈ F ∆T I =
H ∂P K H ∆P K c1.069 × 10 − 9.515 × 10 hPa
S
7
S
6
= 1.702 × 10− 5 ° C Pa = 17.02 ° C MPa
a∂H ∂S f
a∂H ∂Sf
T
P
a f a f a f a f
a f a f a f a f
∂a T , H f ∂a P , S f
µ
=
×
=
= 0.262 ( unitless)
∂ a P , H f ∂a T , S f κ
=
∂ H, T
∂ S,P
∂ H, T
∂ S, P
×
=
×
∂ S ,T
∂ H, P
∂ H, P
∂ S ,T
S
4.2
(a) Start from eqn. 4.4-27
z LMN FH
V
H ( T , P) − H IG (T , P ) = RT ( Z − 1) +
T
V =∞
P=
F ∂P I
H ∂T K
RT
a( T )
− 2
;
V − b V + 2bV − b 2
V
=
z LMN RST
V
V =∞
F
H
OP
Q
− P dV
V
UV
W
OP
Q
R
da dt
RT
a( T )
− 2
−
+ 2
dV
2
V − b V + 2bV − b
V − b V + 2bV − b 2
T
= RT ( Z − 1) + a + T
I
K
R
da dt
− 2
so
V − b V + 2 bV − b 2
H ( T , P) − H IG (T , P )
= RT ( Z − 1) +
dP
dT
da
dT
I
K
z
V
dV
V =∞ V
2
+ 2bV − b2
From integral tables we have
z
dx
a ′x2 + b ′x + c′
In our case
=
1
b ′2 − 4 a ′c′
a′ = 1,
ln
b ′ = 2b ,
(b′)2 − 4a ′c′ = 8b2 = 2 2b .
2a ′x + b ′ − b ′2 − 4 a ′c′
2a ′x + b ′ + b ′2 − 4 a ′c′
for 4a′c′ − b′ 2 < 0
c h
c = −b2 ; so 4 a ′c′ − b ′ 2 = 4 ⋅ 1 ⋅ −b 2 − ( 2b )2 = − 8b2
and
Solutions to Chemical and Engineering Thermodynamics, 3e
H ( T , P) − H IG (T , P )
aa − Tda dTf LMln 2V + 2b − 2 2b
2 2b
MN 2V + 2b + 2 2b
aa − Tda dTf ln V + d1 − 2 ib
= RT ( Z − 1) +
2 2b
V + d1 + 2 ib
= RT ( Z − 1) +
− ln
V
2V + 2b − 2 2b
2V + 2b + 2 2b V = ∞
OP
PQ
or finally
H ( T , P) − H IG (T , P ) = RT ( Z − 1) +
aTda dT − af lnLM Z + d1 + 2 iB OP
2 2b
MN Z + d1 − 2 iB PQ
(b) This part is similar except that we start from eqn. (4.4-28)
S (T , P) − S IG ( T , P) = R ln Z +
z LMNFH
z LMN
V
V =∞
V
= R ln Z +
V =∞
dP
dT
I
K
−
V
OP
Q
R
dv
V
OP
Q
R
da dT
R
− 2
−
dV
2
V − b V + 2 bV − b
V
d i
d i
da dT F Z + d1 + 2 i B I
= R ln( Z − B) +
lnG
J
2 2 b H Z + d1 − 2 i B K
L F ∂ P IJ − P OPdV .
Start with eqn. (4.2-21):
dU = C dT + MT G
MN H ∂ T K PQ
FG ∂U IJ = C + LMTFG ∂ P IJ − POP FG ∂V IJ and
H ∂T K
MN H ∂ T K PQ H ∂T K
FG ∂U IJ − FG ∂U IJ = LMTFG ∂ P IJ − POPFG ∂V IJ .
H ∂ T K H ∂ T K MN H ∂T K PQH ∂T K
(a) Ideal gas PV = RT
F ∂ P IJ − P = 0 ⇒ FG ∂U IJ = FG ∂U IJ
TG
H ∂T K
H ∂ T K H ∂T K
(b) van der Waals gas
L F ∂ P I − POP = a
RT
a F∂ PI
R
P=
−
; G
=
; ⇒ MT G
J
V −b V
H ∂ T K V − b MN H ∂T JK PQ V
V
V −b
da dT V + 1 + 2 b
= R ln Z + R ln
+
ln
V V = ∞ 2 2b V + 1 − 2 b
4.3
V
V
V
P
V
P
P
V
V
V
P
2
Also: dP =
P
V
V
RdT
RT
2a
−
dV + 3 dV
2
V − b (V − b)
V
V
2
V
V =∞
Thus
FG ∂U IJ
H ∂T K
= CV ;
V
Solutions to Chemical and Engineering Thermodynamics, 3e
F ∂V IJ = R (V − b) = LM T − 2a(V − b) OP
⇒G
H ∂T K RT (V − b) − 2a V N (V − b) RV Q
F ∂ u IJ − FG ∂u IJ =
a
⇒G
H ∂ T K H ∂ T K V T (V − b) − 2a(V − b) RV
−1
2
P
P
3
3
2
V
aRV (V − b )
=
RTV 3 − 2 a (V − b) 2
(c) The Virial Equation of State
PV
B C
B
= 1 + + 2 + L = 1 + ∑ ii
RT
V V
i =1 V
RT
B RT
+ ∑ i i +1
V
i =1 V
or P =
Note: This is a total
FG ∂ P IJ = R + ∑ B R + ∑ RT F dB I ← derivative,
since B is
H ∂T K V V
V H dT K
a function of only temperature
F ∂ P IJ − P = ∑ RT dB
⇒ TG
H∂T K
V d ln T
Also need a∂V ∂ T f , but this is harder to evaluate alternatively. Since
FG ∂V IJ FG ∂ P IJ FG ∂T IJ = −1 ⇒ FG ∂V IJ = − a∂ P ∂T f
H ∂T K H ∂V K H ∂ P K
H ∂ T K a∂ P ∂V f
FG ∂ P IJ is given above.
H ∂T K
i
i+1
V
i
i +1
i=1
i
i =1
i
i+1
V
i=1
P
V
P
T
V
P
T
V
FG ∂ P IJ
H ∂V K
⇒
Using
=−
T
FG ∂V IJ
H ∂T K
RT
V
=
2
FG
H
−∑
V i+ 2
i =1
ia
d
V RT V + ∑ Bi RT V i +1 + ∑ RT V i + 1 dBi d ln T
i =1
FG
H
i+ 1
T RT V + ∑ (i + 1) Bi RT V
P
i =1
RT
B RT
= P − ∑ i i +1 , we get
V
i =1 V
FG ∂V IJ
H ∂T K
and
(i + 1) Bi RT
=
P
iadB d ln Tf
T P + ∑ aiB RT f V
d
V P + ∑ RT V
i +1
i
i
i +1
IJ
K
fIJK
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂U IJ − FG ∂U IJ
H ∂T K H ∂T K
P
4.4
LM
MN
d
ia
dBi d ln T
R dBi P + ∑ RT V
=∑ i
V d ln T
P + ∑ iBi RT V i + 1
V
i +1
f OP
PQ
(a) Start from
LM
N
FG IJ OP ⇒ C = − 1 LMV − TFG ∂V IJ OP
H KQ
H∂TK Q
µN
L F ∂V IJ OP = −T FG ∂aV TfIJ and C = T FG ∂ aV T fIJ .
but MV − T G
µ H ∂T K
N H ∂ T K Q H ∂T K
FG ∂ aV T fIJ = µC ; integrate F V I − F V I = z µC dT .
H TK H T K
H ∂T K T
T
1
∂V
V −T
CP
∂T
µ= −
P
P
P
2
2
P
P
P
P
T2 , P
(b)
P
T2 , P
2
2
µCP
T2
T1 , P
1
P=
z
T2 , P
a f a f TT + T
Thus V T2 , P = V T1 , P
4.5
P
2
P
2
T1 , P
T1 , P
dT .
RT
a
− 2
V −b V
FG ∂ P IJ = − RT + 2a
H ∂V K (V − b) V
FG ∂ P IJ = 2RT − 6a
H ∂V K (V − b) V
2
T
(1)
3
2
2
3
(2)
4
T
at the critical point P → PC , T → TC and V → VC
FG ∂ P IJ
H ∂V K
FG ∂ P IJ
H ∂V K
=0⇒
T
2
= 0⇒
2
Dividing (1') by (2') ⇒
a=
Also
a
T
RTC
aV
C
−b
f
2 RTC
aV
C
−b
2
f
3
=
=
2a
(1')
V 3C
6a
V 4C
(2')
f
1
1
V
V −b = VC ⇒b =
from (1')
2 C
3
3
V 3C RTC
a
2 V C −b
f
2
=
(3b )3 RTC
27bRTC 9VC RTC
=
=
8
8
2(3b − b )2
PV
V
a
8a
a
a
=
−
; and PC =
− 2 =
RT V − b RTV
27b( 2b ) 9b
27b 2
⇒ ZC =
=
PC V C
RTC
=
VC
VC − b
−
3 9 3
− = = 0.375
2 8 8
a f
VC
9 8 V C RTC
a
=
−
RTCVC
2 3VC
V C RTC
a f
Solutions to Chemical and Engineering Thermodynamics, 3e
4.6
dS =
For
FG IJ
H K
CP
∂V
dT −
T
∂T
the
dP [eqn. (4.2-20)]
P
ideal
dS
gas
IG
=
i LMN RP − FGH ∂∂VT IJK OPQdP and
d
CP*
R
dT − dP .
T
P
Thus,
at
constant
temperature
d S − S IG =
P
S (T , P ) − S IG (T , P ) − S (T , P = 0) − S IG( T , P = 0 ) =
z RST
P
P= 0
However,
a
FG IJ UVdP
H KW
R
∂V
−
P
∂T
P
S(T , P = 0) − S IG (T, P = 0) = 0 , since all fluids are ideal at
f
PV = Z Tr , Pr RT . Thus
FG ∂V IJ
H ∂T K
RS a
T
FG a
H
f
1
∂ Z Tr , Pr
RZ Tr , Pr + RT
P
∂T
=
P
fIJ UV
KW
P
and
FG IJ
H K
FG IJ
H K
L Z − 1 + T FG ∂ Z IJ OP dP
⇒ S (T , P) − S ( T , P) = − R z M
N P P H ∂T K Q
L ZaT , P f − 1 + T F ∂ Z I OPdP
= −R z M
G J
P H ∂ T K PQ
MN P
R
∂V
−
P
∂T
=−
P
ka
f p
R
RT ∂ Z
Z Tr , Pr − 1 −
P
P ∂T
P
T, P
IG
P
T , P =0
Tr , Pr
Tr , Pr = 0
4.7
r
r
r
r
r
r
r
Pr
(a) Ideal gas
PV = NRT
N =
c
(50 bar ) 100 m 3
−2
( 27315
. + 150) K × 8.314 × 10
h
bar ⋅ m3 kmol K
= 142.1 kmol
Energy balance, closed nonflow system
z
∆U = Q − PdV = Q + W .
However, for ideal gas ∆U = 0 since T is constant (isothermal). Thus
z
z
W = − Q = − PdV = −
NRT
V
P
dV = − NRT ln 2 = − NRT ln 1
V
V1
P2
F 50 I
H 300 K
= −142 .1 kmol × 8.314 J mol K × (27315
. + 150 ) × ln
= 895.9 × 10 3 kJ = 895.9 MJ
Q = −895.9 MJ
Also, by Ideal Gas Law at fixed T and N
P= 0.
Also
Solutions to Chemical and Engineering Thermodynamics, 3e
PV
1 1 = PV
2 2 ⇒ V2 = V1
P1
50
= 100 m 3 ×
= 16.67 m3
P2
300
(b) Corresponding states
Tr
Pr
initial state
150 + 27315
.
= 1.391
304 .2
final state
1.391
50
= 0.679
7376
.
300
= 4.067
73.76
Number of moles of gas = N =
Final volume = V f =
Z
H IG − H
TC
S IG − S
0.94
0.7
0.4
0.765
4.5
2.4
cal
mol K
cal
mol K
PV
142.1
PV
=
= 151.2 kmol ( 142 .1 =
from above)
ZRT
0.94
RT
ZNRT
Pf
0.765 × 151.2 × 8.314 × 10−2 × (273.15 + 150 )
=1356
. m3
300
Energy balance on gas: ∆U = Q + W
=
Q
+ S gen ⇒ S gen = 0 . Therefore
T
Entropy balance on gas processes in gas are reversible: ∆S =
∆S =
Q
or Q = T∆ S
T
c
h mc
h c
∆S = S f − Si = N S f − S i = N S f − S f + S f − S i
IG
IG
IG
P
RS
U
− ( −0.4 × 4.184 )V
P
T
W
300 U
R
= N S−8.368 − 8.314 ln
V = −23.26N J K
50 W
T
= N −2.4 × 4.184 − 8.314 ln
h − cS − S hr
IG
i
i
f
i
Q = T ∆S = ( 27315
. + 150 ) × 151.2 kmol × (− 2326
. ) = −14885
. MJ
W = ∆U − Q = N U f − U i − Q = N H f − H i − N Pf V f − PV
i i −Q
LM d H
MN
= N TC
IG
f
−Hf
TC
i+ c
IG
H f
−H
IG
i
0
h
−TC
dH − H i − Z RT
IG
i
i
TC
f
f
OP
PQ
+ Zi RTi − Q
Since process
is isothermal.
L304 .2(−4.5 − (−0.7)) × 4.184 − 8.314 OP + 14885. MJ
= 151.2 kmolM
N ×(273.15 + 150) × (0.765 − 0.94) Q
= 151.2 × 103 −48365
. + 615.7 J + 14885
. MJ = − 638.2 + 14885
. MJ
= 850.3 MJ
(c) Peng-Robinson E.O.S.
Using the program PR1 with T = 27315
. , P = 1 bar as the reference state, we obtain
T = 150° C , P = 50 bar
Z = 0.9202 ; V = 06475
.
×10−3 m3 mol ; H = 470248
. J mol ; S = −17.57 J mol K .
T = 150° C , P = 300 bar
Solutions to Chemical and Engineering Thermodynamics, 3e
Z = 0.7842 ; V = 0.9197 × 10−4 m3 mol ; H = −60.09 J mol ; S = −4124
. J mol .
N=
V
100 m 3
=
= 154 .44 kmol
V 0.6475 × 10 −3 m 3 mol
Q = TN ∆ S = (273.15 + 150 ) × 154 .44 × (− 4124
. − ( −17.57)) = −1546.9 MJ
W = ∆U − Q = ( H − PV ) f − ( H − PV )i − Q
= N ( H − PV ) f − ( H − PV )i − Q
LM−60.09 − 300 × 0.9197 × 10
= 154.44 M×10 J bar ⋅ m − 4702.48
MMN+50 × 0.6475 × 10 × 10
−4
5
3
−3
5
OP
PP × 10 + 1546.9 × 10
PQ
3
6
= 885.25 MJ
{Note that N, Q and W are close to values obtained from corresponding states.}
4.8
FG ∂ T IJ
H∂ PK
=
∂ (T , S ) ∂ (T , S ) ∂ ( P , T )
∂ ( S , T ) ∂ ( P, T )
=
⋅
=−
∂ ( P, S ) ∂ ( P , T ) ∂ P, S
∂ ( S , P ) ∂ ( T , P)
a f
−a∂ S ∂ Pf
a∂V ∂ Tf V αT
=
a∂ S ∂ Tf = C T = C
S
T
P
P
P
P
and
a fa
a fa
1 V ∂V dP
κS
=
κT
1 V ∂ V dP
f
f
S
=
T
∂ (V , S ) ∂ ( P, S ) ∂ (V , S ) ∂ ( P, T )
=
⋅
∂ (V , T ) ∂ ( P, T ) ∂ (V , T ) ∂ ( P, S )
FG IJ ⋅ FG ∂ T IJ
H K H∂S K
∂ ( S ,V ) ∂ (T , P )
∂S
=
⋅
=
∂ (T , V ) ∂ ( S , P )
∂T
4.9
(a)
V
=
P
CV T
C
⋅
= V
T CP CP
FG ∂ H IJ = ∂( H, T) = ∂(H ,T ) ⋅ ∂ (P ,T ) = FG ∂ H IJ FG ∂ P IJ
H ∂V K ∂ (V ,T ) ∂ (P ,T ) ∂(V , T) H ∂ P K H ∂V K
F ∂ P IJ ≠ 0 (except at the critical point)
Since G
H ∂V K
T
T
T
T
FG ∂ H IJ
H ∂V K
(b)
FG ∂ H IJ = 0
H ∂PK
FG ∂ S IJ = ∂(S, P ) = ∂(S, P) ⋅ ∂(T , P) = FG ∂ S IJ ⋅ FG ∂ T IJ
H ∂V K ∂ (V , P) ∂ (T, P ) ∂(V , P) H ∂T K H ∂V K
C 1 F dT I
C TV
C
F ∂S IJ
=
⋅ ⋅V G
=
=
⇒G
J
H
K
T V
dV
a1 V fa∂V ∂ Tf TV α H ∂V K
= 0 if
T
T
P
P
P
P
P
P
~ α−1
P
P
P
4.10 (a) We start by using the method of Jacobians to reduce the derivatives
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂T IJ
H ∂V K
a f
∂ (T , H ) ∂ (T , H ) ∂ (T , P ) ∂ T ,V
=
⋅
⋅
∂ (V , H ) ∂ (T , P ) ∂ (T ,V ) ∂ (V , H )
=
H
=−
=−
FG IJ a∂ P ∂V f
H K a∂ H ∂T f
∂ ( H , T ) ∂ ( P, T ) ∂ (V , T )
∂H
=−
∂ ( P, T ) ∂ ( H ,V ) ∂ (T ,V )
∂P
a∂ H ∂V f
ad H ∂Tf
T
T
V
T
V
Now from Table 4.1 we have that
FG ∂ H IJ
H ∂PK
=V −T
T
FG ∂V IJ
H ∂T K
FG ∂ H IJ
H ∂T K
and
P
LM
N
= CP + V − T
V
FG ∂V IJ
H ∂T K
P
OP FG ∂ P IJ
QH ∂ T K
V
alternatively, since H = U + PV
FG ∂ H IJ = FG ∂U IJ + FG ∂( PV )IJ
H ∂T K H ∂T K H ∂T K
V
V
= CV + V
V
F dP I
H dT K
V
Thus
FG ∂T IJ
H ∂V K
=
a
f
a
− ∂ P ∂V T V − T ∂V ∂ T
CV + V ∂ P ∂ T V
f
P
=
a
f
F ∂ P IJ FG ∂V IJ = − FG ∂ P IJ
Note: I have used G
H ∂V K H ∂ T K H ∂T K
H
T
FG ∂T IJ
H ∂V K
P
a
f + Ta∂ P ∂Tf
+ V a∂ P ∂ T f
− V ∂ P ∂V
CV
T
V
V
.
V
∂(T , S ) ∂( T , S ) ∂(V , T )
∂( S , T ) ∂(T ,V )
=
⋅
=−
⋅
∂(V , S ) ∂(V , T ) ∂(V , S )
∂(V , T ) ∂( S ,V )
=
S
FG ∂ S IJ FG ∂T IJ
H ∂V K H ∂ S K
=
T
=−
V
FG IJ
H K
T ∂P
CV ∂ T
V
(b) For the van der Waals fluid
FG ∂ P IJ
H ∂T K
=
V
FG IJ
H K
R
∂P
,
V −b
∂V
=
T
− RT
2a
+
(V − b )2 V 3
Thus
FG ∂T IJ
H ∂V K
=
n
s
2
2
− − RTV (V − b ) + 2 a V + RT (V − b )
CV + V R V − b
H
after simplification we obtain
FG ∂T IJ
H ∂V K
− 2 a(V − b ) − RTV b
2
=
H
2
CC (V − b )2 V 2 + R(V − b )V 3
Solutions to Chemical and Engineering Thermodynamics, 3e
and
FG ∂T IJ
H ∂V K
=−
S
RT
CV (V − b )
4.11 There are a number of ways to solve this problem. The method I use is a little unusual, but the
simplest that I know of. At the critical point all three roots of V are equal, and equal to V C .
a
Mathematically this can be expressed as V − V C
f
3
= 0 which, on expansion, becomes
V 3 − 3V C V 2 + 3V 2CV − V 3C = 0
(1)
compare this with
P=
RT
a
RT
a
−
=
− 2
V − b V (V + b ) + b (V − b) V − b V + 2 bV − b 2
which multiplying through by the denominators can be written as
3
V +V
2
F b − RT I + F −3b
H PK H
2
−
I FG
K H
IJ
K
2bRT a
RTb 2 ab
+
V + b3 +
−
=0
P
P
P
P
(2)
Comparing the coefficients of V in Eqns. (1) and (2) gives TC , PC
2
V : b−
(3)
2bRTC a
2
+
= 3V C
PC
PC
(4)
RTC 2 ab
3
b −
= −V C
PC
PC
(5)
V : − 3b 2 −
V 0: b3 +
RTC
= − 3V C
PC
From Eqn. (3)
PCb
−3 PCVC
−1=
= −3 ZC
RTC
RTC
For convenience, let y = 1 − 3ZC or ZC =
PCb
= 1 − 3ZC
RTC
1− y
. Then
3
PCb
=y
RTC
From Eqn. (4)
or
(6)
Solutions to Chemical and Engineering Thermodynamics, 3e
FG P b IJ − 2FG P b IJ + aP
H RT K H RT K aRT f
2
C
−3
C
C
C
C
⇒ −3 y 2 + 2 y +
C
aPC
=
aRT f
C
2
= 3 ZC2
2
3(1 − y)2
9
or expanding and rearranging
aPC
a RT f
C
2
=
c
h
1
10 y 2 + 4 y + 1
3
(7)
Finally from eqn. (5)
FG P b IJ + FG bP IJ − FG P b IJ FG aP IJ = − Z
H RT K H RT K H RT K H aRT f K
1
1
(1 − y)
y + y − y ⋅ c10 y + 4 y + 1h = −
3
27
3
C
C
3
2
C
C
C
C
C
2
C
2
3
C
3
2
or
64 y3 + 6y 2 + 12y − 1 = 0
(8)
This equation has the solution y = 0077796074
.
⇒ b = 0.077796074
a = 0.457235529
RTC
(from Eqn. (6))
PC
aRT f
C
2
(from Eqn. (7))
PC
1− y
= 0.307401309 .
3
Note that we have equated a and b to TC and PC only at the critical point. Therefore these functions
Also ZC =
could have other values away from the critical point. However, as we have equated functions of V ,
we have assumed a and b would only be functions of T. Therefore, to be completely general we
could have
aRT f αFG T IJ
HT K
P
RT F T I
b = 0.077796074
βG J
HT K
P
T
T
FTI
F TI
with αG J → 1 as
→ 1 and βG J → 1 as
→1.
HT K
T
HT K
T
C
a = 0.457235529
2
C
C
C
C
C
C
C
C
C
In fact, Peng and Robinson (and others) have set β = 1 at all temperatures and adjusted α a s a
function of temperature to give the correct vapor pressure (see chapter 5).
4.12 (also available as a Mathcad worksheet)
Solutions to Chemical and Engineering Thermodynamics, 3e
.
.
N1
dN
= N& 1 + N& 2 = 0 ⇒ N& 2 = − N& 1
dt
dU
Q&
= N& 1 H 1 + N& 2 H 2 + Q& = 0 ⇒
= H 2 − H1
dt
N1
M.B.
N2
.
E.B.
Q
Also, now using the program PR1 with T = 27315
. , P = 1 bar reference state we obtain
T = 100° C
P = 30 bar
Z = 0.9032
V = 09340
.
× 10−3 m3 mol
T = 150° C
P = 20 bar
Z = 0.9583
V = 01686
.
× 10−2 m3 mol
H = 3609.72 J mol
H = 679606
. J mol
S = −1584
. J mol K
S = −4.68 J mol
Q&
= 679606
. − 3609.72 = 318634
. J mol
N&
4.13 (also available as a Mathcad worksheet)
Since
process
is
adiabatic
and
reversible
∆S = 0
or
Si = S f ,
i.e.,
S (310 K, 14 bar ) = S (T = ?, 345 bar) . Using the program PR1 with the T = 27315
. K and P = 1
bar reference state we obtain T = 310 K , P = 14 bar , Z = 0.9733 , V = 01792
.
×10−2 m3 mol ,
H = 10908.3 J mol and S = 1575
. J mol K .
By trial and error (knowing P and S , guessing T) we obtain T = 34191
. K , P = 345 bar ,
Z = 0.9717 ,
V = 08007
.
× 10−4 m3 mol ,
⇒ Tf = 34191
. K.
H = 188609
. J mol ,
System = contents of compressor
dN
M.B.:
= 0 = N& 1 + N& 2 ⇒ N& 2 = − N& 1
dt
volume of
compressor
constant
adiabatic
0
dU
dV
= 0 = N& 1 H 1 + N& 2 H 2 + Q& 0 +W&s − P
E.B.:
dt
dt
W&
W&S = − N& 1 H1 + N& 2 H 2 or &S = H 2 − H 1 = 188609
. − 109083
. = 7952.6 J mol
N
4.14 (a)
FG P + a IJ (V − b) = RT ⇒ PV = V − a
H VK
RT V − b RTV
PV
F V − a IJ = 1
i)
lim
= lim G
RT
H V − b RTV K
PV
RV a U
I
ii) B = lim V F
H RT − 1K = lim V STV − b − RTV − 1VW
RV − (V − b) − a UV = lim RS bV − a UV = b − a
= lim V S
T (V − b) RTV W T (V − b) RT W RT
2
P→ 0
V →∞
V →∞
P →0
V →∞
V →∞
V →∞
V →∞
S = 1575
. J mol K
Solutions to Chemical and Engineering Thermodynamics, 3e
iii) C = lim V 2
P →0
V →∞
FG PV − 1 − B IJ = lim V RSbV − b(V − b) UV = lim b V = b
H RT V K
T V −b W V −b
2
V →∞
⇒ C = b2
(b) At the Boyle temperature: lim V
P→ 0
0 =b−
TB =
2
V →∞
F PV − 1I = 0 ⇒ B = 0
H RT K
a
a
9V c RTc
V
, TB =
but a =
, b = c (Eqns. 4.6-3a)
RTB
Rb
8
3
9 8V c RTc 27
=
Tc = 3.375Tc
RVc 3
8
a f
4.15 (a) From table: TB ~ 320 K , i.e., B TB = B(320 K ) = 0
The inversion temperature is the temperature at which
FG ∂ T IJ
H ∂P K
FG ∂ T IJ
H∂P K
F ∂V IJ
V − TG
H ∂T K
= 0= −
H
=
P
P
∂
∂T
P
LM
N
FG IJ OP
H KQ
1
∂V
V −T
CP
∂T
P
LM RT + BOP = R + dB
N P Q P dT
RT
RT
dB
dB
=
+B−
−T
= B− T
P
P
dT
dT
dB
= 0.
dt
Plot up B vs. T, obtain dB dT either graphically, or numerically from the tabular data. I find
Thus, T inv is the temperature at which B − T
T inv ~ 600 K .
Also,
dB dT
decreases
with
increasing
temperature
(i.e.,
dB dT ~ 456
. cm mol K at 87.5 K and 0.027 cm mol K at 650 K. Presumably it is negative
3
3
at even higher temperature!)
(b) Generally
µ=
FG ∂T IJ
H ∂ PK
=−
H
RS
T
FG IJ UV = − 1 RSB − T dB UV
H K W C T dT W
1
∂V
V −T
CP
∂T
P
P
Using the data in the table it is easy to show that for T < T inv , B − T
dB
< 0 ⇒ µ > 0 , while for
dT
dB
> 0⇒ µ< 0 .
dT
(c) Since Fig. 2.4-3 for nitrogen is an H-P plot is easiest to proceed as follows
T > T inv , B − T
−1
−a∂ H ∂ P f
FG ∂T IJ =
=
H ∂ P K a∂ H ∂ Tf a∂ P ∂ Hf a∂ H ∂ Tf
Since a∂ H ∂ T f = C is > 0 and less than ∞ [Except at a phase transition—see Chap. 5 and
Problem 5.1—however, µ has no meaning in the two-phase region], if adT dPf is to be zero,
then ad H dPf must equal zero. That is, an inversion point occurs when isotherms are parallel
T
H
P
P
T
P
P
H
T
to lines of constant H (vertical line). This occurs at low pressures (ideal gas region) and at high
Solutions to Chemical and Engineering Thermodynamics, 3e
pressures (nonideal gas region). See, for example, T = −200° C isotherm near 30 MPa (which
is off the figure).
To identify the inversion temperatures of nitrogen we can use Fig. 2.4-2b, a temperatureentropy diagram. From part a of this problem we note that at T inv
V = T inv
FG ∂V IJ
H∂TK
⇒ P = − T inv
P
FG ∂ P IJ
H ∂T K
P
Thus at each inversion temperatures T inv we can find a density (or pressure) for which this
equation is satisfied. Unfortunately, it is difficult to read the figure.
Solutions to Chemical and Engineering Thermodynamics, 3e
4.16
0.03 m3
18 kg ; V$ =
= 1.667 × 10 −3 m 3 kg . Using Fig. 2.4-2 we find P ≈ 91 bar
18 kg
Using the program PR1, with T = 42315
.
× 10−3 m3 kg , we find, by trial-and-error
. K and V$ = 1667
that P = 108.0 bar .
.
× 10−3 m3 mol ;
4.17 Using the program PR1 we find at 300°C and 35 bar; Z = 0.6853 ; V = 09330
H = 21,033 J mol = 21033
.
kJ mol and S = 7 .06 J mol K .
. bar
To use the principle of corresponding states we will assume the state of T = 16° C and P = 01
is an ideal gas state (i.e., don’t need corrections for nonideality at this condition). At 300°C and 35
bar we have
Tr =
300 + 273.15
= 1.0302
283.2 + 27315
.
35
Pr =
= 0.76754
45.6
We find Z = 0.71; H IG − H TC = 8.37 J mol ; SIG − S = 7113
.
J mol K .
d
i
From Appendix II
C*P = 22.243 + 0.05977T − 3.499 × 10 −5 T 2 + 7.464 × 10 −9 T 3
∆ H IG =
∆S
IG
=
573.15
z
z
CP* dT = 11910 J mol
289 .15
57315
.
CP*
289 .15
T
F 35 bar I = −20.386 J mol K
H 0.1 bar K
dT − R ln
Thus
H (T = 300° C, 35 bar )
= H (16° C, 0.1 bar) + ∆ H
IG
+ TC
FG H − H IJ
H T K
IG
C
300 ° C, 35 bar
= 0 + 11910 − (283.2 + 273.15)(8.37 ) = 7253 J mol
S (T = 300° C, 35 bar) = 0 − 20.386 − 7.113 = 27.499 J mol K
Finally PV = ZRT ; V =
0.71 × 8.314 × 10−5 × 573.15
= 0.9666 × 10 −3 m 3 mol .
35
4.18 Equation of state P (V − b ) = RT
(a)
FG ∂ P IJ
H ∂T K
=
V
R
P
∂V
= ;
V −b T
∂T
FG IJ
H K
=
P
R V −b
∂P
=
; and
P
T
∂V
FG IJ
H K
=−
T
P
V −b
Thus
CP = CV + T
FG ∂V IJ FG ∂ P IJ
H ∂ T K H ∂T K
P
= CV + T
V
for CP ( P, T ) = C*P (T ) , we must have that
R P
⋅ = CV + R
P T
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂ C IJ
H ∂P K
FG ∂ V IJ
H ∂T K
FG ∂ V IJ = 0
H∂T K
∂ F ∂V I
=
G J = ∂∂T
∂T H ∂ T K
2
= −T
P
T
2
2
P
2
P
P
P
R
= 0 ⇒ CP ( T , P) = CP* ( T )
P
P
Similarly, for CV (V , T ) = CV* (T ) , we must have that
FG ∂ P IJ
H ∂T K
2
=
2
V
∂
∂P
∂T V ∂T
FG IJ
H K
=
V
FG ∂ C IJ
H ∂V K
=T
V
T
FG ∂ P IJ
H∂T K
2
= 0.
2
V
∂
R
= 0 ⇒ CV T , V = CV* (T )
∂T V V −b
a f
(b) First case is clearly a Joule-Thomson expansion ⇒ H = constant
FG ∂T IJ
H ∂ PK
=−
H
LM
N
FG IJ OP = − 1 L RT + b − RT O = − b
H K Q C NM P
P QP
C
1
∂V
V −T
CP
∂T
P
P
P
Since CP is independent of P, integration can be done easily
T2
z
CP (T )dT = −b P2 − P1
a
T1
f
to proceed, we need to know how C p depends on T. If C p is independent of T we have
T2 = T1 −
b
P2 − P1
CP
a
f
(1)
Eqn. (1) also holds if CP is a function of T, but then it is the average heat capacity over the
temperature interval which appears in Eqn. (1).
The second expansion is at constant entropy (key words are reversible and adiabatic)
FG ∂ T IJ
H∂ PK
=−
S
a∂ S ∂ Pf
a∂ S ∂ Tf
T
=+
P
a∂V ∂ Tf
P
CP T
=
T R
⇒
CP P
T2
z
T1
P
CP
2
dT
dP
=R
T
P
P
z
1
If CP is independent of T, then
T2 = T1
FG P IJ
HPK
2
R CP
;
(2)
1
more complicated expression arises if CP = CP (T ) .
4.19 (also available as a Mathcad worksheet)
General:
mass balance: N1i = N1f + N 2f
energy balance:
N1i U 1i
=
N1f U 1f
+
N2f U 2f
(1)
(2)
Solutions to Chemical and Engineering Thermodynamics, 3e
state variable constraints: T1 f = T2 f = T f ; P1 f = P2 f = P f ; V1 = V2 ⇒ U 1f = U 2f
(a) Ideal gas solution
Eqns. of state: PV = NRT ; U = CV T − CP TR
P1i 2 P f
=
T1i
T ff
From M.B. get:
From E.B. get: T1i = T f ⇒ T1i = T f = 20°C ; P f =
1 i
P1 = 250 bar .
2
(b) Corresponding states solution
PV = ZNRT or PV = ZRT
IG
IG
IG
U ( T ) = U (T ) + U − U ( T ) = U (T ) + H − H
IG
= U IG
= U IG
d
i
d
(T ) + d H − H i + (1 − Z )( PV )
(T ) + d H − H i + (1 − Z ) RT
IG
i
− PV + ( PV )
IG
T
IG
IG
From the mass balance
P1i
500 × 107
Pf
Pf
5
=
=
1398
.
×
10
=
2
×
;
or
= 6.990 × 104
Z i T1i ( 20 + 27315
. ) × 1.22
ZfT f
ZfTf
where we have used
FG
H
Z i = Z Tr =
IJ
K
.
20 + 27315
5 × 107
. , Pr =
= 1538
= 10.77 = 1.22
190.7
4.64 × 106
From energy balance
N1i U 1i = N1f U 1f + N2f U 2f = N1f + N 2f U
c
h
f
= N1f + N 2f U i ⇒ U f = U i
c
h
where we have used the fact that U 1f = U 2f . Since T1 f = T2 f and P1 f = P2f . But
U 1 −U1 = 0 = U
f
i
IG
bT g − U bT g + c H − H h
IG
f
IG
i
T f ,Pf
+(1 − Z ) RT
T f ,P f
− H−H
c
IG
h
Ti , Pi
− (1 − Z ) RT T i , Pi
and
IG
IG
cT h − U cT h = C cT − T h = 27.25 J mol K cT
dH − H i = −18.0 × 190.7 = −3432.6 J mol
U
f
i
f
i
f
v
− 29315
. K
h
IG
T i , Pi
(1 − Z ) RT T i , P i = − 0.22 × 8.314 × 293.15 = −536.2 J mol
IG
⇒ 0 = 27.25Tf + 293.15{−27.25 + 0.22 × 8.314} + 18.0 × 190.7 + H − H
d
+ (1 − Z )RT
i
T f ,P f
T f ,Pf
27.25T f + H − H IG
d
i
T f ,P f
+ (1 − Z )RT
T f ,Pf
= 4,021
(1)
Solutions to Chemical and Engineering Thermodynamics, 3e
To be solved along with
Pf
= 6.99 × 104
TfZf
(2)
I found the solution by making a guess for T f , using eqn. (2) and Fig. 4.6-3 to find P f , by trial
and error. Then, guessed T f and computed P f were tested in eqn. (1). Solution found:
T f ~ 237 K = −3615
. ° C ; P f ~ 1011
. × 107 Pa = 1011
. bar .
(c) The van der Waals gas
1 f
i
We know that: U i = U f and V = V . To evaluate the final temperature we start from
2
F ∂T IJ
dT = G
H ∂V K
U
F ∂ T IJ
dV + G
H ∂U K
V
dU ⇒ T −
f
V
T i along
path of
const. U
=
f
z FGH
Vi
∂T
∂V
IJ
K
dV
U
but
FG ∂T IJ
H ∂V K
a∂U ∂V f
a∂U ∂T f
=−
=−
T
U
T ∂ P ∂T
a
f
V
+P
CV
V
=−
a
CV V 2
Now, by Eqn. (4.2-36)
FG ∂ C IJ = TFG ∂ P IJ for vdW eos P = RT − a
H ∂V K H ∂T K
V −b V
FG ∂ P IJ = R ; FG ∂ P IJ = 0 ⇒ C is independent of volume ⇒ C
H ∂T K V − b H ∂T K
2
V
2
T
2
V
2
V
2
V
V
V
but C*V = CP* − R = 35565
. − 8.314 = 2725
. J mol K
Vf
⇒T −T = −
f
i
z
V
i
a
dV = −
CV
CVV 2
=−
a
a
2CVV
Thus the first step is to find V i .
i
V
f
z
V
i
dV
V2
since V i =
=+
1 f
V .
2
FG
H
1
1
a
−
CV V f V i
IJ
K
= CV*
Solutions to Chemical and Engineering Thermodynamics, 3e
F a I
GG P + JJ dV − bi = RT
H dV i K
i
i
i 2
⇒ V i = 6.678 × 10 −5 m 3 mol ; V
.
= 1336
× 10−4 m 3 mol .
f
− 0.2283 Pa ⋅ m6 mol2
= −62.73 K
2 × 27.25 J mol K × 6.678 × 10 −5 m3 mol
T f − Ti =
a
f c
h
. − 62.73 = 230.42 K = −42.73° C
⇒ T f = 29315
Pf =
RT
dV
f
f
−b
−
a
=
f 2
i cV h
8.314 × 230 .42
0.2283
−
−4
−5
1.336 × 10 − 4.269 × 10
× 10 −4
.
1336
c
2
h
Pf = 8.286 × 106 Pa = 82.86 bar
(d) Here we will use the program PR1. Using the 273.15 K and 1 bar reference state we find that at
.
, V i = 05365
.
× 10− 4 m3 mol , H i = −328185
. J mol and
the initial conditions Z = 11005
Si = −5912
. J mol K . Therefore
U i = Hi − Pi V i = −328185
. − 500 × 05365
.
×10−4 × 105 = −5964.35 J mol
Now since U f = U i = −5964.35 J mol and
V f = 2V i = 1073
. × 10−4 m3 mol .
We must, by trial-and-error, find the temperature and pressure of the state having these
properties. I find the following as the solution Tf = 230.9 K ; P ~ 99 bar (for which
V = 01075
.
m3 mol and U = −5968.4 J mol ). To summarize, we have the following answers
for the different parts of the problem:
Ideal gas
Corresponding states
van der Waals
Peng-Robinson
Pf
Tf
250 bar
101.1 bar
82.86 bar
99 bar
293.15 K
237 K
230.42 K
230.9 K
Once again, the ideal gas solution is seriously in error.
4.20 Mass balance (system = both tanks): N1i = N1f + N 2f
energy balance (system = both tanks): N1i U 1i = N1f U 1f + N2f U 2f
entropy balance (system = portion of initial contents of tank 1, also in there finally): S i1 = S 1f
Also, P1 f = P2 f = P f ; N1i =
V1
i
V1
(a) Ideal gas solution: obtain
; N1f =
P1i
T1i
=
P1 f
T1 f
V1
f
V1
+
and N2f =
P2f
T2 f
V2
f
V2
from mass balance and
P1i = P1 f + P2 f = 2 P f ⇒ P f = 250 bar = 25
. × 107 Pa from energy balance
Solutions to Chemical and Engineering Thermodynamics, 3e
T1 =
f
R CP
FG P IJ
HPK
f
T1i
FH 1IK
2
⇒ T1 f = (20 + 273.15)
1
i
1
8.314 35.565
= 249 .3 K
= −23.9° C from entropy balance
and
1
T2
=
f
2
T1i
−
1
T1
f
⇒ T2 f = 355.9 K = 82.7° C
Also
N1f
P1 f V1 RT1i 2.5 × 10 7 293.15
=
⋅
=
⋅
= 0.588
N1i
RT1 f P1iV1
249 .3 5.00 × 107
and
FG IJ = 0.412
H K
N2f
N1f
=
1
−
N1i
N1i
(b) Corresponding States Solution:
Initial
Tr =
conditions
29315
.
= 1538
.
;
190 .7
Pr =
5 × 107
= 10.77 ;
4.64 × 106
H IG − H
= 18.0 J mol K ; SIG − S = 96
. J mol K .
TC
Mass balance:
RS
T
P1i
1
1
= Pf
+ f f
i i
f f
Z1 T1
Z1 T1
Z2 T2
UV = 5.0 × 10 = 1.398 × 10
. × 293.15
W 122
7
5
(1)
Entropy balance:
S 1f − S 1i = 0 = S 1 − S 1IG
d
f
i + dS
IG, f
1
i
− S IG,
− S 1 − S IG
1
1
i d
i
i
or
IG f
1
dS − S i
1
+ CP ln
T1 f
Pf
− R ln
= −9 .6
29315
5.0 × 10 7
.
(2)
Energy balance:
N1i U 1i = N1f U 1f + N2f U 2f but N1i = N1f + N 2f ⇒ N1f U 1f − U 1i + N2f U 2f − U i2 = 0
d
or
i
d
i
⇒ Z = 1.22 ;
Solutions to Chemical and Engineering Thermodynamics, 3e
P1 f V1
Z1f RT1 f
f
2
ndH − P V i − dH − P V is + ZP RTV ndH − P V i − dH − P V is = 0
1
ndH − H i + dH − H i − dH − H i − Z RT + Z RT s
Z T
1
+
ndH − H i + d H − H i − d H − H i − Z RT + Z RT s = 0
Z T
f
1
1
i
1
i
1
f , IG
1
f , IG
2
f
2
f
f
2 2
i
1
f , IG
1
f
1
f
f
1 1
f
1
f
2
f
2
i , IG
1
f , IG
2
f
f
2
f
2
f
2
i
1
i
1
i
1
2
i , IG
1
i
1
i , IG
1
i
1
i , IG
1
f
1
f
i
1
1
f
2
f
2
i
1
i
1
i
1
Substituting in the known values gives
1
f , IG
1
f
1
. cT − 29315
. h − 8.314 Z T + 6,406.5s
ndH − H i + 35565
1
+
. cT − 29315
. h − 8.314 Z T + 6 ,406 .5s = 0 (3)
ndH − H i + 35565
Z T
Z1f T1 f
f
f
2 2
f
2
f
f f
1 1
1
f , IG
2
f
2
f f
2 2
Eqns. (1-3) now must be solved. One possible procedure is
i) Guess P f
ii) Use Eqn. (2) to find T1 f
iii) Use Eqn. (1) to find T2 f
iv) Use Eqn. (3), together with T1 f and T2 f to see if guessed P f is correct. If not, go back to
step i.
After many iterations, I found the following solution
P f = 9787
. bar ; T1 f = 2216
. K;
T2 f = 259.4 K ; N1f N1i = 0.645 ; N 2f N1i = 0.355 .
(c) Peng-Robinson equation of state
Here we use the equations
N1i = N1f + N 2f
N1i U 1i
=
+
N1f U 1f
with U = H − PV
N2f U 2f
S i1
=S
f
(4)
(5)
(6)
P1 f = P2 f = P f
and N1i = V1 V 1i ; N1f = V1 V 1f ; N2f = V2 V 2f = V1 V 2f since V1 = V2 (value of V1 cancels out
of problem, so any convenient value may be used). Procedure I used to solve problem was as
follows. From PR1 we know V i1 ⇒ N1i and S i1 given initial conditions. Then
c
h
1.
Guess value of T1 , find P1 = P that satisfies S 1f = S 1i
2.
Use T1 f , P f and V 1f to get N1f ; then N 2f = N1i − N1f so V 2f is known.
3.
From P f and V 2f find (trial-and-error with PR1) T2 f
f
f
f
4. See if eqn. (5) energy balance is satisfied; if not go back to step 1. After a number of
iterations I find
Pf =103.6 bar ; T1 f = 222.3 K ; T2 f = 2555
. K ; N1f N1i = 0.619 ;
N2f N1i = 0381
. .
Solutions to Chemical and Engineering Thermodynamics, 3e
Summary
Pf
T1 f
ideal gas
(part a)
250 bar
249.3 K
Corresponding
states (part b)
97.87
221.6 K
P-R E.O.S.
(part c)
103.6
222.3 K
T2 f
355.9 K
259.4 K
255.5 K
N1f N1i
0.588
0.645
0.619
N2f N1i
0.412
0.355
0.381
Clearly, the ideal gas assumption is seriously in error!
4.21 System = contents of compressor. This is a steady-state, open constant volume system.
dN
= 0 = N& 1 + N& 2
mass balance:
dt
0
dU
dV
= 0 = N& 1 H 1 + N& 2 H 2 + Q& + W& s − P
energy balance:
dt
dt
⇒ 0 = N& 1 H1 − H2 + Q& + W&s
a
entropy balance:
f
dS
Q&
= 0 = N& 1 S 1 + N& 2 S 2 + +
dt
T
Q&
= N& 1 S 1 − S 2 +
T
.
S gen
a
f
0
Thus,
Q& = −TN&1 S 1 − S 2
a
f
Q&
= Q = T S 2 − S1
N& 1
a
f
and
W& S + Q&
= W + Q = H 2 − H1
N& 1
(a) Corresponding states solution
IG
Q = T S 2 − S 1 = T S 2 − S IG
+ S IG
− S 1 − S 1IG
2
2 − S1
a f nd
R
= T SdS − S i − dS
T
2
Now Tr =
IG
2
1
− S 1IG
i d
i d
i − R ln PP UVW
is
2
1
.
37315
1
50
= 0.92 ; Pr , 1 =
= 0.443 . Thus
~ 0.009 ; Pr , 2 =
405.6
112 .8
112.8
Q = − RT ln
RSd
T
50
IG
+ T S2 − S 2
1
i
Pr ,2 = 0.444
Tr = 0.92
IG
− S 1 − S1
d
i
Pr ,1 =0 .009
Tr = 0.92
UV
W
= − 8.314 × 37315
. × ln 50 + 37315
. ( −5.23 − 0) = −14,088.1 J mol
and
W + Q = H 2 − H 1 = ( H 2 − H 2IG ) + ( H 2IG − H 1IG ) − ( H 1 − H 1IG )
0 since
T = constant
Solutions to Chemical and Engineering Thermodynamics, 3e
Thus
W + Q = TC
R|dH
S|
T
2
− H 2IG
TC
i − dH
1
− H 1IG
TC
= − 25472
. J mol
i U|V = 405.6 × (−6.28 + 0)
|W
W = − Q − 25472
. = 11,540.9 J mol
(b) Clausius gas
P (V − b ) = RT ; V =
RT
∂V
+b ;
P
∂T
FG IJ
H K
=
P
R
P
Thus
P2
P2
P2
z FGH ∂∂ SP IJK dP = − z FGH ∂∂VT IJK dP = − R z 1P dP = − R ln PP
F 50I = −12,136.5 J mol
Q = T∆S = − RT ln
H 1K
F ∂ H IJ dP = z LMV − TFG ∂V IJ OPdP = z L RT + b − RT OdP
∆H = z G
P PQ
H ∂PK
N H ∂T K Q MN P
= z bdP = b a P − P f = 182.8 J mol
∆S =
2
T
P1
P
P1
P2
P2
T
P1
1
P1
P2
P
P1
P1
P2
2
1
P1
So W + Q = 182 .8 J mol and
W = −Q + 182 .8 = 12,136.5 + 182.8 = 12,319 .3 J mol
(c) Peng-Robinson equation of state
. and P = 1 bar ideal gas reference state) that
Using the program PR1 we find (for T = 27315
100°C, 1 bar
−1
0.3089 × 10
V
100°C, 50 bar
m mol
0.4598 × 10−3
3
H
3619.67
J/mol
1139.65
S
11.32
J/mol K
–25.94
Note, from PR1, the vapor pressure of NH 3 at 100°C is 62.58 bar. Therefore, use vapor
solution to P-R equation.
. J mol
Then Q = T S 2 − S1 = −13,9036
a
f
W + Q = H 2 − H 1 = 1139.65 − 3619.67 = −24800
. J mol
and W = −Q − 24800
. = 11,423.6 J mol .
Solutions to Chemical and Engineering Thermodynamics, 3e
4.22 (also available as a Mathcad)
Considering the gas that is in the tank finally as the system, this is a closed system undergoing a
reversible, adiabatic expansion. Therefore Si = S f .
FG IJ dP but with P (V − b) = RT or V = RT + b . Then FG ∂V IJ
P
H K
H ∂T K
FG ∂ C IJ = −T FG ∂ V IJ = − T ∂ R = 0 ⇒ C is independent of pressure.
H ∂ P K H∂T K
∂T P
(a) d S =
CP
∂V
dT −
T
∂T
P
=
P
R
; also
P
2
P
P
2
T
Therefore
P
P
CP = CP* . Thus
Tf
0 = ∆S =
z
Ti
Pf
CP*
1
dT − R
dP
T
P
P
z
i
(1)
This has the solution Pf = 1310
. bar . Now to find the initial and final molar volumes we use
V =
RT
+b
P
V i = 0.000709 m3 mol
V
So that
Nf
Ni
=
f
= 0.00197 m 3 mol
Vi
= 0.3595 (or 35.95%)
Vf
(b) Corresponding states
IG
IG
0 = S f − S i = S f − S IG
− S i − S iIG
f + S f − Si
d
i d
i d
i
Initial state
400
= 1.3149
S IG − S = 0.49
304.2
⇒
50
Pr =
= 0.67787 Z = 0.906
73.76
Pf
IG
S IG
= −12.4009 − 8.314 ln
f − Si
50
Tr =
d
i
(As given by eqn. (1) above. Why?)
Guess for final state (use P from part a), then iterate. Final solution is Pf =1152
. bar for which
.
, SIG − S ≅ 029
. and Z f = 0.939 .
Pr = 0156
Nf
Ni
=
cP Z RT h = P
a P Z RT f Z T
f
f
i
i
f
i
f
f
⋅
f
.
400 0.906
Zi Ti 1152
=
×
×
= 0.2964 (or 29.64% )
50
300 0.939
Pi
(c) Peng-Robinson equation of state
Use program PR1 with given heat capacity constants to find a pressure at 300 K which has the
same entropy as the state T = 400 K , P = 50 bar .
By trial-and-error we find that
P = 1337
. bar (somewhat higher than the previous cases). Also, V i = 0.5982 × 10−3 m3 mol
and
Solutions to Chemical and Engineering Thermodynamics, 3e
V
f
Nf
= 0.175 × 10− 2 m3 mol ⇒
Ni
=
Vi
= 0.3416 (or 34.16%)
Vf
4.23 There are two obvious ways to proceed.
1) retain T and P as the independent variables since we have a program, PR1 that calculates
V (T , P) , H ( T , P) and S (T , P) . We can then use
U ( T , P) = H (T , P) − PV = H ( T , P) − ZRT (where Z = Z (T , P ) )
(1)
G (T , P) = H (T , P) − T S ( T , P)
(2)
and
A(T , P ) = G − PV = U − T S = H − PV − T S
Now we will write
T da
H ( T , P) = H IG (T ) + RT ( Z − 1) +
2
and
T
S (T , P ) = S IG (T , P) + R ln( Z − B ) +
a
= H − ZRT − TS
dT − a
f
(3)
X
(4)
ada dTf X
(5)
2b
2 2b
where for convenience, I have used
X = ln
LM Z + d1 + 2 iB OP
MN Z + d1 − 2 iB PQ
Then we find
U ( T , P) = U IG ( T ) +
T da dT − a
a
f
2 2b
X where U IG = HIG − RT
G (T , P) = G IG ( T , P) + RT ( Z − 1) − ln( Z − B ) −
a
X
(6)
(7)
2 2b
and
a
(8)
X
2 2b
Thus we can either use eqns. (1 to 3) and previously calculated values Z, H and S , or modify
PR1 to use Eqns. (6-8) instead of Eqns. (4 and 5).
The second alternative is to take T and V as the independent variables and start from
A(T , P ) = A IG( T , P) − RT ln( Z − B) −
2)
LM FG ∂ P IJ
MN H ∂ T K
dU = CV dT + T
OP
PQ
− P dV and d S =
V
∂P
CV
dT +
∂T
T
to get
T, V
S−S
IG
=
z
T, V =∞
and
T,V
U −U
IG
=
z
T, V =∞
LMF ∂ P I − R OPdV
MNGH ∂T JK V PQ
LMTF ∂ P I − P OPdV
MN GH ∂ T JK PQ
V
V
FG IJ
H K
dV
V
Solutions to Chemical and Engineering Thermodynamics, 3e
Then put in the Peng-Robinson equation of state and from build up a procedure to calculate S ,
H , U , P, A and G with T and V as the independent variables. We will not follow this
alternative further.
4.24 We will do these calculations using
G = H − TS and A = G − PV = H − PV − TS
As an example, consider the T = 0° C isotherm P = 1 bar , H = −742 .14 J mol
G = −742 .14 − 27315
. × ( −2 .59) = − 34.68 J mol ,
S = −2.59 J mol K
V = 22.6800 m3 kmol
A = −3468
. − 1 bar × 0.02268 m3 mol × 105 J = −2302.7 J mol
P = 5 bar
H = −786.05 J mol
S = −16.09 J mol K
⇒
G = 36089
. J mol
V = 0.004513 m3 mol
P = 10 bar
A = 13524
. J mol
H = −840.75 J mol
S = − 2199
. J mol K
⇒
V = 0.002242 m 3 mol
P = 20 bar
H = − 949.56 J mol
S = −28.06 J mol K
⇒
V = 0.001107 m3 mol
P = 40 bar
H = − 116397
. J mol
S = −34.40 J mol K
⇒
V = 0.0005409 m3 mol
P = 60 bar
H = − 137264
. J mol
S = −38.34 J mol K
⇒
V = 0.0003532 m3 mol
P = 80 bar
H = −157376
. J mol
S = −4129
. J mol K
⇒
V = 0.0002602 m 3 mol
P = 100 bar
H = −176561
. J mol
S = −43.69 J mol K
V = 0.0002052 m3 mol
⇒
G = 51658
. J mol
A = 29238
. J mol
G = 67150
. J mol
A = 45010
. J mol
G = 8232.4 J mol
A = 60688
. J mol
G = 9099.9 J mol
A = 69789
. J mol
G = 9704.6 J mol
A = 76230
. J mol
G = 10,168 .3 J mol
A = 8116.3 J mol
Solutions to Chemical and Engineering Thermodynamics, 3e
Similarly, G and A at other points could be computed, though this will not be done here.
ρ=
4.25 (a)
1
V
FG 1 IJ = −V
HV K
∂ρ= ∂
−2
FG ∂ P IJ
H ∂ ρK
∂V ⇒
= −V 2
S
FG ∂ P IJ
H ∂V K
=+
S
V 2 ∂ S ∂V P
∂S ∂ P V
a
a
f
f
by eqn. (4.1-6a)
Now
FG IJ dV ⇒ FG ∂ S IJ = C FG ∂ T IJ
H K
H∂ PK T H∂ PK
C
F ∂V IJ dP ⇒ FG ∂ S IJ = C FG ∂T IJ
dT + G
dS =
H∂TK
H ∂V K T H ∂V K
T
F ∂ P IJ = V C FG ∂ T IJ ⋅ T FG ∂ P IJ = V γ FG ∂ T IJ FG ∂ P IJ
⇒G
H ∂ ρK
H ∂V K H ∂T K
T H ∂V K C H ∂ T K
dS =
CV
∂P
dT +
T
∂T
V
V
V
V
P
P
P
2
P
P
2
P
S
P
V
V
P
= −γV
2
V
FG ∂ P IJ
H ∂V K
( by eqn. ( 4.1- 6a))
Thus vS = −γV 2
FG ∂ P IJ
H ∂V K
= γV 2
T
FG ∂ T IJ FG ∂ P IJ
H ∂V K H ∂T K
P
.
V
CP
C +R
R
= 1+
for the ideal gas CP = CV + R ⇒ γ = V
CV
CV
CV
For the Clausius Gas
(b) γ =
γ=
CP CV + T ∂ V ∂ T
=
CV
CV
with P (V − b ) = RT
a
f a∂ P ∂T f
P
V
= 1+
T ∂V
CV ∂ T
FG IJ FG ∂ P IJ
H K H ∂T K
P
V
T
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂V IJ
H ∂T K
=
P
∂P
R V −b
=
and
∂T
P
T
FG IJ
H K
=
V
R
V −b
Thus
γ = 1+
T V −b
R
R
⋅
⋅
= 1+
CV
T V −b
CV
To show that Cv ≠ Cv (V ) we start from eqn. (4.2-35)
FG ∂ C IJ
H ∂V K
FG ∂ P IJ but FG ∂ P IJ = R ; FG ∂ P IJ = 0 for ideal gas
H ∂T K H ∂ T K V H ∂T K
F ∂ P IJ = R ; FG ∂ P IJ = 0 for Clausius gas]
and G
H ∂ T K V − b H ∂T K
F ∂ C IJ = 0 for the ideal and Clausius Gases
⇒G
H ∂V K
2
=T
V
T
2
2
2
V
V
V
2
2
V
V
V
T
(c) vS ( ideal gas ) = γV
2
T P
= γPV = γRT
V T
T
R
V
=
γRT
V −b V − b V −b
vS ( Clausius gas) = γV 2
=
V
vS (ideal gas )
V −b
at same T and V
4.26 Preliminaries
Pressure = outward force per unit area exerted by gas
Force = tensile force exerted on fiber — at mechanical equilibrium fiber exerts an equal and opposite
inward force
⇒ In all thermodynamic relations replace P by − F A and V by LA, and they will be applicable to
fiber.
In particular, in place of S = S (T ,V ) and dS =
and dS =
FG ∂ S IJ
H ∂T K
dT +
L
FG ∂ S IJ
H ∂T K
dT +
V
FG ∂ S IJ
H ∂V K
FG ∂S IJ dL . Also
H ∂ LK
FG ∂ S IJ = C ⇒ FG ∂ S IJ = C
H ∂T K T H ∂T K T
FG ∂ S IJ = FG ∂ P IJ ⇒ FG ∂ S IJ = −FG ∂ F IJ
H ∂ T K H ∂T K H ∂ L K H ∂ T K
T
V
V
T
L
L
V
T
L
P
F
1
1
dS = dU + dV ⇒ dS = dU = dL
T
T
T
T
dV . We will use S = S T , L
T
a f
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂ S IJ
H ∂T K
(a) From the above dS =
FG ∂ S IJ
H ∂ LK
dT +
L
dL =
T
CL
∂S
dT +
T
∂L
FG IJ dL
H K
and the analog of the
T
Maxwell relation
FG ∂ S IJ = FG ∂ P IJ
H ∂V K H ∂ T K
T
1 ∂S
A ∂L
FG IJ
H K
⇒
V
=−
T
1 ∂F
A ∂T
FG IJ
H K
L
we get
dS =
CL
∂F
dT −
T
∂T
FG IJ
H K
dL =
L
CL
dT − γ L − L0 dL
T
a
f
(b) dU = TdS − PdV ⇒ dU = TdS + FdL
= CL dT − γT L − L0 dL + γT L − L0 dL = CL dT
a
f
a
f
(Note: This is analog of ideal gas expression U = U (T ) or dU = Cv dT )
(c) dS =
CL
dT − γ L − L0 dL
T
a f
RC
U
⇒ S ( L, T ) − S a L , T f = z S dT − γ a L − L fdLV
TT
W
Choosing the path a L , T f → a L , T f → ( L, T ) yields
L, T
L
0
0
0
L0 ,T0
0
0
0
S ( L, T ) − S L0 , T0 =
a
f
T , L0
z
T0 , L0
= αln
T ,L
α + βT
dT − γ
L − L0 dL
T
T ,L
za
f
0
γ
T
+ β T − T0 −
L − L0
T0
2
a
f a
f
2
(d) A reversible (slow), adiabatic expansion ⇒ S L f , Tf − S Li , Ti = 0
h a f
c
0 = S L f , T f − S L0 , T0 − S Li , Ti − S L0 , T0
h a fr k a f a fp
γ
FT I
= αln G J + βcT − T h − c L − L h − a L − L f
2
HT K
mc
2
f
f
i
f
0
2
i
0
i
Need to solve this transcendental equation to find Tf .
(e) dU = CL dT ⇒
FS = −T
FG ∂ S IJ
H ∂LK
FG ∂U IJ
H ∂ LK
= 0 ⇒ FU = 0
T
= − T −γ L − L0
T
k a
fp = γTaL − L f = −T FGH ∂∂FT IJK
0
L
Solutions to Chemical and Engineering Thermodynamics, 3e
4.27 (a) dU = TdS − PdV + GdN ⇒
FG ∂U IJ
H ∂S K
=T ;
V,N
FG ∂U IJ
H ∂V K
= − P ; and
S ,N
FG ∂U IJ
H∂ NK
=G
S ,V
Now equating mixed second derivatives
∂
∂V
∂
∂N
S ,N
S ,V
FG ∂U IJ
H ∂S K
FG ∂U IJ
H ∂S K
=
V ,N
=
V,N
∂
∂ S V ,N
FG IJ
H K
FG ∂U IJ
H∂ N K
∂
∂V
FG ∂U IJ
H∂ NK
∂
∂U
∂ S V , N ∂V
FG ∂T IJ
H ∂V K
F ∂T IJ
⇒G
H∂ NK
⇒
S, N
S ,V
FG ∂ P IJ
H ∂S K
F ∂G IJ
= −G
H ∂S K
=−
S, N
S ,V
(1)
V, N
(2)
V,N
and
∂
∂N
S ,V
FG ∂U IJ
H ∂V K
=
S ,N
S, N
⇒−
S ,V
FG ∂ P IJ
H∂ NK
=
S ,V
F ∂G I
H ∂V K
(3)
S ,N
(b), (c), and (d) are derived in similar fashion.
4.28 (also available as a Mathcad worksheet)
(a) The procedure that will be used is to first identify the temperature at which µ = 0 , and then
show that µ < 0 at larger temperatures, and µ > 0 at lower temperatures. The starting point is,
from Sec. 4.2
µ=
FG ∂T IJ
H ∂ PK
=−
H
V
(1 − Tα)
CP
where, from Illustration 4.2-4, for the van der Waals gas,
α−1 =
R|
S| a
T
TV
2a
V
1
(V − b ) and µ= −
−
1−
V − b RV 2
CP
V V − b − 2a (V − b ) RTV 2
f
U|
V|
W
Simplifying yields
µ= −
a
a
f
f
2
V b V − b − 2a (V − b ) RTV
CP V V − b − 2a (V − b ) RTV 2
2
V RTb (V − b ) − 2 a (V − b) V
=−
CP RTV (V − b ) − 2 a (V − b) V 2
(1)
Now for µ to be zero, either the numerator must be zero, or the denominator infinity. Only the
former is possible. Thus,
T inv =
2 a (V − b) (V − b ) 2 a(V − b )2
⋅
=
the desired expression
b
RV 2
RV 2b
to determine the sign of the Joule-Thomson coefficient in the vicinty of the inversion
temperature, we will replace T in eqn. (1) by T inv + δ , where δ may be either positive or
negative. The result is
Solutions to Chemical and Engineering Thermodynamics, 3e
µ= −
V
Rb (V − b) δ
CP RV (V − b ) δ + 2a (V − b )2 bV 2
It is easily shown that the denominator is always positive.
Thus,
µ is proportional to − δ ⇒ if T > T inv , so that δ > 0 , µ < 0 . Alternatively, if T < T inv , δ < 0
and µ > 0 .
9
V
(b) Using a = V C RTC and b = C
8
3
a
d
f
F I a3V − 1f
H K V
i
9 F V I a 3V − 1f
3a 3V − 1f
=
⋅3 =
=T
4H 3 K V V
4V
2
V −VC 3
V
9
27
T inv = 2 ⋅ V C RTC
=
TC C
2
8
4
3
RV V C 3
T inv
⇒
TC
r
2
r
2
r
2
2
2
C
2
(2)
2
r
2
C
inv
r
2
r
a f
(c) Expression above gives Trinv = Trinv Vr ; what we want is Trinv as a function of Pr . Thus look at
FG P + 3 IJ a3V − 1f = 8T ⇒ P = 8T − 3
H VK
3V − 1 V
r
2
r
r
r
r
r
(3)
2
r
r
Choose Vr as independent variable; use Eqn. (2) to get Trinv , and use Eqn. (3) to get Pr .
Results are tabulated and plotted below.
Vr
Trinv
T inv (K)
Pr
P (bar)
T inv (° C)
0.5
0.625
0.75
1.0
1.25
1.50
1.75
2.0
0.75
1.455
2.048
3.0
3.63
4.083
4.422
4.688
94.65
183.62
262.5
378.6
458.1
515.3
558.1
591.6
0
5.622
7.977
9.0
8.64
8.0
7.344
6.751
0
190.8
270.7
305.5
293.2
271.5
249.3
229.1
–178.55
–89.6
–10.7
105.4
184.9
242.1
284.9
318.4
Solutions to Chemical and Engineering Thermodynamics, 3e
4.29 (also available as a Mathcad worksheet)
4.29
Take the gas to be nitrogen
Enter constants
8.314 .
R
3
Pa .m
.
mole K
Ti
273.15 .K
System is the gas to be compressed. System is closed and isothermal (constant temperature).
Energy balance is U(final) - U(initial) = Q + W = Q - PdV
Initial conditions are 0 C and 1 bar, final conditions are 0 C and 100 bar
a) Ideal gas
R .T
P1( V , T )
Vi
Vf
d
T.
P1( V , T )
dT
V
R .T i
100000 .Pa
Vi =
R .T i
10000000 .Pa
Vf =
0
3
1
0.0227 m mole
2.271 10
4
Vf
P1( V , T i )d V
W
P1( V , T )
Consequently, as we already knew,
the internal energy of an ideal gas is not
a function of pressure or volume, only
temperature.
Initial and final volumes
3
1
m mole
W=
4
1
1.0458 10 mole
Q=
4
1
1.0458 10 mole
joule
U(final)-U(initial)=0, so Q = -W
Vi
Q
W
b) Virial equation of state
B
10.3 .10
6 . m3
mole
C
1.517 .10
9 . m6
2
mole
joule
Solutions to Chemical and Engineering Thermodynamics, 3e
R .T .
P2( V , T )
1
V
T.
d
P2( V , T )
B
C
V
2
V
P2( V , T )
Consequently, the internal energy of this
gas is also not a function of pressure or
volume, only temperature.However, if the
virial coefficient were a function of
temperature (which is the actual case)
then the internal energy of this gas would
be a function of temperature.
0
dT
Guesses for Initial and final volumes
Given
100000 .Pa
Given
10000000 .Pa
Vi
R .T i
1000000 .Pa
find( Vi )
Vi =
3
1
0.0227 m mole
Vf
find( Vf)
Vf =
2.2353 10
P2( Vf, T i )
P2( V , T i )d V
Vf
10000 .Pa
Vi
P2( Vi , T i )
Vf
W
R .T i
W=
4
1
1.0424 10 mole
4
3
1
m mole
joule
Vi
U(final)-U(initial)=0, so Q = -W
Q
Q=
W
4
1
1.0424 10 mole
joule
c) The van der Waals gas
0.1368 .
a
d
6
b
2
mole
R .T
P3( V , T )
T.
Pa .m
V
5 . m3
mole
a
b
P3( V , T )
3.864 .10
2
V
P3( V , T )
.1368 .Pa .
dT
m
=a/V^2
6
Guesses for Initial and final volumes
Given
100000 .Pa
Given
10000000 .Pa
P3( Vi , T i )
P3( Vf, T i )
Vf
P3( V , T i )d V
W
In this case the internal energy is a
function of volume (or pressure)
2 2
mole .V
W=
Vi
R .T i
Vf
10000 .Pa
find( Vi )
Vi =
3
1
0.0227 m mole
Vf
find( Vf)
Vf =
2.132 10
4
1
1.0414 10 mole
Vf
a
2
V
dV
∆U =
joule
635.618 mole
1
joule
Vi
Q
∆U
W
Q=
1000000 .Pa
Vi
Vi
∆U
R .T i
4
1
1.0424 10 mole
joule
4
3
1
m mole
Solutions to Chemical and Engineering Thermodynamics, 3e
d) The Peng-Robinson fluid
0.04
w
b
126.2 .K
0.07780 .R .
b=
5.
Pa
33.94 .10
alf
5.
Pa
33.94 .10
1
1.54226 .w
( 0.37464
alf =
ac .alf
a
T.
d
a=
R .T
P4( V , T )
5
2
2
R .( 126.2 .K )
0.45724 .
ac
2.4051 10
V
0.26992 .w .w ) .
273.15
124.6
b .( V
b)
Pa
P4( V , T ) float , 4
b)
m
.09268 .Pa .
dT
2
mole . V . V
2.405 .10
5 . m3
6
5 . m3 .
V
mole
2.405 .10
mole
2.405 .10
Which shows that the internal energy is a
function of volume (or pressure)
Guesses for Initial and final volumes
Given
100000 .Pa
Given
10000000 .Pa
P4( Vi , T i )
P4( Vf, T i )
Vf
P4( V , T i )d V
W=
Vi
R .T i
Vf
10000 .Pa
R .T i
1000000 .Pa
Vi
find( Vi )
Vi =
0.0227
Vf
find( Vf)
Vf =
2.2006 10
4
1.041 10
mole
1
3
1
m mole
4
3
1
m mole
joule
Vi
273.15 .K
T
Vf
T.
∆U
d
P4( V , T )
P4( V , T ) d V
∆U =
dT
378.0606
Vi
Q
∆U
W
2
a
V .( V
b
P4( V , T )
W
0.5
0.6249
6
2
m mole
0.0927
1
3
1
m mole
Q=
4
1.0788 10
mole
1
joule
mole
1
joule
5 . m3
mole
Solutions to Chemical and Engineering Thermodynamics, 3e
4.30 For an isothermal process involving a fluid described by the Redlich-Kwong equation of state
develop expressions for the changes in
(a) internal energy,
(b) enthalpy, and
(c) entropy
in terms of the initial temperature and the initial and final volumes.
For your information, the Redlich-Kwong equation of state is
P=
RT
a
−
V −b
T ⋅ V ⋅ (V + b )
and
z
F
H
dx
1
x
= ln
x( x + c) c
x+c
I
K
LM F ∂ P I − P OPdV
MN GH ∂ T JK PQ
L R + a1 2fat − RT −
= MT ⋅
N V − b T V (V + b) V − b T
dU = T
V
32
=
12
OP
Q
a
dV
V (V + b )
−adV
2 TV (V + b )
a f a f 2 T z V (VdV+ b) = − 2 aTb lnLMNV V + b V V+ b OPQ
H a T ,V f − H a T ,V f = U aT ,V f − U a T ,V f + P V − PV
a
LV aV + bf OP + RT LM V − V OP
=−
ln M ⋅
2 T b N V aV + b f Q
NV − b V − b Q
a L 1
1 O
−
−
M
V
+
b
V
+ b PQ
T N
F ∂ P IJ dV = LM R + a OPdV
dS =G
H ∂ T K NV − b 2T V (V + b)Q
V −b
a
LV aV + bf OP
S aT ,V f − S aT ,V f = R ln
+
ln M ⋅
V − b 2 T b N V aV + b f Q
G aT ,V f − G aT , V f = H aT ,V f − T S aT ,V f − H aT ,V f − TS aT , V f
a
L V aV + bf OP + RT LM V − V OP
=−
lnM ⋅
2 Tb N V aV + b f Q
NV − b V − b Q
a L 1
−
M − 1 OP
T NV + b V + b Q
V −b
a
LV aV + bf OP
− RT ln
−
ln M ⋅
V − b 2 T b N V aV + b f Q
LV aV + bf OP + RT LM V − V OP
a
=−
ln M ⋅
Tb NV aV + bf Q
NV − b V − b Q
a L 1
1 O
F V + b IJ
−
−
− RT lnG
M
P
HV +b K
T NV + b V + b Q
U T ,V 2 − U T ,V 1 = −
V2
a
2
V1
2
1
2
2
2
1
2
1
1
2
2
1 1
2
2
1
2
1
1
2
1
1
2
1
1
2
32
1
2
1
32
V
2
1
2
2
2
1
1
2
2
1
1
2
1
1
2
2
1
1
1
2
2
1
1
2
2
2
1
1
1
2
2
1
2
1
Solutions to Chemical and Engineering Thermodynamics, 3e
4.31
X
Joule-Thomson Expansion
P1 = 25 bar , T1 = 300 ° C , P2 = 1 bar , T2 = ?
(a) Ideal gas-enthalpy is independent of pressure ⇒ T2 = 300° C
(b) van der Waals gas
IG
H 2 = H 1 ⇒ H2 − H IG
+ H IG
− H 1 − H 1IG = 0
2
2 − H1
c
h c
h c
h
a fL
⇒ 0 = RT a Z − 1f +
MMNTFGH ∂∂ PT IJK − P OPPQdV + C dT − RT aZ − 1f
a fL
−
MMNTFGH ∂∂TP IJK − POPPQ dV = 0
RT
a F ∂ PI
R
F ∂ P IJ − P = RT − RT + a
P=
−
; G
=
; TG
J
V − b V H ∂T K
V −b
H ∂T K
V −b V −b V
z
z
V T2 , P1
2
2
T2
V
V= ∞
z
*
p
1
1
T1
V T1 , P1
V
V =∞
2
2
V
a
f
V
⇒ 0 = RT2 Z2 − 1 + a
a
f
z
z
V2
dV
V =∞ V
V2
0 = RT2 Z2 − 1 + a
dV
V =∞ V
0 = P2V 2 − RT2 − a
2
2
+
z
z
T2
C Vp dT −
a
f
a
f
RT1 Z1 − 1 − a
T1
T2
=
a
V
2
z
V1
dV
V2
V =∞
+ C Vp dT − RT1 Z1 − 1
T1
FG 1 − 1 IJ + ca + bT + cT
HV V K z
T2
2
1
2
h
+ dT 3 dT
T1
− PV
1 1 + RT1
FG 1 − 1 IJ + aaT − T f + b cT
2
HV V K
c
d
+ c T − T h + cT − T h − PV + RT
3
4
0 = P2V 2 − RT2 − a
2
2
3
2
3
1
2
2
1
− T12
h
1
4
2
4
1
1 1
1
Solved together with vdW EOS
T = 575.07 K , ≈ 3019
. °C (T increases?)
(c)
Peng Robinson EOS
Thermodynamic properties relative to an ideal gas at 273.15 K and 1 bar.
H$ (300° C, 25 bar ) = 9.4363 × 103 J mol
After some trial and error
H$ ( 274.1° C, 1 bar) = 9.4362 × 103 J / mol
Close enough
So the solution is T=274.1o C
(d) Steam tables
H$ ( 300° C, 25 bar) = 30088
. kJ kg
H$ ( T = ?, 1 bar ) = 30088
. kJ kg
H$ ( T = 250 ° C, 1 bar ) = 2974.3
H$ ( T = 300 ° C, 1 bar ) = 3074.3
⇒ T ≅ 267 ° C = 540 K
4.32 Note error in first printing. The problem statement should refer to
Solutions to Chemical and Engineering Thermodynamics, 3e
Problem 4.13, not the previous problem. The solution is available only
as a Mathcad worksheet.
4.33
LM f (T ) V Za N V , Tf OP
N!
N
Q
N
= − NRT ln f (T ) − NkT lnV − NkT ln Z F , T I + kT ln N !
HV K
Stirling’s approximation ln N ! = N ln N − N
N
A( N ,V , T ) = − NkT ln f (T ) − NkT lnV − kT ln Z F , T I + NkT ln N − NkT
HV K
FG ∂ A IJ = − P = − NkT − NkT ∂ln Za N V ,T f ∂a N V f
H ∂V K
V
∂a N V f
∂V
NkT N kT ∂ ln Z a N V , T f
P=
−
V
V
∂a N V f
FG ∂A IJ = −S = − Nk ln f (T) − NkT d ln f (T ) − Nk lnV − k ln ZF N , TI
HV K
H∂TK
dT
∂ ln Z a N V , T f
− NkT
+ Nk ln N − kT
N
A( N ,V , T ) = −kT ln Q ( N ,V , T ) = − kT ln
T ,N
N
N
T
T
2
2
T
V ,N
∂T
N ,V
F I
H K
F
H
d ln f (T )
N
N
− Nk ln
+ k ln Z
,T
dT
V
V
∂ ln Z N V , T
+ NkT
− kT
∂T
N ,V
S = Nk ln f ( T ) + NkT
a
G molecule =
FG ∂ A IJ
H∂ NK
G = N ⋅ G molecule
f
= − kT ln f ( T ) − kT ln V − kT
T ,V
a
f
I
K
a f ∂a N V f
a f ∂N
∂ ln Z N V , T
∂ NV
V
− NkT ln Z N V , T + kT ln N + kT − kT
NkT ∂ ln Z N V , T
= − NkT ln f ( T ) − NkT ln V −
V
∂ NV
+ NkT ln N − NkT ln Z
a f
a f
F N ,TI
HV K
As a check
F N , TI + NkT ln N
HV K
L NkT − NkT ∂ ln Za N V , Tf OPV
− NkT + M
N V V ∂a N V f Q
N
= − NkT ln f (T ) − NkT ln V − NkT ln Z F , T I + NkT ln N
HV K
a
f
∂
−
+
−
∂a
f
N
= − NkT ln f ( T ) − NkT ln V − NkT ln Z F , T I + NkT ln N
HV K
NkT ∂ ln Z a N V , T f
−
V
∂a N V f
G = A + PV = − NkT ln f (T ) − NkT ln V − NkT ln Z
2
NkT
NkT
NkT
V
ln z N
N
T
V , T
V
T
V
Solutions to Chemical and Engineering Thermodynamics, 3e
which checks!
∂
A
U
=− 2
∂T V, N T
T
F I
H K
∂
LM− Nk ln f (T) − Nk lnV − Nk ln Z F N , TI + Nk ln N − Nk OP
=
HV K
∂T
N
Q
L d ln f (T) − Nk F ∂ ln Za N V , TfI OP
= M− Nk
GH ∂ T JK P
dT
MN
Q
d ln f ( T )
F ∂ ln Za N V ,T fIJ
U = NkT
+ NkT G
dT
H ∂T K
dU I
d ln f (T )
d ln f (T )
C =F
H dT K = 2 NkT dT + NkT dT
F ∂ ln Za N V , TfIJ + NkT FG ∂ ln Za N V ,T fIJ
+ 2 NkT G
H ∂T K
H ∂T K
V, N
N,V
2
2
N, V
2
2
V
2
N, V
2
2
N, V
H = U + PV
= NkT 2
+
FG
H
a
d ln f ( T )
∂ ln Z N V , T
+ NkT 2
dT
∂T
a f
a f
NkT NkT ∂ ln Z N V , T
−
V
V
∂N V
N,V
fIJ
K
N ,V
T
etc.
4.34
LM FG ∂V IJ OP dP − C dT
N H ∂T K Q
L F ∂V IJ OPdP
= c C − C hdT + MV − T G
N H ∂T K Q
L F ∂ P IJ − P OPdV
dU = c C − C hdT + MT G
N H∂T K Q
C
F ∂ P IJ dV ⇒ d S = C − C dT + LMFG ∂ P IJ − P OPdV
dS =
dT + G
H ∂T K
T
T
MNH ∂T K T PQ
LF ∂V IJ − V OP dP
C −C
or d S =
− MG
T
NH ∂ T K T Q
d G = d H − d c TS h = d H − Td S − S dT
L F ∂V IJ OPdP − cC − C hdT − LMTFG ∂V IJ − V OPdP
= c C − C hdT + MV − T G
N H∂TK Q
N H ∂T K Q
d H res = d H − d H IG = CPdT + V − T
*
P
P
*
P
P
P
res
*
V
V
V
res
V
V
*
V
V
res
V
*
V
V
P
res
res
P
res
res
res
*
p
P
P
− S resdT
dG res = S resdT
res
*
P
P
Solutions to Chemical and Engineering Thermodynamics, 3e
z LMMN FGH
V
H ( T , P) − H IG (T , P ) = RT ( Z − 1) +
T
V =∞
P=
RT
a
− 2
V −b V
FG ∂ P IJ = R
H ∂T K V − b
F ∂ P IJ − P = RT − RT + a
TG
H∂T K
V −b V −b V
∂P
∂T
IJ
K
OP
PQ
− P dV
V
V
=
2
V
IG
H ( T , P) − H (T , P ) = RT ( Z − 1) +
= RT ( Z − 1) −
a
V2
z
V
a
V =∞ V
2
dV = RT ( Z − 1) − a
LM 1 − 1 OP
NV ∞ Q
a
RT a P
= RT ( Z − 1) −
⋅
V
P V RT
aP
ZRT
aP
IG
res
H ( T , P) − H (T ) = RT ( Z − 1) −
=H
ZRT
= RT ( Z − 1) −
a f
d
U res = U ( T , P) − U IG (T ) = H T , P − PV − H IG − PV IG
i
a f
PV PV IG
+
RT = H res T , P + RT (1 − Z )
RT
RT
aP
= H res(T , P) − RT ( Z − 1) =
ZRT
= H res(T , P) − RT
S
res
IG
= S (T , P ) − S (T , P) = −
z LMN
V
=−
V =∞
S res(T , P) = R ln
z LMMNFGH
V
V =∞
OP
Q
∂P
∂T
IJ
K
−
V
OP
PQ
R
dV
V
R
R
(V − b )
V
−
dV = − R ln
+ R ln
(V → ∞ ) − b
(V → ∞ )
V −b V
V
Z
Pb
= R ln
; B=
V −b
Z− B
RT
4.35 a) The Soave-Redlich-Kwong equation of state is
P=
RT
a (T )
−
V −b V V −b
a
f
Rewrite this in the power series of V
V3 −
FH
IK
RT 2
RT
a (T )
a (T )b
V + − b2 −
b−
V−
=0
P
P
P
P
Notice that the three roots of volume at the critical point are identical so we can write
bV − V g
3
C
=0
or
V 3 − 3V C V 2 + 3V 2C V − V 3C = 0
Solutions to Chemical and Engineering Thermodynamics, 3e
At critical point, the second and fourth equations must be satisfied simultaneously. Consequently, the
coefficients of each power of V must be the same. Thus,
RTC
PC
3V C =
3V 2C = - b 2 V 3C =
a f
a f
RTC
a TC
b−
PC
PC
and
a TC
b
PC
Solving the above three equations together for a(T), b and Vc, we get
VC =
b=
c
3
RTC
3 PC
h
2 − 1 V C = 0.08664
a f
a TC =
PC V 3C
b
= 0 .42748
RTC
PC
a RT f
C
a f
a (T ) = a TC α( T ) = 0.42748
ZC =
2
PC
Also
b)
and
aRT f α(T )
C
2
PC
PCV C PCV C RTC 1
=
=
RTC
RTC 3 PC 3
4.36 (also available as a Mathcad worksheet)
RT
a( T )
P=
−
V − b V (V + b )
z LMMN FGH
V
H ( T , P) − H IG (T , P ) = RT ( Z − 1) +
T
V =∞
∂P
∂T
IJ
K
OP
PQ
− P dV
V
FG ∂ P IJ = R − 1 da(T)
H ∂ T K V − b V (V + b) dT
F ∂ P IJ − P = RT − T da(T) − RT + a
TG
H ∂T K
V − b V (V + b) dT
V − b V (V + b)
1
La − T da(T )OP
=
V (V + b) MN
dT Q
V
V
So the integral to be done is
z
V
z
V
LM
N
1
dV
1
1⋅ V + b
dV =
=
ln
V
(
V
+
b
)
V
(
V
+
b
)
(
−
b
)
V
V =∞
V =∞
LM
N
OP
Q
LM
N
1 V +b
1
V +b
= − ln
+ ln
b
V
b
V
So
OP
Q
V =∞
OP
Q
V
V =∞
LM
N
1 V +b
= − ln
b
V
OP
Q
Solutions to Chemical and Engineering Thermodynamics, 3e
z LMMN FGH
V
H ( T , P) − H ( T , P) = RT (Z − 1) +
T
IG
V =∞
IJ
K
OP
PQ
− P dV
V
f FG IJ
H K
T a da dT f − a L Z + a Pb RT f O
= RT (Z − 1) +
ln
b
NM Z QP
T a da dT f − a L Z + B O
= RT (Z − 1) +
ln M
b
N Z QP
(T , P) = R ln Z +
z LMMNFGH ∂∂ PT IJK − VR OPPQdV
= RT (Z − 1) −
a
∂P
∂T
a − T da dT
V +b
ln
b
V
V = ZRT P
S (T , P ) − S IG
FG ∂ P IJ
H ∂T K
z
V =∞
−
V
V = ZRT P
V =∞
= R ln
V
R
R
1
da R
=
−
−
V V − b V (V + b ) dT V
LM R − 1 da − R OPdV
NV − b V (V + b) dT V Q
V −b
V
da 1 V + b
− R ln
+
ln
V − b V →∞
V V →∞ dT b
V
LM a fOP
N
Q
Z
−
B
da
1
Z
+
B
F I + lnLM OP
= R ln
H Z K dT b N Z Q
Z − B I da 1 L Z + B O
S (T , P ) − S (T , P ) = R ln Z + R ln F
H Z K + dT b lnMN Z PQ
da 1 L Z + B O
= R ln( Z − B ) +
ln
dT b MN Z PQ
= R ln
V − b da 1
Z + Pb RT
+
ln
V
dT b
Z
IG
G
res
=H
res
− TS
res
= RT (Z − 1) −
aP
Z
− RT ln
zRT
Z− B
Redlich-Kwong
P=
RT
a
−
V −b
TV (V + b )
FG ∂ P IJ = R + a1 2fa
H ∂ T K V − b T V (V + b)
F ∂ P IJ − P = RT + a1 2fa − RT +
TG
H ∂T K
V −b
TV (V + b ) V − b
a3 2fa = 3a
=
V
32
V
TV (V + b )
2 TV (V + b )
a
T V (V + b )
Solutions to Chemical and Engineering Thermodynamics, 3e
z
V
H ( T , P) − H IG (T , P ) = RT ( Z − 1) +
3a
dV
(
V
V
+ b)
V =∞ 2 T
3a
= RT ( Z − 1) +
ln
V
V +b
2b T
3a
Z
= RT ( Z − 1) +
ln
Z
+
bP
RT
2b T
3a
Z
= RT ( Z − 1) +
ln
2b T Z + B
3a
Z
U ( T , P) − U IG (T , P ) =
ln
Z+ B
2b T
a
z LMMNFGH
z LMN
V
S (T , P) − S IG (T , P ) = −
V =∞
V
=−
V =∞
∂P
∂T
IJ
K
−
V
OP
PQ
f
R
dV
V
a f
OP
Q
R
12a
R
+
−
dV
V − b T 3 2V (V + b ) V
a f
= R ln
V
12a
V
−
ln
V − b T 3 2b V + b
= R ln
Z
a
Z
ln
−
32
Z − B 2T b Z + B
Solutions to Chemical and Engineering Thermodynamics, 3e
4.37 (also available as a Mathcad worksheet)
Problem 4.37
Critical properties and heat capacity for oxygen:
154.6 . K
Tc
Pc
25.460 .
Cp ( T )
6
5.046 . 10 . Pa
joule
mole . K
ω
2
1.519 . 10 . T .
0.021
joule
Tref
298.15 . K
5 2
0.715 . 10 . T .
mole . K
2
5
10 . Pa
Pref
joule
9 3
1.311 . 10 . T .
mole . K
3
joule
mole . K
4
Soave-Redlich-Kwong Constants:
R
joule
8.31451 .
K . mole
a( T )
2
2
2
R . Tc
0.42748 .
Pc
a1
a1. α ( T )
b
α ( T)
R. Tc
0.08664 .
Pc
1
1.574 . ω
0.480
b = 2.207 10
5
2
0.176 . ω . 1
3
1
m mole
(Initial guess
for solver)
Temperature and Pressure:
173.15 . K
T
P
5
1 . 10 . Pa
4 3
10 . m . mole
V
1
Solve block for Volume:
Given
R. T
P
a( T )
.
V (V b)
V b
V
Find( V)
Calculation of Compressibility:
P .V
R. T
Z
Calculation of Enthalpy and Entropy:
d
a( T ) . T
dT
. ln V
b
V b
a( T )
HDEP
SDEP
R. T . ( Z
R. ln
1)
Z .( V b )
d
a( T )
dT
.
V
b
T
HIG
Cp ( T ) d T
Tref
T
ln
V
Cp ( T )
SIG
V b
T
Tref
H
HIG HDEP
S
Final Results:
T = 173.15
K
5
P = 1 10
Pa
Z = 0.9952
3
1
m mole
V = 0.014327
3
H = 3.60273 10
S = 15.62
1
mole
joule
1
1
K mole
joule
SIG SDEP
dT
R. ln
P
Pref
T
Tc
Solutions to Chemical and Engineering Thermodynamics, 3e
4.38 (also available as a Mathcad worksheet)
4.38 with the SRK equation
Property Data
(T in K, P in bar):
Tc
R
126.2
Pc
33.94
0.00008314
kap
om
0.04
Cp 1
Initial Conditions (Vt = total volume, m^3):
Ti
0.08664 .
b
170
Pi
R. Tc
100
Vt
0.42748 .
ac
0.15
T
2.
R Tc
Pc
Initial temperature
4.2 . 10
Cp 2
1.574 . om 0.176 . om. om
0.480
SRK Constants:
27.2
Pc
Ti
Note that these are being defined as a
function of temperature since we will need to
interate on temperature later to obtain the final
state of the system
2
alf( T )
1. 1
T
kap . 1
R. Ti
V
Pi
Given
R. T
Pi
a( T )
V. ( V b )
V b
Vi = 1.02 10
Initial molar volume and
number of moles
Entropy departure at the
initial conditions
d
a( T )
dT
Da ( T )
Start with initial guess for volume, m3 /mol
Solve SRK EOS for initial volume
ac . alf( T )
a( T )
Tc
Find initial molar volume and number of moles
DELSi
R. ln ( Vi
4
Vi
Vt
Ni
b).
Pi
R. T
Da ( T ) .
ln
Vi
b
Nf
Ni
Find( V)
3
Ni = 1.471 10
Vi
Now consider final state
b
. 105
Vi
0.15
10 . 50 Vf
V
Vf
Nf
Type out final number of
moles and specific volume
Nf = 971.269
Final pressure, will change in course
of solving for the final temperature
Entropy departure
at final conditions
T
Type out solution
Pf( T )
R. T
V b
DELS( T )
Solve for final
temperature using
S(final) - S(initial) = 0
2
R. ln ( V b ) .
Pf( T )
R. T
Da ( T ) .
b
Vf = 1.544 10
4
a( T )
V. ( V b )
ln
V b
. 105
V
GIVEN
0 Cp 1 . ln
T
Cp 2 . ( T
Ti
Ti)
Pf( T )
5
R. 10 . ln
Pi
DELS( T )
DELSi
FIND( T )
V = 1.544 10
4
T = 131.34
Pf( T ) = 37.076
3
Solutions to Chemical and Engineering Thermodynamics, 3e
4.39 (also available as a Mathcad worksheet)
Problem 4.39
Critical properties for carbon dioxide:
Tc
304.2
K
7.376 . 10
6
Pc
ω
Pa
0.225
Soave-Redlich-Kwong Equation of State and Constants:
R
3
joule
Pa . m
or
K . mol
( K . mol)
8.31451
0.42748 .
a1
2
2
R . Tc
b
0.08664 .
Pc
R. Tc
Pc
2
α ( T)
1
1.574 . ω
0.480
R. T
P( V, T )
0.176 . ω
2
. 1
50 . 10
Pa
T
a( T )
Tc
a1. α ( T )
a( T )
.
V (V b)
V b
Data given in the problem:
T
( 150
273.15 ) K
R.
V
T
5
P1
P2
300 . 10
5
Pa
V1total
3
100 m
(Initial guess needed for solver)
P1
Solving for the initial molar volume and the number of moles of carbon dioxide:
P1 P( V, T )
Given
V1molar
Find( V)
V1total
N
5
N = 1.518 10
V1molar
V1molar = 6.587 10
Solving for the final molar volume and the final total volume:
( R. T )
V
P2
P2 P( V, T )
Given
V2molar
Find( V)
V2total
V2molar. N
V2molar = 9.805 10
(i) V2total = 14.885
Calculating the amount of work done to compress the gas:
Work
N.
V2molar
8
( ii) Work = 8.823 10
P( V, T ) d V
joule
V1molar
Since the temperature is constant, the change in enthalpy, H(T, P2) - H(T,P1), is just equal
to Hdep(T,P2) - Hdep(T,P1) :
Hdep( T , P )
H( T , P )
Q
P .V
R. T .
R. T
a( T )
1
8
Q = 3.837 10
d
a( T )
dT
.
ln
b
V
V b
3
H( T , P2 ) = 4.382 10
Hdep( T , P )
N . ( ( H( T , P2 )
T.
H( T , P1 ) )
joule
( P2 . V2molar P1 . V1molar) )
3
H( T , P1 ) = 7.314 10
Work
5
4
Solutions to Chemical and Engineering Thermodynamics, 3e
4.40 (also available as a Mathcad worksheet)
Problem 4.40
Critical properties and heat capacity for ethylene:
Tc
282.4
K
Ca
3.950
Cb
5.036 . 10
6
Pc
15.628 . 10
2
ω
Pa
Cc
0.085
8.339 . 10
5
17.657 . 10
Cd
9
Soave-Redlich-Kwong Equation of State and Constants:
J
R
T2
a1
a1. α ( T )
a( T )
T1
mol. K
8.31451
( 100
b
0.08664 .
α ( T)
1
R. Tc
K
273.15 )
K
P1
30 . 10
P2
20 . 10
5
5
Pa
( R. T1)
V
Pa
0.480
P( V, T )
Pc
273.15 )
( 150
2
2
2
R . Tc
0.42748 .
Pc
1.574 . ω
R. T
V b
0.176 . ω
2
. 1
a( T )
.
V (V b)
(Initial guess for solver)
P1
Solving for the initial and final molar volume (Only one root is possible for each volume because
both temperatures are above the critical temperature):
Given
V1
V1 = 9.501 10
Find( V)
Given
V2
P1 P( V, T1)
4
3
m
P2 P( V, T1)
V2 = 1.467 10
Find( V)
3
3
m
Defining the reference state as P=1 bar, and T=300 K :
T
Hig ( T )
Ca
Cb . TT
Cc . TT
2
Cd . TT d TT
a( T )
T.
3
300
Hdep( T , P , V)
H( T , P , V)
Q
R. T .
R. T
Hdep( T , P , V)
H( T2, P2 , V2 )
3
Q = 2.64 10
P .V
1
b
Hig ( T )
H( T1, P1 , V1 )
joule
d
a( T )
dT
.
ln
V
V b
T
Tc
Solutions to Chemical and Engineering Thermodynamics, 3e
4.41 (also available as a Mathcad worksheet)
Problem 4.41 Soave-Redlich Kwong EOS with MATHCAD
mol
1
101300 . Pa
bar
Property Data
(T in K, P in bar):
Tc
8.314 .
RE
304.2 . K Pc
73.76 . bar
joule
mol. K
om
0.225
kap
0.480
Initial Conditions:
Ti
400 . K
SRK Constants:
b
0.08664 .
RG
0.00008314 .
T
50 . bar
Pi
RG. Tc
ac
0.42748 .
2
2
RG . Tc
Pc
Ti
Note that these are being defined as a
function of temperature since we will need to
interate on temperature later to obtain the final
state of the system
joule
Heat capacity Cp1 22.243 .
Cp2
mol. K
constants
2
alf( T )
1. 1
kap . 1
Da ( T )
Cp4
Given Pi
a( T )
2 joule
5.977 . 10 .
mol. K
Find initial molar volume and number of moles
Start with initial guess for volume, m^3/mol
Solve SRK EOS for initial volume
T
Tc
5 joule
3.499 . 10 .
mol. K
Cp3
V
RG. Ti
V = 6.6512 10
Pi
RG. T
a( T )
V b
V. ( V b )
Vi
Entropy departure at the
initial conditions
DelHi
RE. T . ( Zi
1
DELSi
Zi
T.
a( T )
d
a( T )
dT
Pi
RG. T
d
a( T )
dT
4
3
m
Find( V)
Da ( T ) . Vi b
ln
b . RG
Vi
4
3
m
. RE
Zi = 0.91653
Zi
. ln
b
b).
Pi . Vi
RG. T
joule
1)
3
DelHi = 1.24253 10
ln ( Vi
ac . alf( T )
9 joule
7.464 . 10 .
mol. K
Vi = 6.09602 10
DELSi = 2.37467 K
3
1.574 . om 0.176 . om. om
Pc
Initial temperature
bar . m
mol. K
b . Pi
RG. T
Zi
joule
Final temperature is 300 K, and final pressure is unknown; will be found by equating the initial and final
entropies. Guess final temperature is 10 bar
Pf
10 . bar
T
300 . K
V
RG. T
Pf
V = 2.4942 10
3
3
m
Solutions to Chemical and Engineering Thermodynamics, 3e
Temperature part of ideal gas entropy change
ti
tf
Cp1. ln
DSidealT
400
Cp2. ( tf
tf
Cp3 .
ti)
ti
DSidealT = 11.24332 K
300
2
Cp4 .
2
tf
ti
2
1
3
3
tf
ti
3
joule
Note: To use the given and find commands for variable with different dimensions such as P and V, will have
to convert to dimensionless variables so as not to have a units conflict. Define x=V/b and y=P/Pc
initial guess
V
x
x = 83.95847
b
Given
y . Pc
RG. 300 . K
x. b b
RE. ln
0 DSidealT
y . Pc
FIND( x, y )
Y=
ln ( x. b
b).
y . Pc
RG. 300 . K
Da ( 300 . K ) . x 1
ln
b . RG
x
Vf = 1.59393 10
Pf
Y1 . Pc
6
Pf = 1.46849 10 Pa
bar
DELSi
0.19654
Y0 . b
= 14.49643
. RE
53.6541
Vf
Pf
0.5
a( 300 . K )
.
x b . ( x. b b )
Pi
Y
y
3
3
m
Vi
Fraction mass remaining in tank =
Vf
= 0.38245
4.42 (also available as a Mathcad worksheet. In fact, this file contain graphs and other
information.)
∂ CV
∂2 P
∂ CP
∂2V
Easier to work with
=T
than with
= −T
.
2
∂V T
∂T V
∂P T
∂ T2 P
FG
H
IJ
K
FG
H
IJ
K
FG
H
V =∞
= 0.75
a
T3 2
V
V +b
ln
V V =∞
FG IJ
H K
FG ∂ P IJ dV
H∂T K
a
F V + bIJ
= 0.75
ln G
HV K
T
z
V
CV (V , T ) − CV (V = ∞ , T ) = CV (V , T ) − CV* (T ) =
IJ
K
2
T
2
32
Solutions to Chemical and Engineering Thermodynamics, 3e
So CV (V , T ) = CV* (T ) + 0.75
a
32
T b
FG V + b IJ .
HV K
Clearly as V → ∞ (ideal gas) we
ln
recover CV = C*V .
Procedure:
Choose collection of V ’s
Calculate CV − CV* for given V and T
Calculate P from RK EOS get CV − CV* vs. P
Next use
CP = CV − T
FG ∂V IJ FG ∂ P IJ
H ∂ P K H ∂T K
T
2
= CV − T
V
a∂ P ∂T f
a∂ P ∂V f
2
V
T
to convert from CV to CP . Have done both parts using MATHCAD. See the
MATHCAD F file.
4.43 (also available as a Mathcad worksheet)
a) P =
Z=
RT
a
PV
B C
−
;
= 1 + + 2 + L= Z
V −b
V V
T V (V + b ) RT
PV
V
V
=
−
RT V − b RT
a
T V (V + b )
=
V
a
−
32
V − b RT (V + b)
1
aV
1
a /V
−
=
−
1 − b / V RT 3 2 (1 + b / V ) 1 − b / V RT 3 2 (1 + b / V )
Now expanding in a power series in 1V
=
Z = 1+ b /V −
B =b−
a
RT 3 2
F
H
I
K
a /V
a
1
3 2 = 1+ b −
32
RT
RT
V
a
; B = 0; b =
RTB3 2
F I
H K
a
a 23
⇒ TB =
= 876.5 K
bR
bR
b) Using the Redlich-Kwong parameters
TB3 2 =
TB =
FaI
H bR K
23
= 876.5 K
4.44 (also available as a Mathcad worksheet)
Z=
V
a
−
V − b RT 3 2 (V + b )
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂ Z IJ
H∂ PK
=
T
FG IJ
H K
1
∂V
V −b ∂ P
−
T
RT
a
P=
−
V −b
T V 2 + bV
c
V
(V − b )
2
FG ∂V IJ
H ∂ PK
+
T
a
RT
32
(V + b)
2
P→ 0
V →∞
2
2
2
P→ ∞
V →∞
T
T
2
P →0
V→∞
T
h
FG ∂ P IJ = − RT + a (2V + b)
H ∂V K (V − b) T dV + bV i
F ∂ P IJ = 0 ; lim FG ∂ P IJ = ∞
lim G
H ∂V K
H ∂V K
F ∂ Z IJ = 0 = LM 1 − V + a OPFG ∂V IJ
lim G
H ∂P K
NV − b (V − b) RT (V + b) QH ∂ P K
FG ∂ Z IJ = 1 (V − b) − mV (V − b) r + ma RT (V + b) r
H ∂ P K m− RT (V − b) r + {a T cV + bV h }(2V + b)
F ∂Z IJ = V − b − V (V − b) = b
lim G
H ∂ P K − RT (V − b)
RT
F ∂ Z IJ = 1 V − 1 V + a cRT V h
lim G
H ∂ P K − RT (V ) + a d T V i ⋅ 2V
a d RT V i
aV
a
=
=−
=−
T
FG ∂V IJ
H ∂P K
2
32
2
T
2
32
T
T
2
2
2
2
P →∞
V →b
2
T
32
P→ 0
V →∞
2
T
2
32
4
2
− RT V 2
RT 3 2V 2
RT 3 2
4.45 a) The Redlich-Kwong equation of state is
RT
a
−
V −b
T V (V + b )
P=
which we rewrite as follows
PV
V
a
=
−
RT V − b RT 1.5(V + b )
so that
PV
V
a
b
a
−1=
−1−
=
−
RT
V −b
RT 1.5 (V + b) V − b RT 1.5 (V + b)
PV
V
a
V
V
−1 = b
−
15
.
RT
V − b RT V + b
F
H
Lim P→0
I
K
PV
I
VF
H RT − 1K = Lim
V→∞
V
and
F PV − 1I = Lim LMb V − a
H RT K
N V − b RT
V →∞
1.5
OP
Q
V
a
= b−
= B( T )
V +b
RT 1.5
Solutions to Chemical and Engineering Thermodynamics, 3e
To proceed further, we now need to have expressions for a and b in terms of the critical
properties. To obtain these we proceed as in Problem 4.35 and rewrite this in the power
series of V
V3 −
F
H
I
K
RT 2
RT
a
ab
V + −b 2 −
b−
V−
=0
P
P
TP
TP
Notice that the three roots of volume at the critical point are identical so we can write
bV − V g
3
=0
C
or
V 3 − 3V C V 2 + 3V 2C V − V 3C = 0
At critical point, the second and fourth equations must be satisfied simultaneously.
Consequently, the coefficients of each power of V must be the same. Thus,
3V C =
RTC
PC
2
3V C = -b 2 -
RTC
a
b−
PC
TC PC
a
VC =
3
TC PC
and
b
Solving the above three equations together for a(T), b and Vc, we get
VC =
b=
d
3
RTC
3 PC
i
2 − 1 V C = 0.08664
a = 0.42748
So B( T ) = b −
RTC
PC
and
R2 TC2.5
PC
a
RTC
RTC2.5
=
0
.
08664
−
0
.
42748
RT 1.5
PC
PC T 1.5
For n-pentane, TC=469.6 K and PC = 33.74 bar
The resulting virial coefficient as a function of temperature is shown below.
Solutions to Chemical and Engineering Thermodynamics, 3e
0
.0
B( T )
0.001
0.002
200
200
4.46
400
600
800
1000
1200
1400
FG ∂ T IJ = − V 1 − Tα ⇒ αT = 1
H ∂P K C
1 F ∂V I
F ∂V IJ FG ∂ P IJ FG ∂T IJ = −1 by triple product rule.
α= G
but G
J
H ∂T K H ∂V K H ∂ P K
V H ∂T K
a∂ P ∂T f
FG ∂V IJ =
−1
=−
H ∂T K a∂ P ∂V f a∂ T ∂ Pf a∂ P ∂V f
Inversion temperature µ = 0 =
H
P
P
T
P
V
V
P
(a) vdW EOS; P =
T
FG IJ
H K
RT
a
∂P
−
;
V −b V2
∂T
V
=
V
R
V −b
FG ∂ P IJ = − RT + 2a
H ∂V K (V − b) V
− R (V − b)
FG ∂V IJ =
H ∂T K n(− RT ) (V − b) s + 2a V
1 F ∂V I
k− R V (V − b) p
α= G
=
J
V H ∂T K
n− RT (V − b) s + 2a V
2
T
3
2
P
2
P
Tα = 1 ;
n− RT
3
− RT V (V − b )
s
(V − b) 2 + 2 a V 3
=1
3
1600
3
1.5 .10
T
T
Solutions to Chemical and Engineering Thermodynamics, 3e
RT
RT
2a
=
− 3
2
V (V − b ) (V − b )
V
TR
LM 1 − 1 OP = − 2a
N V (V − b) (V − b) Q V
2
3
TR
2a
V −b −V = − 3
2
(
)
V V −b
V
bRT
2a
=
V (V − b )2 V 3
FG
H
IJ
K
2
2 a (V − b) 2 2a
b
=
1−
2
bR V
bR
V
RT
a
also P =
−
V −b V2
Choose V
Calculate Tinv from Eqn. (1)
Calculate P from Eqn. (2)
Solution done with MATHCAD (see MATHCAD worksheet).
T=
(1)
(2)
(b) RK EOS
RT
a
P=
−
V −b
TV (V + b )
FG ∂ P IJ
H ∂V K
FG ∂ P IJ
H ∂T K
α=
=−
T
=
V
RT
a
a
+
+
2
2
2
(V − b)
T V (V + b )
TV (V + b )
a f
R
12 a
+ 32
V − b T V (V + b )
FG IJ
H K
1 ∂V
V ∂T
=−
P
a
a
1 ∂ P ∂T
V ∂ P ∂V
l
f
f
V
T
a f
q
− R (V − b ) + 1 2 a T 3 2V (V + b )
1
=
V − RT (V − b )2 + a TV 2 (V + b ) + a T V (V + b) 2
m
r n
s n
s
Tα = 1
− RT
12a
RT
a
a
−
=−
+
+
V (V − b ) T 1 2V 2 (V + b )
(V − b) 2
T V 2 (V + b )
TV (V + b )2
a f
LM
OP
a
LM 1 + 1 + 1 2 OP
N
Q TV (V + b) NV V + b V Q
RT LV − b − V O
RTb
a
LM a3 2f(V + b) + V OP
−
=
=
M
P
V − b N V (V − b ) Q V (V − b)
TV (V + b ) N V (V + b)
Q
−
RT 1
1
−
=
V − b V V −b
2
Solutions to Chemical and Engineering Thermodynamics, 3e
RTb
V (V − b )
2
RTb
(V − b )
2
T3 2 =
T=
4.47
=
=
5V + 3b
2 TV (V + b ) V (V + b)
a
a(5V + 3b)
2 T V (V + b )2
a (5V + 3b ) (V − b )2
2
Rb
2V (V + b )
RS a L5 + 3 b O (V − b) UV
T 2Rb MN V PQ (V + b) W
2
23
2
Sorry, in the first printing this problem was misplaced. It is Problem 5.47, and
the solution appears in Chapter 5 of the solution manual.
The replacement problem is
“Repeat the calculation of Problem 4.31 with the Soave version of the RedlichKwong equation.”
The solution is as follows:
Using a Mathcad program for the Soave-Redlich-Kwong EOS we find
H ( 300° C, 25 bar ) = 9 .45596 × 103 J mol
(relative to ideal gas at 273.15 K and 1 bar).
Now by trial and error until enthalpies match
H ( 274.5° C, 1 bar ) = 9.45127 × 103
⇒ T = 274.6o C
H ( 274.6° C, 1 bar ) = 9 .45486 × 103
Close enough
Note that this solution is only very slightly different from that obtained with the
Peng-Robinson equation (274.1oC compared to 274.6oC obtained here).
4.48
(also available as a Mathcad worksheet)
Problem 4.48 Peng-Robinson EOS with MATHCAD
mol
1
bar
Property Data
(T in K, P in bar):
101300 . Pa
Tc
5.19 . K
RE
Pc
2.27 . bar om
8.314 .
joule
mol. K
RG
0.387
kap
Initial Conditions and total volume Vt:
Ti
298 . K
Peng-Robinson Constants:
b
0.07780 .
Cp
0.37464
Pi
RG. Tc
0.00008314 .
T
3
2.5 . RE
1.54226 . om 0.26992 . om. om
400 . bar
ac
0.045 . m
3
Vt
0.45724 .
Pc
Initial temperature
bar . m
mol. K
2
2
RG . Tc
Pc
Ti
Note that these are being defined as a
function of temperature since we will need to
interate on temperature later to obtain the final
state of the system
2
alf( T )
1. 1
kap . 1
T
Tc
a( T )
Da ( T )
ac . alf( T )
d
a( T )
dT
Solutions to Chemical and Engineering Thermodynamics, 3e
Find initial molar volume and number of moles
Start with initial guess for volume, m^3/mol
Solve P-R EOS for initial volume
RG . Ti
V
Given
Pi
RG. T
Pi
a( T )
.
V ( V b ) b .( V b )
V b
Initial molar volume and
number of moles
Vi = 7.10667 10
Entropy departure at the
initial conditions
DELSi = 0.0151 K
RE. T . ( Zi
DelHi
1
DELSi
joule
Zi
T.
a( T )
1)
d
a( T )
dT
ln ( Vi
b).
Pi . Vi
RG. T
3
m
N
Vt
Da ( T )
2 . 2 . b . RG
5
3
m
Vi
Find( V)
N = 633.20762
Vi
Pi
RG. T
. ln Vi
1
2 .b
Vi
1
2 .b
. RE
Zi = 1.14736
Zi
1
b . Pi
2 .
RG. T
Zi
1
b . Pi
2 .
RG. T
. ln
2. 2 .b
5
V = 6.19393 10
DelHi = 274.27781 joule
Final pressure is 1.013 bar, and final temperature is unknown; will be found by equating the initial and final
entropies. Guess final temperature is 30 K
Pf
1.013 . bar
25.7 . K
T
V
RG. T
V = 2.10928 10
Pf
3
3
m
Note: To use the given and find command for variables with different dimensions such as T and V, will have
to convert to dimensionless variables so as not to get a units conflict. Define x=V/b, y=T/Tc
initial guess
x
Given
RG. y . Tc
x. b b
Pf
0 Cp . ln
y . Tc
RE. ln
Ti
Y
V
10
Pf
y
5
a( y . Tc )
x. b . ( x. b b ) b . ( x. b
RE. ln ( x. b
Pi
FIND( x, y )
Y0 . b
Y=
Pf . V
RG. T
Pf
RG. y . Tc
Da ( y . Tc ) . RE .
2 . 2 . b . RG
ln
x. b
x. b
1
2 .b
1
2 .b
150.89161
5.27265
V = 2.2315 10
Zf
b).
b)
3
3
m Tf
Zf = 1.05794
Y1 . Tc
Tf = 27.36506 K
Final temperature
DELSi
Solutions to Chemical and Engineering Thermodynamics, 3e
Tf .
RE. Tf . ( Zf
DelHf
1)
d
a( Tf )
d Tf
a( Tf )
Zf
1
Zf
1
b . Pf
2 .
RG. Tf
b . Pf
2 .
RG. Tf
. ln
2. 2 .b
DelHf = 11.61232 joule
Uf - Ui = (Hf - Zf*R*Tf) - (Hi - Zi*R*Ti) = Hf - Hi - Zf*R*Tf + Zi*R*Ti
= (Hf-HfIG) + HfIG -(Hi - HiIG) -HiIG - Zf*R*Tf + Zi*R*Ti
= DelHf - DelHi + Cp*(Tf-Ti) - Zf*R*Tf + Zi*R*Ti
DelU
( DelHf
Cp . ( Tf
DelHi
6
DelU = 2.08062 10
TNTeq = 0.45231 kg
joule
4600000 .
kg
4.49
Zi. RE. Ti) . N
joule
DelU
TNTeq
Zf . RE. Tf
Ti)
(also available as a Mathcad worksheet)
PENG-ROBINSON EQUATION OF STATE CALCULATION
Nitrogen
The Property Data should be as follows
Tc (in K), Pc (in bar), omega, Tb (in K)
Cp1, Cp2, Cp3, Cp4
(In eqn Cp=Cp0+Cp1*T+Cp2*T^2+Cp3*T^3)
Tref (in K), Pref (in bar)
(reference conditions)
Note that in the 1st and 2nd printings, carbon dioxide was used as the fluid. This gave
unreasonable answers when this problem was revisited with the Peng-Robinson eqn.
of state, as both the initial and final states were found to be in the liquid state. Therefore
from the 3rd printing on, the fluid has been changed to nitrogen.
i
0 , 1 .. 3
Cp 0
Tc
Trs
R
28.883
126.2
273.15
0.00008314
0.157 . 10
Cp 1
Pc
33.94
Prs
om
1.0
Peng-Robinson Constants:
2
Cp 2
0.808 . 10
5
Cp 3
2.871 . 10
9
0.04
kap
b
0.37464
0.07780 .
1.54226 . om 0.26992 . om. om
R. Tc
Pc
ac
0.45724 .
2
2
R . Tc
Pc
Solutions to Chemical and Engineering Thermodynamics, 3e
Input temperature and pressure of calculation
T
2
alf( T )
1. 1
T
kap . 1
ac . alf( T )
a( T )
Tc
A
CA ( T , P )
B
CB( T , P )
A .B
P .b
R. T
CB( T , P )
B
Vector of coefficients in the PR equation
in the form
0=-(A*B-B^2-B^3)+(A-3*B^2-2*B)*Z-(1-B)*Z^2+Z^3
2.B
2
(1
2
( R. T )
bar
3
B
3.B
A
V
2
a( T ) . P
CA ( T , P )
140
d
a( T )
dT
Da ( T )
Z( T , P )
K, P
298.15
B)
1
ZZ
Solution to the cubic
polyroots ( V)
for i ∈ 0 .. 2
ZZi
0
if
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ2
Im ZZi
0
ZZ0 < 10
5
ZZ2 < 10
5
ZZ0 if
Set any imaginary roots to zero
Sort the roots
Set the value of any imaginary roots
to value of the real root
ZZ
Calculate molar volumes
Z( T , P ) 0 . R . T
VL( T , P )
. 103
P
VV( T , P )
Z( T , P ) 2 . R . T
. 103
P
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
fv ( T , P )
phil( T , P )
fugl( T , P )
Z( T , P ) 0
Z( T , P ) 2
1
1
exp ( fl( T , P ) )
P . phil( T , P )
ln Z( T , P ) 0
ln Z( T , P ) 2
CB( T , P )
CB( T , P )
phiv ( T , P )
fugv( T , P )
CA ( T , P )
. ln
2 . 2 . CB( T , P )
CA ( T , P )
2 . 2 . CB( T , P )
exp ( fv ( T , P ) )
P . phiv ( T , P )
. ln
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Solutions to Chemical and Engineering Thermodynamics, 3e
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
R. ln Z ( T , P )
DELSL( T , P )
R. ln Z( T , P ) 2
0
DELSV( T , P )
Da ( T ) .
CB( T , P )
ln
2. 2 .b
Da ( T ) .
CB( T , P )
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
R . T . Z( T , P ) 0
DELHL( T , P )
1
a( T ) .
ln
2. 2 .b
R . T . Z( T , P ) 2
DELHV( T , P )
T . Da ( T )
1
T . Da ( T )
a( T ) .
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Ideal gas properties changes relative to the reference state
Cp 0 . ( T
DELHIG( T )
Cp 0 . ln
DELSIG( T , P )
Trs )
Cp 1 . T
2
Trs
2
Cp 2 . T
2
T
Trs
Cp 1 . ( T
3
Trs
3
Cp 3 . T
3
Trs )
Cp 2 . T
2
4
Trs
4
4
Trs
2
Cp 3 . T
2
3
3
Trs
3
P
5
R. 10 . ln
Prs
Total entropy and enthalpy relative to ideal gas reference state
SL( T , P )
HL( T , P )
DELSIG( T , P )
DELHIG( T )
DELSL( T , P )
DELHL( T , P )
SV( T , P )
HV( T , P )
SUMMARY OF RESULTS
T = 298.15
K
Pressure, bar
P = 140
DELSIG( T , P )
DELSV( T , P )
DELHIG( T )
DELHV( T , P )
Solutions to Chemical and Engineering Thermodynamics, 3e
LIQUID
VAPOR
Compressibility
Z( T , P ) 0 = 0.99907
Z( T , P ) 2 = 0.99907
Enthalpy, J/mol
HL( T , P ) = 96.20674
HV( T , P ) = 96.20674
Entropy, J/mol K
SL( T , P ) = 41.04818
SV( T , P ) = 41.04818
Fugacity coefficient
phil( T , P ) = 0.97024
phiv ( T , P ) = 0.97024
Fugacity, bar
fugl( T , P ) = 135.83299
fugv( T , P ) = 135.83299
Volume, m^3/kmol
VL( T , P ) = 0.17689
VV( T , P ) = 0.17689
Number of moles initially
N
Tf
100
Pf
2
3.1416 . ( .01 ) . .06 .
1000
VV( T , P )
N = 0.10656
1.103
Given
SV( Tf , Pf )
41.04818
Tf = 69.36841K
Tf
Pressure, bar
find( Tf )
Tf = 69.36841
Pf = 1.103
LIQUID
VAPOR
3
Compressibility
Z( Tf , Pf ) 0 = 5.61855 10
Enthalpy, J/mol
4
HL( Tf , Pf ) = 1.16967 10
3
HV( Tf , Pf ) = 5.96312 10
Entropy, J/mol K
SL( Tf , Pf ) = 114.62977
SV( Tf , Pf ) = 41.04818
Fugacity coefficient
phil( Tf , Pf ) = 0.31698
phiv ( Tf , Pf ) = 0.94395
Fugacity, bar
fugl( Tf , Pf ) = 0.34963
fugv( Tf , Pf ) = 1.04117
Volume, m^3/kmol
VL( Tf , Pf ) = 0.02938
VV( Tf , Pf ) = 4.91926
U( T , P )
HV( T , P )
U( Tf , Pf )
W
G
140 .
100
HV( Tf , Pf )
N . ( U( Tf , Pf )
0.17689
1.013 .
4.91926
100
U( T , P ) )
U( T , P ) = 96.45439
3
U( Tf , Pf ) = 5.96317 10
W = 625.17152
W
4600
G = 0.13591
grams of TNT
Z( Tf , Pf ) 2 = 0.94081
Solutions to Chemical and Engineering Thermodynamics, 3e
PENG-ROBINSON EQUATION OF STATE CALCULATION
Carbon dioxide
The Property Data should be as follows
Tc (in K), Pc (in bar), omega, Tb (in K)
Cp1, Cp2, Cp3, Cp4
(In eqn Cp=Cp0+Cp1*T+Cp2*T^2+Cp3*T^3)
Tref (in K), Pref (in bar)
(reference conditions)
Note that in the 1st and 2nd printings, carbon dioxide was used as the fluid. This gave
unreasonable answers when this problem was revisited with the Peng-Robinson eqn.
of state, as both the initial and final states were found to be in the liquid state. Therefore
from the 3rd printing on, the fluid has been changed to nitrogen.
i
0 , 1 .. 3
Cp 0
Tc
Trs
R
22.243
0.00008314
5.977 . 10
304.2
Cp 1
Pc
273.15
73.76
Prs
om
2
Cp 2
3.499 . 10
5
Cp 3
7.464 . 10
0.225
1.0
kap
Peng-Robinson Constants:
b
1.54226 . om 0.26992 . om. om
0.37464
0.07780 .
R. Tc
ac
0.45724 .
Pc
Input temperature and pressure of calculation
T
1. 1
T
kap . 1
a( T )
ac . alf( T )
Tc
Da ( T )
Z( T , P )
A
CA ( T , P )
B
CB( T , P )
A .B
298.15
a( T ) . P
( R. T )
2
140
bar
CB( T , P )
P .b
R. T
d
a( T )
dT
B
Vector of coefficients in the PR equation
in the form
0=-(A*B-B^2-B^3)+(A-3*B^2-2*B)*Z-(1-B)*Z^2+Z^3
2.B
2
(1
CA ( T , P )
K, P
3
B
3.B
A
V
2
2
2
R . Tc
Pc
2
alf( T )
9
B)
1
ZZ
Solution to the cubic
polyroots ( V)
for i ∈ 0 .. 2
ZZi
0
if
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ2
Im ZZi
0
ZZ0 < 10
5
ZZ2 < 10
5
ZZ0 if
Set any imaginary roots to zero
Sort the roots
Set the value of any imaginary roots
to value of the real root
ZZ
Calculate molar volumes
VL( T , P )
Z( T , P ) 0 . R . T
P
. 103
VV( T , P )
Z( T , P ) 2 . R . T
P
. 103
Solutions to Chemical and Engineering Thermodynamics, 3e
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl ( T , P )
fv ( T , P )
phil( T , P )
Z( T , P )
0
Z( T , P ) 2
ln Z( T , P ) 0
1
ln Z( T , P ) 2
1
exp ( fl( T , P ) )
fugl( T , P )
. ln
2 . 2 . CB( T , P )
CA ( T , P )
CB( T , P )
phiv ( T , P )
P . phil( T , P )
CA ( T , P )
CB( T , P )
. ln
2 . 2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
exp ( fv ( T , P ) )
P . phiv ( T , P )
fugv( T , P )
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
DELSL( T , P )
DELSV( T , P )
R. ln Z( T , P ) 0
Da ( T ) .
R. ln Z( T , P ) 2
CB( T , P )
ln
2. 2 .b
Da ( T ) .
CB( T , P )
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
DELHL( T , P )
DELHV( T , P )
R . T . Z( T , P ) 0
1
T . Da ( T )
a( T ) .
ln
R . T . Z( T , P ) 2
2. 2 .b
1
T . Da ( T )
a( T ) .
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Ideal gas properties changes relative to the reference state
DELHIG( T )
DELSIG( T , P )
Cp 0 . ( T
Cp 0 . ln
Trs )
Cp 1 . T
2
Trs
2
Cp 2 . T
2
T
Trs
Cp 1 . ( T
3
Trs
3
Cp 3 . T
3
Trs )
Cp 2 . T
2
2
Trs
4
Trs
4
4
2
Cp 3 . T
3
3
Trs
3
P
5
R. 10 . ln
Prs
Solutions to Chemical and Engineering Thermodynamics, 3e
Total entropy and enthalpy relative to ideal gas reference state
SL( T , P )
HL( T , P )
DELSIG( T , P )
DELHIG( T )
DELSL( T , P )
SV( T , P )
DELHL( T , P )
DELSIG( T , P )
DELSV( T , P )
DELHIG( T )
DELHV( T , P )
HV( T , P )
SUMMARY OF RESULTS
T = 298.15
K
P = 140
Pressure, bar
LIQUID
VAPOR
Compressibility
Z( T , P ) 0 = 0.29126
Z( T , P ) 2 = 0.29126
Enthalpy, J/mol
4
HL( T , P ) = 1.03464 10
4
HV( T , P ) = 1.03464 10
Entropy, J/mol K
SL( T , P ) = 67.27151
SV( T , P ) = 67.27151
Fugacity coefficient
phil( T , P ) = 0.365
phiv ( T , P ) = 0.365
Fugacity, bar
fugl( T , P ) = 51.09983
fugv( T , P ) = 51.09983
Volume, m^3/kmol
VL( T , P ) = 0.05157
VV( T , P ) = 0.05157
Number of moles initially
N
Tf
50
Pf
2
3.1416 . ( .01 ) . .06 .
1000
VL( T , P )
N = 0.36552
1.013
Given
SL( Tf , Pf ) SL( T , P )
Tf = 277.04181
K
Tf
Pressure, bar
find( Tf )
Tf = 277.04181
Pf = 1.013
LIQUID
VAPOR
3
Compressibility
Z( Tf , Pf ) 0 = 2.52794 10
Z( Tf , Pf ) 2 = 0.99308
Enthalpy, J/mol
4
HL( Tf , Pf ) = 1.10995 10
HV( Tf , Pf ) = 93.98326
Entropy, J/mol K
SL( Tf , Pf ) = 67.27151
SV( Tf , Pf ) = 0.2929
Fugacity coefficient
phil( Tf , Pf ) = 26.04907
phiv ( Tf , Pf ) = 0.99311
Fugacity, bar
fugl( Tf , Pf ) = 26.3877
fugv( Tf , Pf ) = 1.00603
Volume, m^3/kmol
VL( Tf , Pf ) = 0.05748
VV( Tf , Pf ) = 22.58025
Solutions to Chemical and Engineering Thermodynamics, 3e
U( T , P )
HV( T , P )
U( Tf , Pf )
W
G
4
U( T , P ) = 1.03464 10
HV( Tf , Pf )
N . ( U( Tf , Pf )
U( Tf , Pf ) = 93.98326
3
W = 3.81613 10
U( T , P ) )
W
G = 0.82959
4600
grams of TNT
Note that this answer does not make sense. The reason is that with carbon
dioxide, but the initial and final states would be liquid. Therefore, carbon
dioxide is a poor choice of fluid for this problem, and also problem 3.44. In
the third and later printings, nitrogen is used.
4.50
We start from
d S = CV dT +
F ∂P I
H ∂T K
dV
V
Since the entropy at 0 K is not a function of temperature, it follows that CV = 0.
Also, since the entropy is not a function of specific volume, it follows that
F ∂P I
H ∂T K
V
=0
However, by the triple product rule
F ∂P I F ∂V I FG ∂T IJ = −1 or
H ∂T K H ∂P K H ∂V K
F ∂P I = −FG ∂P IJ F ∂V I = 0
H ∂T K H ∂V K H ∂T K
V
T
V
P
P
T
but from the thermodynamic stability condition
FG ∂P IJ
H ∂V K
<0
T
which implies that
F ∂V I
H ∂T K
4.51
= 0 and α =
P
F I
H K
1 ∂V
V ∂T
=0
P
Rewrite the Clausius equation as
RT
∂V
R
V=
+ b Then
= ;
P
∂T P T
FG ∂ V IJ
H ∂T K
F I
H K
2
2
= 0 (which m eans CP is independent of pressure and equal to CP* )
P
and V − T
F ∂V I
H ∂T K
a) Therefore
=b
P
Solutions to Chemical and Engineering Thermodynamics, 3e
z
z
T2
P2
T1
P1
∆H = H (T2 , P2 ) − H ( T1 , P1 ) = CP* ( T )dT + bdP = 0 or
β 2
γ
( T2 − T12 ) + (T23 − T13 ) + b ( P2 − P1 ) = 0
2
3
is the line of constant enthalpy.
b)
α( T2 − T1 ) +
∆S = S ( T2 , P2 ) − S ( T1 , P1 ) =
z
T2
T1
=
z
T2
z FH
P
2
C*P (T )
∂V
dT −
T
∂T
P
1
C*P (T )
T1
dT −
T
z
P2
P1
I
K
dP = 0
P
R
dP = 0
P
or
αln
T2
γ
P
+ β(T2 − T1 ) + ( T22 − T12 ) − R ln 2 = 0
T1
2
P1
is the line of constant entropy.
c) For the fluid to have a Joule-Thomson inversion temperature
∂T
µ=
must undergo a sign change. However
∂P H
F I
H K
∂T
µ= F I
H ∂P K
LMV − TF ∂V I OP RT + b − T R
H ∂T K Q = − P
P =− b
=−N
P
H
CP
CP
CP
This is always negative, so the Clausius does not have a Joule-Thomson
inversion temperature.
Solutions to Chemical and Engineering Thermodynamics, 3e
5
5.1
(also available as a Mathcad worksheet)
(a) G$ = H$ − TS at P = 25
. MPa and T = 22399
. ° C = 49714
. K
equal with
G$ V = H$ V − TS$ V = 28031
. − 497.14 × 6.2575 = −307 .8 J g
the accuracy
G$ L = H$ L − TS$ L = 962.11 − 497 .14 × 2.5547 = − 307.9 J g
of tables
V
V
L
(b) T ( ° C) T ( K )
$
$
$
H
−
TS
G
|UV
|W
225
498.15 28063
. − 498.15 × 6.2639 =
−314 .1 J g
250
52315
.
28801
. − 523.15 × 6.4085 =
−472 .5
300
57315
.
30088
. − 573.15 × 6.6438 =
−799 .1
350
62315
.
31263
. −
62315
. × 6.8403 =
−11362
.
400
67315
.
32393
. − 673.15 × 7 .0148 = − 14827
.
(Note: All Gibbs free energies are relative to the internal energy and entropy of the liquid phase
$ L − TS$L , we have that G$ L = 0 at
being zero at the triple point. Since H$ L ~ U$ L , and G$ L = H
the triple point.)
(c) T (° C) T ( K)
H$ L
−
TS$ L
G$ V
160
43315
.
675.55 −
43315
. × 19427
.
=
−165.9 J g
170
44315
.
719 .21 − 44315
. × 2.0419 =
− 185.7
180
45315
.
763.22
− 453.15 × 2.1396 =
− 206.3
190
46315
.
807.62 − 46315
. × 2.2359 =
−227 .9
200
47315
.
852 .45 − 47315
. × 2.3309 =
−250.4
48315
.
897.76 − 48315
. × 2.4248 = −273.8
RESULTS
(d)
T (° C)
150
160
180
200
220
3
$
V m kg 0.001091 0.001102 0.001127 0.001157 0.001190
210
c
h
224
0.001197
to
0.07998
T (° C)
$
V m3 kg
c
h
225
250
300
350
400
0.08027 0.08700 0.09890 010976
.
012010
.
(e) Will compute CP from CP ~
FG ∆H$ IJ
H ∆T K
=
P
H$ (T + ∆ T ) − H$ ( T )
∆T
Solutions to Chemical and Engineering Thermodynamics, 3e
T (° C)
a
CP kJ kg K
150
f
170
180
190
200
210
224
4.6225
4.328 4.392 4.430 4.472 4.518 4 .572
to
3.200
T (° C)
a
f
250
300
350
400
CP kJ kg K 2.952 2 .574 2.350 2.260
These results are plotted below.
5.2
dU
dV
= Q& − P
dt
dt
&
dS Q &
Closed system entropy balance:
= + Sgen
dt T
(a) System at constant volume and constant entropy
Closed system energy balance:
dV
dS
= 0 and
=0
dt
dt
dU
Q&
⇒
= Q& and 0 = + S&gen ⇒ Q& = −TS&gen
dt
T
Solutions to Chemical and Engineering Thermodynamics, 3e
and
⇒
dU
= − TS&gen ; T > 0 ; S&gen ≥ 0
dt
dU
≤ 0 or U = minimum at equilibrium at constant V and S.
dt
(b) System at constant entropy and pressure again Q& = − TS&gen .
Now
dP
dV
d
= 0⇒ P
= ( PV ) . Thus
dt
dt
dt
dU
dV
d
= Q& − P
= − TS& gen − ( PV )
dt
dt
dt
and
dU d
d
dH
+ ( PV ) = (U + PV ) =
= − TS&gen ≤ 0
dt dt
dt
dt
Therefore, enthalpy is a minimum at equilibrium at constant S and P.
5.3
(a) The condition for equilibrium at constant T and V is that the Helmholtz free energy A shall be a
minimum.
i) Equilibrium analysis (following analysis in text)
dAI =
FG ∂ A IJ
H ∂T K
I
dT I +
I
V ,M
FG ∂ A IJ
H ∂V K
I
dV I +
I
T ,M
FG ∂ A IJ
H∂M K
I
dM I
I
T ,V
but dTI = 0 , since temperature is fixed, and
FG ∂ A IJ
H ∂V K
I
= − P I and
I
T, M
FG ∂ A IJ
H∂ M K
I
= G$ I
I
T ,V
Thus, following the analysis in the text, we obtain
c
h
c
h
dA = − P I − PII dV I + G$ I − G$ II dM I ⇒ PI = P II and G$ I − G$ II
ii) Stability analysis:
Here again we follow analysis in Sec. 5.2—and find
1 2
2
2
d A = AVV ( dV ) + 2 AVM (dV )( dM ) + AMM ( dM ) ≥ 0
2
This can be rewritten as
a f
a f
1 2
2
d A = θ1 dx1 + θ2 dx2
2
2
≥0
where
θ1 = AVV ; θ2 =
2
AMM AVV − AVM
A2
= AMM − VM
AVV
AVV
Solutions to Chemical and Engineering Thermodynamics, 3e
and
dx1 = dV +
AVM
dM , dx2 = dM
AVV
Thus, θ1 ≥ 0 and θ2 ≥ 0
θ1 =
FG ∂ A IJ
H ∂V K
2
=
2
T, M
or
∂
∂V
FG ∂ A IJ
H ∂V K
T ,M
FG ∂ P IJ
H ∂V K
=
T, M
FG ∂ P IJ
H ∂V K
T ,M
≥0
T, M
≤ 0 as previously found
T ,M
∂
∂M
T ,V
2
AVM
≥0
AVV
FG IJ = −FG ∂ P IJ
H K
H∂ M K
FG ∂ A IJ = FG ∂ G$ IJ
H∂M K H∂ MK
; AVM =
AMM =
FG ∂ P IJ
H ∂V K
( − P) ⇒ −
T, M
θ2 = A MM −
AVV = −
∂
∂V
∂ ∂A
∂ M ∂V
T ,M
T ,V
;
T ,V
T ,V
Now be Eqn. (4.8-17) on a mass basis
LMFG ∂G IJ
MNH ∂ M K
F ∂G IJ
$
Also, dG = VdP − SdT + GdM
⇒G
H∂ MK
FG ∂ G$ IJ
H∂ MK
=
T ,V
1
M
T ,V
T, V
FG ∂ G$ IJ
H∂ MK
=
T, V
FG
H
1
∂P
V
M
∂M
IJ
K
T, V
OP
PQ
F ∂ P IJ
= VG
H∂ M K
− G$
FG
H
∂P
= V$
∂M
IJ
K
+ G$ and
T ,V
= AMM
T, V
so
AMM −
FG IJ − − a∂ P ∂ M f
H K − a∂ P ∂V f
F ∂ P IJ LMV$ + FG ∂V IJ FG ∂ P IJ OP
=G
H∂ M K N H∂P K H∂ M K Q
F ∂ P IJ LMV$ − FG ∂V IJ OP by the triple product
=G
H ∂ M K N H ∂ M K Q rule; Eqn. (4.1 - 6)
2
AVM
∂P
= V$
AVV
∂M
2
T ,V
T ,V
T, M
T ,V
T, M
T ,V
T, P
T ,V
Since
FG ∂V IJ
H∂ MK
T ,P
A2
= V$ ⇒ AMM − VM = θ2 = 0!
AVV
(b) The Gibbs free energy must be a minimum for a system constrained at constant T and P
Solutions to Chemical and Engineering Thermodynamics, 3e
i)
Equilibrium analysis
dG I =
FG ∂G IJ
H ∂T K
I
dT I +
I
P, M
Since T and P are fixed,
dG I =
Thus
FG ∂ G IJ
H∂P K
I
I
FG ∂ G IJ
H∂M K
I
T, M
FG ∂G IJ
H∂M K
I
dM I
I
T ,P
dM I = G$ IdM I
I
c
dPI +
T ,P
h
dG = G$ I − G$ II dM I = 0 and G$ I = G$ II
ii) Stability analysis
FG
H
1 2
∂ 2G
d G = G MM ( dM )2 =
2
∂M2
IJ
K
(dM ) 2 > 0
T ,P
Now
FG ∂ G IJ
H∂ M K
2
=
2
T ,P
FG ∂G$ IJ
H∂ MK
Eqn.
⇒
T ,P
4.9 -10
1
M
LMFG ∂G IJ
MNH ∂ M K
OP
PQ
− G$ =
T ,P
1 $ $
G −G
M
Thus G MM ≡ 0 , and stability analysis gives no useful information.
5.4
(a) At constant M, T and V, A should be a minimum. For a vapor-liquid mixture at constant M, T
and V we have:
A = AL + AV
and at equilibrium dA = 0 = dAL + dAV . Thus
m
r m
dA = 0 = − P L dV L − S L dT L + G$ L dM L + − PV dV V − S V dT V + G$ V dM V
but
M = constant ⇒ dM L + dM V = 0 or dM L = −dM V
V = constant ⇒ dV L + dV V = 0 or dV L = − dV V
T = constant ⇒ dT L + dT V = 0
⇒ dA = − P L − P V dV L + G$ L − G$ V dM L = 0
c
h
c
h
Since dV L and dM L are independent variations, we have that
P L = P V ; and G$ L = G$ V
also T L = T V by constraint that T is constant and uniform.
(b) At constant M, T, and P, G = minimum or dG = 0 or equilibrium.
m
r m
dG = V L dPL − S LdT L + G$ LdM L + V V dP V − S V dT V + G$ V dM V
r
r
Solutions to Chemical and Engineering Thermodynamics, 3e
and M = constant ⇒ dM L = −dM V
P = constant ⇒ dPL = dP V = 0
T = constant ⇒ dT L = dT V = 0
⇒ dG = G$ L dM L + G$ V dM V = G$ V − G$ L dM L = 0
c
$L
$V
or G = G
h
for vapor-liquid equilibrium at constant T and P.
(Also, T and P are uniform—this is implied by constraints.)
5.5
From Sec. 4.2 we have
CP = CV − T
FG ∂V IJ FG ∂ P IJ
H ∂ P K H ∂T K
T
2
FG ∂V IJ FG ∂ P IJ
H ∂ T K H ∂V K
2
= CV − T
V
P
T
a
It is the last form of the equation which is useful here now T > 0 and ∂V ∂ T
f
2
P
≥ 0 . However
FG ∂ P IJ RS< 0
H ∂V K T= 0 at critical point or limit of stability
T
Thus CP > CV in general; except that CP = CV
i) at the critical point or limit of stability of a single phase.
ii) For the substances with zero valuer (or very small value) of the coefficient of thermal expansion
α = 1 V ∂V ∂ T P such as liquids and solids away from the critical point.
a fa
5.6
f
Stability conditions for a fluid are
FG ∂ P IJ
H ∂V K
T
FG ∂ F IJ
H ∂ LK
T
CV > 0 and
<0
for a fiber these translate to
CL > 0 and
>0
Now CL = α + βT ; if CL > 0 for all T, then CL > 0 at T ⇒ 0 implies α > 0 ; CL > 0 as T → ∞
a
f
implies β > 0 . Also, ∂ F ∂ L
5.7
T
= γT > 0 since T > 0 , this implies γ > 0 .
1
P
d U + dV
T
T
1
∂S
P
=
and
= . These relations, together with the equation
T
∂V U T
dU = Td S − PdV ⇒ d S =
Thus
FG ∂ S IJ
H ∂U K
V
FG IJ
H K
S = S o + αln
U
V
+ β ln o
Uo
V
will be used to derive the required equation.
(1)
Solutions to Chemical and Engineering Thermodynamics, 3e
[Note that Eqn. (1), which is of the form S = S (U ,V ) is a fundamental equation of state , in the
sense of Sec. 4.2.]
(a)
(b)
FG ∂ S IJ
H ∂U K
FG ∂ S IJ
H ∂V K
FG U IJ
HU K
=α
V
=
U
−1
o
1
U
=
o
α 1
= ⇒ U = αT
U T
(2)
P
Vo 1
β
= β ⋅ o = . Thus PV = βT .
T
V V
V
(3)
[Clearly, the fluid with an equation of state given by (1) is an ideal gas with constant heat
capacity]
(c) Stability criteria:
FG ∂ S IJ < FG ∂ S IJ FG ∂ S IJ
H ∂U∂V K H ∂V K H ∂U K
2
2
2
2
2
U
FG ∂ S IJ
H ∂V K
2
2
and
V
2
<0;
U
FG ∂ S IJ
H ∂U K
2
2
<0
V
Now
FG ∂ S IJ
H ∂U K
FG ∂ S IJ
H ∂V K
2
V
2
2
U
U|
|V
||
W
∂
∂U
α
α
= − 2 < 0 ⇒ α > 0 for fluid
U
U
V
to be
∂
β
β
=
= − 2 < 0 ⇒ β > 0 stable
∂V U V
V
=
2
[Note: α, β > 0 by problem statement.]
and
∂ 2S
∂
=
∂ U ∂V ∂U
V
FG ∂S IJ
H ∂V K
=
U
∂
∂U
V
β
≡0
V
Thus, the stability criteria yield
α=
FG ∂U IJ
H ∂T K
= CV > 0 since α and β are positive constants
V
from
Eqn. (2)
and
β= −
FG IJ
H K
V 2 ∂P
T ∂V
>0⇒
T
FG ∂ P IJ
H ∂V K
<0
T
Thus, fluid is always stable and does not have a first order phase transition.
5.8
a
At limit of stability ∂ P ∂V
f
T
= 0 for the van der Waals equation:
FG P + a IJ (V − b) = RT
H VK
2
So that at limit of stability
FG ∂ P IJ
H ∂V K
2a
V
3
=
RT
(V − b )
2
=0=
T
=
− RT 2 a
+
= 0 ; or
V − b V3
FG
H
IJ
K
FG
H
1
RT
1
a
=
P+ 2
V −b V −b
V −b
V
IJ
K
Solutions to Chemical and Engineering Thermodynamics, 3e
Thus P = a(V − 2b) V 3 ; or using
a = 3PCVC2 and b =
VC
3
⇒
Pr =
3Vr − 2
Vr3
To obtain the envelope, we compute Pr for various values of Vr
Vr
10
2
1
0.8
0.7
Pr
0.028
0.5
1.0
0.781
0.0343
a
f
Notice, that the critical point Vr = 1, Pr = 1 is the upper limit of metastability (i.e., Pr ≤ 1 ), as well
as the limit of single phase stability.
5.9
T and P will be taken as the independent variables at a second order phase transition
FG ∂ G IJ
H ∂T K
F ∂ G IJ
, where V = G
H ∂P K
G I = G II ; S I = SII , since S = −
Then
and
V I = V II
P
T
and, of course, T = T and P = P .
From S I = SII we have that along the 2nd order phase transition curve that d SI = dS II or
I
II
FG ∂ S IJ
H ∂T K
I
II
FG ∂ S IJ dP = FG ∂S IJ dT + FG ∂ S IJ
H ∂ P K H ∂T K H ∂ P K
F ∂V IJ dP = C dT − FG ∂V IJ dP
C
⇒
dT − G
T
H ∂T K
T
H ∂T K
I
I
dT +
II
P
T
I
P
II
P
I
II
P
dP
T
II
P
P
Thus
FG ∂ P IJ
H ∂T K
along
transition
curve
=
CPI − CPII
od
T ∂V I ∂ T
i − d∂V
II
P
∂T
it
(1)
P
Similarly, equating dV I = dV II yields
FG ∂V IJ
H ∂T K
I
dT +
P
FG ∂V IJ
H ∂PK
T
FG ∂V IJ
H ∂T K
od
d∂V
i − d∂V
∂ Pi − d∂V
I
dP =
II
dT +
P
FG ∂V IJ dP
H ∂P K
II
T
Thus
F dP I
H dT K
along
transition
curve
=
− ∂V I ∂ T
I
II
P
II
T
it
∂ Pi
∂T
P
(2a)
T
However, since V I = V II , we can divide numerator and denominator by V and obtain
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂ P IJ
H ∂T K
=
along
transition
curve
αI − αII
κ IT − κIIT
(2b)
Note: The Clausius-Clapeyron equation is
T
FG ∂ P IJ
H∂T K
along
transition
curve
=
H I − H II
(3)
V I − V II
However, this form is indeterminate for a 2nd order phase transition. Applying L’Hopital’s rule to
eqn. (3) taking derivatives of numerator and denominator with respect to T at constant P
T
FG ∂ P IJ
H∂T K
=
along
transition
curve
CPI − CPII
d∂V
i d
∂ T − ∂V II ∂ T
I
i which is eqn. (1)!
Similarly, applying L’Hopital’s rule, but now taking derivatives with respect to P at constant T.
F ∂ P IJ
TG
H∂T K
c∂ H
=
c∂V
along
transition
curve
=
VI
h − c∂ H ∂ P h
∂ P h − c∂V ∂ Ph
− T c ∂V ∂ T h − V + T c∂ V
c∂V ∂ Ph − c∂V ∂ P h
∂P
I
II
T
T
I
II
T
T
I
II
II
P
I
∂T
h
P
II
T
T
but V I = V II so that
F ∂ P IJ
⇒ TG
H ∂T K
along
transition
curve
d∂V
= −T
d∂V
I
I
i − d∂V
∂ P i − d∂V
∂T
II
P
II
T
i
∂ Pi
∂T
P
T
which is eqn. (2a)!
5.10 (a)
F dP I
H dT K
along
transition
curve
=
∆H
T ∆V
⇒
dP
H$ − H$ S
335
. × 105 J kg
= L
=
d ln T
V$L − V$S
−0.000093 m3 kg
= −361
. × 103 J m3
dP
T
= − 3.61 × 109 J m3 = −3.61 × 109 Pa ⇒ P2 − P1 = − 3.61 × 109 ln 2
d ln T
T1
RS −2.985 × 10 a P − P fUV
T Pa
W
−10
or T2 = T1 exp
(b)
dP
∆H
=
dT along
T
∆V
transition
curve
but
∆V ≈ V
2
V
~
RT
P
1
Solutions to Chemical and Engineering Thermodynamics, 3e
dP ∆ H P d ln P ∆ H
=
;
=
dT
RT 2
dT
RT 2
If we assume that ∆H is constant, then
ln
FG
H
IJ
K
LM
N
P2
∆H 1
1
1
R
P
=−
−
⇒ T2 =
−
ln 2
P1
R T2 T1
T1 ∆ H P1
OP
Q
−1
(c) Denver: P2 = 846
. × 105 Pa
m
c
T2F = 273.15 exp −2.985 × 10− 10 8.46 × 104 − 1.013 × 105
hr
= 273.15 K = 0° C ( freezing point essentially unchanged )
T2B =
LM 1 − 8.314 × 10 ln 8.46 × 10 OP
.
2.255 × 10 × 18 1.013 × 10 Q
N 37315
3
6
4
5
−1
= 368 .08 K = 94.9° C
Solutions to Chemical and Engineering Thermodynamics, 3e
5.11 This problem involves the application of the Clausius-Clapeyron equation. We will assume that the
heats of fusion, sublimation and vaporization are all constant. Thus we will use
P
∆H 1
1
ln 2 = −
−
in all cases. Now ∆ H vap = ∆ H sub − ∆ H fus . To calculate ∆H sub we
P1
R T2 T1
will use the following sublimation data:
FG
H
IJ
K
State 1: Triple point; T = 112.9° C = 386.05 K ; P = 1157
. × 104 Pa
3
State 2: T = 1054
. °C = 378.55 K ; P = 8.00 × 10 Pa
⇒ ln
FG 1157
.
× 10 I
∆H F 1
1 I
= 0.369 = −
−
J
H
H 8.00 × 10 K
R
386 .05 378 .55K
4
sub
3
⇒ ∆ H sub = 5.980 × 10 4 J mol
∆ H fus = 1527
.
× 104 J mol ⇒ ∆ H vap = (5.980 − 1527
. ) × 104
= 4.453 × 104 J mol
and ∆H vap R = 5356 K . To find the normal boiling temperature we again use Clausius-Clapeyron
equation.
FG 1013
.
× 10 I − 4.453 × 10 F 1
J = 8.314 GH T − 3861.05IJK
H 1157
.
× 10 K
5
ln
4
4
2
State 1 = T. P.
State 2 = N.B. P.
⇒ T2 = 457 .7 K = 184 .5° C
Experimental value = 183° C ; difference due to assumption that ∆H vap is a constant.
5.12 (a) At equilibrium P sat (ice) = Psat (water )
Equating the ln P sat ’s gives
28.8962 −
614.01
54328
.
= 26.3026 −
T
T
⇒
T = 273.1° C
and
ln Psat (ice) = 288962
.
−
(b) ln P = A −
61401
.
= 6.4096 ⇒ P sat = 607.7 Pa
2731
.
B
d ln P ∆ H
and
=
also
T
dT
RT 2
d ln P
B
= + 2 ⇒ ∆H = B ⋅ R
dT
T
Thus
∆H
R
∆H
R
∆H
ice→ vapor
water → vapor
fus
= 61401
. and ∆H sub = 5105
. × 104 J mol
= 54328
. and ∆H sub = 4517
. × 104 J mol
= ∆ H ice→ water = ∆ H sub − ∆ H vap = 5.880 × 10 3 J mol
Solutions to Chemical and Engineering Thermodynamics, 3e
5.13 (Also available as a Mathcad worksheet. The Mathcad solution includes graphs.)
(a) Use the Clausius-Clapeyron equation
a
f
∆H vap ln P2 P1
=
R
1 T1 − 1 T2
1
and graphically taking slope, I find ∆H vap ~ 42700 J mol .
T
(b) The vapor pressure is low enough that the ideal gas approximation should be valid—thus
Plotting ln P vs.
d ln Pvap ∆ ln P vap ∆ H vap
=
=
dT
∆T
RT 2
either graphically or analytically, we find
∆H vap ~ 313,600 J mol
5.14 (a) Start with Eqn. (5.4-6)
f = P exp
but
RS 1 FV − RT I dPUV ⇒ ln f = F PV − 1I dP
T RT z H P K W P z H RT K
P
0
0
1
d ( PV ) dV dZ dV
dP =
−
=
−
so
P
PV
V
Z
V
z
z
Z
ln
P
V
F
H
z FGH
I
K
IJ
K
V
f
dZ
PV
dV
Z
P
1
=
( Z − 1)
−
−1
= ( Z − 1) − ln −
−
dV
P Z =1
Z V = ∞ RT
V
1 V = ∞ RT V
or
ln
(b) Z = 1 +
f
1
= ( Z − 1) − ln Z +
P
RT
FG
H
B(T )
RT
B
and P =
1+
V
V
V
ln
FG
H
IJ
K
=
z
V =∞
IJ
K
RT
− P dV
V
(Eqn. 5.4-8)
IJ
K
f
B
B
1
= − ln 1 +
+
P V
V
RT
V
z FGH
V
z LMNFGH
V
V =∞
RT RT RT B
−
−
⋅
V
V
V V
B
1
ZB
− ln Z + B
dV =
− ln Z
2
V
V
V =∞ V
IJ OP dV
KQ
Solutions to Chemical and Engineering Thermodynamics, 3e
(c)
vdW e.o.s.
P=
z FGH
V
V =∞
RT
a
PV
V
a
−
; Z vdW =
=
−
V −b V2
RT V − b V 2 RT
z FGH
IJ
K
IJ
K
V
V
RT
RT
RT
a
V
a
− P dV =
−
+ 2 dV = RT ln
−
V
V
V −b V
V − b V V =∞
V =∞
= RT ln
V
a
−
V −b V
so
vdW
f
ln
P
V
a
−
+ ( Z − 1) − ln Z
V − b RTV
Pb
a RT P
= ln Z − ln Z −
−
⋅
⋅
+ ( Z − 1) − ln Z
RT
RT PV RT
= ln
F
H
= ( Z − 1) −
aP
I
K
A
− ln( Z − B)
Z
Pb
.
RT
( RT )
(d) Peng-Robinson equation of state. Start with
where A =
2
z
V
and B =
a
FG RT − PIJ dV = z LM RT − RT +
OPdV
HV K
N V V − b V (V + b) + b(V − b)Q
L V + d1 + 2 ib OP
V
a
= RT ln
−
lnM
V − b 2 2 b NM V + d1 − 2 ib QP
LV + d1 + 2 ib OP
Z
a
= RT ln
−
lnM
Z − B 2 2 b MNV + d1 − 2 ib PQ
V
V =∞
V =∞
[See solution to Problem 4.2 for integral]. Therefore
ln
f
PR
P
= (Z − 1) − ln Z + ln
= (Z − 1) − ln( Z − B ) −
5.15 (a)
LM d
MN d
i OP
i PQ
V + 1+ 2 b
Z
a
−
ln
Z − B 2 2 bRT
V + 1− 2 b
a
2 2bRT
ln
d i
V + d1 − 2 ib
V + 1+ 2 b
a f
fHliq2 S = f Hvap
; f vap = P f P , where the fugacity coefficient,
2S
corresponding states.
20
= 0.2237
89.42
255
. + 273.15
TC , H 2 S = 373.2 K ⇒ Tr =
= 0.8002
373.2
ZC , H 2 S = 0.284, which is reasonably close to 0.27
PC , H 2 S = 89.42 bar ⇒ Pr =
f P will be gotten from
Solutions to Chemical and Engineering Thermodynamics, 3e
f
= 0.765 , fH 2 S = 20 × 0765
.
= 153
. bar .
P
(b) For a liquid, from Eqn. (5.4-18)
Fro m Fig. 5.4-1,
f = P vap
FfI
H PK
LM
MN z
P
exp
sat
P vap
V
dP
RT
OP
PQ
Since Pvap = 6.455 × 103 Pa at the temperature of interest, we will assume that
Also, we will consider the liquid to be incompressible. Thus
z
P
P vap
V
V
dP =
RT
RT
z
P
dP =
c
V P − P vap
P vap
h
RT
and
fH 2 S = P vap exp
LMV c P − P
MN RT
vap
h OP = 6455 expL 0.018( P − 6455) O Pa
MN 8.314 × 10 × 310.6PQ
PQ
3
so that
Pressure, Pa
P = 10
. × 10
50
. × 107
10
. × 108
7
fH 2 S, Pa
6,921
9,146
12,960
Reported
6,925
9,175
12,967
5.16 (also available as a Mathcad worksheet)
(a) There are (at least) two ways to solve this problem. One way is to start from
f = P exp
RS 1 FV − RT I dP UV
T RT z H P K W
P
0
or
RT ln
f
=
P
z FH
P
V−
0
I
K
RT
dP
P
RT 8.314 × 10−6 MPa ⋅ m3 mol K × ( 27315
. + 400 )K 0.310748
=
=
m3 kg
−3
P
P(MPa ) × 18.01 g mol × 10 kg g
P
a f Pf
sat
~ 1.
Solutions to Chemical and Engineering Thermodynamics, 3e
From Steam Tables T = 400° C
P MPa
V$ m3 kg
V$ − RT P
0.01
0.05
0.10
0.2
0.3
0.4
0.5
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
31.063
6.029
3.103
1.5493
1.0315
0.7726
0.6173
0.5137
0.3843
0.3066
0.2548
0.2178
0.19005
0.16847
0.15120
–0.0118
–0.00596
–0.00448
–0.00444
–0.00433
–0.00427
–0.00420
–0.00421
–0.00413
–0.00415
–0.00416
–0.00416
–0.00417
–0.00417
–0.00417
By numerical integration of this data we find that
f
~ −0.0084 MPa ⋅ m3 kg
P
f
−0.0084 MPa ⋅ m3 kg
⇒ ln =
= −0.027032
P 0.310748 MPa ⋅ m3 kg
RT ln
so f P = 0.97333 and f = 1947
.
MPa . A second way to use the steam tables is to assume that
steam at 400°C and 0.01 MPa is an ideal gas. From the steam tables, at these conditions, we have
H$ = 3279.6 kJ kg ; S$ = 9.6077 kJ kg K
⇒ G$ = H$ − TS$ = 3279.6 − 673.15 × 9.6077 = −3187.8 kJ kg
= −57412.7 kJ kmol = G ( 400° C, 0.01 MPa) = G IG( 400° C, 0.01 MPa)
Also
G IG( T = 400° C, 2 MPa) = −574127
. kJ kmol +
z
2 MPa
0.01
z
2 MPa
G IG( T = 400° C, 2 MPa) − G IG( T = 400° C, 0.01 MPa ) =
V IGdP
P = 0.01 MPa
RT
dP
P
= −57412.7 + 8.314 ln 200 = −277603
. kJ kmol
Also, from steam tables
G (T = 400° C, 2 MPa) = (3247.6 − 673.15 × 7 .1271) × 18.01
= − 2791563
. kJ kmol
FG
H
IJ
K
F
H
f
G − G IG
−2791563
. + 277603
.
= exp
= exp
P
RT
8.314 × 67315
.
= 0.9726
f = 0.9726 × 2 MPa = 1.945 MPa
I
K
Solutions to Chemical and Engineering Thermodynamics, 3e
a
(b) Corresponding states TC = 647.3 K, PC = 22.048 MPa, w = 0.344
Tr =
f
400 + 273.15
2
= 1.04 ; Pr =
= 0.0907
647 .3
22.048
From corresponding states chart (actually from Table in Hougen, Watson and Rogatz, Vol. II, p.
601) we have
f
= 0.983 ⇒ f = 1.966 MPa
P
(c) Using the program PR1 we find
f = 19.40 bar = 1940
.
MPa
Comment: The steam table results are probably the most accurate, and the corresponding states
results the least accurate. Note that with the availability of the computer program PR1, the P-R
e.o.s. is the easiest to use. The results would be even more accurate if the PRSV equation was
used.
5.17 (a)
PV
B C
RT BRT CRT
= 1 + + 2 +L ; ⇒ V =
+
+
+L
RT
V V
P
V P V 2P
Thus
PV
B
= 1+
2
RT
RT P + BRT V P + CRT V P + L
+
=1+
C
2
RT P + BRT V P + CRT V P + L
a
B P RT
k
p
1 + B RT P + L
−1
k
f
p
+ C RT P + L
−2
+L
+
+L
a
Now keeping terms of order 1, B, B 2 and C only yields
F I F1 − B P + LI + CF P I
H K H RT K H RT K
BP
F P I +L
= 1+
+ cC − B h
H RT K
RT
PV
P
= 1+ B
RT
RT
2
2
2
(b) V =
c
RT
+ B + C − B2
P
hFH RTP IK + L and V
IG
=
RT
. Therefore
P
hFH RTP IK + L and
R 1 L B + cC − B h P + LOdPUV
f
= expS
P
RT
QP W
T RT z NM
c
V − V IG = B + C − B 2
P
2
0
or
f
2
C P RT
1 + 2B V + L
Solutions to Chemical and Engineering Thermodynamics, 3e
R|
S|
T
c
2
f
BP C − B
= exp
+
P
RT
2
hF P I
H RT K
2
U|
V|
W
+L
We will consider a number of alternatives for using the virial coefficient data. The first is to
start with Eqn. (5.4-6a)
RS
T
f
1
= exp
P
RT
z FH
P
V−
0
I UV
K W
RS
T
RT
1
dP = exp
P
RT
P
V − V IG dP
0
zd
P
thus we need to evaluate the integral 1 RT
i UVW
zd
(1)
i
V − V IG dP . Since the truncated virial equation
0
P RT = 1 V + B V 2 + C V 3 can not easily be solved for V as a function of T and P, the
following procedure will be used:
i) Choose values of V and compute
P=
FG
H
RT
B C
1+ + 2
V
V V
IJ
K
and PIG =
RT
V
ii) Plot P and P IG as a function of V
iii) Use these two plots to obtain V IG and V (real gas) at the same value of P, also compute
V (T , P )− V IG (T , P )
iv) Finally, use a numerical or graphical integration scheme to get f P as a function of P
Same representative values of V − V IG are given below
c h
−dV − V i
P 10 6 Pa
IG
1
2
3
4
5
6
7
8
0.187
0.180
0.187
0.223
0.206
0.211
0.201
0.180
9
10
11
12
13
14
15
m3 kmol
c h
−dV − V i
P 10 6 Pa
IG
0.1573 0.1384 0.1215 0.1069 0.0944 0.0834 0.0739
3
m kmol
Using the data and performing the integration we obtain
c
P 10 6 Pa
f P
h
1
3
5
7
9
10
11
13
15
0.939 0.822 0.703 0.602 0.527 0.499 0.475 0.439 0.412
FG ∂ P IJ dV and, with
H ∂V K
RT F
P=
G 1 + B + C IJ , that FGH ∂∂VP IJK = FGH − V1 − V2B − V3C IJK RT
V H V V K
An alternative is to note that dP =
T
2
T
2
3
4
Solutions to Chemical and Engineering Thermodynamics, 3e
and
z
z
z
cV − V hdP = RT1 V FGH ∂∂VP IJK dV − RT1 RTP dP
L F 1 + 2B + 3C I dV − dP OP
= lim M−
PP
MN a f GH V V V JK
Q
L
F 1 − 1 IJ + 3C FG 1 − 1 IJ OP
= lim M− ln PV + ln P V a P f + 2 BG
H V ( P) V a P f K 2 H V ( P) V a P f K Q
N
1
RT
V(P)
P
0
IG
P
T
V (0 )
z
P0 → 0
0
z
V (P )
P
2
3
V P0
P0
0
P0 → 0
0
2
2
0
2 B 3C
= − ln PV + ln RT +
+
V
2V 2
[Note:
0
a f
lim V P0 = ∞ ]
P0 → 0
Thus
RS
T
f
RT 2 B 3C
= exp ln
+
+ 2
P
PV V
2V
UV = RT expRS 2B + 3C UV
W PV T V 2V W
2
but PV RT = 1 + B / V + C / V 2 and
m
exp 2 B V + 3C 2V
f
=
P
1+ B V +C V 2
r = expm−0.3326 V + 0.01938 V r
2
2
1 − 01663
.
V + 0.012921 V 2
(2)
for V in m3 kmol .
The use of this equation leads to results that are somewhat more accurate than the graphical
integration scheme. Still another possibility is to use the results of part (a) which yields
m
r
f
= exp −0.00619 P − 1.0207 × 10−5 P2 for P in bar
P
(3)
The results of using this equation are listed below.
Finally we can also compute f P using corresponding states (Figure 5.4-1). For methyl
a f
fluoride TC = 317.7 K and PC = 5.875 MPa (note
ZC is unknown is V C has not been
measured). Thus, Tr = ( 27315
. + 50) 317 .7 = 1.017 , and for each pressure Pr can be computed,
and f P found from Fig. 5.4-1.
The results for each of the calculations are given below:
P (bar)
eqn. (1) f P
10
30
50
70
100 130 150
0.939 0.822 0.703 0.602 0.499 0.439 0.412
eqn. (2) f P
0.939 0.822 0.710 0.607 0.503 0.442 0.416
eqn. (3) f P
Corresponding
states f P
0.939 0.823 0.715 0.617 0.486 0.376 0.314
0.96
0.85
0.72
0.60
0.47
0.39 0.345
Note that at low pressure, all the results for f P are similar. At high pressures, however, the
results differ. Equation (3) is approximate, and probably the least accurate. Equation (2) should
be the most accurate, except that there is a question as to how accurate it is to use an equation of
state with only the second and third virial coefficients for pressures as high as 150 bar.
Solutions to Chemical and Engineering Thermodynamics, 3e
5.18 (a) Assume the vapor phase is ideal, and that ∆H vap is approximately constant (or an average
∆H vap can be used).
ln
⇒ ln
⇒
FG
H
P2
∆ H vap 1
1
=−
−
P1
R
T2 T1
IJ
K
(1)
F 2.026 I = − ∆ H F 1 − 1 I
H 1013
K R H 222 .0 + 27315
.
.
178.0 + 27315
. K
vap
vap
∆H
R
= 3.52 × 103 K
∆ H vap = 2.93 × 104 J mol
(b) ∆H vap(T) = H(sat. vap,T )− H(sat. liq, T)
IG
IG
= H (sat. vap, T ) − H ( T ) − H (sat. liq. , T ) − H (T )
⇒ ∆ H vap( T ) = TC
LMF H − H I
MNGH T JK
IG
C
−
sat. vap, T
FG H − H IJ
H T K
IG
sat. liq., T
T
200 + 27315
.
(c) Tr =
=
= 0.851
TC 2831
. + 27315
.
FG H − H IJ
H T K
IG
C
= 5.06 J mol K and
sat. vap
Tr = 0 .851
FG H − H IJ
H T K
IG
C
OP
PQ
= 44.69 J mol K
sat. liq.
Tr = 0 .851
∆H vap(T) = 556.45 K 44.69 − 506
. = 2205
. × 104 J mol
(d) The reason for the discrepancy is probably not the inaccuracy of corresponding states (since
ZC = 0.272 which is close to 0.27) but rather the assumption of an ideal vapor phase in the
Clausius-Clapeyron equation. We correct for gas-phase nonideality below.
at T = 178° C , Tr = 0.811 , Z = 082
.
T = 222° C , Tr = 0.890 , Z = 0.71
The average value of the compressibility is
Z=
1
(0.82 + 0.71) = 0.765
2
We now replace eqn. 1 with
vap
ln
P2 −∆ H
=
P1
ZR
FG 1 − 1 IJ ⇒ ∆ H
HT T K
2
vap
c
h
= 0.765 × 2.93 × 104 J mol
1
= 2.24 × 104 J mol
which is in much better agreement with the result of part (c). A better way to proceed would be
to compute the compressibility as a function of temperature, i.e., find Z = Z (T , P ) and then
integrate
dP
∆ H vap P
=
dT Z (T , P ) RT 2
Solutions to Chemical and Engineering Thermodynamics, 3e
rather than use an average value of Z, i.e. Z .
5.19 Basis: vessel volume 1 m3 (cancels out of problem)
x = fraction of vessel filled with liquid water
x (1 − x)
N = total number of moles of water = L + V (per 1 m3 of vessel)
V
V
1) Total number of moles same at all conditions
2) x is the same at initial loading and at critial point
State 1—low pressure V V >> V L
⇒ N1 =
State 2—Critical point V
V
=V
L
x
L
V
+
(1 − x )
V
= V C ⇒ N2 =
N1 = N2 ⇒
x
V
L
=
V
≈
x
VL
1− x
x
1
+
=
but
VC VC VC
1
VL
or x =
VC
VC
(a) Using steam tables
V L (25° C) = 0001003
.
m3 kg ; V C = 0.003155 m3 kg
x=
0.001003
= 0.3179
0.003155
⇒ Initial fill should contain 31.79% of volume with liquid (which was reported in the Chemical
and Engineering News article).
(b) Peng-Robinson equation of state
V L (25° C) = 0.2125 × 10−4 m3 mol
and the P-R equation of state predicts ZC = 0.3074 (solution to Problem 4.11b) so
VC =
ZC RTC 0.3074 × 8.314 × 10− 6 MPa m3 mol K × 647.3 K
=
PC
22.048 MPa
= 0.75033 × 10− 4 m3 mol ⇒ x = 0.2832
or an initial fill of 28.32% of volume with liquid.
5.20 (a) One theory for why ice skating is possible is because ice melts due to the pressure put upon it
under the ice skates, and then refreezes when skate leaves and the pressure is released. Skate
actually moves over a film of water on the sheet of ice. To find the lowest temperature we use
the Clapeyron equation to calculate the change in freezing point as a result of the applied
pressure. Properties of ice:
∆H$ fus = ∆H$ sub
ρ = 090
. g cc ⇒ V S = 111
. cc g
vap
$
− ∆H = 28348
. − 25013
. = 3335
. J g (at 0.01°C)
(Appendix III)
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂ P IJ
H ∂T K
=
sat
∆H$
∂P
333.5 J g
⇒
=
∂ T 27315
. K × (1 − 111
. ) cc g
T ∆V$
= − 1110
. J cc K = − 111.4 bar K
or
FG ∂T IJ ~ − 0.0090 K bar ⇒ ∆T = −0.009 K bar (∆P )
H ∂ PK
Assume 70 kg person on 0.6 cm2 skate area (well sharpened)
70
= 116.7 kg cm2 × 0.9807 bar kg cm2 = 114.4 bar
0.6
⇒ ∆T = −103
. °C
∆P =
assuming skate makes complete contact with ice. If the surface is irregular (as it is) maybe
contact only over 10% of area. In this case ∆T = −93
. ° C . My observation in Minnesota was
that it was possible to skate down to ~ −20° C (5% contact area??). Of course, the
thermodynamic model for this process may be incorrect. Other possibilities include the melting
of ice as a result of friction, or by heat transfer from the skater’s foot to the ice. I believe the
thermodynamic theory to be the a reasonable explanation of the phenomena.
(b) Since ∆H$ fus > 0 and ∆V$fus > 0 for CO 2 and most other materials, freezing point will be
elevated not depressed. Liquid film can not form and ice skating is impossible.
(c) More difficult to quantify. Similar to (a) for freezing point depression, which on release of
pressure causes refreezing and formation of snowball; but in this case there is also considerable
heat transfer from the hands to the surface of the snowball that causes melting.
Solutions to Chemical and Engineering Thermodynamics, 3e
8
−1
V
3
−
3
c h
V 2
=
8
L
3
−1
−
3
,
c h
L 2
or, solving for the reduced temperature
{c
h c h}
h c
hr
2
2
1 V − 1 L
3
=
8 1 3V V − 1 − 1 3V L − 1
mc
ln
R| c3
S|c3
T
+
6
V
V
V
iii) Use
L
and
and
−1
V
L
. Using Eqn. (3) in Eqn. (2) gives'
L
(4)
L
+
L
V
(c) Procedure used in solution
i) Guess (or choose) a value of
ii) Use
and
h U|V + c3 − 1h − c3 − 1h
− 1h |
W
RS 1 − 1 UV = 0
T3 − 1 3 − 1 W
−1
L
V
V
,
This is an independent equation between
(3)
L
V
L
, compute
so obtained to compute
from Eqn. (3).
=
and Eqn. (1) to get
which satisfies Eqn. (4).
vap
. [Note: This calculation was done on a
digital computer.] Results are at end of problem solution.
(d) The Clausius-Clapeyron equation can be written as
coexistance
curve
=
∆
∆
vap
=
=
∆
vap
∆
or
vap
Since
=3
2
vaporization, ∆
;
vap
=
∆
vap
1
∆
=
∆
vap
∆
∆
vap
=
∆
9 ∆
vap
∆
has units of (energy )−1 , so define a reduced heat of
= 3 . Now 9
, to be 9
2
3
vap
=
. Thus
vap
∆
Thus ∆ vap can be computed by taking derivatives of results of part (c). (This was done
graphically). The results are as shown below.
V
L
∆
vap
3 × 103
0.3690
2.617 × 10−4
0.2948
1 × 10
0.3745
−4
0.3300
3
8 × 10
2
5 × 102
8772
.
×10
0.3758
1123
. ×10
−3
0.3382
0.3789
1894
.
× 10−3
0.3571
vap
0.3124
9.1
0.3341
8.69
0.3477
8.69
0.3686
8.51
Solutions to Chemical and Engineering Thermodynamics, 3e
0.3829
33503
.
× 10−3 0.3802
15
. × 10
0.3896
7.299 × 10
−3
100
0.3947
0.01153
0.4423
75
0.3985
0.01597
0.4621
60
0.4020
0.02056
0.4787
50
0.4052
0.02527
0.4931
40
0.4095
0.03254
0.5119
30
0.4158
0.04505
0.5383
20
0.4268
0.07112
0.5799
15
0.4365
0.09810
0.6129
10
0.4538
0.1534
0.6649
8
0.4658
0.1954
0.6965
6
0.4849
0.2650
0.7402
5
0.4998
0.3198
0.7697
4
0.5226
0.3996
0.8072
3
0.5596
0.5240
0.8573
2
0.6410
0.7332
0.9270
1.8
0.6710
0.7899
0.9437
1.5
0.7364
0.8830
0.9697
1.0
1.0
1.0
1.0
3 × 102
2
Results are plotted below.
0.3986
8.77
0.4297
8.87
0.4522
8.70
0.4704
8.52
0.4859
8.54
0.5025
8.62
0.5251
8.53
0.5591
8.85
0.5964
8.01
0.6389
7.81
0.6807
7.51
0.7184
7.33
0.7550
6.97
0.7885
6.64
0.8323
5.86
0.8922
4.89
0.9354
3.94
0.9567
3.22
0.9849
1.43
0.4171
Solutions to Chemical and Engineering Thermodynamics, 3e
The reduced vapor pressure and heat of vaporization for the van der Waals fluid.
5.30 (a) Restricted form of Gibbs Phase Rule:
=3−
⇒
must be 3 or less
⇒ quaternary point can not exist in a 1-component fluid.
(b) 2 phases ⇒ = 3 − 2 = 1 degree of freedom.
Thus, if any property of
of the phases is specified, this is sufficient to fix all of the
thermodynamic properties of both phases! However, if only the total molar volume of the twophase system (or some other two-phase property), i.e., = I I ( , )+ II II ( , ) , this is not
sufficient to solve for I , I and II . That is, many choices of and can yield the same
value of the total molar volume by varying the mass distribution between the two phases.
Consider now the situation in which the total molar volume and total molar enthalpy is
specified. In this case we have
1=
I
+
=
I
I
=
I
and
II
( , )+
I
( , )+
II
II
II
( , )
II
( , )
The unknowns here are I , II and either or (note that since the system has only one
degree of freedom, either or , but not both are independent variables). Thus, given equation
and
to and , and the two-phase coexistance curve), the
of state information (relating
equations above provide 3 equations to be solved for the 3 unknowns.
5.31 (a)
=
FG ∂ IJ
H∂ K
+
FG ∂
H∂
FG ∂
H∂
IJ
K
IJ
K
. Thus
sat
curve
=
FG ∂ IJ + FG ∂ IJ FG ∂ IJ
H∂ K H∂ K H ∂ K
sat
curve
Solutions to Chemical and Engineering Thermodynamics, 3e
but
FG ∂ IJ
H∂ K
and, by the Maxwell relations
Thus
sat
FG ∂ IJ
H∂ K
=
=
P
;
FG ∂ IJ
H∂ K
FG ∂ IJ
H∂ K
=−
=
−α
sat
curve
sat
curve
FG ∂ IJ
H∂ K
P
∆
∆
=
=
P
−α
∆
vap
∆
vap
=
vap
=
vap
P
vap
vap
=− α
and
sat
∆
∆
−
P
−
FG ∂ IJ
H∂ K
FG ∂ IJ
H∂ K
∆
∆
∆
vap
∆
vap
vap
vap
;
where denotes the phase.
(b) For the liquid
∆
For
a∂
[
the
f
∂
P
~
a
and ∂
vap
vapor
>>
L
we
a f− a f a
2
V
sat
=
2
∂
f
, and α ~ 0 ⇒
1
will
−
1
f and
V
P
− αV
L
sat
~
use
V
∆
vap
∆
vap
P
V
P
F I
H K
F I
H K
~
a f− a f a
2
2
2776.4 − 26761
.
= 2.006 kJ kg K ;
50
19364
.
− 1.6958
=
= 0.00481 m3 kg K
50
=
V
P
= 14.6 kJ kg K and
= 2 .006 kJ kg K − 0.00481 m3 kg K ×
= −4.488 kJ kg K
at 370°C
2
−
1
f
and
to be the next higher temperature in
= 0.000141 m3 kg K . Thus, at 100°C
V
sat
1
will be evaluated using finite differences above and steam tables. In each
case 1 will be taken as the saturation temperature, and
the steam tables.]
Thus, at 100°C (0.1 MPa)
at 370°C (21.0 MPa)
L
P
22570
. kJ kg
1.6719 m 3 kg
Solutions to Chemical and Engineering Thermodynamics, 3e
V
sat
= 14.6 kJ kg K − 0.000141 m3 kg K ×
441.6 kJ kg
0.0027 m 3 kg
= −8.359 kJ kg K
5.32 (a) Multiply out the terms, and this is easily proved.
∂2
∂ ∂
=
(b)
=
∂
∂
,
=
but
∂2
∂ ∂
=
FG ∂ IJ
H∂ K
1
∂
∂
=−
F 1 I = − 1 FG ∂ IJ
H K
H∂ K
,
−
F− I
H K
1 F∂ I
GH ∂ JK
⇒
=−
,
=
−
=
2
,
+
=
Now
∂
∂
+
FG ∂ IJ
H∂ K
FG
H
FG ∂
H∂
1 ∂
∂
2
IJ
K
IJ
K
,
FG ∂ IJ
H∂ ∂ K
2
=−
so that
,
FG ∂ IJ
H∂ K
1 F∂ I F ∂ I
GH ∂ JK GH ∂ JK
−
2
,
,
=−
+
2
; thus
FG ∂ IJ
H∂ K
=
FG ∂ IJ
H∂ K
−
and
=−
R|S FG ∂ IJ
T| H ∂ K
1
V
but
=−
(c)
+
=
=
2
2
FG ∂ IJ
H∂ K
−
V
+
+
2
V
. Thus
=−
2
V
,
V
,
2
2
FG ∂ IJ
H∂ K
FG ∂ IJ
+
H∂ K
IJ FG ∂ IJ +
K H∂ K
2
,
,
but
U|V +
W|
FG ∂ IJ +
H∂ K
∂ ∂
∂ F I
1F∂ I
=
=
= G
∂ ∂
∂ H K
H ∂ JK −
∂ ∂
F ∂a fIJ = − 1 FG ∂ IJ
=
= −G
H ∂ K
H∂ K
∂ ∂
1 F∂ I
=−
GH ∂ JK + FGH ∂∂ IJK = − 1 FGH ∂∂
F ∂ IJ 1 LM − FG ∂ IJ OP
=G
H∂ K
MN H ∂ K PQ
1
FG ∂ IJ
H∂ K
−
,
2
,
,
2
FG ∂ IJ
H∂ K
Solutions to Chemical and Engineering Thermodynamics, 3e
=
LM FG ∂ IJ
MN H ∂ K
+
V
OP
PQ
−
⇒
FG ∂ IJ
H∂ K
LM FG ∂ IJ
MN H ∂ K
1
=−
V
OP
PQ
−
Also
FG ∂ IJ
H∂ K
f ∂ a , f ∂a , f
f = ∂a , f ⋅ ∂a , f
a∂ ∂ f a∂ ∂ f − a∂ ∂ f a∂
=
a∂ ∂ f
F ∂ IJ − a∂ ∂ f ;
=G
H ∂ K a∂ ∂ f
F ∂ IJ = FG ∂ IJ − and FG ∂ IJ
; G
H∂ K H∂ K
H∂ K
=
a
a
∂
∂
,
,
∂
f
=
FG ∂ IJ
H∂ K
V
=
−
Thus,
FG ∂ IJ = FG ∂ IJ − − a∂
H∂ K H∂ K
a∂ ∂
FG ∂ IJ = − − FG ∂ IJ + +
H∂ K
H∂ K
R|S FG ∂ IJ − U|VR|S − FG ∂ IJ U|V
|T H ∂ K |W|T H ∂ K |W
|R F ∂ IJ |UV|RS FG ∂ IJ − |UV +
=S − G
T| H ∂ K W|T| H ∂ K W|
−
f
∂
f
V
−
a∂
V
a∂
f
∂
∂
−
f
FG ∂ IJ
H∂ K
V
so that, finally,
=
FG IJ
H K
1 ∂
∂
−
,
|RS
|T
1
=−
2
V
=
(d)
FG ∂ IJ = ∂
H∂ K ∂
RSFG ∂ IJ −
1 |
−
|TH ∂ K
=
2
A4-10
Now
−
2
,
eqn.
,
=
−
+
2
,
FG ∂
H∂
FG ∂
H∂
RS−
T
|UV +
|W
⇒
IJ
K
IJ |UV|RS FG ∂ IJ
K |W|T H ∂ K
,
|UV −
|W
UV = − 1 FG ∂ IJ + FG ∂
H∂
W H∂ K
FG ∂ IJ
H∂ K
−
2
,
FG ∂ IJ
H∂ K
IJ
K
,
2
FG ∂ IJ
H∂ K
,
=
,
FG ∂ IJ
H∂ K
−
,
FG ∂ IJ
H∂ K
+
,
Solutions to Chemical and Engineering Thermodynamics, 3e
⇒
=−
FG ∂ IJ
H∂ K
=−
FG ∂ IJ
H∂ K
+
FG ∂ IJ + FG ∂ IJ
H∂ K
H∂ K
FG ∂ IJ = − FG ∂ IJ + FG ∂ IJ
H∂ K
H∂ K
H∂ K
+
,
2
,
2
,
,
,
,
,
but from above
1
2
FG ∂ IJ
H∂ K
FG ∂ IJ
H∂ K
=
V
,
−
2
V
and, from equating the two expressions for
FG IJ
H K
1 ∂
∂
=
−
a∂
f
∂
+
V
,
V
FG ∂ IJ
H∂ K
FG ∂ IJ
H∂ K
2
−
Putting these expressions together yields
=
2
V
FG ∂ IJ
H∂ K
2
2
−
2
−
a∂
∂
V
f
2
2
+
V
FG ∂ IJ
H∂ K
(e) It is now simple algebra to combine the expressions above, and those in Sec. 5.2, and show that
θ3 =
−
2
−
a
b
−
f
g
2
−
2
is exactly zero!!
5.33 (Mathcad worksheets in the Mathcad Utilities Directory are also available to do these calculations)
Students in my thermodynamics courses have produced thermodynamics diagrams for many fluids
using the program PR1 and following the methods in illustrations of Chapters 4 and 5. The
following figures are examples of some of these diagrams. It should be noted that all of these
diagrams are in qualitative, but not quantitative agreement with thermodynamic diagrams generated
using more accurate equations of state. In particular, liquid densities are not predicted very
accurately from the Peng-Robinson e.o.s. so that the location of the two-phase dome is somewhat
shifted as are the other thermodynamic properties. Diagrams for other substances will be found in
the file named “Other figs”
Thermodynamic properties of nitrogen by Tom Petti
Solutions to Chemical and Engineering Thermodynamics, 3e
Pressure-volume diagram for nitrogen Peng-Robinson eos.
Solutions to Chemical and Engineering Thermodynamics, 3e
Solutions to Chemical and Engineering Thermodynamics, 3e
5.34 Thermodynamic properties of water (steam) by Allen Donn.
Pressure-volume diagram of steam computed with the Peng-Robinson equation of state
Solutions to Chemical and Engineering Thermodynamics, 3e
Temperature-entropy diagram of steam computed with the Peng-Robinson equation of state.
Solutions to Chemical and Engineering Thermodynamics, 3e
T= 425°C
T= 300°C
T= 200°C
T= 150°C
Pressure-enthalpy diagram of steam computed with the Peng-Robinson equation of state.
5.35(a) This would be a difficult problem if it were not for the availability of the program PR1. Using this
. K,
program, the critical properties and the heat capacity data in the text, and the = 27315
= 1 bar reference state (which cancels out of the problem) we find for ethylene
85 bar and 25° C = 29815
. K
= −6388 J mol
= −52.79 J mol K
By trial and error, using guessed values of
= 22135
. K;
L
V
= −29.44 J mol K ;
= −7871
. J mol K ;
L
until we obtain
V
vap
= 10 bar , we obtain
= 01536
.
×10−2 m3 mol ;
= 05454
.
× 10− 4 m3 mol .
Solutions to Chemical and Engineering Thermodynamics, 3e
Now considering the fluid initially in the tank that will be in the tank finally as the system we
have
=
=
and
(b) Now there can not be only vapor in the tank (entropy too high) or only liquid (entropy too low),
so there must be two phase mixture. Let L = mass (or mole) fraction of liquid. Thus:
L
L
L
L
c
+ 1−
L
c
h
( −78.71) + 1 −
=
V
=
L
= −52.79 J mol K
h(− 29.44) = −52.79
− 52.79 + 29.44
= 0.474 ;
−78.71 + 29.44
V
= 0.526
Thus, 47.4 wt % of fluid in tank is liquid, and 52.6% is vapor. Based on 1 mole in tank we have
= 0.474 × 05454
.
× 10 −4 + 0.526 × 0.1536 × 10 −2 = 8.338 × 10−4 m3 mol
volume % liquid =
0.474 × 0.5454 × 10−4
8.338 × 10−4
volume % vapor = 96.9%
× 100 = 31%
.
Solutions to Chemical and Engineering Thermodynamics, 3e
5.36
(also available as a Mathcad worksheet)
FG ∂ P IJ
H ∂T K
=
sat
Assume V
∆H
T ∆V
V
L
>> V ⇒ ∆V ~ V
FG ∂ P IJ
H ∂T K
V
=
=
sat
ZRT
'
P
FG
H
∆H
∂ ln P
⇒
2
ZRT P
∂T
IJ
K
=
sat
∆H
ZRT 2
but
FG ∂ ln P IJ
H ∂T K
=
sat
F
H
∂
5622.7
43552
.
−
− 4 .70504 ln T
∂T
T
=+
I
K
56227
.
4.70504
1
−
= 2 (5622.7 − 4.70504T )
2
T
T
T
Thus
∆H
1
=
(5622.7 − 4 .70504T )
ZRT 2 T 2
or
∆H = ZR (56227
. − 4.70504T ) = 31,602 J mol at 75°C
Z=
31,602
= 0.9539
8.314 × (5622.7 − 4.70504 × (273.15 + 75))
but
PV
B
= 1 + = Z = 0.9539
RT
V
so
B
= 0.9539 − 1 = −0.04607 ; B = −0.04607V
V
Then V = 0.9539 RT P . To find P use
ln Pvap = 43552
.
−
5622.7
− 4 .70504 ln( 27315
. + 75)
(273.15 + 75)
Pvap = 0.8736 bar
V=
0.9539 × 8.314 × 10−5 × (273.15 + 75)
= 31606
.
× 10 −2 m3 mol
0.8736
and
B = −1456
. × 10−3 m3 mol
Solutions to Chemical and Engineering Thermodynamics, 3e
5.37
We start with Eqn. (5.7-4), the Clapeyron equation
FG ∂ P IJ
H ∂T K
sat
=
G I =G II
∆H
T∆V
[Note: Error in problem statement of 1st printing. Disregard comment that the volume change on
fusion is zero.]
From the problem statement ∆H = 48.702 kJ mol , but no data on ∆V is given. Also Psat = 1013
.
bar at T = 185
. ° C = 29165
. K . Based on other hydrocarbons, we can guess that
∆V fus ~ 1 to 2 × 10−4 m3 kg
We will use this as an estimate and determine the effect on Tm . Also, the molecular weight of
hexadecane is 226.45. Thus
dP
48.702 kJ mol × 1000 J kJ
=
d ln T δ × 10 −4 m3 kg × 226 .54 g mol × 1 kg 1000 g
=
214 .98
× 107 J m3
δ
[where δis 1 or 2]
214.98 × 107
214 .98 × 10 2
J m3 × 10 −2 bar ⋅ m 3 kJ × 10−3 kJ J =
bar
δ
δ
21498
21498
T2
dP =
d ln T ⇒ ( P − 1.013 bar ) =
ln
δ
δ
291.65 K
294.36 if δ = 1
(200 − 1013
. )×δ
T2 = 291.65 exp
=
21498
297.10 if δ = 2
=
OP RS
Q T
LM
N
So the freezing point is raised between 2.7 and 5.5 K, depending on the (unknown) value of ∆V fus .
5.38 (also available as a Mathcad worksheet)
This is a one-component adiabatic flash process. I will assume that only vapor + liquid are present,
and then show that this is indeed the case.
There are two ways to solve this problem. One is to calculate all the thermodynamic properties, and
the second is to use the steam tables. Both methods will be considered here
(1) Calculating all thermodynamic properties, and assuming the vapor phase is ideal.
energy balance: 10. U$ L (T = 95° C) = (10 − x)U$ L (T) + xU$ V (T)
equilibrium requirements: T L = T V ; P L = P V ; and G L = GV ⇒ P = P vap
Also, using data supplied earlier,
F
H
Pvap = exp 14.790 −
and by the ideal gas law
54328
.
T
I
K
Solutions to Chemical and Engineering Thermodynamics, 3e
P=
N V RT x 18 mol × 8.314 × 10− 5 (bar ⋅ m 3 mol K ) T
=
VV
1 × 10−3 − (10 − x) 106
volume taken up
by liquid
Equating P and P vap we have
F
H
I
K
x 18 × 8.314 × 10−5 T
5432.8
x
8314
. T
= exp 14.790 −
=
⋅
1 × 10 −3 − (10 − x) 10 6
T
18 1000 − (10 − x)
Also we have for the internal energies
U$ L (T = 0° C) = 0 reference state
U$ L (T = 95° C) = 95° C × 4184
.
J g° C = 397 .48 J g
L
$
U (T ) = (T − 273.15) × 4184
.
assuming C = constant
v
8.314 × 273.15
U (T = 0° C) = ∆H − RT = 2260 −
= 213383
. J g
18
U$ V (T ) = 213383
. + (T − 27315
. ) × 2.09
$V
$ vap
so that
10 × 397 .48 = (10 − x) × 4 .184 × (T − 27315
. ) + x 213383
. + (T − 27315
. ) × 2.09
I find that the solution to these equations is
T = 352.68 K and P = 05411
.
bar
x = 03289
.
g
This is so far above the melting point of water, that the presence of an ice phase is impossible.
(2) Using the steam tables
energy balance:
10 ⋅ U$ L ( T = 95° C) = 10 × 397 .88 = (10 − x )U$ L ( T ) − xU$ V ( T )
1442443
both at saturation
also P = P sat (T) and
V = 0.001 m3 = (10 − x)V$ L (T) + xV$ V (T)
Procedure
i) Guess T, get Psat (T) , V$L (T ) , V$ V (T) , U$ L (T) and U$ V (T ) from steam tables
ii) See if Eqns. (1) and (2) are satisfied by using Eqn. (2) to get x, and then seeing if Eqn. (1) is
satisfied.
For example, guess T = 80° C :
V$ L = 1029
.
×10−6 m3 g
V$ V = 3407
.
× 10−3 m3 g
P = 4739
. kPa
U$ L = 334.86 J g
U$ V = 2482.2 J g
⇒ x( eqn. (2 )) = 0.29058 ; x( eqn. (1)) = 0.29348
Solutions to Chemical and Engineering Thermodynamics, 3e
by iteration and interpolation
T = 794
. ° C , P = 04739
.
bar and x = 029
. grams .
Difference between this solution and the previous one is due to the inaccuracies of the
approximate vapor pressure equation in Part 1, and the assumption of constant heat capacities.
5.39 All the P-V data for this problem was obtained with a simple basic language program written for this
problem. Calculations were done for n-butane as a representative fluid. The van der Waals loop
region is shown on the diagram. What is interesting is that, in addition to the van der Waals loop,
there is much structure in the P-V plot. Much of it occurs in the region of b > V and V < 0 , so that
it has no relevance to our calculations. In the region V > b there is only the van der Waals loop
behavior at low reduced temperatures, and the hyperbolic behavior ( PV = RT ) at very high
temperatures. The main point is that the cubic equations we use exhibit quite complicated P- V
behavior, but only relatively simple behavior in the region of interest to us, which is V > b .
P- V diagram for n-butane calculated with the Peng-Robinson equation of state for realizable (V > b ) and
physically unrealizable ( V < b ) regions.
Solutions to Chemical and Engineering Thermodynamics, 3e
5.40 Let TE = the equilibrium transition temperature when both solid phases are stable.
d G = VdP − SdT
Also dH = TdS +VdP so at constant pressure
FG ∂ S IJ
H∂TK
=
P
FG IJ
H K
1 ∂H
T ∂T
=
P
Cp
T
⇒ phase with higher heat capacity will have a higher entropy since the entropy of both phases are
zero at 0 K.
Then, again at constant pressure
FG ∂ G IJ
H ∂T K
= −S
P
Since both phases have the same Gibbs free energy at the temperature, T, this implies that the
substance with the larger entropy (which arises from larger heat capacity) will have the lower Gibbs
free energy, and therefore be the stable phase.
5.41
PV
B( T )
RT B(T ) RT
=1+
; P=
+
RT
V
V
V2
FG ∂ P IJ
H ∂V K
=−
T
2 B(T )
3
>−
RT
V
1
2
2
;
−2 B(T ) RT
V3
; B( T ) > −
V
V
Back to virial eq.
<0
V3
2
V 2
=−
V
V
; B( T ) > −
2
2
a
PV 2
+1 ± 1 + 4 PB RT
− V − B( T ) = 0 ; V =
RT
2 P RT
V=
RT RT
4 PB
±
1+
2P 2P
RT
B >−
LM
N
OP LM
QN
1 RT
4 PB
1± 1+
2 2P
RT
B
a
1 ± 1 + 4 PB RT
In fact, B( T ) > −
f
>−
f
OP
Q
RT
fluid will be stable
4P
V
is sufficient since B(T ) << V in all conditions where second virial coefficient is
2
used.
Approximation B( T ) > −
V
RT
⇒ B(T ) ~> −
for stability.
2
2P
Solutions to Chemical and Engineering Thermodynamics, 3e
5.42 Easy way
dU
dV dS Q& &
= Q& − P
;
= + Sgen
dt
dt dt T
System of constant entropy Q& = − TS&gen
Also constant pressure
dU
dV
d
= − TS&gen − P
= − TS&gen − ( PV )
dt
dt
dt
dU d
d
dH
+ ( PV ) = (U + PV ) =
= − TS&gen ≤ 0
dt dt
dt
dt
⇒ H = maximum at equilibrium
dH = 0
d2H ≥ 0 ⇒
dH = TdS + VdP
FG ∂ T IJ (dS) + FG ∂ T IJ dSdP + FG ∂V IJ
H∂S K
H∂ PK
H ∂S K
F ∂ T IJ a∂ S f ≥ 0 ⇒ FG ∂ T IJ ≥ 0
d H=G
H∂S K
H ∂SK
FG ∂ T IJ = T ≥ 0 ⇒ C > 0
H ∂SK C
2
d2H =
P
S
dPdS +
P
FG ∂V IJ (dP)
H ∂ PK
2
S
2
2
P
P
P
P
P
More theoretically correct way
Equilibrium criterion for a closed system at constant entropy and pressure.
dU
dV dS Q& &
= Q& − P
;
= + Sgen
dt
dt dt T
dS
Q& = T
− TS&gen
dt
dU
dS
dV
=T
−P
− TS&gen
dt
dt
dt
dS
dV
d
Constant entropy
= 0 ; constant pressure P
= ( PV )
dt
dt
dt
dU
d
d
(U + PV ) = −TS&gen
⇒
= − ( PV ) − TS&gen ;
dt
dt
dt
dH
dH
= − TS&gen ≤ 0 ⇒
≤ 0 ⇒ H = minimum stability
dt
dt
d 2 H > 0 but dH = TdS + VdP + G i dN
c
hc h + cH + H hcdN h
+ 2c H + H h dS dN > 0
N +N
=
N H cdS h + 2 N H dS dN
N N
I
II
d 2 H = HSS
+ HSS
dS I
I
SN
I
II
SN
II
I
II
2
I
I
NN
I
I
SS
I 2
II
NN
I
I 2
Making a transformation of variables
H
dx1 = dS I + SN dN ; dx2 = dN I
H NN
I
I
SN
I
I
c h
I
+ N I H NN
dN I
2
>0
Solutions to Chemical and Engineering Thermodynamics, 3e
θ1 = NHSS ; θ2 =
cNH
SS NH NN
2
− N 2 HSN
h
NHSS
As a check
θ1dx12 + θ2 dx22
FG
H
HSN
dN I
HSS
= NH SS dS I +
= NH
+
SS
NH
dS
I2
+ 2 NH
I2
NN dN
SS
2
+
H SN
2
2
NHSS NH NN − N 2 HSN
dN I
NHSS
I
dS dN
+ NH
I
H SS
2
SN
NH
−
⋅
IJ
K
H SS
⋅
SS
2
H SN
dN
2
H SS
I2
I2
dN
which is correct so
θ1dx12 + θ2 dx22 ≥ 0 ⇒ θ1 > 0; θ2 > 0
FG ∂ H IJ = N ∂ FG ∂ H IJ = N FG ∂T IJ > 0
H ∂ S K ∂S H ∂S K H∂S K
C
F ∂V IJ ; FG ∂ S IJ = C ⇒ FG ∂ S IJ = NC
but d S =
dT − G
H ∂T K H ∂ T K T H ∂ T K T
T
θ1 = NHSS = N
2
2
P
P
P
P
P
P
P
P
P
T
⇒N
> 0 ; N > 0, T > 0 ⇒ CP > 0
NCP
Second criterion
FG
H
2
NHSS NHNN − N 2 HSN
∂2 H
; H NN =
NHSS
∂ N2
HSN =
NH NN
IJ
K
=
S, P
∂
(G )
∂N
S,
FG ∂ H IJ = ∂ (T) = FG ∂T IJ
H ∂S K ∂ N
H∂ NK
H
F ∂ G IJ − N a∂T ∂ N f
−N
= NG
H
H∂ NK
a∂ T ∂ S f
∂
∂N
S, P
S ,P
P, N
S ,P
SS
P
FG ∂G IJ
H∂ NK
S, P
S ,P
2
S, P
2
SN
=
=?
P, N
5.43 (also available as a Mathcad worksheet)
5.43
Cp
0
R
ISENTHALPIC CLAUSIUS EQUATION OF STATE CALCULATION
20.97
8.314 . 10
bb 0
4.28 . 10
5
bb 1
1.35 . 10
7
b( T )
bb 1 . T
bb 0
5
Input initial temperature and pressure of calculation
Input final pressure
Pf
Initial state calculations
Ti
bar
10
T
Ti
P
Pi
Vi
120
273.15 Pi
( R. Ti)
50
b ( Ti)
Pi
Zi
( Pi . Vi )
R. Ti
Zi = 1.1467
DELHin
R. T . ( Zi
1)
Vi
DELHin = 214
Guess for final state
Residual enthalpy (DELHF)
2
R. Ti . bb 1
T
0.8 . Ti
V( T , P )
P
( R. T )
P
. 105
b ( Ti)
Pf
b( T )
Z( T , P )
( P . V( T , P ) )
R. T
bar
Solutions to Chemical and Engineering Thermodynamics, 3e
R .T .( Z ( T , P )
DELHF ( T , P )
2
R. T . bb 1
1)
V( T , P )
. 105
b( T )
DELHIG( T , P )
Ideal gas properties changes relative to the initial state
Cp 0 . ( T
Ti)
Solve for the exit temperature
Given
δH
DELHF( T , P )
DELHF( T , P )
DELHIG( T , P )
DELHIG( T , P )
DELHin 0
DELHin
T
DELHF( T , P )
HF
T = 401.314
find( T )
DELHIG( T , P )
SUMMARY OF RESULTS
FEED
EXIT
Temperature, K
Ti = 393.15
T = 401.314
Pressure, bar
Pi = 50
P = 10
Zi = 1.1467
Compressibility
Enthalpy
(relative to the feed)
Z( T , P ) = 1.0291
HF = 214
0
13
δH = 3.1264 10
Symbolic determination of enthalpy departure function for the
Clausius equation of state
b ( T , bb )
bb
0
File: 5-43 symbolic
bb 1 . T
( R. T )
P( T , V, R, bb )
V b ( T , bb )
R
d
P( T , V, R, bb )
dT
der( T , V, R, bb )
V bb 0
bb 1 . T
T
R.
V bb 0
bb 1 . T
. bb
2
1
d
P( T , V, R, bb )
dT
Int( T , V, R, bb )
T . der( T , V, R, bb )
Int( T , V, R, bb )
T.
P( T , V, R, bb )
R
V bb 0
Upon simplification
bb 1 . T
T
R.
V bb 0
2
R. T .
bb 1 . T
. bb
2
1
R.
V bb 0
bb 1
V bb 0
bb 1 . T
T
2
bb 1 . T
Solutions to Chemical and Engineering Thermodynamics, 3e
V
DelH ( T , V , R , bb )
Int( T , S , R, bb ) d S
15
MATHCAD has trouble with an infinite lower
limit, so use a very large number instead
10
bb 1
2
R. T .
DelH( T , V, R, bb )
V bb 0
1000000000000000
bb 0
bb 1 . T
This term can be neglected, would
be zero if an infinite lower limit
could be used
bb 1
2
R. T .
DelH( T , V, R, bb )
bb 1 . T
bb 1
2
R. T .
bb 1 . T
V bb 0
Final result
5.44 Clausius EOS: P =
RT
V − b( T )
Condition for stability is
FG ∂ P IJ
H ∂V K
<0
T
For the Clausius equation
∂P
RT
=−
Since R > 0 , T > 0 and (V − b)2 > 0 .
∂V T
(V − b) 2
FG IJ
H K
F ∂ P IJ must be negative or
Then G
H ∂V K
FG ∂ P IJ < 0 ⇒ Single phase is stable at all conditions.
H ∂V K
T
T
5.45 See solution to Problem 5.41. If fluid is unstable, then a vapor-liquid phase
transition can occur.
5.46 Redlich-Kwong equation of state
ln
f
1
=
P RT
z
V = ZRT P
V =∞
FG RT − PIJ dV − ln Z + (Z − 1)
HV K
Solutions to Chemical and Engineering Thermodynamics, 3e
z FGH RTV − PIJKdV = z FGH RTV − VRT− b + V(aV(T+)b)IJKdV
V
V =∞
z
V
= RT ln
V
V −b
dV
− RT ln
+ a (T )
V →∞
(V − b)V → ∞
V (V + b)
V =∞
F I FG
H K H
F I
H K
F I
H K
F
H
IJ
K
F
H
I
K
V
1
V +b
Z
a
Z+ B
+ a − ln
= RT ln
− ln
V −b
b
V
Z −B b
Z
f
Z
a
Z +B
ln = ln
−
ln
− ln Z + ( Z − 1)
P
Z − B bRT
Z
a
Z+B
= − ln( Z − B ) −
ln
+ ( Z − 1)
bRT
Z
a
Z + Pb RT
= ( Z − 1) − ln(Z − B) −
ln
bRT
Z
aP
bP
A=
; B=
RT 2
RT
f
A
Z +B
ln = ( Z − 1) − ln( Z − B) − ln
P
B
Z
Using the same analysis for the Soave-Redlich-Kwong equation of state leads to
the following
f
Pb
a (T )
Z + Pb RT
ln = ( Z − 1) − ln Z −
−
ln
P
RT
RTb
Z
a (T )
Z +B
= ( Z − 1) − ln( Z − B) −
ln
RTb
Z
= RT ln
F
H
F
H
I
K
I
K
LM a
N
LM OP
N Q
I
K
fO
QP
5.47 (also available as a Mathcad worksheet)
See Mathcad for the graphs.
Problem 5.47
R .T
P( V, T , R , a , b )
R. T
b
T
126.2 . K
0.5 .
V. ( V b )
R. T . ln ( V)
P( V, T , R, a , b ) d V
V
Tc
V
a
a
. ln ( V)
a
1. .
T .b
6
3.396 . 10 . Pa
Pc
1. . ln ( V 1. . b ) . R. T
8.314 .
R
Pa . m
mole . K
3
a
0.42748 .
. ln ( V b )
T .b
2
2.5
R . Tc
b
0.08664 .
R. Tc
Pc
Pc
110 . K
T
0 .. 100
i
Pi
1.5 . b
V0
R. T
Vi
fopi
Vi . 1.001
i
1
V0 = 4.015 10
a
b
ln
Vi
T
0.5 .
Vi
Vi
b
Vi . Vi
ln Zi
Zi
b
Zi
a
1
T
1.5 .
R. b
. ln
Pi . Vi
R. T
Vi
Vi
b
5
3
1
m mole
V100 = 5.655 10
3
3
1
m mole
Solutions to Chemical and Engineering Thermodynamics, 3e
fi
Pi . exp fopi
150
100
P
i
50
5
10 .Pa
0
50
0
0.5
1
1.5
log
2
2.5
V
i
b
6
2.5 10
6
2 10
6
1.5 10
f
i
6
1 10
5
5 10
0
0
0.5
1
log
T
150 . K
i
0 .. 100
V0
1.5 . b
Vi
Vi . 1.001
i
1
1.5
2
V
i
b
V0 = 4.015 10
5
3
1
m mole
2.5
Solutions to Chemical and Engineering Thermodynamics, 3e
R. T
Pi
Vi
fopi
fi
a
b
ln
T
0.5 .
Vi . Vi
Vi
Vi
b
Zi
b
ln Zi
Zi
a
1
T
1.5 .
R. b
. ln
Pi . Vi
V100 = 5.655 10
R. T
3
3
1
m mole
Vi
Vi
b
Pi . exp fopi
500
400
300
P
i
5
10 .Pa
200
100
0
0
0.5
1
1.5
log
2
2.5
V
i
b
7
2.5 10
7
2 10
7
1.5 10
f
i
7
1 10
6
5 10
0
0
0.5
1
log
1.5
V
i
b
2
2.5
Solutions to Chemical and Engineering Thermodynamics, 3e
5.48 (also available as a Mathcad worksheet)
Problem 5.48
a) If ethanol is an ideal gas, the f = P, so that the fugacity of ethanol is 505 kPa
b) Starting from eqn. 5.48 we have that
ln (f/P)=(B/V) - ln Z +(Z-1)
505000 . Pa
P
T
mol
126 ) . K
( 273.15
V = 6.571 10
P
( P . V)
1
R. T
Given
Z
( P . V)
R. T
fsat
8.314 .
R
( R. T )
V
1
B
V
find( V)
V
Pa . m
mol. K
3
3
3
B
6 m
523 . 10 .
mol
3
m
V = 5.998 10
3
3
m
Z = 0.913
P . exp
B
ln ( Z )
(Z
1)
5
fsat = 4.592 10 Pa
V
5.49 (also available as a Mathcad worksheet)
Problem 5.49
The density of ethanol is 0.789 g/cc at 20 C which we will also use at 126 C, and its molecular
weight is 46.07. Therefore its liquid molar volume is
kPa
3 kg
46.07 . 10 .
mol
Vl
Vl = 5.839 10
3 kg
0.789 . 10 .
5
3
10 . Pa
MPa
6
10 . Pa
3
m
3
m
a)
f
fsat . exp
( ( 25 . MPa
505 . kPa ) . Vl )
R. T
5
f = 7.066 10
Pa
b)
VV( p )
f
Vl . 1
fsat . exp
1 .
R. T
6
1
1.09 . 10 . kPa . ( 101.3 . kPa
p)
5
VV( 25 . MPa ) = 5.681 10
25 . MPa
505 . kPa
VV( p ) d p
5
f = 7.024 10
Pa
3
m
Solutions to Chemical and Engineering Thermodynamics, 3e
5.50 (also available as a Mathcad worksheet)
5.50 FUGACITY CALCULATION USING SRK EQUATION
Read in properties for Pentane
Tc
kappa calculation
469.6
kap
S-R-K Constants:
R
Pc
0.480
0.00008314
33.74
om
0.251
1.574 . om 0.176 . om. om
0.08664 .
b
R. Tc
0.42748 .
ac
2
2
R . Tc
Pc
Note that these are being defined as a
function of temperature.
Pc
2
alf( T )
1. 1
T
kap . 1
ac . alf( T )
a( T )
Tc
CA ( T , P )
Z( T , P )
A
CA ( T , P )
B
CB( T , P )
A .B
A
V
2
B
a( T ) . P
CB( T , P )
2
( R. T )
P .b
R. T
Da ( T )
d
a( T )
dT
Vector of coefficients in the SRK equation
in the form
0=-A*B+(A-B^2-B)*Z-*Z^2+Z^3
B
1
1
ZZ
polyroots ( V)
Solution to the cubic
for i ∈ 0 .. 2
ZZi
0
if
Im ZZi
0
Set any imaginary roots to zero
Sort the roots
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ0 < 10
5
ZZ2
ZZ0 if
ZZ2 < 10
5
Set the value of any imaginary roots
to value of the real root
ZZ
Enter temperature T, and pressure P.
T
100 C
T
273.15
T K
P
50
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
fv ( T , P )
Z( T , P ) 0
Z( T , P ) 2
1
1
ln Z( T , P ) 0
ln Z( T , P ) 2
CB( T , P )
CB( T , P )
CA ( T , P ) .
CB( T , P )
ln
CA ( T , P ) .
CB( T , P )
ln
Z( T , P ) 0
CB( T , P )
Z( T , P ) 0
Z( T , P ) 2
CB( T , P )
Z( T , P ) 2
Solutions to Chemical and Engineering Thermodynamics, 3e
Fugacity
fugl
fugv
P . exp ( fl( T , P ) )
Fugacity coefficient
fugl = 6.49272
P . exp ( fv ( T , P ) )
fl( T , P ) = 2.04134
fugv = 6.49272
phil
fugl
phil = 0.12985
P
fv ( T , P ) = 2.04134
fugv
phiv
P
phiv = 0.12985
SUMMARY OF RESULTS
T = 373.15
K
Vapor pressure, bar P = 50
LIQUID
Compressibility
VAPOR
Z( T , P ) 0 = 0.23249
Z( T , P ) 2 = 0.23249
Fugacity coefficient
phil = 0.12985
phiv = 0.12985
Fugacity, bar
fugl = 6.49272
fugv = 6.49272
Read in properties for Benzene
Tc
kappa calculation
S-R-K Constants:
kap
R
562.1
Pc
0.480
0.00008314
48.94
om
0.212
1.574 . om 0.176 . om. om
b
0.08664 .
R. Tc
ac
0.42748 .
Pc
Note that these are being defined as a
function of temperature.
2
2
R . Tc
Pc
2
alf( T )
1. 1
kap . 1
T
a( T )
ac . alf( T )
Tc
CA ( T , P )
a( T ) . P
2
( R. T )
CB( T , P )
P .b
R. T
Da ( T )
d
a( T )
dT
Solutions to Chemical and Engineering Thermodynamics, 3e
dT
Z( T , P )
A
CA ( T , P )
B
CB( T , P )
A .B
A
V
2
B
Vector of coefficients in the SRK equation
in the form
0=-A*B+(A-B^2-B)*Z-*Z^2+Z^3
B
1
1
ZZ
Solution to the cubic
polyroots ( V)
for i ∈ 0 .. 2
ZZi
0
if
Im ZZi
0
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ0 < 10
5
ZZ2
ZZ0 if
ZZ2 < 10
5
Set any imaginary roots to zero
Sort the roots
Set the value of any imaginary roots
to value of the real root
ZZ
Enter temperature T, and pressure P.
100 C
T
T
273.15
T K
P
50
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
Z( T , P ) 0
fv ( T , P )
Z( T , P ) 2
1
1
ln Z( T , P ) 0
CB( T , P )
ln Z( T , P ) 2
CB( T , P )
CA ( T , P ) .
CB( T , P )
ln
CA ( T , P ) .
CB( T , P )
ln
Z( T , P ) 0
CB( T , P )
Z( T , P ) 0
Z( T , P ) 2
CB( T , P )
Z( T , P ) 2
Fugacity
fugl
fugv
P . exp ( fl ( T , P ) )
fugl = 2.01968
P . exp ( fv ( T , P ) )
fugv = 2.01968
Fugacity coefficient
fl( T , P ) = 3.20908
fv ( T , P ) = 3.20908
phil
fugl
phil = 0.04039
P
phiv
fugv
P
phiv = 0.04039
SUMMARY OF RESULTS
T = 373.15
K
Vapor pressure, bar P = 50
LIQUID
Compressibility
Z( T , P ) 0 = 0.17187
VAPOR
Z( T , P ) 2 = 0.17187
Fugacity coefficient
phil = 0.04039
phiv = 0.04039
Fugacity, bar
fugl = 2.01968
fugv = 2.01968
Solutions to Chemical and Engineering Thermodynamics, 3e
5.51 (Solution using Mathcad worksheet)
T = -200o C
Pvap = 0.10272 bar
Z V = 0.99512
Z L = 4 .414 × 10 −4
H V = − 55795
.
× 103
H L = −1.2994 × 104
S V = 17.372
S L = −118.74
T = -180o C
Pvap = 1.348 bar
Z = 0.96359
Z L = 4 .955 × 10−3
H V = − 51095
.
× 103
H L = − 11948
.
× 104
S V = −32.734
S L = −106.15
V
T = -160o C
Pvap = 6.750 bar
Z V = 0.8810
Z L = 0.02307
H V = − 4.7953 × 103
H L = −1.0805 × 104
S V = −42.099
S L = −95.210
T = -140o C
Pvap = 20.676 bar
Z V = 0.73096
Z L = 0.07305
H V = − 4.7988 × 103
H L = − 9.4328 × 103
S V = −49.6785
S L = −84.481
T = -130o C
Pvap = 32.310 bar
Z V = 0.61800
Z L = 0.12528
H V = − 5.0406 × 10 3
H L = −8.5449 × 103
S V = −53.938
S L = −78.418
T = -125o C
Pvap = 39.554 bar
Z = 0.54226
Z L = 016843
.
H V = − 52985
.
× 10 3
H L = − 7.9739 × 10 3
S V = −56.685
S L = −74.744
V
T = -120o C
Pvap = 47.848 bar
Z V = 0.42788
Z L = 0.24887
H V = − 58378
.
× 10 3
H L = − 71281
.
× 103
S V = −61034
.
S L = −69.459
Solutions to Chemical and Engineering Thermodynamics, 3e
The Mathcad worksheet for this file is shown below.
Solutions to Chemical and Engineering Thermodynamics, 3e
Solutions to Chemical and Engineering Thermodynamics, 3e
5.52 (also available as a Mathcad worksheet)
5.52 Pure component properties calculation using the SRK equation of state
Read in properties for oxygen
Tc
154.6
Pc
50.46
om
0.021
Heat capacity constants
Cp 0
25.460
1.519 . 10
Cp 1
2
Cp 2
0.715 . 10
5
1.311 . 10
Cp 3
9
Reference state and kappa calculation
Trs
298.15
S-R-K Constants:
Prs
1.0
kap
R
0.480
0.00008314
1.574 . om 0.176 . om. om
b
0.08664 .
R. Tc
ac
0.42748 .
Pc
Note that these are being defined as a
function of temperature for convenience.
2
2
R . Tc
Pc
2
alf( T )
1. 1
kap . 1
T
a( T )
ac . alf( T )
Tc
CA ( T , P )
a( T ) . P
2
( R. T )
CB( T , P )
P .b
R. T
Da ( T )
d
a( T )
dT
Solutions to Chemical and Engineering Thermodynamics, 3e
CA ( T , P )
Z( T , P )
A
CA ( T , P )
B
CB( T , P )
A .B
2
A
V
B
a ( T ) .P
( R .T )
P .b
R. T
CB( T , P )
2
Da ( T )
d
a( T )
dT
Vector of coefficients in the SRK equation
in the form
0=-A*B+(A-B^2-B)*Z-*Z^2+Z^3
B
1
1
ZZ
polyroots ( V)
Solution to the cubic
for i ∈ 0 .. 2
ZZi
0
if
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ2
ZZ0 if
Im ZZi
0
ZZ0 < 10
5
ZZ2 < 10
5
Set any imaginary roots to zero
Sort the roots
Set the value of any imaginary roots
to value of the real root
ZZ
Enter temperature T, and pressure P.
T
125 C
T
273.15
T K
P
100
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
Z( T , P ) 0
fv ( T , P )
1
Z( T , P ) 2
1
ln Z( T , P ) 0
ln Z( T , P ) 2
CB( T , P )
CB( T , P )
CA ( T , P ) .
CB( T , P )
ln
CA ( T , P ) .
CB( T , P )
ln
Z( T , P ) 0
CB( T , P )
Z( T , P ) 0
Z( T , P ) 2
CB( T , P )
Z( T , P ) 2
Fugacity
fugl
fugv
P . exp ( fl( T , P ) )
P . exp ( fv ( T , P ) )
Fugacity coefficient
fugl = 34.64672
fl( T , P ) = 1.05997
phil
fugv = 34.64672
fv ( T , P ) = 1.05997
phiv
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
fugl
P
fugv
P
phil = 0.34647
phiv = 0.34647
Solutions to Chemical and Engineering Thermodynamics, 3e
R. ln Z ( T , P )
DELSL
0
R. ln Z( T , P ) 2
DELSV
Da ( T ) .
CB( T , P )
Z( T , P ) 0
ln
Da ( T ) .
Z( T , P ) 2
ln
. 105
Z( T , P ) 0
b
CB( T , P )
CB( T , P )
CB( T , P )
. 105
Z( T , P ) 2
b
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
DELHL
DELHV
R . T . Z( T , P ) 0
1
R . T . Z( T , P ) 2
1
T . Da ( T )
a( T ) .
ln
Z( T , P ) 0
Z( T , P ) 0
b
T . Da ( T )
a( T ) .
ln
CB( T , P )
Z( T , P ) 2
. 105
CB( T , P )
. 105
Z( T , P ) 2
b
Ideal gas properties changes relative to the reference state
DELHIG
DELSIG
Cp 0 . ( T
Cp 0 . ln
Trs )
Cp 1 . T
2
Trs
2
Cp 2 . T
2
T
Trs
Cp 1 . ( T
3
Trs
3
Cp 3 . T
3
Trs )
Cp 2 . T
2
Trs
4
Trs
4
4
2
Cp 3 . T
2
3
Trs
3
3
P
5
R. 10 . ln
Prs
Total entropy and enthalpy relative to ideal gas reference state
SL
DELSIG DELSL
V0
SV
2 T
Z( T , P ) 0 . 8.314 . 10 .
P
DELSIG DELSV
HL
V0 = 0.04081
DELHIG DELHL
V2
HV
DELHIG DELHV
2 T
Z( T , P ) 2 . 8.314 . 10 .
P
V2 = 0.04081
SUMMARY OF RESULTS
T = 148.15
K
Vapor pressure, bar P = 100
LIQUID
Compressibility
Z( T , P ) = 0.3313
VAPOR
Z( T , P ) = 0.3313
Solutions to Chemical and Engineering Thermodynamics, 3e
LIQUID
VAPOR
Compressibility
Z( T , P ) 0 = 0.3313
Z( T , P ) 2 = 0.3313
Volume, m^3/kmol
V0 = 0.04081
V2 = 0.04081
Enthalpy, J/mol
3
HL = 9.29109 10
3
HV = 9.29109 10
Entropy, J/mol K
SL = 83.19194
SV = 83.19194
Fugacity coefficient
phil = 0.34647
phiv = 0.34647
Fugacity, bar
fugl = 34.64672
fugv = 34.64672
Some representative results are shown below.
T (C)
P=1 bar
Z
V
H
S
-125
-150
-175
-200
0.9923
12.2227
-4301.41
-19.97
0.9872
10.1072
-4994.48
-25.1
0.9766
7.9693
-5684.02
-31.35
0.9505
5.7804
-6375.9
-35.49
P=10 bar
Z
V
H
S
0.9193
1.1323
-4561.07
-40.28
0.8565
0.877
-5357.77
-46.18
0.03572
0.02914
-12395.6
-106.11
0.04292
0.0261
-13706.2
-121.5
P=50 bar
Z
V
H
S
0.1946
0.04795
-8938.78
-79.34
0.1647
0.03373
-10919.2
-93.79
0.17634
0.02878
-12338.7
-106.71
0.21349
0.02597
-13628.9
-121.86
0.3313
0.04081
-9291.09
-83.19
0.318
0.03256
-10896.8
-95.02
0.34788
0.02839
-12261.8
-107.39
0.42446
0.02581
-13530.9
-122.29
P=100 bar
Z
V
H
S
.
5.53 (also available as a Mathcad worksheet)
5.53 Pure component properties calculation using the SRK equation of state
Read in properties for Water
Tc
647.3
Pc
220.48 om
0.344
Heat capacity constants
Cp 0
32.218
Cp 1
0.192 . 10
2
Cp 2
1.055 . 10
5
Cp 3
3.593 . 10
9
Solutions to Chemical and Engineering Thermodynamics, 3e
Reference state and kappa calculation
Trs
373.15
Prs
1.013
kap
S-R-K Constants:
R
0.480
0.00008314
1.574 . om 0.176 . om. om
0.08664 .
b
R. Tc
0.42748 .
ac
2
2
R . Tc
Pc
Note that these are being defined as a
function of temperature for convenience.
Pc
2
alf( T )
1. 1
T
kap . 1
ac . alf( T )
a( T )
Tc
CA ( T , P )
Z( T , P )
A
CA ( T , P )
B
CB( T , P )
A .B
A
V
2
B
a( T ) . P
( R. T )
CB( T , P )
2
P .b
R. T
Da ( T )
d
a( T )
dT
Vector of coefficients in the SRK equation
in the form
0=-A*B+(A-B^2-B)*Z-*Z^2+Z^3
B
1
1
ZZ
polyroots ( V)
Solution to the cubic
for i ∈ 0 .. 2
ZZi
0
if
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ2
ZZ0 if
Im ZZi
0
ZZ0 < 10
5
ZZ2 < 10
5
Set any imaginary roots to zero
Sort the roots
Set the value of any imaginary roots
to value of the real root
ZZ
Enter temperature T, and pressure P.
T
50
C
T
273.15
T K
P
0.15
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
fv ( T , P )
Z( T , P ) 0
Z( T , P ) 2
1
1
ln Z( T , P ) 0
ln Z( T , P ) 2
CB( T , P )
CB( T , P )
CA ( T , P ) .
CB( T , P )
ln
CA ( T , P ) .
CB( T , P )
ln
Z( T , P ) 0
CB( T , P )
Z( T , P ) 0
Z( T , P ) 2
CB( T , P )
Z( T , P ) 2
Solutions to Chemical and Engineering Thermodynamics, 3e
Fugacity
fugl
fugv
Fugacity coefficient
P . exp ( fl ( T , P ) )
fugl = 0.09983
fl( T , P ) = 0.40721
P . exp ( fv ( T , P ) )
fugv = 0.14972
3
fv ( T , P ) = 1.83629 10phiv
fugl
phil
phil = 0.6655
P
fugv
P
phiv = 0.99817
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
DELSL
DELSV
R. ln Z( T , P ) 0
CB( T , P )
Da ( T ) .
R. ln Z( T , P ) 2
CB( T , P )
Da ( T ) .
Z( T , P ) 0
ln
CB( T , P )
. 105
Z( T , P ) 0
b
Z( T , P ) 2
ln
CB( T , P )
. 105
Z( T , P ) 2
b
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
DELHL
DELHV
R . T . Z( T , P ) 0
1
R . T . Z( T , P ) 2
1
T . Da ( T )
a( T ) .
ln
Z( T , P ) 0
Z( T , P ) 0
b
T . Da ( T )
a( T ) .
ln
CB( T , P )
Z( T , P ) 2
. 105
CB( T , P )
. 105
Z( T , P ) 2
b
Ideal gas properties changes relative to the reference state
DELHIG
DELSIG
Cp 0 . ( T
Cp 0 . ln
Trs )
Cp 1 . T
2
Trs
2
Cp 2 . T
2
T
Trs
Cp 1 . ( T
3
Trs
3
Cp 3 . T
3
Cp 2 . T
Trs )
2
Trs
4
Trs
4
4
Cp 3 . T
2
2
3
Trs
3
3
P
5
R. 10 . ln
Prs
Total entropy and enthalpy relative to ideal gas reference state
SL
DELSIG DELSL
SV
DELSIG DELSV
HL
DELHIG DELHL
HV
DELHIG DELHV
SUMMARY OF RESULTS
T = 323.15
K
Vapor pressure, bar P = 0.15
LIQUID
VAPOR
4
Compressibility
Z( T , P ) 0 = 1.35706 10
Enthalpy, J/mol
4
HL = 4.74037 10
3
HV = 1.71382 10
Entropy, J/mol K
SL = 127.05678
SV = 10.96203
Fugacity coefficient
phil = 0.6655
phiv = 0.99817
Fugacity, bar
=
Z( T , P ) 2 = 0.99816
=
Solutions to Chemical and Engineering Thermodynamics, 3e
5.54 (also available as a Mathcad worksheet)
5.54 ISENTHALPIC PENG-ROBINSON EQUATION OF STATE CALCULATION
Tc
154.6
Cp 0
Pc
25.46
50.46
om
1.591 . 10
Cp 1
2
Peng-Robinson Constants:
0.021
0.7151 . 10
Cp 2
R
kap
1.54226 . om 0.26992 . om. om
0.37464
5
1.311 . 10
Cp 3
0.00008314
0.07780 .
b
9
R. Tc
0.45724 .
ac
Pc
Input initial temperature and pressure of calculation
Input final pressure
Pf
Ti
T
Ti
P
T
kap . 1
a( T )
ac . alf( T )
Tc
Da ( T )
Z( T , P )
A
CA ( T , P )
B
CB( T , P )
A .B
2
(1
bar
30
CA ( T , P )
a( T ) . P
( R. T )
2
P .b
R. T
CB( T , P )
d
a( T )
dT
3
B
3.B
A
V
2
Pi
Pi
2
1. 1
Pc
bar
3.0
Initial state calculations
alf( T )
K,
120.
B
Vector of coefficients in the PR equation
in the form
0=-(A*B-B^2-B^3)+(A-3*B^2-2*B)*Z-(1-B)*Z^2+Z^3
2.B
B)
1
ZZ
Solution to the cubic
polyroots ( V)
for i ∈ 0 .. 2
ZZi
0
if
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ2
ZZ0 if
2
2
R . Tc
Im ZZi
0
ZZ0 < 10
5
ZZ2 < 10
5
Set any imaginary roots to zero
Sort the roots
Set the value of any imaginary roots
to value of the real root
ZZ
Calculate inital properties
Calculate initial molar volume
and enthalpy and entropy
departure
Zf( T , P )
VL
Z( T , P )
Z( T , P ) 0 . R . T
P
0.0888
. 103
Z( T , P ) =
0
0.0888
Solutions to Chemical and Engineering Thermodynamics, 3e
R . T . Z( T , P ) 0
DELHin
a( T ) .
ln
2. 2 .b
R. ln Z( T , P ) 0
DELSin
T . Da ( T )
1
Da ( T ) .
CB( T , P )
3
DELHin = 5.9875 10
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
. 105
. 105
DELSin = 40.1647
Guess for final state
0.8 . Ti
T
P
Pf
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
fv ( T , P )
Given
Z( T , P ) 0
Z( T , P ) 2
fl( T , P )
1
1
ln Z( T , P ) 0
ln Z( T , P ) 2
fv ( T , P ) 0
CA ( T , P )
CB( T , P )
2 . 2 . CB( T , P )
CA ( T , P )
CB( T , P )
T
. ln
. ln
2 . 2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
T = 101.906
find( T )
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
DELSL( T , P )
DELSV( T , P )
R. ln Z( T , P ) 0
R. ln Z( T , P ) 2
CB( T , P )
CB( T , P )
Da ( T ) .
ln
2. 2 .b
Da ( T ) .
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
DELHL( T , P )
DELHV( T , P )
R . T . Z( T , P ) 0
R . T . Z( T , P ) 2
1
T . Da ( T )
a( T ) .
ln
2. 2 .b
1
T . Da ( T )
a( T ) .
2. 2 .b
ln
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Solutions to Chemical and Engineering Thermodynamics, 3e
Ideal gas properties changes relative to the initial state
DELHIG( T , P )
Cp 0 . ( T
DELSIG( T , P )
T
Cp 0 . ln
Ti
Ti)
Cp 1 . T
2
Cp 2 . T
2
Ti
2
Cp 1 . ( T
Find vapor-liquid split
3
Cp 3 . T
3
Ti
3
Ti)
Cp 2 . T
2
4
Ti
4
2
Ti
Cp 3 . T
2
x
4
3
3
Ti
P
5
R. 10 . ln
Pi
3
0.5
Given
x. DELHV( T , P )
x
x) . DELHL( T , P )
(1
DELHIG( T , P ) DELHin
x = 0.1618
find( x)
Fraction vapor
HV
DELHV( T , P )
DELHIG( T , P )
SV
DELSV( T , P )
DELSIG( T , P )
HL
DELHL( T , P )
DELHIG( T , P )
SL
DELSL( T , P )
DELSIG( T , P )
δH
x. HV ( 1
x) . HL DELHin
δS
x. SV ( 1
x) . SL DELSin
SUMMARY OF RESULTS
Temperature, K
FEED
Ti = 120
Pressure, bar
Pi = 30
VAPOR
T = 101.906
P=3
P=3
x = 0.1618
Vapor-liquid split
Compressibility
LIQUID
T = 101.906
Z( Ti, Pi ) 0 = 0.0888
Z( T , P ) 0 = 9.3464 10
Enthalpy, J/mol
(relative to feed)
0
3
HL = 7.0203 10
Entropy, J/mol K
(relative to feed)
0
SL = 48.8038
3
Enthalpy change
J/mol
δH = 0
Entropy change
J/mol K
δS = 1.4957
Z( T , P ) 2 = 0.9309
HV = 635.2466
SV = 13.8527
Solutions to Chemical and Engineering Thermodynamics, 3e
5.55 (also available as a Mathcad worksheet)
ISENTROPIC PENG-ROBINSON EQUATION OF STATE CALCULATION
Tc
154.6
Cp 0
Pc
25.46
50.46
om
1.591 . 10
Cp 1
2
Peng-Robinson Constants:
0.021
0.7151 . 10
Cp 2
R
kap
1.54226 . om 0.26992 . om. om
0.37464
5
1.311 . 10
Cp 3
0.00008314
0.07780 .
b
9
R. Tc
ac
0.45724 .
2
2
R . Tc
Pc
Input initial temperature and pressure of calculation
Input final pressure
Pf
Ti
T
Ti
P
T
kap . 1
a( T )
ac . alf( T )
Tc
Da ( T )
Z( T , P )
A
CA ( T , P )
B
CB( T , P )
A .B
2
(1
CA ( T , P )
a( T ) . P
( R. T )
2
P .b
R. T
CB( T , P )
d
a( T )
dT
3
B
3.B
A
V
2
bar
30
Pi
2
1. 1
Pi
bar
3.0
Initial state calculations
alf( T )
K,
120.
Pc
B
Vector of coefficients in the PR equation
in the form
0=-(A*B-B^2-B^3)+(A-3*B^2-2*B)*Z-(1-B)*Z^2+Z^3
2.B
B)
1
ZZ
Solution to the cubic
polyroots ( V)
for i ∈ 0 .. 2
ZZi
0
if
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ2
ZZ0 if
Im ZZi
0
ZZ0 < 10
5
ZZ2 < 10
5
Set any imaginary roots to zero
Sort the roots
Set the value of any imaginary roots
to value of the real root
ZZ
Calculate inital properties
Calculate initial molar volume
and enthalpy and entropy
departure
Zf( T , P )
VL
Z( T , P )
Z( T , P ) 0 . R . T
P
0.0888
. 103
Z( T , P ) =
0
0.0888
Solutions to Chemical and Engineering Thermodynamics, 3e
R . T . Z( T , P ) 0
DELHin
a( T ) .
ln
2. 2 .b
R. ln Z( T , P ) 0
DELSin
T . Da ( T )
1
Da ( T ) .
CB( T , P )
3
DELHin = 5.9875 10
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
. 105
. 105
DELSin = 40.1647
Guess for final state
0.8 . Ti
T
P
Pf
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
fv ( T , P )
Given
Z( T , P ) 0
Z( T , P ) 2
fl( T , P )
1
1
ln Z( T , P ) 0
ln Z( T , P ) 2
fv ( T , P ) 0
CA ( T , P )
CB( T , P )
2 . 2 . CB( T , P )
CA ( T , P )
CB( T , P )
T
. ln
. ln
2 . 2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
T = 101.906
find( T )
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
DELSL( T , P )
DELSV( T , P )
R. ln Z( T , P ) 0
R. ln Z( T , P ) 2
CB( T , P )
CB( T , P )
Da ( T ) .
ln
2. 2 .b
Da ( T ) .
ln
2. 2 .b
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
DELHL( T , P )
DELHV( T , P )
R . T . Z( T , P ) 0
R . T . Z( T , P ) 2
1
T . Da ( T )
a( T ) .
ln
2. 2 .b
1
T . Da ( T )
a( T ) .
2. 2 .b
ln
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 0
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
Z( T , P ) 2
1
2 . CB( T , P )
. 105
. 105
Solutions to Chemical and Engineering Thermodynamics, 3e
Ideal gas properties changes relative to the initial state
DELHIG( T , P )
Cp . ( T
DELSIG( T , P )
T
Cp 0 . ln
Ti
0
Ti)
Cp 1 . T
2
Cp 2 . T
2
Ti
2
Cp 1 . ( T
Find vapor-liquid split
x
Ti)
3
Cp 3 . T
3
Ti
3
Cp 2 . T
2
2
Ti
4
4
Ti
4
Cp 3 . T
2
3
3
Ti
P
5
R. 10 . ln
Pi
3
0.5
Given
x. DELSV( T , P )
x
(1
x) . DELSL( T , P )
DELSIG( T , P ) DELSin
x = 0.1379
find( x)
Fraction vapor
HV
DELHV( T , P )
DELHIG( T , P )
SV
DELSV( T , P )
DELSIG( T , P )
HL
DELHL( T , P )
DELHIG( T , P )
SL
DELSL( T , P )
DELSIG( T , P )
δH
x. HV ( 1
x) . HL DELHin
δS
x. SV ( 1
x) . SL DELSin
SUMMARY OF RESULTS
Temperature, K
FEED
Ti = 120
Pressure, bar
Pi = 30
VAPOR
T = 101.906
P=3
P=3
x = 0.1379
Vapor-liquid split
Compressibility
LIQUID
T = 101.906
Z( Ti, Pi ) 0 = 0.0888
Z( T , P ) 0 = 9.3464 10
3
Z( T , P ) 2 = 0.9309
Enthalpy, J/mol
(relative to feed)
0
3
HL = 7.0203 10
HV = 635.2466
Entropy, J/mol K
(relative to feed)
0
SL = 48.8038
SV = 13.8527
Enthalpy change
J/mol
Entropy change
J/mol K
δH = 152.4165
δS = 0
Solutions to Chemical and Engineering Thermodynamics, 3e
5.56 (also available as a Mathcad worksheet)
5.56 ISENTHALPIC S-R-K EQUATION OF STATE CALCULATION
Tc
154.6
Cp 0
Pc
25.46
50.46
om
1.591 . 10
Cp 1
2
S-R-K Constants:
0.021
0.7151 . 10
Cp 2
R
kap
1.574 . om 0.176 . om. om
0.480
5
1.311 . 10
Cp 3
0.00008314
0.08664 .
b
9
R. Tc
ac
0.42748 .
2
2
R . Tc
Pc
Input initial temperature and pressure of calculation
Input final pressure
Pf
Ti
T
Ti
P
T
kap . 1
a( T )
ac . alf( T )
CA ( T , P )
Tc
A
CA ( T , P )
B
CB( T , P )
A .B
A
V
2
B
bar
30
a( T ) . P
( R. T )
2
P .b
R. T
CB( T , P )
d
a( T )
dT
Da ( T )
Z( T , P )
Pi
Pi
2
1. 1
K,
120.
bar
3.0
Initial state calculations
alf( T )
Pc
Vector of coefficients in the S-R-K equation
in the form
0=-A*B+(A-B^2-B)*Z-Z^2+Z^3
B
1
1
ZZ
polyroots ( V)
Solution to the cubic
for i ∈ 0 .. 2
ZZi
0
if
Im ZZi
0
Set any imaginary roots to zero
Sort the roots
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ0 < 10
5
ZZ2
ZZ0 if
ZZ2 < 10
5
Set the value of any imaginary roots
to value of the real root
ZZ
Zf( T , P )
Calculate inital properties
Calculate initial molar volume
and enthalpy and entropy
departure
DELHin
R . T . Z( T , P ) 0
1
VL
Z( T , P )
Z( T , P ) 0 . R . T
0.1004
. 103
Z( T , P ) =
P
T . Da ( T )
b
a( T ) .
ln
0
0.1004
Z( T , P ) 0
CB( T , P )
Z( T , P ) 0
. 105
Solutions to Chemical and Engineering Thermodynamics, 3e
R. ln Z ( T , P )
DELSin
Da ( T ) .
0
CB( T , P )
3
DELHin = 6.0618 10
ln
Z( T , P ) 0
CB( T , P )
Z( T , P ) 0
b
. 105
DELSin = 40.9502
Guess for final state
0.8 . Ti
T
P
Pf
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
fl( T , P )
fv ( T , P )
Given
Z( T , P ) 0
Z( T , P ) 2
fl( T , P )
1
1
ln Z( T , P ) 0
CA ( T , P ) .
CB( T , P )
ln Z( T , P ) 2
CB( T , P )
CA ( T , P ) .
CB( T , P )
fv ( T , P ) 0
T
Z( T , P ) 0
ln
Z( T , P ) 0
Z( T , P ) 2
ln
CB( T , P )
CB( T , P )
CB( T , P )
Z( T , P ) 2
T = 102.0671
find( T )
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
DELSL( T , P )
R. ln Z( T , P ) 0
CB( T , P )
Da ( T ) .
DELSV( T , P )
R. ln Z( T , P ) 2
CB( T , P )
Da ( T ) .
Z( T , P ) 0
ln
CB( T , P )
. 105
Z( T , P ) 0
b
Z( T , P ) 2
ln
CB( T , P )
. 105
Z( T , P ) 2
b
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
DELHL( T , P )
R . T . Z( T , P ) 0
1
DELHV( T , P )
R . T . Z( T , P ) 2
1
T . Da ( T )
a( T ) .
ln
Z( T , P ) 0
a( T ) .
ln
. 105
Z( T , P ) 0
b
T . Da ( T )
CB( T , P )
Z( T , P ) 2
CB( T , P )
. 105
Z( T , P ) 2
b
Ideal gas properties changes relative to the initial state
DELHIG( T , P )
Cp 0 . ( T
Ti)
Cp 1 . T
2
2
2
Ti
Cp 2 . T
3
3
3
Ti
Cp 3 . T
4
4
4
Ti
Solutions to Chemical and Engineering Thermodynamics, 3e
T
Cp . ln
DELSIG( T , P )
0
Cp 1 . ( T
Ti
Find vapor-liquid split
Cp 2 . T
Ti)
2
Cp 3 . T
2
Ti
2
x
3
3
Ti
P
5
R. 10 . ln
Pi
3
0.5
Given
x. DELHV( T , P )
x
x) . DELHL( T , P )
(1
DELHIG( T , P ) DELHin
x = 0.1661
find( x)
Fraction vapor
HV
DELHV( T , P )
DELHIG( T , P )
SV
DELSV( T , P )
DELSIG( T , P )
HL
DELHL( T , P )
DELHIG( T , P )
SL
DELSL( T , P )
DELSIG( T , P )
δH
x. HV ( 1
x) . HL DELHin
x. SV ( 1
δS
x) . SL DELSin
SUMMARY OF RESULTS
Temperature, K
FEED
Ti = 120
Pressure, bar
Pi = 30
LIQUID
T = 102.0671
P=3
P=3
x = 0.1661
Vapor-liquid split
Z( Ti, Pi ) 0 = 0.1004
Compressibility
5.57
VAPOR
T = 102.0671
Enthalpy, J/mol
(relative to feed)
0
Entropy, J/mol K
(relative to feed)
0
Z( T , P ) 0 = 0.0106
Z( T , P ) 2 = 0.934
3
HL = 7.1435 10
HV = 630.1699
SL = 49.936
SV = 13.8781
Enthalpy change
J/mol
δH = 0
Entropy change
J/mol K
δS = 1.6121
(also available as a Mathcad worksheet)
5.57 ISENTROPIC S-R-K EQUATION OF STATE CALCULATION
Tc
Cp 0
154.6
25.46
Pc
50.46
Cp 1
om
1.591 . 10
2
S-R-K Constants:
Cp 2
R
0.021
kap
0.7151 . 10
1.574 . om 0.176 . om. om
0.480
5
1.311 . 10
Cp 3
0.00008314
b
0.08664 .
9
R. Tc
ac
0.42748 .
Pc
Input initial temperature and pressure of calculation
Input final pressure
Initial state calculations
Pf
bar
3.0
T
Ti
P
Pi
Ti
120.
K,
2
2
R . Tc
Pc
Pi
30
bar
Solutions to Chemical and Engineering Thermodynamics, 3e
Initial state calculations
T
Ti
P
Pi
2
alf ( T )
1. 1
T
kap . 1
a( T )
ac . alf( T )
Tc
A
CA ( T , P )
B
CB( T , P )
A .B
A
V
2
B
P .b
R. T
CB( T , P )
2
( R. T )
d
a( T )
dT
Da ( T )
Z( T , P )
a( T ) . P
CA ( T , P )
Vector of coefficients in the S-R-K equation
in the form
0=-A*B+(A-B^2-B)*Z-Z^2+Z^3
B
1
1
ZZ
polyroots ( V)
Solution to the cubic
for i ∈ 0 .. 2
ZZi
0
if
Im ZZi
0
Set any imaginary roots to zero
Sort the roots
ZZ
sort ( ZZ )
ZZ0
ZZ2 if
ZZ0 < 10
5
ZZ2
ZZ0 if
ZZ2 < 10
5
Set the value of any imaginary roots
to value of the real root
ZZ
Zf( T , P )
Calculate inital properties
Calculate initial molar volume
and enthalpy and entropy
departure
DELHin
DELSin
R . T . Z( T , P ) 0
R. ln Z( T , P ) 0
1
VL
Z( T , P ) 0 . R . T
0.1004
. 103
Z( T , P ) =
P
T . Da ( T )
a( T ) .
ln
CB( T , P )
Da ( T ) .
ln
CB( T , P )
Z( T , P ) 0
Z( T , P ) 0
CB( T , P )
Z( T , P ) 0
b
DELSin = 40.9502
T
0.8 . Ti
P
0
0.1004
Z( T , P ) 0
b
3
DELHin = 6.0618 10
Guess for final state
Z( T , P )
Pf
Fugacity expressions [actually ln(f/P)] for the liquid fl and vapor fv
. 105
. 105
Solutions to Chemical and Engineering Thermodynamics, 3e
fl ( T , P )
Z( T , P )
fv ( T , P )
Z( T , P ) 2
Given
fl( T , P )
1
0
ln Z( T , P ) 0
ln Z( T , P ) 2
1
CA ( T , P ) .
CB( T , P )
CB( T , P )
CA ( T , P ) .
CB( T , P )
fv ( T , P ) 0
T
ln
CB( T , P )
Z( T , P ) 0
Z( T , P ) 2
ln
CB( T , P )
Z( T , P ) 0
CB( T , P )
Z( T , P ) 2
T = 102.0671
find( T )
Residual entropy for liquid (DELSL) and vapor (DELSV) phases
DELSL( T , P )
R. ln Z( T , P ) 0
CB( T , P )
Da ( T ) .
DELSV( T , P )
R. ln Z( T , P ) 2
CB( T , P )
Da ( T ) .
Z( T , P ) 0
ln
CB( T , P )
. 105
Z( T , P ) 0
b
Z( T , P ) 2
ln
CB( T , P )
. 105
Z( T , P ) 2
b
Residual enthalpy for liquid (DELHL) and vapor (DELHV) phases
DELHL( T , P )
R . T . Z( T , P ) 0
1
DELHV( T , P )
R . T . Z( T , P ) 2
1
T . Da ( T )
a( T ) .
Z( T , P ) 0
ln
a( T ) .
. 105
Z( T , P ) 0
b
T . Da ( T )
CB( T , P )
Z( T , P ) 2
ln
CB( T , P )
. 105
Z( T , P ) 2
b
Ideal gas properties changes relative to the initial state
DELHIG( T , P )
Cp 0 . ( T
DELSIG( T , P )
Cp 0 . ln
Ti)
Cp 1 . T
2
2
Ti
Cp 2 . T
2
T
Ti
Find vapor-liquid split
Cp 1 . ( T
x
Ti)
3
3
Ti
3
Cp 2 . T
2
(1
x) . DELSL( T , P )
4
4
Ti
4
2
Ti
2
0.5
Given
x. DELSV( T , P )
Cp 3 . T
DELSIG( T , P ) DELSin
Cp 3 . T
3
3
3
Ti
P
5
R. 10 . ln
Pi
Solutions to Chemical and Engineering Thermodynamics, 3e
x
x = 0.1408
find ( x )
Fraction vapor
HV
DELHV( T , P )
DELHIG( T , P )
SV
DELSV( T , P )
DELSIG( T , P )
HL
DELHL( T , P )
DELHIG( T , P )
SL
DELSL( T , P )
DELSIG( T , P )
δH
x. HV ( 1
x) . HL DELHin
δS
x. SV ( 1
x) . SL DELSin
SUMMARY OF RESULTS
Temperature, K
FEED
Ti = 120
Pressure, bar
Pi = 30
LIQUID
T = 102.0671
P=3
P=3
x = 0.1408
Vapor-liquid split
Z( Ti, Pi ) 0 = 0.1004
Compressibility
Z( T , P ) 0 = 0.0106
Z( T , P ) 2 = 0.934
Enthalpy, J/mol
(relative to feed)
0
3
HL = 7.1435 10
HV = 630.1699
Entropy, J/mol K
(relative to feed)
0
SL = 49.936
SV = 13.8781
Enthalpy change
J/mol
δH = 164.5454
Entropy change
J/mol K
5.58
VAPOR
T = 102.0671
δS = 7.1054 10
15
This problem was solved using the attached Mathcad worksheet. The results are
T(o C)
Pvap with α(T)
Pvap with α=1
273.15
283.15
293.15
303.15
323.25
343.15
373.15
393.15
423.25
448.15
474.15
523.15
0.3137
0.5529
1.697
3.208
9.994
26.681
92.355
186.67
463.23
886.08
1599.4
4065.2
166.57
221.329
288.55
369.83
580.97
867.65
1467.0
1997.1
3016.5
4094.2
5456.5
8759.0
(P in kPa)
Solutions to Chemical and Engineering Thermodynamics, 3e
623.15
643.15
16744
21060
18865
As can be seen, the S-R-K equation is of comparable accuracy to the P-R
equation. In both cases if the α parameter is set to one, the results are not very
good, indeed quite bad at low temperatures.
The Mathcad worksheet used in solving this problem is given below.
Solutions to Chemical and Engineering Thermodynamics, 3e
5.59
5.59
(also available as a Mathcad worksheet)
The solution is that the final temperature is 131.34 K, and the final pressure is
37.036 bar.
Using SRK EOS with the approximate two-constant heat capacity expression
Property Data
(T in K, P in bar):
Tc
R
126.2
Pc
0.00008314
33.94
kap
om
0.480
Initial Conditions (Vt=total volume, m^3):
Cp1
b
27.2
Cp2
0.0042
1.574 . om 0.176 . om. om
Ti
Peng-Robinson Constants:
Initial temperature
0.04
0.08664 .
170
Pi
R. Tc
ac
100
Vt
0.42748 .
0.15
R Tc
Pc
T
2.
2
Pc
Ti
Note that these are being defined as a
function of temperature since we will need to
interate on temperature later to obtain the final
state of the system
Find initial molar volume and number of moles
Start with initial guess for volume, m^3/mol
2
alf( T )
1. 1
kap . 1
T
Tc
ac . alf( T )
a( T )
d
a( T )
dT
Da ( T )
V
R. Ti
Pi
Solutions to Chemical and Engineering Thermodynamics, 3e
Solve P-R EOS for initial volume
Given
Pi
R .T
V
Vi = 1.02 10
Initial molar volume and
number of moles
Entropy departure at the
initial conditions
DELSi
R. ln ( Vi
a( T )
( V.( V
b
4
Vi
b))
Vt
Ni
3
Ni = 1.471 10
Vi
b).
Now consider final state
Pi
R. T
Da ( T ) .
ln
Vi
b
Nf
Ni
Find ( V )
b
. 105
Vi
0.15
10 . 50 Vf
V
Vf
Nf
Type out final number of
moles and specific volume
Nf = 971.269
Final pressure, will change in course
of solving for the final temperature
Entropy departure
at final conditions
Solve for final
temperature using
S(final)-S(initial)=0
DELS( T )
GIVEN
0 27.2 . ln
T
Ti
T
R. ln ( V b ) .
0.0042 . ( T
Pf( T )
R. T
Ti)
R. T
a( T )
V b
V. ( V b )
Da ( T ) .
ln
b
Pf( T )
5
R. 10 . ln
Pi
V b
. 105
V
DELS( T )
T = 131.34
a) At a given temperature, the stability limit of a fluid is determined by the
following criterion (Note that this leads to the spinodal curve)
∂P
=0
∂V T
FG IJ
H K
For the given EOS, the stability limit of a fluid undergoing a pressure change at
constant temperature is
∂P
RT
BRT
CRT
= − 2 − 2 3 −3 4 = 0
∂V T
V
V
V
FG IJ
H K
or V 2 + 2 BV + 3C = 0
In order to have a phase transition, there must be two distinct stability limits,
i.e., the above quadratic equation must have two different roots of V. Therefore,
( 2 B ) 2 − 4 × 1 × (3C) > 0
or B 2 > 3C
4
DELSi
FIND( T )
Type out solution
5.60
Pf( T )
Vf = 1.544 10
Pf( T ) = 37.076
Solutions to Chemical and Engineering Thermodynamics, 3e
b) According to Illustration 4.2-1
LM F ∂P I
N H ∂T K
dU = CV dT + T
V
OP
Q
− P dV
LM F ∂P I
N H ∂T K
But for the given EOS T
V
OP
Q
−P =0
Therefore,
∆U = dU = CV (V , T ) dT
z z
Since
FG ∂C IJ
H ∂V K
=T
V
T
FG ∂ P IJ
H ∂T K
2
2
= 0 (Because B and C are not functions of T)
V
Therefore Cv = Cv (T) = CV* = a + bT
and
z
T
2
b
∆U = ( a + bT ) dT = a ( T2 − T1 ) + ( T22 − T12 )
2
T
1
The internal energy change is the same for an ideal gas.
c) According to Eqn 4.2-19
C
∂P
d S = V dT +
dV so that
T
∂T V
F I
H K
∂P
FG RT + BRT + CRT IJ
TF I
H
K
FG ∂T IJ = − ∂T = P = H V V V K
H ∂V K
C
a + bT
a + bT
V
S
V
For an ideal gas,
FG ∂T IJ
H ∂V K
S
FG IJ
H K
RT
P
V
RT
=
=
=
(
a + bT a + bT V a + bT )
2
3
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Solutions to Chemical and Engineering Thermodynamics, 3e
6
6.1
(a) By Eqn. (6.2-3)
FG ∂ H IJ
H∂N K
i
= Gi ;
P , S , N j≠ i
but Gi = Hi − TSi . Thus
FG ∂ H IJ
H∂N K
i
f
F ∂U IJ
dU = G
H ∂S K
= Hi − TSi
P , S , N j≠ i
a
(b) Since U = U S ,V , N
FG ∂U IJ dV + ∑ FG ∂U IJ
H ∂V K
H∂ N K
F ∂U IJ dN
= TdS − PdV + ∑ G
H ∂N K
dS +
V,N
i
S ,N
i
dN i
S ,V , N j≠i
(1)
i
i
i
S ,V , N j≠i
However, we also have U = H − PV ; dU = dH − PdV − VdP , and, by Eqn.
(6.2-3)
dU = VdP + TdS + ∑ Gi dNi − PdV −VdP = TdS − PdV + ∑ Gi dNi
Equating (1) and (2) shows that Gi =
A = A( T ,V , N )
⇒ dA =
FG ∂ A IJ
H ∂T K
dT +
V,N
FG ∂U IJ
H∂N K
FG ∂ A IJ
H ∂V K
i
T, N
(2)
. Next we start from
S ,V , N j ≠i
dV + ∑
i
FG ∂ A IJ
H∂ N K
i
dN i
T ,V , N
j≠ i
or
dA = − SdT − PdV + ∑
i
However, we also have that A = U − TS ;
FG ∂ A IJ
H∂N K
i
dN i
T ,V , N
j≠i
(3)
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
dA = dU − TdS − SdT = TdS − PdV + ∑ Gi dNi − TdS − SdT
or
dA = −SdT − PdV + ∑ Gi dNi
(4)
Comparing (3) and (4) yields
Gi =
6.2
FG ∂ A IJ
H∂N K
i
T ,V , N j≠i
a
(a) General: θ = ∑ Ni θi where θi = ∂θ ∂ Ni
f
and
T , P , N j≠ i
d θ = ∑ θi dN i + ∑ Ni d θi
(1)
However, we also have that
dθ =
FG ∂θ IJ
H ∂TK
dT +
V,N
FG ∂θ IJ
H ∂V K
T, N
dV + ∑
FG ∂θ IJ
H∂N K
i
dN i
(2)
T ,V , N j ≠i
Subtracting (2) from (1) yields
FG ∂θ IJ
H ∂T K
0=−
dT −
V ,N
FG ∂θ IJ
H ∂V K
T ,N
LM F ∂θ I
MN GH ∂ N JK
dV + ∑ θi −
i
OP
PQdN + ∑ N dθ
i
T ,V , N
j≠ i
i
i
At constant T and V
LM FG ∂θ IJ
MN H ∂ N K
0 = ∑ θi −
OP dN + ∑ N dθ
PQ
i
i T ,V , N
i
i
(general equation)
For θ = A ,
θi −
FG ∂θ IJ
H∂N K
i
θi = Ai
and
FG ∂θ IJ
H∂N K
i
=
T ,V , N
j≠i
FG ∂ A IJ
H∂N K
i
= Gi .
T ,V , N j≠i
= Ai − Gi = − PVi and
T ,V , N j≠i
∑ NidAi T , V
= P∑ Vi dNi T , V specific equation for θ = A
(b) Following the analysis above, we also get
Thus,
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂θ IJ
H ∂U K
0=−
dU −
V ,N
FG ∂θ IJ
H ∂V K
U,N
LM F ∂θ I
MN GH ∂ N JK
dV + ∑ θi −
i U ,V , N j ≠i
OP
PQdN + ∑ N dθ
i
i
i
and, at constant U and V
OP dN + N dθ
PQ ∑
G
F ∂S IJ
Now, choosing θ = S , and using that G
=−
, which is easily
H∂N K
T
0=
LMθ − F ∂θ I
∑ M GH ∂ N JK
N
i
i
i
i
i
U ,V , N j≠i
i
i U ,V , N j≠i
derived, yields
−T ∑ Ni dSi U ,V = ∑ Hi dN i U ,V
(c) Following a similar analysis to those above, we obtain
FG ∂θIJ
H ∂SK
0=−
dS −
V ,N
FG ∂θ IJ
H ∂V K
S, N
LM F ∂θ I
MN GH ∂ N JK
dV + ∑ θi −
i
OP
PQdN + ∑ N dθ
i
S ,V , N j ≠i
i
which, at constant V and S, reduces to
LM F ∂θ I
MN GH ∂ N JK
Finally, using θ = U , and a∂ U ∂ N f
0 = ∑ θi −
i
OP
PQdN + ∑ N dθ
i
S ,V , N j ≠i
i S ,V , N j≠i
i
i
= Gi yields
∑ NidU i S ,V = ∑ l− PVi + TSiqdNi S ,V
6.3
(a) At constant U and V, S = maximum at equilibrium
C
C
i =1
i =1
S = S I + S II = ∑ NiI Si I + ∑ NiII Si II
but
FG ∂ S IJ
H ∂U K
F ∂ S IJ
+G
H ∂U K
dS = 0 =
I
I
FG ∂ S IJ
H ∂V K
F ∂ S IJ
+G
H ∂V K
dU I +
V ,N
II
dU II
II
V ,N
I
I
U ,N
FG ∂ S IJ
H∂ N K
F ∂ S IJ
+ ∑G
H∂ N K
dV I + ∑
II
dV II
II
U ,N
Since U = U I + U II = constant, dU II = −dU I
I
I
i U ,V , N
dN iI
j≠ i
II
II
i
dNiII
U ,V , N j ≠i
i
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
Since V = V I +V II = constant, dV II = −dV I
and since Ni = NiI + NiII = constant, dNiII = −dNiI
Also,
FG ∂ S IJ
H ∂U K
=
V ,N
FG IJ
H K
1
∂S
;
T
∂V
=
U ,N
FG
H
P
∂S
and
T
∂ Ni
IJ
K
=−
U ,V , N j≠i
Gi
T
(see previous problem)
Thus
dS = 0 =
F 1 − 1 I dU + FG P
HT T K H T
I
I
I
II
I
−
IJ
K
FG
H
IJ
K
P II
G1I G1II
I
dV
−
−
dN iI
∑
T II
T I T II
i
⇒ T I = T II ; P I = PII ; and Gi I = Gi II
for equilibrium in a closed system at constant U and V.
(b) For a closed system at constant S and V, U has an extremum. Thus
FG ∂U IJ
H ∂S K
F ∂U IJ
+G
H ∂S K
I
dU = 0 =
I
FG ∂U IJ
H ∂V K
F ∂U IJ
+G
H ∂V K
I
dS I +
V,N
II
dS II
II
V,N
I
S, N
FG ∂U IJ
H ∂N K
F ∂U IJ
+ ∑G
H∂ N K
I
dV I + ∑
I
i
i
II
dV II
II
S ,V , N j ≠i
II
II
i U ,V , N j ≠i
i
S ,N
dNiI
dN iII
but S, V and N j , j = 1, L, C are constant. Thus
c
h
c
h
c
h
dU = 0 = T I − T II dS I + P I − PII dV I + ∑ Gi I − Gi II dN iI
i
⇒ T I = T II , P I = PII and Gi I = Gi II
for equilibrium in a closed system at constant S and V.
6.4
(a) For a closed system at constant T and V, A is a minimum at equilibrium; thus
dAV , T = 0 . From Eqn. (6.2-5)
dA = − PdV − SdT + ∑ Gi dNi or dA V , T = ∑ Gi dNi
But, Ni = Ni , 0 + ν i X . Thus dN i = νi dX and
dA V , T =
b∑ ν G gdX = 0 or FGH ∂∂ XA IJK
i
i
V ,T
= ∑ νi Gi = 0 .
i
(b) For a closed system at constant U and V, S = maximum, or dS U ,V = 0 . From
Eqn. (6.2-4) dS =
1
P
1
dU + dV − ∑ Gi dN i ; thus
T
T
T
Solutions to Chemical and Engineering Thermodynamics, 3e
dS U ,V = −
1
1
Gi dNi or dS U ,V = −
∑
T
T
b∑ G ν gdX
i i
and
∂S
∂X
6.5
=−
U ,V
1
T
∑νiGi = 0
i
Let mi = molecular weight of species i. Multiplying Eqn. (6.3-2a) by mi and
summing over all species i yields, for a closed system
∑ mi Ni = total mass in system = ∑ mi Ni,0 + X ∑νi mi
total mass in
system initially
However, since the total mass is a conserved quantity,
∑ mi Ni = ∑ mi Ni,0 ⇒ X ∑ν i mi = 0 , where X can take on any value.
Consequently, if this equation is to be satisfied for all values of X, then
∑ ν imi = 0 !
M
Similarly, in the mu lti-reaction case, starting from Ni = Ni ,0 + ∑ νij X j , we get
j =1
C
C
C
M
C
M
M
C
i =1
i= 1
i =1
j =1
i =1
j =1
j =1
i =1
∑ mi Ni = ∑ mi Ni,o + ∑ mi ∑νij X j ⇒ ∑ mi ∑νij X j = 0 = ∑ X j ∑νijmi
Since the X j ’s are not, in general, equal to zero, we have
C
∑ νijmi = 0
i =1
a f
a f
In particular, for the reaction H 2 O = H 2 + 1 2 O 2 , or H 2 + 1 2 O 2 − H 2 O = 0 , we
have
∑ νijmi = (+ 1)(2) + FH 2IK (32) + ( −1)(18) = 0 .
1
i
6.6
From Eqns. (6.6-4) we have
V1 = V 1 + ∆V mix + x2
and
a
∂ ∆V mix
∂ x1
f
(1)
T ,P
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
V2 = V 2 + ∆V mix + x1
a
f
∂ ∆V mix
∂ x1
(2)
T ,P
Now since T, P and X, are the independent variables, we have that
0 since pure component volume is a function of
dV1
T, P
=
=
=
dV 1
T, P
a
∂ ∆V mix
∂x 1
a
T and
P only
+ d ∆V
mix
a
f
+
T , P
f
∂ 2 ∆V mix
∂ x 12
x 2
f
+d
T, P
a
f
∂ ∆V mix
∂x 1
LMN ∂ a∆ f OPQ
∂
V mix
x 2
T ,
x1
∂x 2
∂x 1
P
T , P
dx 1
+
x2
a
∂ 2 ∆ V mix
∂ x 12
f
dx 1
T, P
∂x 2
= −1
∂x 1
dx 1 since
T , P
Similarly
a
∂ 2 ∆V mix
dV2 T , P = − x1
∂ x12
f
dx1
T ,P
Thus
∑ xiαVi T , P = x1x2
a
∂2 ∆V mix
∂ x12
f
dx1 − x2 x1
a
∂2 ∆V mix
T, P
∂ x12
f
dx1 ≡ 0
T, P
Thus, V1 and V2 given by equations (1) and (2) identically satisfy the GibbsDuhem equation
∑ xidθ i T , P = 0 .
A similar argument applies for the partial molar enthalpies of Eqn. (6.6-9).
6.7
(also available as a Mathcad worksheet)
The students can solve this problem by drawing tangent lines to the
∆V mix
curves. Polak and Lu smoothed their data using the Redhich-Kister equation (see
Eqn. (6.6-5a)). That is, they fitted their data to
a
n
∆V mix = x1 x2 ∑ C j x2 − x1
j =1
Now
a
f a
1
a
= x1 1 − x1
f∑ C (1 − 2 x)
j
j −1
1
1
mix
j −1
j
1
j −1
j
1
2
1
mix
1
and
j −1
f∑ C a1 − 2x f
− x ∑ C a1 − 2 x f − 2 x a1 − x f∑ C ( j − 1)a1 − 2 x f
∂ a ∆V f
V − V = a ∆V f − x
= a1 − x f k A − 2 x Bp
∂x
∂ ∆V mix
= 1 − x1
∂ x1
Thus
f
1
j
2
1
j −2
1
1
(1)
Solutions to Chemical and Engineering Thermodynamics, 3e
a
f
V2 − V 2 = ∆V mix − x1
∂ ∆V mix
= x12 A + 2 x2 B
∂ x1
∑ Cj a1 − 2 x1f j −1
B = ∑ C j ( j − 1) 1 − 2 x1
a
f
k
p
(2)
where
A=
n
a
n
and
j =1
j =1
f
j −2
Taking species 1 to be methyl formate, Polak and Lu found
C1
methyl formate - Methanol
C2
C3
C4
− 0.33259 − 010154
.
− 0.0516 0.0264
methyl formate - Ethanol
0.81374
−0.00786
0.0846
−3
3
[units are cc/mol; multiply by 10
0.0448
to get m kmol ]
I have used the equations above and the constants given to find V1 − V 1 and
V2 − V2 , since this leads to more accurate results than the graphical method.
The results are tabulated and plotted below.
Methyl formate - Methanol
xMF
0
0.1
0.2
0.3
0.4
0.5
0
–0.039
–0.065
–0.080
–0.085
–0.083
V1 − V 1
–0.459
–0.329
–0.225
–0.148
–0.093
–0.058
V2 −V 2
0
–0.007
–0.025
–0.051
–0.080
–0.109
0.6
0.7
0.8
0.9
1.0
–0.075
–0.063
–0.047
–0.027
0
V1 − V 1
–0.035
–0.021
–0.011
–0.004
0
V2 −V 2
–0.136
–0.162
–0.192
–0.236
–0.309
∆V mix
acc molf
xMF
∆V mix
acc molf
b
g
Thus VMF = 6278
. + V1 −V 1 cc/mol or 10−3 m3 kmol .
b
g
VM = 4073
. + V2 −V 2 .
Methyl formate - Ethanol
xMF
0
0.1
0.2
0.3
0.4
0.5
0
0.080
0.136
0.174
0.196
0.203
V1 − V 1
0.935
0.682
0.507
0.381
0.285
0.205
V2 −V 2
0
0.013
0.043
0.085
0.137
0.201
0.6
0.7
0.8
0.9
1.0
0.196
0.174
0.134
0.077
0
V1 − V 1
0.138
0.081
0.037
0.010
0
V2 −V 2
0.284
0.390
0.522
0.680
0.861
∆V mix
acc molf
xMF
∆V mix
acc molf
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
b
g
Thus VMF = 6278
. + V1 −V 1 cc/mol. Multiply by 10−3 for m3 kmol .
b
VE = 5868
. + V2 − V 2
6.8
g
This problem is similar to the last one, and will be treated in a similar fashion.
Fenby and Ruenkrairergasa give their data in the form
a
f a
∆H mix J mol = x2 1 − x2
f∑ C a1 − 2 x f
n
j =1
j
j −1
(1)
2
where component 2 is the fluorobenzene.
The constants given in the
aforementioned reference and Fenby and Scott J. Phys. Chem 71, 4103 (1967) are
given below
System
C1
C2
C3
C4
C6 H 6 − C6 F5Cl
–2683
929
970
0
C6 H 6 − C6 F5Br
–3087
356
696
0
C6 H 6 − C6 F5I
–4322
–161
324
0
C6 H 6 − C6 F6
–1984
+1483
+1169
0
230
+578
+409
+168
C6 H 6 − C6 F5H
Solutions to Chemical and Engineering Thermodynamics, 3e
If we replace x2 with 1 − x1 in Eqn. (1), we regain the equation of the previous
illustration, except for a factor of (−1) j −1 in the sum and the corresponding places
in the other equations.
xC 6 H 6
∆H mix
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0
–252
–463
–609
–679
–671
–590
–453
–284
–119
0
bH − Hg
C 6H6
bH − Hg
C 6 F5Cl
–2642
–2171
–1790
–1466
–1175
–903
–646
–409
–205
–57.8
0
0
–39.2
–130
–242
–349
–439
–506
–555
–601
–666
–784
xC 6 F5Cl
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
[Note: J/mol]
C6 H 6 − C6 F5Br
xC 6 H 6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
∆H mix
0
–263
–488
–654
–751
–772
–717
–595
–420
–212
0
bH − H g bH − H g
C6 H 6
C6 F5 Br
–2747
–2248
–1829
–1469
–1149
–861
–600
–370
–181
–50.0
0
0
–42.9
–153
–306
–486
–683
–893
–1120
–1374
–1671
–2035
C6 H 6 − C6 F5I
∆H mix
0
–359
–657
–883
–1026
–1081
–1042
–910
–688
–382
0
bH − H g bH − H g
C6 H 6
C6 F5 I
–3837
–3119
–2489
–1937
–1456
–1040
–689
–402
–187
–48.9
0
0
–52.1
–200
–431
–740
–1121
–1572
–2095
–2695
–3379
–4159
xC 6 F5 x
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
C6 H 6 − C6 F6
6.9
xC 6 H 6
∆H mix
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0
–218
–392
–502
–536
–496
–394
–253
–108
–4.5
0
bH − H g bH − H g
C6 H 6 − C6 F5H
∆H mix
bH − H g bH − H g
C6 H 6
C6 F6
–2298
–1899
–1590
–1332
–1097
–867
–637
–413
–212
–60.9
0
0
0
61.0
0
–31.2
–2.2
36.2
–1.1
–93.0
–3.9
–2.8
+6.8
–146
13.5
–42.3
+37.4
–162
31.4
–72.3
+100
–125
57.5
–87.0
+202
–28.9
86.9
–84.5
344
+121
110
–66.7
524
+308
116
–39.4
+737
+503
85.9
–12.6
+973
+688
0
0
1217
↑
↑
↑
Note: Changes in sign in column
C6 H 6
xC 6 F5 x
C6 F5 H
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
(a) Gibbs Phase Rule: F = C − M − P + 2
P = 2 , C = 2 , M = 0 ⇒ F = 2 − 0 − 2 + 2 = 2 degrees of freedom
Thus can fix two variables, usually from among T, P, x and y.
(b) P = 1 , C = 3 and M = 1 ⇒ F = 3 − 1 − 1 + 2 = 3 degrees of freedom
Thus, we can fix 3 variables, for example, T, P and xH 2 .
(c) Formation reactions
C + 2O → CO 2
C + O → CO
2H → H 2
C + 4H → CH 4
2H + O → H 2 O
Use O = CO − C and H =
1
H 2 to eliminate O and H from the set so that
2
C + 2( CO − C) → CO 2
a
a
f
2CO → CO 2 + C
C + 4 1 2 H 2 → CH 4 ⇒ C + 2H 2 → CH 4
f
2 1 2 H 2 + ( CO − C) → H 2 O
H 2 + CO → H 2 O + C
Thus we have found a set (there is no unique set) of three independent
reactions among the six species. Consequently, C = 6 , M = 3 , P = 2 (solid
carbon + gas phase).
F = C − M − P + 2 = 6 − 3 − 2 + 2 = 3 degrees of freedom. As a check:
c
# of unknowns = 8 T S , PS , T V , P V , xCO 2 , xCO , xH 2 , xCH 4
Note: xH 2 O = 1 − xCO 2 − xCO − xH 2 − xCH 4
h
Solutions to Chemical and Engineering Thermodynamics, 3e
Relations among the unknowns T S = TV , PS = P V , no phase equilibrium
relations, but 3 chemical equilibrium relations of the form ∑ νij Gi = 0 .
8 unknowns− 5 eqns. = 3 unspecified unknowns or
3 degrees of freedom
a
f
6.10 (a) In general, for a binary, two-phase mixture C = 2, M = 0, P = 2
F = C − M − P + 2 = 2 − 0 − 2 + 2 = 2 degrees of freedom.
However, for an azeotrope there is the additional restriction x1 = y1 , which
eliminates one degree of freedom. Thus, there is only 1 degree of freedom for a
binary, azeotropic system.
(b) In osmotic equilibrium P I ≠ PII , since the membrane is capable of supporting
a pressure difference, and G2I ≠ G2II , where 2 is the species which does not
pass through the membrane. Therefore, the independent unknowns are T I ,
PI , x1I , T II , P II and x1II . [Note, x2I and x2II are not independent unknowns
since x2I = 1 − x1I and x2II = 1 − x1II ].
There are two equilibrium relations
between these six unknowns: viz. T = TII and G1I = G1II . Consequently,
I
there are four degrees of freedom … that is, as we shall see in Sec. 8.7, if T, PI ,
P II and x1I are specified, x1II will be fixed.
(c) Case I: M = 0, C = 2, P = 2 ⇒ F = 2 − 0 − 2 + 2 = 2
Case II: M = 0, C = 2, P = 3 ⇒ F = 2 − 0 − 3 + 2 = 1
6.11 (a) Gibbs Phase Rule: F = C − M − P + 2
C = 2 , M = 0 ⇒ F = 2 − 0 − P + 2 = 4 − P degrees of freedom.
Therefore, a maximum of 4 phases can exist at equilibrium (for example a solid,
two liquids and a vapor, or two solids, a liquid and a vapor, etc.)
(b) Gibbs Phase Rule: F = C − M − P + 2
C = 2 , M = 1 ⇒ F = 2 − 1 − P + 2 = 3 − P degrees of freedom.
Therefore, a maximum of e phases can exist at equilibrium (for example a two
liquids and a vapor, or a solid, a liquid and a vapor, etc.)
6.12 (a)
dNi
= N& i + N& i ,rxn
dt
dU
0
dV
= ∑ N& i Hi + Q& − W s −P
dt
dt
&
dS
Q
= ∑ N& i Si + + S&gen
dt
T
dS
T
− T ∑ N& i Si − TS&gen = Q&
dt
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
dU
dS
dV
= ∑ N& i Hi + T
− T ∑ N& i S i − TS&gen − P
dt
dt
dt
dU
dV
dS
+P
−T
= ∑ N& i Hi − TSi − TS&gen
dt
dt
dt
dU
dV
dS
dNi
dX
+P
−T
= ∑ N& i µi − TS&gen = ∑
− νi
µi − TS&gen
dt
dt
dt
dt
dt
General expression
Now
System is only permeable to species 1
dU
dV
dS
dN 1
dX
+P
−T
−
− ν1
µ1 = − TS& gen ≤ 0
dt
dt
dt
dt
dt
When T and P constant
d
d
(U + PV − TS ) −
N1 − ν1 X µ1 ≤ 0
dt
dt
d
G − N1 − ν1 X µ1 ≤ 0
dt
⇒ G − N1 − ν1 X µ1 = minimum at equilibrium
(b) When T and V are constant
d
d
(U − TS ) −
N1 − ν1 X µ1 ≤ 0
dt
dt
⇒ A − N1 − ν1 X µ1 = minimum at equilibrium
b
g
F
H
6.13 (a)
f
f
a
a
f
I
K
I
K
a
a
a
F
H
f
f
2N → N 2
2O → O 2
2N + O → N 2 O
2N + 2O → 2NO
2N + 4O → N 2 O 4
1
N2 + O2 → N2O
2
N 2 + O 2 → 2NO
N 2 + 2O2 → N 2 O 4
2N + 4O → 2NO 2
N 2 + 2O2 → 2NO 2
5
2N + 5O → N 2 O 5
N2 + O2 → N2O5
2
⇒ 5 independent reactions
(b) F = C − M − P + 2 = 7 − 5 − 1+ 2 = 9 − 6 = 3
F = 3 degrees of freedom
(c) 1 degree of freedom used in O 2 : N 2 ratio ⇒ 2 degres of freedom
6.14 Mass balance: M1 + M2 = M f
Molecular weight H 2O = 18.02 g mol
Energy balance: M1U$ 1 + M2U$ 2 = M f U$ f
In each case the system is M1 kg of solution 1 + M2 kg of solution 2.
Since Q = 0 , Ws = 0 (adiabatic mixing)
For liquids U$ ≡ H$ . Thus we have
M H$ + M2 H$ 2
H$ f = 1 1
M1 + M2
Solutions to Chemical and Engineering Thermodynamics, 3e
c
h
1 $
when M1 = M2 ; H$ f =
H1 + H$ 2 .
2
(a) Read from Figure 6.1-1
H$ 1 = 6.9 × 103 J kg
H$ = −6.1 × 103 J kg
2
c
h
1
Thus H$ f = 5.410 × 104 = 2 .705 × 10 4 J kg
2
To find the composition, so a sulfuric acid balance
1
ρ1 M1 + ρ2 M 2 = ρ f M f ⇒ ρ f =
ρ1 + ρ2
2
where ρi = weight percent of ith flow stream.
a
Thus ρf =
f
since
M1 = M2
1
(10 + 90) = 50 wt % sulfuric acid. From Figure 6.1-1
2
50 wt % H2SO 4
⇒ Tf ~ 110° C
H$ = U$ = 2 .705 × 10 4 J kg
(b) Here H$1 = 69
. × 103 J kg ,
1
H$ 2 = −3186
.
× 105 J kg ⇒ H$ f = ( 6.9 − 318.6) × 103 = −156
. × 105 J kg
2
ρ1 = 10 wt % , ρ2 = 60 wt % ⇒ ρ f = 35 wt % . Using
and
Figure 6.1-1, Tf ~ 22° C .
Notice that there is a balance between the energy released in mixing, ∆H$ mix ,
and the energy absorbed in heating the mixture, CP ∆T . In case (a), ∆H$ mix is
very large, and Tf > T1 or T2 , while in case (b) ∆H$ mix is smaller, so that
Tf ~ T1 .
6.15 (a) MW H 2 O = 18.02 g mol ; MW H 2SO 4 = 98.08 g mol
100 g H 2 O = 555
. mol
100 g H 2 SO 4 = 1.02 mol
Note: When these are mixed, a solution containing
5.44 mol H 2 O /mol acid is formed. ∆H s for such a solution is –58,390 J/mol
acid. Thus,
a
f
total heat released = 1.02 mol acid × −58,390 J mol acid = −59,558 J
(Negative sign means that heat is released!)
(b) Adding another 100 grams of water produces a solution which contains 10.88
mol H 2 O /mol acid. From the graph ∆H s = −64,850 J mol acid . However, –
58,390 J/mol of acid were released in preparing the first solution, so that only –
6,460 J/mol acid, or 6,590 J, are released on this further dilution.
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
H 2 SO 4 ⇒
(c) 60 wt %
40 18.02
= 3.629 moles H 2 O moles acid
60 98.08
for which
∆H s = −52,300 J mol acid , and
60 mol aci d
= −31,990 J
98.08
Note: Enthalpy of 60 WT% solution is –31,990 J relative to pure components
at
the
same
temperature.
Similarly
25 wt % H 2SO 4 ⇒ 16.27 mol H 2 O mol acid ,
∆H s ~ −68,830 J mol acid
∆Hs = −52,300 J mol acid ×
and
0.25 × 75
= −13,160 J
98.08
Final solution =175 grams ; 78.75 grams acid = 0803
.
mol,
∆Hs = −68,830 J mol acid ×
96.25 grams water = 5347
.
mol ⇒ 6.66 mol H 2 O mol acid . So that
∆ H s = −60,670 J mol acid
∆H s = −48,720 J
Thus, enthalpy change on mixing, ∆H mix is
∆Hmix = −48,720 − ( −31,990 − 13160
, ) = −3570 J
Thus, 3570 J = 357 kJ must be removed to keep solution isothermal!
N
(d) For 1 mole of solute: 1 + N 2 H mix = H 1 + N2 H 2 + 1 ⋅ ∆ H s 2 (argument of
N1
a
FG IJ
H K
f
∆H s ) and for N1 moles of solute and N2 moles of solvent.
a N + N fH
1
2
mix
FG N IJ = H
HN K
F N IJ + N ∂a∆ H f ⋅ ∂aN N f
+ ∆H G
H N K ∂a N N f ∂ N
= N1 H 1 + N 2 H 2 + N1∆ H s
2
mix
1
Now
H1 =
FG ∂ H IJ
H ∂N K
= H1
mix
1
T ,P
2
s
1
s
1
2
1 T, P
2
or
FG N IJ − N LM ∂ ∆ H aN N f OP
H N K N N ∂a N N f Q
F ∂ H IJ we obtain
Similarly, starting from H = G
H ∂N K
∂ ∆H a N N f
H −H =
∂a N N f
H1 − H 1 = ∆ H s
2
2
1
1
2
s
2
1
T, P
1
a
T ,P
f
∂ N 2 N1
N
= − 22
∂ N1
N1
mix
2
2
2
T ,P
2
s
1
2
2
(e) 50 wt % acid ⇒
since
1
1
1
T ,P
50 18.02
= 5.443 mol H 2 O mol acid
50 98.08
∆H s (5.443) = − 58,370 J mol and, from the accompanying graph
a
∂ ∆H s N2 N1
∂ N2 N1
a
f
f
=
at N 2 N1 =5.443
( −91,630 ) − (− 46,030)
= −2,280 J mol
20
so that H2 − H 2 = −2,280 J mol and
Solutions to Chemical and Engineering Thermodynamics, 3e
H1 − H1 = (−58,370) − 544
. (−2,280) = −45,967 J mol .
6.16 To get partial molar properties it is easiest to first convert all data in problem to
mole fractions and properties per mole.
xCCl 4 =
a
wt % CCl4 153.84
wt % CCl4 153.84 + 100 − wt % CCl4 7811
.
f a
f
where MWCCl 4 = 15384
. ; MWC6 H 6 = 7811
. .
CP ( mole mixture ) = CP (grams mixt ure) × ( MW of mixture )
c
h
= CP × xCCl 4 × 153.84 + 1 − xCCl 4 × 7811
.
∑ xiCP,i , where CP,i = heat capacity of pure
∆CP , mix = CP (mixture) − ∑ xi CP, i . Results are given below:
also, compute
species i and
a
Wt % CCl4
xCCl 4
CP J mol K
∑ xiCP,i
∆CP ,mix J mol K
0
10
20
30
40
50
60
70
80
90
100
0
0.0534
0.1126
0.1787
0.2529
0.3368
0.4323
0.5423
0.6701
0.8205
1
137.90
133.91
129.55
124.45
118.85
113.98
111.29
110.48
110.59
114.44
124.15
137.90
137.17
136.35
135.44
134.42
133.72
131.96
130.44
128.69
126.62
124.15
0
–3.26
–6.80
–10.99
–15.67
–19.74
–20.67
–19.96
–18.10
–12.18
0
f
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
Using these data, and the graphical procedure introduced in Sec. 6.6, we obtain
the following results.
xCCl 4
bC
P
− CP
CP, CCl 4
bC
P
− CP
g
CCl 4
g
C 6H6
CP, C6 H 6
xCCl 4
bC
P
− CP
CP, CCl 4
bC
P
− CP
CP, C6 H 6
g
CCl 4
g
C 6H6
0
0.1
0.2
0.3
0.4
0.5
–71
–60.7
–58.0
–44.5
–27.5
–17.5
53.15
63.5
66.2
79.7
96.7
106.7
0
–0.5
–1.3
–6.7
–16.0
–24.0
137.9
137.4
136.6
131.2
121.9
113.9
0.6
0.7
0.8
0.9
1.0
–11.8
–8.7
–4.1
–1.2
0
112.4
115.45
120.1
123.0
124.15
–30.8
–36.7
–49.8
–67.5
–80.5
107.1
101.2
88.1
70.4
57.4
Solutions to Chemical and Engineering Thermodynamics, 3e
An alternate solution to this problem follows.
Alternate Solution to Problem 6.16
Instead of using Equations (6.6-10a and b) and ∆CP , mix data, Equations (6.611a and b) and the heat capacity data for the mixture can be used. Since
Equations (6.6-11a and b) are very similar to Equations (6.6-10a and b) [of
which Equations (6.6-4a and b) and (6.6-9a and b) are special cases], it follows
that the graphical construction discussed in Sec. 6.6 can be used. The
difference, however, is that the tangents to the CP, mix vs. mole fraction curve
will give CP, CCl4 and CP ,C6 H6
bC
P
− CP
g
C6 H6
as before.
directly, rather than
bC
P
− CP
g
CCl4
and
An illustrative graph, and the numerical results
obtained using a much larger graph are given below:
xCCl 4
0
0.1
0.2
0.3
0.4
0.5
CP, CCl4
3.0
63.0
63.0
82.0
97.0
106.2
CP ,C6 H6
137.9
137.9
137.9
130.7
122.9
114.7
xCCl 4
0.6
0.7
0.8
0.9
1.0
CP, CCl4
113.0
117.0
120.3
123.2
124.15
CP ,C6 H6
105.7
97.6
87.3
70.8
55.9
Chapter 6
Solutions to Chemical and Engineering Thermodynamics, 3e
Note that these results differ from previous results by small amounts. Previous results are
probably more accurate since the curvature of ∆CP , mix vs. xCCl 4 is greater than that of
CP, mix vs. xCCl 4 , so tangents are found with greater accuracy.
6.17
Let
UV
W of 60 WT% solution
x = lbs. of 20 WT% solution used to make 1 lb.
y = lbs. of pure acid
(a) Total mass balance: x + y = 1
Species mass balance on acid: 0.2 x + y = (0.6 )(1)
⇒ 0.2 x + (1 − x ) = 0.6 or x = 05
. kg 20 WT% solution, y = 05
. kg pure acid.
(b) From Figure 6.1-1
H$ ( 20 wt%, 5° C) = −1.22 × 105 J kg
H$ (100 wt%, 50° C) = 7 .10 × 10 4 J kg
H$ ( 60 wt%, 70° C) = −159
. × 105 J kg
H$ ( 60 wt%, boiling point) = H$ ( 60 wt%, 143° C) ~ 0 J kg
Using the change over a time interval form of the energy balance equation,
considering the initial state to be two 0.5 lbs. of separated 20 WT% and pure
acid solutions, and the final state to be 1 lb. of mixed solution, and neglecting
the difference between H$ and U$ for these liquids, yields
c
h
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Solutions to Chemical and Engineering Thermodynamics, 3e
at Tf = 70° C
b
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6.18 Suppose there was enough information available on ∆θmix , where θ is any
extensive thermodynamic property of a mixture, as a function of the three mole
numbers N1 , N2 , and N3 , that the data could be fitted to a polynomial expression
in x1 , x2 and x3 or, equivalently, in N1 , N2 and N3 where N = ∑ Ni . The
i
partial molar properties could then be obtained by differentiation of the polynomial
expression for ∆θmix . That is since
a
3
θ = Nθ = ∑ Ni θi + ∆θmix N1, N 2 , N 3
i= 1
θi =
∂θ
∂ Ni
= θi +
T , P , N j≠ i
a
∂ ∆θmix
∂ Ni
f
f
T , P, N
j≠ i
so that
θi − θi =
a
∂ ∆θmix
∂ Ni
f
T , P , N j≠ i
Alternately, graphical methods could be developed for finding θi − θi along paths
where Ni is varied, and other mole numbers are fixed (i.e., xi is varied, while the
mole ratios of the other species in the mixture are fixed.)
Since it is unlikely that enough information will be available for any mixing property
to obtain ∆θmix as an explicit function of mole fractions or species mole numbers
for ternary, quaternary, etc. mixtures, it is not surprising that there is little
information on partial molar properties in such systems.
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Solutions to Chemical and Engineering Thermodynamics, 3e
6.29
The mass balance of the acetic acid-pyridine streams at steady-state is
kmol
0 = N& P + N& A + N& mix ⇒ − N& mix = N& P + N& A ⇒ N& mix = -2
min
The energy balance is
0 = N& P H P + N& A H A + N& mix H mix + Q&
= N& P H P + N& A H A + N& mix xP H P + xA H A + ∆ H mix + Q&
a
a
f a
f
f
a
f
= 1 ⋅ H P − H P + 1 ⋅ H A − H A − 2 ⋅ ∆ H mix xP = 0.5 + Q&
so
Q& = 2 ⋅ ∆ H
mix
ax
P
f
= 0.5
Now from the table
a
f
∆ H a x = 0.5029f = −4765 J / mol
By interpolation
∆ H a x = 0.5f ≈ −4773 J / mol
∆ H mix xP = 0.4786 = −4833 J / mol
mix
P
mix
P
and
kmol
J
1 kJ 1000 mol
Q& = 2
⋅ ( −4773)
⋅
⋅
min
mol 1000 J
kmol
kJ
= −9546
min
Negative sign means that heat must be removed (or cooling supplied) to keep
the process at a constant temperature. Since ethylene glycol has a value of CP =
2.8 kJ/kg K. From an energy balance we have that
kJ
kJ
2.8
× 20 K × M& = 9546
kg ⋅ K
min
Therefore
& =
M
6.30
9546
kJ
min
kJ
2.8
× 20 K
kg ⋅ K
= 170.5
kg ethylen e glycol
min
(also available as an Mathcad worksheet)
Problem 6.30
x
0
0
H0
0
x1
0.0371
H1
1006
x2
0.0716
H2
1851
x3
0.1032
H3
2516
x4
0.1340
H4
3035
x5
0.1625
H5
3427
x6
0.1896
H6
3765
x7
0.2190
H7
4043
x8
0.2494
H8
4271
x10
0.3006
H10
4571
x11
0.3234
H11
4676
x9
0.2760
H9
4440
x12
0.3461
H12
4760
x13
0.3671
H13
4819
x14
0.3874
H14
4863
x15
0.3991
H15
4832
x16
0.4076
H16
4880
x17
0.4235
H17
4857
x18
0.4500
H18
4855
x19
0.4786
H19
4833
x20
0.5029
H20
4765
Solutions to Chemical and Engineering Thermodynamics, 3e
18
18
19
19
20
20
x21
0.5307
H21
4669
x22
0.5621
H22
4496
x23
0.5968
H23
4253
x24
0.6372
H24
3920
x25
0.6747
H25
3547
x26
0.7138
H26
3160
x27
0.7578
H27
2702
x28
0.8083
H28
2152
x29
0.8654
H29
1524
x30
0.9277
H30
806
x31
1.0
H31
i
0
0 , 1 .. 31
0
2000
H
i
4000
6000
0
0.5
1
x
i
One-constant Margules fit
f( x)
x. ( 1
x)
4
S = 1.961 10
linfit ( x, H , f )
S
Two-constant Margules fit
f( x)
x. ( 1
x. ( 1
x)
.
x) ( 2 . x 1 )
SS
linfit ( x, H , f )
4
1.893 10
SS =
3
8.068 10
Three-constant Margules fit
f( x)
x. ( 1 x)
x. ( 1 x) . ( 2 . x 1 )
x. ( 1
HH( x)
x) . ( 2 . x 1 )
4
1.88 . 10 . ( x. ( 1
dHH( x)
d
HH( x)
dx
∆ H1( x)
HH( x)
(1
4
1.88 10
SS
linfit ( x, H , f )
2
SS =
3
7.983 10
3
1.143 10
x) )
3
7.983 . 10 . ( x. ( 1
x) . dHH( x)
∆ H2( x)
x) . ( 2 . x 1 ) )
HH( x)
3
1.143 . 10 . x. ( 1
x. dHH( x)
x) . ( 2 . x 1 )
2
Solutions to Chemical and Engineering Thermodynamics, 3e
Solutions to Chemical and Engineering Thermodynamics, 3e
Solutions to Chemical and Engineering Thermodynamics, 3e
∆ H1 xi
HH xi
∆ H2 xi
2.793 . 10
0
970.622
1.917 . 10
0
64.168
230.38
462.512
753.807
1.7 . 10
1.074 . 10
4
2.45 . 10
4
3
1.76 . 10
2.16 . 10
4
2.393 . 10
3
3
2.931 . 10
3
3.363 . 10
3.718 . 10
3
4.045 . 10
3
3
4.322 . 10
4.517 . 10
3
4.66 . 10
3
3
4.762 . 10
3
4.835 . 10
4.88 . 10
3
4
3
4
1.516 . 10
1.418 . 10
1.355 . 10
1.828 . 10
1.195 . 10
2.288 . 10
1.044 . 10
2.716 . 10
9.243 . 10
3.127 . 10
8.227 . 10
3
3.52 . 10
7.36 . 10
3
3.92 . 10
6.565 . 10
4.296 . 10
5.886 . 10
4.663 . 10
5.281 . 10
4.875 . 10
4.953 . 10
5.03 . 10
4.724 . 10
5.319 . 10
4.317 . 10
5.8 . 10
3.696 . 10
6.315 . 10
3.102 . 10
6.747 . 10
2.653 . 10
7.231 . 10
2.2 . 10
7.763 . 10
4
4
4
4
3
3
3
3
4.902 . 10
3
4.906 . 10
4.905 . 10
3
4.895 . 10
3
3
4.853 . 10
4.777 . 10
3
4.688 . 10
3
3
4.561 . 10
4.388 . 10
3
4.162 . 10
3
3
3.86 . 10
3.545 . 10
3
3.186 . 10
3
3
2.751 . 10
3
2.218 . 10
1.581 . 10
859.046
0
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
1.759 . 10
8.327 . 10
1.349 . 10
963.67
683.156
458.239
274.249
135.442
48.028
8.49
0
8.946 . 10
3
3
3
3
9.48 . 10
3
9.989 . 10
3
1.05 . 10
4
1.1 . 10
4
1.144 . 10
4
1.177 . 10
4
1.196 . 10
4
Solutions to Chemical and Engineering Thermodynamics, 3e
6.31
Starting by writing the equation for the formation of each of the six compounds
present from their elements
C + 4H = CH4
2O = O2
C + 2O = CO2
C + O = CO
2H = H2
2H + O = H2 O
(1)
(2)
(3)
(4)
(5)
(6)
Now using eqn. (2) to eliminate the oxygen atom, and eqn. (5) to eliminate the
hydrogen atom. We obtain
C + 2H2 = CH4
C + O2 = CO2
C + 1/2O2 = CO
H2 + 1/2O2 = H2O
Thus from the Denbigh method, we find there are four independent reactions.
One such set is listed above.
6.32 (a) N 2 ( g ) + H 2 ( g ) = 2NH 3 ( g )
∆Hrxn = 2 × (− 461
. ) = −92.2 kJ mol
∆Grxn = 2 × (− 165
. ) = −33.0 kJ mol
(b) C3 H 8 ( g ) = C2 H 4 ( g ) + CH 4 ( g )
∆H rxn = 52.5 − 74.5 − ( −104 .7) = 82.7 kJ mol
∆Grxn = 685
. − 505
. − ( − 24.3) = 42.3 kJ mol
(c) CaCO3 ( s) = CaO( s) + CO 2 ( g )
∆Hrrxn = −635.1 − 393.5 − ( −1206.9) = 178.3 kJ mol
∆Grxn = −604 .0 − 394.4 − ( −11288
. ) = 130.4 kJ mol
(d) 4CO( g ) + 8H 2 ( g ) = 3CH 4 ( g ) + CO 2 ( g ) + 2 H 2 O( g )
∆H rxn = 3 × ( −74.5) + ( −393.5) + 2 × ( −241.8) − 4 × ( −110.5) = −658.6 kJ mol
∆Grxn = 3 × (− 505
. ) + ( −394 .4 ) + 2 × (− 228.6 ) − 4 × ( −137 .2 ) = −454 .3 kJ mol
6.33
Buckmasterfullerene C60 (BF) + 60 O2 = 60 CO2 for which
∆Hcomb = 26,033 kJ/mol= 26,033 kJ/60 mols C
Graphite 60C + 60 O2 = 60CO2 for which ∆Hcomb = 60×393.513= 23,611 kJ/60
mols C
For these reactions sice only carbon, carbon dioxide and oxygen are involved,
∆H f = -∆Hcomb
Subtracting the first chemical reaction above from the second yields
60C > C60 (BF) => -26033 – (-23611) = -2422 kJ/mol C60 (BF)
Solutions to Chemical and Engineering Thermodynamics, 3e
6.34 (also available as an Mathcad worksheet)
Problem 6.34 Partial molar enthalpies
x0
0
H0
File: 6-34.MCD
0
x1
0.0120
H1
68.8
x2
0.0183
H2
101.3
x3
0.0340
H3
179.1
x4
0.0482
H4
244.4
x5
0.0736
H5
344.6
x6
0.1075
H6
451.1
x7
0.1709
H7
565.3
x8
0.1919
H8
581.0
x10
0.2636
H10
566.1
x11
0.2681
H11
561.9
x13
0.3073
H13
519.6
x14
0.3221
H14
508.0
x9
0.2301
x12
0.2721
x15
H9
585.0
H12
557.8
0.3486
H15
468.5
x16
0.3720
H16
424.4
x17
0.3983
H17
369.1
x18
0.4604
H18
197.1
x19
0.4854
H19
135.4
x20
0.5137
H20
66.1
x21
0.5391
H21
1.9
x22
0.5858
H22
117.1
x23
0.6172
H23
186.5
x24
0.6547
H24
266.9
x25
0.7041
H25
360.3
x26
0.7519
H26
436.6
x27
0.7772
H27
470.5
x28
0.7995
H28
495.9
x29
0.8239
H29
510.0
x30
0.8520
H30
515.8
x31
0.8784
H31
505.3
x32
0.8963
H32
486.0
x33
0.9279
H33
420.5
x34
0.9532
H34
329.2
x35
0.9778
H35
184.7
x36
0.9860
H36
123.3
x37
0.9971
25.1
x38
1.0
H38
0.0
12
12
i
13
13
14
H37
14
0 , 1 .. 38
1000
500
H
i
0
500
1000
0
0.5
x
i
One-constant Margules fit
f( x)
x. ( 1
HH( x)
x)
SS
SS0 . x. ( 1
linfit ( x, H , f )
SS = 528.45491
x)
dHH( x)
d
HH( x)
dx
PH1( x)
HH( x)
(1
x) . dHH( x)
PH2( x)
HH( x)
x. dHH( x)
1
Solutions to Chemical and Engineering Thermodynamics, 3e
Two-constant Margules fit
x. ( 1
f ( x)
HH( x)
PH1( x)
x. ( 1
x)
x) . ( 2 . x 1 )
x. ( 1
x) . SS0
HH( x)
(1
linfit ( x, H , f )
SS
SS1 . ( 2 . x 1 )
x) . dHH( x)
PH2( x)
SS =
dHH( x)
HH( x)
337.24041
5707.44046
d
HH( x)
dx
x. dHH( x)
Solutions to Chemical and Engineering Thermodynamics, 3e
Three-constant Margules fit
x. ( 1 x)
f ( x)
x. ( 1
x) . ( 2 . x 1 )
x. ( 1
2
x) . ( 2 . x 1 )
HH( x)
SS0 . ( x. ( 1
dHH( x)
d
HH( x)
dx
x) )
SS
488.57112
linfit ( x, H , f )
SS =
5672.45617
970.2807
SS1 . ( x. ( 1
PH1( x)
x) . ( 2 . x 1 ) )
HH( x)
(1
SS2 . x. ( 1
x) . dHH( x)
x) . ( 2 x 1 )
PH2( x)
2
HH( x)
x. dHH( x)
Solutions to Chemical and Engineering Thermodynamics, 3e
Four-constant Margules fit
x. ( 1
f ( x)
HH( x)
dHH( x)
x)
x. ( 1
x) . ( 2 . x 1 )
x. ( 1
2
x) . ( 2 . x 1 )
x. ( 1
3
x) . ( 2 . x 1 )
x. ( 1
x) . SS0
d
HH( x)
dx
SS
SS1 . ( 2 . x 1 )
PH1( x)
linfit ( x, H , f )
2
SS2 . ( 2 . x 1 )
HH( x)
(1
3
SS3 . ( 2 . x 1 )
x) . dHH( x)
PH2( x)
HH( x)
x. dHH( x)
Solutions to Chemical and Engineering Thermodynamics, 3e
Solutions to Chemical and Engineering Thermodynamics, 3e
HH x
i
0
75.89284
112.97104
197.51978
264.87861
365.61955
465.07305
563.45181
575.52278
576.41179
558.17713
554.54726
551.10303
512.60889
492.43244
451.15379
409.80542
358.55983
221.37494
161.16559
90.55115
25.5922
95.51229
176.23404
269.19976
380.26243
466.63265
500.06403
520.4105
530.95404
525.02317
498.49631
467.10138
380.96262
279.46841
148.70801
97.36045
21.20369
0
PH1 xi
6620.61619
6037.1938
5746.57438
5066.79383
4503.47003
3607.09619
2606.80402
1219.73582
871.2077
347.75299
14.29521
57.06648
94.0214
379.84612
481.19324
638.98059
756.15737
866.44984
1052.07298
1101.2544
1140.97708
1162.73326
1169.06642
1148.81031
1098.66681
990.04823
841.94903
748.83482
660.0669
557.74661
436.6149
323.9931
250.96992
135.3294
62.11408
15.16587
6.1957
0.27562
0
PH2 xi
0
3.48837
7.95429
26.13746
50.23256
108.09291
207.10545
428.17388
505.3063
644.75104
763.09798
778.58558
792.26028
908.52549
955.04467
1034.5447
1100.47128
1169.46453
1307.91205
1351.95197
1391.46839
1415.53852
1422.80738
1391.88005
1303.49667
1070.73515
670.81348
367.73066
36.47376
405.60256
1033.96809
1759.05242
2335.16915
3542.17012
4706.43722
6030.57744
6517.96343
7216.85219
7407.10877
Solutions to Chemical and Engineering Thermodynamics, 3e
6.35 Assume reactor operates in steady state. Then the mass balance is
0 = Ni ,in − Ni ,out + νi X
or
N i ,in = + N i, out − νi X
and the energy balance is
0=
∑N
i ,in
∑N
H i ,in −
i
i ,out
H i , out +Q
i
or
∑ N H +∑ N H
= −∑( N
− ν X ) H +∑ N H
= ∑ N bH
− H g + ∆ H (T ) X
Q=−
i ,in
i, out
i ,in
i
i, out
i
i, out
i
i ,out
i ,in
i
i ,out
i
i ,out
i ,out
i, in
rxn
in
i
Using a Mathcad worksheet, the heats of formation and heat capacities in the
appendices of the textbook, we find that 59.93 kJ must be supplied per mole of N2
entering reactor. See Mathcad worksheet for this problem.
6.36 (also available as an Mathcad worksheet)
6.36
x25
0
H250
x500
0
H500
x251
0.027
H251
223.16
x501
0.031
H501
76.20
x252
0.034
H252
290.15
x502
0.043
H502
121.84
x253
0.054
H253
329.50
x503
0.082
H503
97.55
x254
0.094
H254
384.25
x504
0.098
H504
52.75
x255
0.153
H255
275.07
x505
0.206
H505
125.60
x256
0.262
H256
103.41
x506
0.369
H506
370.53
x257
0.295
H257
81.22
x507
0.466
H507
435.43
x258
0.349
H258
11.35
x508
0.587
H508
473.11
x259
0.533
H259
133.98
x509
0.707
H509
460.55
0
0
0
x2510
0.602 H2510
168.31
x5010
0.872 H5010
238.23
x2511
0.739 H2511
177.94
x5011
0.9999 H5011
0.0
x2512
1.0
0.0
x5012
1.0
0.0
i
0 .. 12
H2512
H5012
Nomenclature
H25 = enthalpy at 25C
H50 = enthalpy at 50C
HH25 and HH50 are
correlated enthalpies.
∆H125 = difference
between partial molar
and pure component
enthalpies of species
1 at 25C,
etc.
Solutions to Chemical and Engineering Thermodynamics, 3e
x. ( 1 x)
x. ( 1 x) . ( 2 . x 1 )
Three-constant Margules fit
f( x)
x. ( 1
s25
x) . ( 2 . x 1 )
2
851.268
linfit ( x25, H25 , f )
s25 =
3
1.543 10
3
5.027 10
s50
3
1.95 10
linfit ( x50, H50 , f )
s50 =
3
1.443 10
3
2.099 10
HH25( x)
s250 . x. ( 1
x)
s251 . x. ( 1
x) . ( 2 . x 1 )
s252 . x. ( 1
2
x) . ( 2 . x 1 )
HH50( x)
s500 . x. ( 1
x)
s501 . x. ( 1
x) . ( 2 . x 1 )
s502 . x. ( 1
x) . ( 2 . x 1 )
2
Solutions to Chemical and Engineering Thermodynamics, 3e
dHH25 ( x )
d
HH25 ( x )
dx
dHH50 ( x )
d
HH50 ( x )
dx
∆ H150( x50)
HH50( x50)
(1
x50) . dHH50( x50)
∆ H125( x25)
HH25( x25)
(1
x25) . dHH25( x25)
∆ H250( x50)
HH50( x50)
x50. dHH50( x50)
∆ H225( x25)
HH25( x25)
x25. dHH25( x25)
Solutions to Chemical and Engineering Thermodynamics, 3e
6.37 (also available as an Mathcad worksheet)
6.37
x0
0.2108
H0
738
x1
0.2834
H1
900
x2
0.3023
H2
933
x3
0.4285
H3
1083
x4
0.4498
H4
1097
x5
0.5504
H5
1095
x6
0.5562
H6
1096
x7
0.6001
H7
1061
x8
0.6739
H8
976
x9
0.7725
H9
780
x10
H10
622
i
0.8309
0 .. 10
Hxxi
Hi
xi . 1
xi
600
4420
800
4430
H
i
Hxx
i
1000
4440
1200
4450
0.2
0.4
0.6
x
i
0.8
1
0.2
0.4
0.6
x
i
0.8
1
Solutions to Chemical and Engineering Thermodynamics, 3e
4430 . 1
∆ H1i
xi
∆ H1i
2
∆ H2i
4.436 . 10
3
196.854
355.798
404.837
813.402
896.278
3
2.275 . 10
3
2.156 . 10
3
1.447 . 10
3
4.432 . 10
3
4.424 . 10
3
4.422 . 10
3
3
4.433 . 10
3
4.425 . 10
3
1.342 . 10
1.341 . 10
895.481
872.526
708.446
471.092
229.28
126.675
3
2
Hxxi
∆ H2i
2.759 . 10
4430 . xi
3
1.37 . 10
4.44 . 10
3
1.595 . 10
3
4.421 . 10
3
2.012 . 10
3
4.441 . 10
3
2.644 . 10
3
4.438 . 10
3
3.058 . 10
3
4.427 . 10
3
1
6.38 (a) C8 H 18 + 12 O 2 + 47.02N2 → 8CO2 + 9H 2 O + 47.02N2
2
∆Hrxn = 8∆ H f , CO 2 + 9 ∆ H f , H 2 O − ∆H f , C 8 H 18
= 8(− 393.5) + 9( −2418
. ) − (− 208.4 ) = −51158
. kJ
∆U rxn = ∆ Hrxn − ∆ NRT = − 51158
. kJ(17 − 135
. ) × 8.314 ×
298.15
kJ
1000
= − 51158
. − 8.7 kJ = −5124.5 kJ
CP of mixture = 8 × 5125
. + 9 × 39.75 + 47.02 × 32.43 = 2292.61 J mol K
. J mol K
CV of mixture = CP − NR = 2292.61 − 64.02 × 8.314 = 176035
A
# of
moles
∆U rxn
51245
. × 10 3
= 298 .15 +
= 298.15 + 291107
. = 3209.2 K
CV
1760.35
by ideal gas law
N f Tf
N f Tf
NT
PV = NRT ⇒ i i =
; Pf = Pi
Pi
Pf
Ni Ti
Tfinal = Tin +
64.02 32092
.
×
= 11386
.
bar
60.52 29815
.
(b) Adiabatic expansion
Pf = 1 bar
Solutions to Chemical and Engineering Thermodynamics, 3e
CP ( per mole) =
FPI
T = TG J
HPK
229261
.
229261
.
=
= 3581
.
8 + 9 + 47.02
64.02
R CP
= 3209.2 ×
2
2
1
1
F 1 I
H 11386
K
.
8 .314 3581
.
T2 = 3209.2(0087827
.
)0 .23217 = 18245
. K
0
dU
dS
= W& ;
= 0 + S gen
dt
dt
W = CV ∆T = −2.438 × 106 J mol of octane
(c)
This is like Carnot cycle with a varying upper T
0
dU
dT dS Q&
C dT
= Q& + W& = CV
;
=
+ S gen
= P
; T2 = 150° C
dt
dt dt TL
T dt
dU
dS &
= TL
+W
dt
dt
∆U − TL ∆S = W
or
W = CV∆T − TL ⋅ CP ln
TL
TH
W = 176035
. (423.15 − 18245
. ) − 42315
. × 2292.61ln
42315
.
18245
.
= −1049
.
× 106 J mol of octane
dT
C dT &
dT
C dT
= TL P
+ W ⇒ W& = CV
− TL P
dt
T dt
dt
T dt
TL
WS = CV TL − TH − TL CP ln
TH
dU
= ∑ N& i ,in Hi ,in − ∑ N& i ,out H i ,out + Q& + W& = 0
dt
0
dS
Q&
= ∑ N& i , in Si ,in − ∑ N& i ,out S i ,out +
+ S gen
=0
dt
Tamb
CV
a
f
(for maximum work)
Q& = −Tamb ∑ N& i ,in Si , in − ∑ N& i , out Si , out
b
bH
bH
g
b
g − ∑ N& bH − T
g − ∑b N& + Xν gbH
g
g
0 = ∑ N& i , in Hi ,in − Tamb Si , in − ∑ N& i , out Hi ,out − Tamb Si , out + W&
−W& = ∑ N& i , in
= ∑ N& i , in
i ,in
− Tamb Si , in
i ,in
− Tamb Si , in
i , out
i ,out
i , in
i
amb Si , out
− Tamb Si ,out
i ,out
Absolute maximum work Tin = Tout = Tamb
Ideal gas Hi = H i ; Gi , in = Gi ,out ; Si = Si − R ln xi
−W& =
∑N
i ,in
G i , in − RTamb
i
RTamb
∑N
i
∑N
i ,in
i ,in
∑N G +
+ X ∑ ν bG − RT ln x g
ln xi, in −
i
ln xi , out
i, in
i ,in
i
i
i
i
amb
i ,out
g
Solutions to Chemical and Engineering Thermodynamics, 3e
−W& = RTamb
∑N
= RTamb
∑N
i , in
ln
i ,in
ln
xi , in
xi ,out
xi ,in
xi ,out
+X
∑ ν G − RT ∑ ln x
i
i
+ X∆Grxn − RTamb
νi
i ,out
amb
∑ ln x
νi
i ,out
6.39
C is the number of components, and M is the number of phases.
Then the unknowns are
N iK (number of moles of species i in phase K)
= C × P unknowns
P (pressure in phase K) = P unknowns
K
T K (temperature in phase K) = P unknowns.
Total number of unknowns is C × P + P + P = P × (C + 2 )
Then restrictions are that
T is the same in all phases, i.e., T I = T II = T III =.....
P - 1 restrictions
P is the same in all phases, i.e., P I = PII = PIII =.....
P - 1 restrictions
K
Gi
must be the same for species i in all phases
= C × ( P - 1) restrictions
In addition we have the stoichiometric relation for each species that
K
M
k=1
j =1
N i = ∑ Nik = Ni , o + ∑ νij X j which provides an addition C restrictions.
Therefore the number of degrees of freedom F are
F = (C + 2) ⋅ P - (C + 2) ⋅ ( P -1) - C = C + 2 - C = 2
independent of the number of components, phases or independent chemical reactions.
Therefore Duhem's theorem is valid.
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7
7.1
a
f
PV T , P, N1 , N2 K = ∑ Ni RT
a
f
⇒ Vi ( T , P, x) = V i (T , P )
U T , P, N 1 , N 2 K = ∑ N i U i ( T , P) ⇒ U i ( T , P, x ) = U i ( T , P)
Also Si (T, P , x) = S i (T, P) − R ln xi
∆U mix = ∑ xi U i ( T , P, x) − U i ( T , P) = ∑ xi 0 = 0
i
i
∆V mix = ∑ xi Vi ( T , P, x) − V i (T , P ) = ∑ xi 0 = 0
i
i
∆ H mix = ∆U mix + P∆V mix = 0
∆S mix = ∑ xi Si (T , P, x )− S i (T, P) = ∑ xi −R ln xi = −R∑ xi ln xi
b
g a
∆G mix = ∑ xi Gi ( T , P, x) − G i (T , P) = ∑ xi Hi − TSi − H i − T S i
b
g
b
= ∑ xi Hi − H i − T ∑ xi Si − S i
g
f
= ∆ H mix − T ∆S mix = RT ∑ xi ln xi
Similarly
∆ Amix = ∆U mix − T∆ S mix = RT∑ xi ln xi
.
7.2
The picture of the process here is as follows
Mixture ∑ N i Moles of gas at T and V !
1 T ,V , P1
B
2 T ,V , P2
Mix keeping
 
→ T ,V , Pf
T and V fixed
M
C T , V , PC
(a)
Let
Pi = initial pressure of species i (pressure in unmixed state)
P = final pressure of mixed gas
Pi = xi P = partial pressure of species i in final state
want to show that Pi = Pi
P = NRT V =
Pi = N i RT V
and
b∑ N gRT V
i
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
↓ initial pressure of pure i
Pi = xi P = xi NRT V = Ni RT V = Pi
↑ partial pressure of species i
Q.E.D.
(b) Since the internal energy of an ideal gas is independent of total pressure, it follows that
U i IGM ( T , x ) = U IG
i ( T ) for mixing at constant total pressure or constant partial pressure. Thus,
∆U IGM
mix = 0 .
Next,
Vi IGM (T , P , x) =
RS∑ N RT UV = RT = RTx = RTx
T P W P Px P
RT
V aP , T f
=x
= x V aT , P f
∂V
∂ Ni
= xi
j
i
i
T ,P
IG
i
Pi
i
i
Ni
IG
i
i
i
i
a f
Vi IGM (T , P, x) = xiV iIG T , Pi
Thus
∆θIGM
mix
Can now define two
m
r
a fr
m
∆θ IGM
= ∑ xi θi (T , P, x )− θ i (T , P ) and ∆θ IGM
= ∑ xi θ i (T, P , x) − θ T, Pi
mix
mix
1
2
∆θ IGM
was computed in Section 7.1 and will not be considered here. We will be concerned
mix
1
with ∆θ IGM
!
mix
2
∑ x lV (T , P, x ) − V aT , P fq = ∑ x kx V aT , P f − V aT , P fp
V
V
V
= ∑ x a x − 1f
= ∑ x a x − 1f
=
∑ a x − 1f
N
xN N
∆V mix =
IGM
i
i
IGM
i
i
i
i
i
i
i
i
C
i
i
i
i
i
=
i
i
i =1
V
V (1 − C )
(1 − C) = C
N
Ni
∑
i =1
C
Note:
∑1 = C
i =1
where C = number of components.
For enthalpy we have
a f
Hi IGM (T , P, x) = U iIGM ( T , P, x) + PVi IGM ( T , P, x) = U iIG (T , P) + PxiV iIG T , Pi
=
Thus
U IG
i (T , P ) +
d
H iIG( T ,
P)
i
∆H IGM
= ∑ xi Hi IGM (T, P , x) − H IG
mix
i (T , P ) = 0
1
and
PVi (T , P) =
IG
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
a fi
d
= ∑ x nU ( T , P) + PV (T , P ) − U ( T , P) − PV aT , P fs
= ∑ x k0 + PV ( T , P) − PV aT , P fp = ∑ x {RT − RT } = 0
C
IGM
∆H mix
2
= ∑ xi H iIGM (T , P , x ) − H 1 T , Pi
IG
i =1
IG
i
i
IG
i
i
i
i
i
i
i
i
i
i
i
i
To compute ∆S IGM
mix we use the same sort of argument as in Section 7.1, but noting here that the
volume occupied by each gas in the initial and final states are the same. Therefore
SiIGM ( T , P, x) = S IG
i T , Pi . Since T and V of each species is unchanged (see eqn. (3.4-2)).
a f
Therefore
a fs = 0
n
∆S IGM
= ∑ xi Si IGM (T , P , x) − S IG
mix
i T , Pi
2
i
For the Helmholtz free energy we note that
A = U − TS ⇒ Ai IGM (T , P , x) = U iIGM ( T , P, x) − TSi IGM( T , P, x)
a f
a f
= U iIG( T ) − TS IG
T , Pi = AIG
T , Pi
i
i
Thus, ∆ AIGM
= 0 . Finally G = A + PV , ⇒
mix
2
a f
a f
IG
Gi IGM (T , P , x) = Ai IGM( T , P, x) + PVi IGM ( T , P, x) = AIG
i T , Pi + Pxi V i T , Pi
=
AiIG
aT, P f + PV aT, P f = aT, P f
i
i i
IG
G IG
i
i
i
So that
∆GIGM
=0
mix
2
7.3
Generally mixing at constant T and P and mixing at constant T and V are quite different.
However, for the ideal gas we have
PVi = N i RT (pure fluids) and PV = ∑ Ni RT (mixtures)
Thus for the pure fluids (same T and P)
N RT
N RT
V1 = 1
and V2 = 2
P
P
N1 RT N2 RT
RT
⇒ V1 + V2 =
+
= N1 + N2
=V
P
P
P
So for the ideal gas the mixing process described in problem statement is also a mixing process at
constant T and P and Table 7.1-1 applies here also.
a
7.4
f
a
f
We have the following properties for a mixture for mixing at constant T and P:
U ( T , P, x) = ∑ Ni U i (T , P)
V (T , P , x) = ∑ Ni V i (T , P )
S (T , P , x) = ∑ Ni S i (T , P) − R∑ Ni ln xi
0
and S i = S i + CV ,i ln
Ui
U 0i
+ R ln
Vi
V i0
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
S 0i , U 0i , V 0i are at some reference state.
(a) Find Vi , U i , S i and G i in terms of S 0i , U 0i , V 0i , CV, i , R, T, and P . Need U i , V i .
∂S
∂U
We know dU = Td S − PdV →
1 ∂S
;
T ∂V
=
V
=
U
P
U
V
and S i = S i0 + CV,i ln i0 + R ln 0i for
T
Ui
Vi
pure component i.
∂S
1
1
= = CV ,i
→ U i = CV ,i T
∂U V T
Ui
∂S
1
∂S
∂U
=R
and
⋅
∂V U
Vi
∂V U ∂ S
∂U
∂S
=T
V
∂V
∂U
RT
P
∂U
So U i =
∂ Ni
=−
S
1
∂S
⇒
P
∂V
⋅
V
∂V
∂U
= T⋅
U
= −1
S
−1
∂S
= −1 →
P
∂V
=
U
R
P
=
Vi T
⇒Vi =
=
T , P , N j ≠i
∂
∂Ni
∑ Ni CV,iT = CV,i T = U i
U i = CV,i T
Vi =
7.5
∂V
∂ Ni
=
T , P , N j≠ i
∂
∂ Ni
∑ Ni V i =
RT
= Vi = V i
P
(a) Start with eqn. 7.2-13
ln
fi
1
= ln φi =
xi P
RT
P=
∂P
∂ Ni
N
∂P
∂ Ni
=
V
V
z
V = ZRT P
V =∞
LM RT − N FG dP IJ
MN V H ∂N K
i V
OPdV − ln Z
PQ
RT
a
NRT
N 2a
NRT
∑ ∑ Ni N jaij
− 2 =
− 2 =
−
V −b V
V − Nb V
V − ∑ Ni bi
V2
RT
V − ∑ Ni bi
−
2∑ N a
−b f −
a
V
bV − ∑ N b g
NRT
j ij
2
i i
=
RT
NRTbi 2 ∑ N j aij
+
−
V − Nb (V − b) 2
V2
=
2 ∑ x j a ij
RT
RTbi
+
−
2
V
(V − b )
V2
i
2
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
f
1
ln φi = ln i =
xi P RT
z
ZRT P
LM
N
= ln
2 ∑ x j a ij
i
2
V =∞
1
V
=
RT ln
RT
V −b
LM RT RT RTb
MM V − V − b − (V − b)
N
OP
Q
ZRT P
V =∞
Z
Bi
+
−
Z − B ( Z − B)
ZRT P
RTbi
+
−
(V − b ) V = ∞
j
+
V2
2 ∑ x j a ij
OP
PPdV − ln Z
Q
ZRT P
j
− ln Z
V2
V =∞
2 ∑ x j a ij
j
F where B = Pb I
H
RT K
− ln Z
RTV
f
Bi
⇒ ln φi = ln i =
− ln (Z − B) −
xi P Z − B
i
i
2∑ x j aij
j
RTV
(b) For a pure van der Waals fluid (Eqn. 5.4-13)
ln
a
f
fi
a
= (Z − 1) − ln Z − Bi − i
P
RTV
and, by definition of the activity coefficient
fi = xi f iγ i ⇒
R|
2∑ x A U
|V
B
f = x P exp S
− ln( Z − B ) −
Z
|T Z − B
|W
f = P exp R
ST(Z − 1) − lnaZ − B f − AZ UVW
j
i
i
ij
j
i
i
mixture
ii
i
pure fluid i
so
fi
=
xi f i
RS
|T
LMF 2∑ x A I
JK
NGH
expk( Z − 1) − lna Z − B f − a A f Z p
exp Bi ( Z − B ) − ln( Z − B) −
j
Z
ij
j
i
ii
OP UV
Q |W
mixture
= γi
pure fluid
Note that the compressibilities in pure fluid and mixture will generally be different at the same T
and P.
7.6
As a preliminary note that, from Eqns. (4.4-27 and 28)
z
LMTF ∂ P I − P OPdV
MN GH ∂ T JK PQ
LMFG ∂ P IJ − R OPdV
MNH ∂ T K V PQ
V = ZRT P
H ( T , P) − H IG (T , P ) = RT ( Z − 1) +
V =∞
and
z
V = ZRT P
S (T , P ) − S IG (T , P) = R ln Z +
V =∞
V
V
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
vdw E.O.S. P =
RT
a
− 2 so
V −b V
FG ∂ P IJ = R ; T FG ∂ P IJ
H ∂T K V − b H∂T K
FG ∂ P IJ − R = R − R ;
H ∂ T K V (V − b) V
V
RT
RT
a
a
−
+ 2 = 2
V − b (V − b) V
V
−P=
V
V
z
V = ZRT P
⇒ H ( T , P) − H IG (T , P ) = RT ( Z − 1) +
a
V2
V =∞
and
dV = RT ( Z − 1) −
z
V = ZRT P
S (T , P) − S IG ( T , P) = R ln Z +
V =∞
a
RTA
= RT (Z − 1) −
V
Z
LM R − R OPdV
NV − b V Q
ZRT P
(V − b)
= R ln Z + R ln
= R ln( Z − B )
V V =∞
Now on to solution of problem.
RT
ex
(a) V = V mix − ∑ xi V i =
Z mix − ∑ xi Zi = ∆V mix
P
Zmix = compressibility of mixture at T and P
b
g
Zi = compressibility of pure fluid i at T and P
Will leave answer to this part in this form since the analytic expression for Zi and Zmix
(solution to cubic) is messy. Though it can be analytically and symbolically with a computer
algebra program such as Mathcad, Mathematica, Maple, etc.)
RT
RTAmix
RTAi
(b) H ex = H mix −
xi H i =
−
−
xi RT Zi − 1 −
Z mix − 1
Z mix
Z mix
i
L
∑ MN a
∑
FG x A IJ − RTA
HZ K Z
F x A − A IJ
= RT b Z − ∑ x Z g + RT G ∑
H Z Z K
= aH
− PV f − ∑ x a H − PV f
= bH
− ∑ x H g − PbV
−∑xV g
F x A − A IJ − RT bZ
= RT b Z − ∑ x Z g + RT G ∑
H Z Z K
F x A − A IJ
= + RT G ∑
H Z Z K
b
g
= RT Z mix − ∑ xi Zi + ∑ RT
U
ex
mix
mix
mix
i
i
mix
i
i
i
mix
i
i
mix
i
i
i
mix
i
i
i
i
mix
i
i i
mix
i
i
i i
mix
i
mix
mix
OP
Q
f
i
mix
mix
mix
− ∑ xi Zi
g
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
(c)
S mix − ∑ xi Si − R ∑ xi ln xi
=S
ex
a
= R lna Z
= R lna Z
f
a f
f − R∑ x lnaZ − B f + ln x
f − R x ln x aZ − B f
= P ln Z mix − Bmix − R X i ln Zi − Bi − R∑ xi ln xi
= R ln
mix
− Bmix
mix − Bmix
i
i
i
i
i
i
i
i
Zmix − Bmix
a
Π Z i − Bi
i
f
xi
b
g
(d) G ex = H ex − TS ex = RT Zmix − ∑ xi Zi + RT
− RT ln
i i
mix
i
mix
Z mix − Bmix
a
Π Zi − Bi
i
f
xi
A ex = U ex − TS ex = + RT
7.7
FG ∑ x A − A IJ
H Z Z K
FG ∑ x A − A IJ − RT ln Z − B
H Z Z K
Πa Z − B f
i i
mix
mix
i
mix
i
i
mix
xi
i
(a) Start from eqn. (7.2-13)
ln φi = ln
fi
1
=
yi P RT
z
V = ZRT P
V =∞
LM RT F ∂P I
MN V − N GH ∂N JK
i
T ,V , N j≠i
OP
PQdV − ln Z
but
PV
B
∑ ∑ yi y j Bij
= 1 + mix = 1 +
RT
V
V
P=
RT Bmix RT
+
=
V
V2
∂P
RT
=
+
∂ Ni
V
=
V2
j
V2
2 N ∑ N j Bij RT
⇒
z
V
2 ∑ N j Bij RT
∂P
NRT
N
=
+
∂ Ni
V
f
1
ln i =
yi P RT
∑ Ni RT + RT ∑ ∑ Ni N j Bij
V = ZRT P
V =∞
j
V
2
=
RT 2 ∑ x j Bij
+
RT
V
V2
LM RT RT 2 RT ∑ x B OP
MM V − V − V PPdV − ln Z
N
Q
2
∑ x j Bij − ln Z
V
j ij
j
2
(eqn. 7.4 - 6)
Note also that
PV
B
PV 2
PV 2
=1+
or
=V + B
−V − B = 0
RT
V
RT
RT
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
⇒V =
1 ± 1 + 4 PB RT
PV
1
or
= Z = 1 ± 1 + 4 PB RT
2 P RT
RT
2
d
i
as P → 0, Z → 1 (ideal gas limit) so only + sign allowed
Z=
d
1
1 + 1 + 4 PB RT
2
i
Note that at low pressures we can obtain a simpler expression.
At low pressures
PV
B
B P
= 1 + mix ≈ 1 + mix
RT
V
RT
Then
m
RT
RT
+ Bmix =
+ y12 B11 + 2 y1 y2 B12 + y22 B22
P
P
RT
RT
V − V IG =
+ Bmix −
= ∑ ∑ yi y j Bij = V ex
P
P
V=
r
Also
a
V = N V = N1 + N 2
f RTP + N +1 N mN
1
2
1 B11 +
N22 B12 + 2 N1 N2 B22
2
r
2 ∑ N j Bij
∂
RT
1
V1 =
V=
−
∑ ∑ Ni N j Bij + Nj + N
∂ N1
P
N1 + N 2
i
1
2
=
=
=
=
V1 − V1 IGM =
ln
f1
1
=
y1 P RT
RT
P
RT
P
RT
P
RT
P
RT
P
− ∑ ∑ yi y j Bij + 2 ∑ y j Bij
+ 2 y1B11 + 2 y2 B12 − y12 B11 − 2 y1 y2 B12 − y22 B22
a f
+ y a2 − y f B
+ y a2 − y f B
0
f
1
1
11 +
2 y22 B12 − y22 B22
1
1
11 +
2 y22 B12 − y22 B22 −
zc
P
a
+ y1 2 − y1 B11 + 2 y2 1 − y1 B12 − y22 B22
a
h
RT
P
f
V1 − V1 IGM dP = y1 2 − y1 B11 + 2 y22 B12 − y22 B22
RS a
T
f
f
P
⇒ 1 = exp y1 2 − y1 B11 + 2 y22 B12 − y22 B22
y1 P
RT
P
RT
UV
W
This is an alternate, some approximate expression that we will use in what follows. Also, for
the pure component we have
LM
N
f1
B P
= exp 11
P
RT
OP
Q
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
Note that these expressions are slightly easier to use then the full expressions since we don’t
have to solve for V (or Z) first
l a
f
a
exp y1 2 − y1 B11 + 2 y22 B12 − y22 B22 P RT
f1
= γ1 =
y1 f1
exp B11 P RT
RS b
T
= exp LM y a 2 B
N
k
g
p
= exp − 1 − 2 y1 + y12 B11 + 2 y22 B12 − y22 B22
2
2
12
− B11 − B22
f RTP OPQ = γ
P
RT
fq
UV
W
1
δ12 y22 P
where δ12 = 2 B12 − B11 − B22
RT
(b) Repeating the argument for a ternary mixture
or ln γ 1 =
ln γ 1 =
a
f
P 2
y2 δ12 + y2 y3 δ12 + δ13 − δ22 + y32δ13
RT
By simple generalization
ln γ 1 =
7.8
P
2 RT
∑ ∑ yi y jcδ1i + δij − δijh
Note: δii = 0
i
j
i ≠1 j ≠1
i≠ j
a
k
fp
N N R
= NG =
S A + BFGH NN −+ NN IJK UVW
N +N T
F ∂ G IJ = N RS A + BFG N − N IJ UV
=G
H ∂ N K N + N T H N + N KW
N N
RS A + B aN − N fUV
−
aN + N f T N + N W
N N R B
Ba N − N f U
+
−
S
N + N TN + N
aN + N f VW
= x k A + Ba x − x fp − x x k A + Ba x − x fp + x x Ba1 − x + x f
(a) G ex = x1 x2 A + B x1 − x2
G ex
ex
1
1
2
2
1
2
1
2
ex
G1ex
2
T ,P
1
1
1
2
2
1
2
1
2
1
2
1
2
2
1
1
1
2
2
1
2
1
2
1
2
2
2
2
1
2
1 2
1
2
2
3
2
3
= Ax2 + Bx2 − 2 Bx2 + 2 Bx2 − 2 Bx2
Thus G1ex = RT lnγ 1 = ( A + 3B )x22 − 4 Bx32 .
2
1 2
1
2
Now by repeating the calculation, or by using the symmetry of G ex and replacing B by −B and
interchanging the subscripts 1 and 2 we obtain
G2ex = RT lnγ 2 = ( A − 3B )x12 + 4 Bx13
2 RTa12 x1q1 x2 q 2
⇒
x1q1 + x2q 2
After some algebra
(b) G
ex
=
G ex = N G
ex
=
2 RTa12 q1q2 N1 N2
N1q1 + N2 q 2
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
G1ex
=
d
∂ N G ex
∂ N1
⇒ ln γ 1 =
Similarly ln γ 2 =
(c)
i
2 RTa12 q1
=
a
1 + x1q1 x2 q 2
T , P, N 2
G1ex
2a12 q1
=
RT
1 + x1 q1 x2 q 2
a
β
a
1 + βx2 αx1
f
f
=
2
f
2
a
α
1 + αx1 βx2
f
2
.
2
G ex
= 2 a12 z1 z2 + 2a13 z1z3 + 2a 23z2 z3 ;
RT
zi =
where
xi qi
∑ x jq j
and
j
FG ∑ N q IJ ∑ ∑ a
H
K
−1
G ex = N G = RT
ex
k
k
k
k
kj
Nk N j q k q j ; thus
j
G ex ∂
ln γ i = i
G ex
=
RT ∂ Ni
T , P , N j ≠i
− qi ∑ a kjq k q j xk x j
j, k
FG ∑ x q IJ
H
K
+
2
2 ∑ aij qi q j x j
j
∑ xk q k
k
k k
k
Now setting
α12 = 2q1 a12 , β12 = 2 q2 a 21 = 2q 2 a12 = α21 ⇒ q 2 =
q1 β13
, etc. For the case i = 1
α13
similarly, q 3 =
ln γ 1 =
nx α aβ
2
2
12
12
α12
q1 β12
;
α12
f + x α aβ α f s + x x aβ α faβ α f α
kx + x aβ α f + x aβ α fp
2
2
3
2
13
13
13
1
2 3
2
12
12
12
12
3
13
13
13
2
12
a
f
a
f
a
f
+ α13 − α23 α12 β12
13
Interchanging indices 1 and 2
ln γ 2
nx β aα
=
2
1 12
12
β12
f
2
a
f s + x x aα β faβ α f β
kx + x aα β f + x aβ α fp
2
+ x32α23 β23 α23
2
1 3
1
12
12
12
12
3
23
23
23
2
+ α13 − α23 β12 α12
12
23
Finally, interchanging indices 1 and 3 in the original equation yields
ln γ 3 =
nx β aα
2
2 23
23
f
2
a
f s + x x aα β faα β f β
kx + x aα β f + x aα β fp
β23 + x12 β13 α13 β13
3
7.9
2
1 2
2
23
23
(also available as a Mathcad worksheet)
b
Using Eqns. (7.6-b) yields: α = 1
RT
LM a
Nb
1
1
23
a
− 2
b2
23
1
OP
Q
2
13
13
13
2
23
+ β13 − β12 β23 α23
13
b
and β = 2
RT
LM a
Nb
1
1
a2
−
b2
OP
Q
2
. From Section 4.6,
we have three different expressions relating the a and b parameters to the critical properties:
27 R 2 TC2
RTC
and b =
. Eqn. (4.6-4a)
64 PC
8 PC
1)
a=
2)
a = 3 PC V 2C and b =
VC
. Eqn. (4.6-4b)
3
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
9V C RTC
V
and b = C . Eqn. (4.6-3a)
8
3
Since ZC is not equal to 3/8 for the fluids under consideration, each set of relations will give a
3)
a=
different pair of values for a and b. Generally, set 1 is used, since V C is known with less accuracy
than PC and TC . All three sets of parameters will be considered here.
VC = MW ρC
Set 1
a
Benzene
2,2,4-trumethyl pentane
0.2595 m3 kg
0.4776 m3 kg
c h
0.119 c m h
c h
0.232 c m h
2
1.875 × 106 m3 Pa kmol2
b
3 2
3 2
kmol
α
β
Set 2
a
kmol
0.480
0.937
9 .945 × 105 m3 Pa kmol2
c h
1.698 × 106 m3 Pa kmol2
c h
0.0865 m3 kmol
0159
.
m3 kmol
2
b
α
β
Set 3
2
3.609 × 10 6 m3 Pa kmol2
2
0.353
0.658
b g
c h
a
1.3655 × 106 m3 Pa kmol2
2 .476 × 106 m3 Pa kmol2
b
0.0865 m3 kmol
0159
.
m3 kmol
2
2
α
β
Set 4
0.433
0.807
Example 7.5-1
Fitting the van Laar equation
α = 0.415
β = 0.706
Example 7.6-1
α=
Set 5
Regular Solution Theory
a
f
V1
2
δ1 − δ2 = 0.703
RT
V
β = 2 α = 1304
.
V1
The 5 sets of results are plotted below.
Numbers in circles denote parameter sets used.
Parameter set 4 , fitted to the experimental data, should be the most accurate. Parameter set
obtained using V C and TC data should be reasonably good, also.
3
,
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
B
7.10 i)
One-constant Margules equation
a
RT lnγ 1 = Ax22
Thus
RT ln γ *1 = RT ln
or
γ *1 = exp
f
RT ln γ 1 x1 = 0 = A
af
γ 1 x1
= Ax22 − A = − A 1 − x22
γ 1 x1 = 0
a
LM − Ac1 − x hOP
MN RT PQ
2
2
c
f
γ1 =
and
h
LM c
MN
− A 1 − x22
x1 ⋅ 1000
exp
ms M1
RT
ii) Two-constant Margules equation
af
RT ln γ a x = 0f = α + β
γ ax f
RT ln γ = RT ln
= −α c1 − x h − β c1 − x h
γ a x = 0f
L −α c1 − x h − β c1− x hOP
γ = exp M
RT
MN
PQ
L −α c1 − x h − β c1 − x hOP
x ⋅ 1000
γ =
exp M
mM
RT
MN
PQ
RT ln γ 1 x1 = α1 x22 + β1 x23
1
*
1
1
1
1
1
2
2
2
2
1
1
s
1
1
1
1
*
1
and
1
1
1
Thus
1
3
2
1
2
2
1
3
2
3
2
hOP
PQ
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
iii) van Laar equation
a f
a
f ; ln γ a x = 0f = α
α
ln γ a x f = ln γ a x f − ln γ a x = 0f =
−α
1 + aαx βx f
α
ln γ 1 x1 =
*
1
1
1
1
1
1
2
1 + αx1 βx2
1
1
2
1
2
or, upon rearrangement,
LM −α x a2βx + αx f OP
N aβx + αx f Q
L −α x a2βx + αx f OP
x ⋅ 1000
=
exp M
mM
N aβx + αx f Q
2
γ *1 = exp
1
2
1
2
2
and
γ1
1
2
1
1
2
1
2
s
1
2
1
iv) Regular Solution Theory
a f
a
a
f
a f
a f c ha
LV cφ − 1haδ − δ f OP and
γ = exp M
RT
MN
PQ
LV cφ − 1haδ − δ f OP
x ⋅ 1000
γ =
exp M
mM
RT
MN
PQ
f
RT ln γ 1 x1 = V 1φ22 δ1 − δ2
af
2
a
f
and RT ln γ 1 x1 = 0 = V 1 δ1 − δ2
2
x2V 2
⇒ 1 as x1 → 0 )
x1V 1 + x2 V 2
(since φ2 =
Thus
f
RT ln γ *1 x1 = RT ln γ 1 x1 − RT lnγ 1 x1 = 0 = V 1 φ 22 − 1 δ 1 − δ 2
1
*
1
2
2
2
1
s
2
2
2
1
1
1
2
2
1
2
1
v) UNIQUAC Model
LM
θτ
∑ x l − q M1 − ln ∑ cθ τ h − ∑ θ τ
∑
NM
ln γ a x → 0f = lim a ln γ f
φ z
θ
φ
ln γ i = ln i + q i ln i + li − i
xi 2
φi
xi
j ij
j j
i
j ij
j
j
k
kj
k
i
i
OP
PP
Q
i
xi → 0
Now consider xi → 0 , then θ j →1 and θi → 0
a
f
a
f
FrI z Fq r I r L
τ O
ln γ = lnG J + q lnG ⋅ J + l − l − q M1 − lncτ h −
P
τ Q
Hr K 2 Hq r K r N
γ
F φ r IJ + z q lnFG θ ⋅ q r IJ − φ ∑ x l + r l
ln
= ln γ = ln G
γ ax → 0f
H x r K 2 Hφ q r K x
r
LM
O
θτ
τ P
− q − ln ∑ θ τ − ∑
+ ln τ +
MM
τ P
∑θ τ
PQ
N
φi
xi ri
1
ri
φ
r
=
⋅ =
, i = i ;
xi
x1r1 + x2 r2 xi x1r1 + x2 r2 xi rj
i
i
j
1
1
i
i
j
j
i
i j
*
1
i
i
i
ii
1
θi xi qi x1q1 + x2 q2
q rj
=
= i
φi
xi ri x1r1 + x2 r2
ri q j
j
j
i
i
j i
i
i
i j
i
j ij
i
j
j ij
j
k
k
j j
ij
kj
ij
ij
ij
jj
jj
i
j
i
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
γ1 =
x1 ⋅ 1000 *
γ 1 where γ 1* is as given above.
ms M1
7.11 An ideal gas constrained to remain at constant volume and T, is also a system at constant internal
energy and volume, since U is only a function of temperature for the ideal gas. Consequently, at
equilibrium, the entropy should be a maximum.
Suppose there were N1 molecules and N lattice sites. For any distribution of the N1 molecules
among the N lattice sites there will be N1 lattice sites with molecules, and N 2 = N − N1 empty
lattice sites. Thus we can consider the “lattice gas” to be a mixture of N1 molecules and N2 holes,
and the entropy of various configurations of this binary system can be computed. Following the
analysis of Appendix 7.1, it is clear that the random mixture, or uniform distribution of gas
molecules, is the state of maximum entropy. A completely ordered state (for example, the first N1
lattice sites filled, and the next N 2 = N − N1 lattice sites empty) is an especially low entropy
configuration.
7.12 The principle of corresponding states, and the pseudo-critical constant concept will be used first,
then the Peng-Robinson equation of state (program PR1)
290
800
(a) O 2 : TC = 154 .6 K ; PC = 50.46 bar ; Tr =
= 1876
.
; Pr =
= 1585
. .
154.6
50.46
f
f
From Figure 5.4-1:
= 1025
.
⇒ f =P
= 820 bar .
P
P
Using the P-R e.o.s.. and the program PR1 fO 2 = 7351
. bar .
F I
H K
(b) N 2 :
TC = 126.2 K ;
PC = 3394
. bar ; Tr =
290
= 2.298 ;
126.2
Pr =
800
= 23.57 ;
33.94
fN 2 = 1088 bar .
Using the P-R e.o.s. fN 2 = 1043 bar .
(c) Lewis-Randall Rule
Corresponding states
P-R e.o.s.
fO 2
0.3 × 820 = 2460
. bar
0.3 × 7351
. = 220.5 bar
fN 2
0.7 ×1088 = 7616
. bar
0.7 × 1043 = 7301
. bar
(d) Kay’s Rule
TCM = 0.3 × 154 .6 + 0.7 × 126.2 = 134.72 K
PCM = 0.3 × 50.46 + 0.7 × 38.94 = 38.90 bar
ψO 2 = −
T
2
TCM
cT
C, O 2
h
− TCM = −
290
134 .72 2
(154 .6 − 134.72) = −0.318
290
(126.2 − 134.72) = +0136
.
134 .72 2
P PC, O 2 − PCM
=−
= −6112
.
2
PCM
ψN 2 = −
ψO2 2
c
h
ψN2 2 = 2.622
290
800
TrM =
= 2.1526 ; PrM =
= 20.565
134.72
38.90
f
≅ 1.36 ;
P
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
f
H − H IG
17
.
~ 1.23 ;
≅−
= − 0.856 ; ZM ~ 1.6
P
RTC
1.987
ln
f N2
xN 2 P
= ln 123
. −
( 0136
. )( −0.856 )
( 2.1526)2
+
(1.6 − 1)(2.622 )
20.565
= 0.2070 + 0.0251 + 0.0765 = 0.3086
f N2
⇒
ln
f O2
xO 2 P
xN 2 P
= 13615
.
; f N 2 = 0.7 × 800 × 1.3615 = 762.4 bar
= ln 1.23 −
fO 2
⇒
(− 0.318 )(− 0856
. )
(1.6 − 1)( −6112
. )
= − 0.02929
20.65
= 0.97113; fO 2 = 0.97113 × 0.3 × 800 = 2331
. bar
xO 2 P
(e) Prausnitz-Gunn Rule
PCM = R ∑ xi Z C,i ∑ xi TC,i /
b
+
(2 .1526)2
gb
= Z CM
= TCM
g c∑ x V h
i C,i
=V CM
ZCM = 0.3 × 0.288 + 0.7 × 0.290 = 0.2894
TCM = 134 .72 ( see part d )
V CM = 0.3 × 0.0732 + 0.72 × 0.0895 = 0.08461
0.08314 × 0.2894 × 134.72
= 38.31 bar {vs. 38.90 bar in part d}
0.08461
and PrM are so close to results in (d) that ZM , f P ; H − H IG are all the same. Also, ψ’s
PCM =
TrM
are the same.
ψ2 2 =
RSF
TH
I F
K H
I F
K H
I UV
KW
ψO2 2
RSF
TH
I F
K H
I F
K H
I UV
KW
N
ln
fN2
xN 2 P
800
0.0895 − 0.0846
126 .2 − 134 .72
0.290 − 0.2894
−
−
38.31
0.0846
134 .72
0.2894
800
=
{0.05792 + 0.06324 − 0.00207} = 2.487
38.31
800
0.0732 − 0.0846
154 .6 − 134 .72
0.288 − 0.2894
=
−
−
38.31
0.0846
134 .72
0.2894
= −5.794
= 0.2070 + 0.0251 +
φN 2 =
ln
f N2
xN 2 P
f O2
xO2 P
= 1.3547
0.6 × 2.487
= 0.30357
2088
.
f N 2 = 758.6 bar
= −0.01819 ; φO 2 =
fO 2
xO 2 P
= 0.9820 ; fO 2 = 235.7 bar
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
(f)
Using the program PR1we find fO 2 = 224 .9 bar and fN 2 = 732.3 bar .
Lewis-Randall
with corresponding states
with Peng-Robinson eos
Kay’s Rule
Prousnitz-Gunn
Peng-Robinson e.o.s. directly
(program PR1)
SUMMARY
fN 2
fO 2
761.6 bar
246.0 bar
730.1
762.4
758.6
732.3
220.5
233.1
235.7
224.9
7.13
This problem was solved using the program UNIFAC. To present the extent of nonideality, two
measures will be used. One is the infinite dilution activity coefficients, and the other is G ex(max) ,
that is, the maximum value of the excess Gibbs free energy. The results appear below for the case of
T = 50° C
1
water
ethanol
benzene
2
ethanol
benzene
toluene
benzene
toluene
toluene
γ ∞1
γ ∞2
G ex (J/mol)
2.7469
304.0
446.5
8.8774
8.1422
0.9650
7.2861
1867.7
8776.2
4.5590
5.4686
0.9582
829.6
3507.9
3765.4
1162.0
1177.7
–26.1
These results were obtained treating toluene as 5 ACH groups + 1 ACCH 3 group.
An alternative is to consider toluene to be 5 ACH groups, 1 AC group and 1 CH 3 group. We do this
just to demonstrate that there can be a number of possible group assignments, each of which will
result in somewhat different activity coefficients.
1
water
ethanol
benzene
2
toluene
toluene
toluene
γ ∞1
γ ∞2
G ex (J/mol)
340.1
9.928
1.0058
6162.0
5.966
1.0080
3685.0
1269.2
4.5
We see, from the results (independent of which group assignment is used for toluene) that the
benzene-toluene mixture, which contains chemically similar species, is virtually an ideal solution.
The water-toluene and water-benzene mixtures consist of very dissimilar species and, therefore, the
mixtures are very nonideal. Ethanol contains a hydrocarbon end and a polar -OH end.
Consequently, it is almost equally compatible (or incompatible) with both water and hydrocarbon
solvents and forms only moderately nonideal mixtures with both this behavior is predicted above.
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.14 Regular solution theory should not be used with hydrogen-bonded solvents such as
water and ethanol. However, merely for demonstration, we will use R.S.T. for
these compounds.
The “Handbook of Chemistry and Physics” reports
δ H 2 O = 9.9 .
δEtOH = 10.0
and
Also,
V H 2 O = 18 cc mol
and
V EtOH = 58.4 cc mol .
In regular solution theory
a
RT lnγ i = V i φ 2j δ 1 − δ 2
f
2
so γ ∞i = exp
LMV aδ − δ f OP
N RT Q
2
i
1
2
so
γ ∞1
γ ∞2
ethanol
1.000
1.001
δEtOH = 10.0
benzene
1.014
1.070
δbenz = 9.2
toluene
1.028
1.181
δtol = 8.9
1
2
Water
Ethanol
benzene
1.060
1.093
toluene
1.116
1.223
Benzene toluene
1.013
1.015
Since the solubility parameters of all the components are similar, regular solution
theory predicts essentially ideal solution behavior, even though, for example, the
water-aromatic hydrocarbon mixtures are highly nonideal. This is an example of
how bad the regular solution theory predictions can be when used for mixtures for
which it is not appropriate.
This example should serve as a warning about the improper use of thermodynamic
models.
7.15 Start from
LM
N
∂ G − G IGM
∂ Tr
RT
In general
LM
OP
N
Q
OP = T ∂ LM G − G OP
Q R ∂T N RT Q
G( T , P, x) − G IGM (T , P, x)
H − H IG
=
− Tr , M S − S IG .
TCM
Tc
IGM
CM
F I
H K
∂ G
H
= − 2 ; using this above yields
∂T T
T
LM
N
OP
Q
LM
N
∂ G − G IGM
T
1
H − H IGM
= − CM2 H − H IGM = −
2
∂ Tr
RT
RT
RTrm
TCM
Also
LM
N
IGM
∂ G−G
∂ Pr
RT
OP = P ∂ dG − G
Q RT ∂ P
P F PV PV
=
G − RT
P H RT
CM
IGM
OP
Q
i = PRT dV − V i
IJ = 1 aZ − 1f
K P
(7.7-12)
IGM
CM
IGM
CM
Using these equations in Eqns. (7.7-9 and 11) gives
r
M
(7.7-13)
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
ln
FG f IJ − T dH − H i N FG ∂ T IJ
H PK
x P
RT
HNK
F∂ P I
1
+ a Z − 1f N G
P
H N JK
F f I dH − H i ψ + aZ − 1f ψ
= lnG J −
H P K RT T
P
IGM
fj
CM
= ln
rm
2
j
j
T , P , N i≠ j
r, M
M
r
j
T , P , Ni ≠ j
IGM
j
1
2
r ,M CM
j
2
M
r
(7.7-14)
For Kay’s rules, TCM = ∑ xi TC ,i and PCM = ∑ xi PC, i , we have
F ∂ T I = N ∂ RS T UV = − NT ∂ T = − NT ∂ RS ∑ N T UV
GH ∂ N JK ∂ N TT W T ∂ N T ∂ N T N W
NT R T
∑ N T U = − T mT − T r
=−
−
S
VW T
T T N
N
ψ1j ( K ) = N
r
i C ,i
CM
j
j
C, j
2
CM
CM
i C,i
2
2
CM
2
CM
j
C, j
2
CM
j
CM
Since the “combining rule” for PCM is the same as for TCM (for Kay’s rule), it
follows that
ψ2j ( K ) = −
m
P
PC, j − PCM
2
PCM
r
For the Prausnitz-Gunn rule, TCM = TCM ( Kay ) , so ψ 1j (PG ) = ψ 1j (K) . However
R
FG ∑ x Z IJ FG ∑ x T IJ
H
KH
K
i C ,i
i
PCM =
i C,i
i
∑ xiV C,i
i
Thus,
ψ2j ( PG ) = N
F ∂P I
GH ∂ N JK
r
k p
RS R ∑ N Z ∑ N T UV
|T ∑ N ∑ N V |W
2
PCM
j T , P, N
i≠ j
=−
NP ∂
2
∂Nj
PCM
i C, i
i C ,i
i
i
C, i
RS RZ ∑ N T + RT ∑ N Z − P LM 1 − V OPUV
T|∑ N ∑ N V ∑ N ∑ N V
N ∑ N ∑ N V QW|
T
V
UV
NP R Z
1
=−
+
−
−
S
P |
T∑ N Z ∑ N T N ∑ N V |W
T
Z U
P R V
=
1+
−
−
S
V
P T V
T
Z W
−V I F T − T I F Z − Z I O
P LF V
ψ ( PG ) =
G
JK − GH T JK − GH Z JK PQ
M
P NH
V
=−
NP
2
PCM
C, j
i
i C ,i
i
CM
CM
C, j
i
C,i
C, j
Thus
∂
PCM
∂Nj
NP
=−
C, j
C, j
CM
CM
CM
C, j
CM
CM
i
C,i
i
C, j
i C,i
C, j
CM
i
C, j
CM
C, j
i C,i
j
2
i C ,i
i
C, j
CM
CM
C,i
C, j
CM
CM
C ,i
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.16
T, P
x1 moles pure 1
Process
T, P
x1 , x2
T, P
x2 moles pure 2
System: 1 mole of initial mixture
system is closed, isothermal and isobaric
Mass balance: x1 + x2 = 1
dU
dV
Energy balance:
= Q& − P
+ W&s
dt
dt
dS Q& &
Entropy balance:
= + Sgen
dt T
Thus
dU &
dV
W&s =
−Q+ P
dt
dT
dS
Q& = T
− TS&gen
dt
dU
dS
dV
⇒ W&s =
−T
+ TS&gen + P
dt
dt
dt
a) Since both P and T are constant, we can write
dU d
d
W&s =
+ ( PV ) − ( TS ) + TS&gen
dt dt
dt
d
dG
= (U + PV − TS ) + TS&gen =
+ TS&gen
dt
dt
dG
Clearly, for W&s to be a minimum, S&gen = 0 , and W&smin =
.
dt
Wsmin
FG per mole of IJ = G
H initial mixtureK
f
− G i = x1 G 1 + x2 G 2 − x1G1 − x2 G2
b
g b
= x1 G 1 − G1 + x2 G 2 − G2
= x1 RT ln
f1(T , P)
f 1 (T , P , x)
g
+ x2 RT ln
b) Now for either ideal mixtures or Lewis-Randall mixtures,
f 2 ( T , P)
f 2 ( T , P, x )
fi (T , P, x )
= xi .
fi (T , P )
Therefore,
Wsmin = RT −x1 ln x1 − x2 ln x2 ≥ 0 , so work must be added!
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
dS
dS
Q& = T
− TS&gen ; thus Q& max = T
, which occurs when Ws = Wsmin .
dt
dt
Following same analysis as above leads to
[Note:
l
Q max = T x1 S 1( T , P) − S1( T , P, x) + x2 S 2 (T , P ) − S 2 ( T , P, x)
q
c) Using the ideal gas or ideal mixture assumption, since isomers can be
expected to form ideal mixtures, and the result above
J
× 300 K × − 0.5 ln 0.5 − 0.5 ln 0.5
mol ⋅ K
J
J
= −8.314
× 300 K × ln 0.5 = −17288
.
mol ⋅ K
mol of feed
Wsmin = RT − x1 ln x1 − x2 ln x2 = 8.314
7.17
LM a
N a
G1 =
f OP
f Q
2
N1 N 2
C N1 − N 2
A+
2
N1 + N2
N1 + N2
NG =
a
f
∂( N G )
= x2 A + C x1 − x2
∂ N 1 T , P, N 2
a
f
− x1 x2 A + C x1 − x2
2
2
LM a N − N f − 2Ca N − N f OP
N a N + N f aN + N f Q
= x a1 − x f A + Ca x − x f + x x 2 Ca x − x f − 2 Ca x − x f
= x A + Ca x − x f + 2Cx x a x − x f 1 − x + x
= x A + Ca x − x f + 4Cx x a x − x f = RT ln γ
2
+ x1 N2 2C
1
2
1
2
3
2
1
2
1
2
2
2
1
2
1 2
1
2
1
2
2
2
2
2
1
2
2
2
2
G2 =
1
1
2
1 2
1
2
2
1 2
1
2
a
∂( N G )
= x1 A + C x1 − x2
∂ N2 T , P , N1
f
1
2
1
a
− x1 x2 A + C x1 − x2
2
fLMN −a2Na N+ −N Nf f − 2Ca NaN+−NNf f OPQ
= x a1 − x f A + Ca x − x f − 2 x x Ca x − x f 1 + x − x
= x A + Ca x − x f − 4Cx x a x − x f = RT ln γ
a
f
2
2
+ x1 x2 N1 + N2
1
2
2
1
1
2
2
3
1
2
2
1
2
2
1
or
1
2
1 2
2
1
2
2 1
2
1
1
2
2
1
2
2
a f + 4Cx ax − x f
A + Ca x − x f − 4 Cx a x − x f
RT ln γ 1 = x22 A + C x1 − x2
2
RT ln γ 2 = x12
2
1
1
2
2
1
1
2
2
7.18 (a) i) One constant Margules equation.
a
f
RT ln γ 1 = A 1 − x1 2 ; RT
a
∂ ln γ 1
= −2 A 1 − x1
∂ x1
f
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
a
or
∂ ln γ 1 −2 A 1 − x1
=
∂ x1
RT
lim
x1 → 1
f
a
f
∂ ln γ 1
2 A 1 − x1
= lim
=0
x
→
1
∂ x1
RT
1
ii) Two constant Margules equations
a f + β a1 − x f
1
=
n−2α a1 − x f − 3β a1 − x f s → 0 as x → 1
RT
2
RT ln γ 1 = α1 1 − x1
∂ ln γ 1
∂ x1
3
1
1
1
1
2
1
1
1
iii) van Laar Equation
ln γ 1 =
α
a
1 + αx1 βx2
f
2
Thus
a f
a f
∂ ln γ 1
2α2 β2 1 − x1
=
∂ x1
β 1 − x1 + αx1
→ 0 as x1 → 1
3
iv) Regular Solution Theory expression
R.S.T. has the same form as the van Laar Equation, so that proof follows
from (iii) above.
(b) Starting from the Gibbs-Duhem Equation, Eqn. (7.3-16)
C
0 = ∑ xi
i =1
F ∂ lnγ I
GH ∂ x JK
i
j
T, P
we obtain
FG ∂ln γ IJ
H ∂x K
0 = x1
+ x2
1
1
T, P
FG ∂ lnγ IJ
H ∂x K
2
1
T ,P
Alternatively, since dx2 = −dx1 , we have
x1
FG ∂ lnγ IJ
H ∂x K
1
Now lim x1 = 1 and lim
1
1
T, P
2
2
(*)
T ,P
T ,P
=0
1
FG ∂ln γ IJ
H ∂x K
=0
1
FG ∂ln γ IJ
H ∂x K
⇒ lim x1
x1 → 1
T ,P
FG ∂ lnγ IJ
H ∂x K
x1 → 1
x1 → 1
= x2
1
⇒
lim x2
x1 → 1
FG ∂ln γ IJ
H ∂x K
=0
2
2
T ,P
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
which also implies that lim x2
x2 → 0
FG ∂ lnγ IJ
H ∂x K
= 0 or, more generally
2
2
lim xi
xi → 0
T ,P
FG ∂ln γ IJ
H ∂x K
i
i
=0
T ,P
Thus we have
F ∂ lnγ IJ
limG
H ∂x K
ln γ 1 departs, with
= 0 ⇒ zero slope, from its
i
xi →1
i
value of 0 at xi = 1
T, P
and
ln γ 1 rises less rapidly
1
= 0 ⇒ than
as xi → 0. Thus
i
xi → 0
x
i
i
T ,P
xi ln γ i is bounded!!!
7.19 Let M = molality of salt in solution.
i) For KCl: z+ = 1 , z− = 1 , M K = M , MCl = M ;
F ∂ln γ IJ
lim x G
H ∂x K
i
1
1
zi2 Mi = (1 × M + 1 × M ) = M
∑
2
2
z+ = 3
z− = −1
1 2
ii) For CrCl 3 :
I =
3 × M + 1 × 3M = 6 M
M + = M M− = 3 M
2
I=
a f
iii) For Cr2 SO 4 3 :
UV
W
z+ = 3
z− = −2
M+ = 2 M
M− = 3M
UV
W
m
r
I =
1 2
3 × 2 M + 4 × 3 M = 15 M
2
m
Now, the Debye-Hückel expression is lnγ ± = −α z+ z −
and Equation (7.11-18)
ln γ ± =
i)
KCl
M
0.1
0.2
0.3
0.5
0.6
0.8
1.0
−α z+ z−
1+ I
ln γ ±
experiment
0.770
0.718
0.688
0.649
0.637
0.618
0.604
I
+ 0.1 z + z− I ;
I
α = 1178
.
Debye-Hückel
ln γ ± = −1178
.
M
0.689
0.590
0.525
0.435
0.402
0.349
0.308
r
F molI
H lit K
12
at 25°C
Eqn. (7.11-18)
ln γ ± =
−1178
.
M
1+ M
0.761
0.709
0.679
0.645
0.635
0.621
0.613
+ 0.1 M
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
ii)
CrCl 3
ln γ ±
M
0.1
0.2
0.3
0.5
0.6
0.8
1.0
a f
iii) Cr2 SO 4
M
0.1
0.2
0.3
0.5
0.6
0.8
1.0
ln γ ± = −8657
.
M
0.331
0.298
0.294
0.314
0.335
0.397
0.481
0.065
0.021
87
. × 10−3
2.2 × 10−3
12
. × 10−3
4.3 × 10−4
17
. × 10−4
ln γ ±
ln γ ± = −27374
.
M
ln γ ± =
− 8.657 M
1 + 2.449 M
0.256
0.226
0.227
0.262
0.291
0.373
0.492
+ 18
. M
3
0.0458
0.0300
0.0238
0.0190
0.0182
0.0185
0.0208
174
. ×10−4
4.82 × 10−6
3.08 × 10 −7
392
. × 10−9
618
. × 10−10
2.33 × 10−11
129
. ×10−12
ln γ ± =
−27.374 M
1 + 3.873 M
0.0502
0.0113
0.122
0.508
1.104
5.559
29.44
+9 M
Thus the Debye-Hückel and extended Debye-Hückel (with a fixed value of
the δ parameter) theories are not very accurate. However, if the δ parameter is
adjusted, much better agreement with experimental data can be achieved. This
is left to the student to prove.
7.20 (also available as a Mathcad worksheet)
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
lngDH2
i
1.178 . M i
1
Mi
0.5
0.5
0.1 . M i
lngDH3i
1.178 . M i
1
Mi
0.5
0.5
0.30 . M i
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.21
The Gibbs-Duhem equation, written in terms of molalities and using the mean
ionic activity coefficient is as follows:
MSd G S + M E d G E = 0 where S is solvent and E is electrolyte
but
c
+ RT lna x γ f
G E = G oE + RT ln γ ν± M ν+ + M ν− −
G oS
h
GS =
S S
So that
MSd G S + M E d G E = 0
a f
c
M d lna x γ f + M d lncγ
h
MSd ln xSγ S + M E d ln γ ν± M +ν + M−ν − = 0
S
S S
E
ν
ν
± M±
h = 0 = M d lnax γ f + M νd lndγ M i
S
S S
E
±
±
This is the Gibbs-Duhem equation for the solute-electrolyte system.
For HCl
M + = M HCl
ν+ = 1
M − = M HCl
M±2
=
M1HCl
ν− = 1
⋅
M1HCl
=
2
MHCl
1000
= 55.56
18
MS
5556
.
xS =
=
MS + M E 55.56 + M E
MS =
a
f
MSd ln γ S xS + M HCl (1 + 1) d ln γ ± M HCl = 0
a
f
a
f
d ln γ S xS =
−2 M HCl d ln γ ± MHCl
MH 2 O
d ln γ S xS =
−2 M HCl d ln γ ± MHCl
5556
.
∆ ln γ S xS =
−2 M HCl
∆ ln γ ± M HCl
5556
.
a
f
This can now be used as a basis for numerical integration with the activity
coefficient expression from Illustration 7.11-1. Or proceeding further,
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
d ln γ S xS =
a
−2 MHCl d ln γ ± M HCl
M H2O
f
a f
a
f
2 M HCl d ln γ ±
−2 M HCld ln M HCl
+
5556
.
5556
.
5556
.
2 M HCl d ln γ ±
−2 M HCld ln M HCl
d ln γ S + d ln
=−
+
5556
. + M HCl
5556
.
5556
.
d ln γ S + d ln xS = −
FG
H
IJ
K
a f
a
f
From Illustration 7.11-1
ln( γ ± ) =
−1178
.
M HCl
1+
FG
H
d ln(γ ± ) = −
d ln γ S −
MHCl
+ 0.3 M HCl
1
1178
.
2 M HCl 1 + M HCl
+
IJ
K
1
1178
.
+ 0.3 dMHCl
2 (1 + MHCl ) 2
dM HCl
5556
. + MHCl
FG
H
IJ
K
2 M HCl
1
1178
.
1
1178
.
2
−
+
+ 0.3 dM HCl −
dM HCl
5556
.
55.56
2 M HCl 1 + M HCl 2 (1 + M HCl ) 2
d ln γ S
=−
=
FG 1
. +M
H 5556
+
HCl
1178
.
M HCl
5556
. ⋅ (1 + M HCl )
−
1178
.
M HCl
5556
. ⋅ (1 + M HCl )
2
−
IJ
K
0.6 M HCl
2
−
dM HCl
5556
.
5556
.
This can only be solved by numerical integration. (See MATHCAD file for this
problem).
1
1
gamma ( M ) 0.5
4.5185 .10
3
0
0
0
10
20
M
30
30
7.22 (a) The two-constant Redlich-Kister expansion, which leads to the two-constant
Margules equation is
k
a
G ex = x1 x2 A + B x1 − x2
fp
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
Thus
a
f
G ex
= A + B 2 x1 − 1
x1x2
(1)
Which is a linear function of x.
The form of the Wohl Equation which leads to the van Laar Equation is
G ex =
2 RTa12 x1q1 x2 q2
x1q1 + x2 q 2
which can be rearranged to
x1x2
G ex
=
a
f
x1q1 + 1 − x1 q2
2 RTa12 q1q2
(2)
which is also a linear function of x. Equations (1) and (2) provide the
justification for the procedure.
(b) The figure below is the required plot. Clearly, neither equation is an accurate
fit of the data. [The 2-constant Wohl (or van Laar) equation plot of the data,
i.e., the form of Eq. (2), is closest to being linear, and therefore should be the
better of the two-constant fits of the data. The data can, however, be fit quite
well with a 3-constant Redlich-Kister expansion—See Illustration 8.1-4]
7.23
Expression for G ex in this problem is the same as that of Eq. (7.6-6). If we
recognize that A and B in Eq. (7.6-6) is replaced by ART and BRT here. Also,
since 1 − 2 xAr = xCH 4 − xAr , species 1 is methane and species 2 is argon.
a
f c
h
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
(a)
Therefore
RT lnγ 1 = ( ART + 3BRT )x22 − 4 BRTx23
and
RT lnγ 2 = ( ART − 3BRT )x22 + 4 BRTx23
At x1 = x2 = 0.5
1
1
( A + B ) = (0.2944 + 0.0118) = 0.0766 ; γ CH 4 = 10796
.
4
4
1
1
ln γ 2 = ( A − B) = (0.2944 − 0.0118) = 0.0706 ; γ Ar = 10732
.
4
4
ln γ 1 =
a
a
fk
(b) G ex = RTxAr 1 − xAr A + B 1 − 2 xAr
G
ex
ax
Ar
f
= 0.5 =
fp at x
Ar
= 05
.
a
f
ART
G ex xAr = 0.5
AT
and
=
A
R
4
Thus
a
ax
f
G
= 0.5f
0.2944
( at T = 112 K ) = 112 ×
= 8.2432 K
R
4
G a x = 0.5f
0.2804
(at T = 115.74 K ) = 115.74 ×
= 81134
.
K
G ex xAr = 0.5
0.3036
(at T = 109 K ) = 109 ×
= 8.2731 K
R
4
ex
Ar
ex
Ar
R
4
Now replacing differentials with finite differences
G ex
R
∆T
T
109
∆
FG G IJ
HRK
G ex
RT
ex
8.2731
–0.0299
ex
–0.0023
8.2432
0.0736
3.74
–0.1298
115.74
FG G IJ
H RT K
0.0759
3
112
∆
–0.0035
8.1134
0.0701
Next using
a
∂ ∆G RT
∂T
Thus
∆T .
c
∆G
ex
∆T
RT
f
=
P
LM c
MN
∂ ∆ G RT
−∆ H
2 and
RT
∂T
h × b− RT g = ∆ H
2
mix
ex
h OP
PQ
=
P
− H ex − ∆ H mix
=
RT 2
RT 2
where T = average temperature over
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
⇒ ∆H mix ≅ −8.314 × (112 K )2 ×
(c) From
FG ∂G IJ
H∂TK
= − S we obtain
P
S ex = −
d
R∆ G ex / R
RS
T
UV
W
1 −0.0023 −0.0035
+
= 888
. J mol
2
3
3.74
FG IJ
H K
∂ G ex
∂T R
=−
P
S ex
. Therefore,
R
i = −8.314 J mol K × 1 RS− 0.0299 + −0.1298UV
2T 3
3 W
∆T
= 0.2213 J mol K
Also
∆S mix = − R ∑ xi ln xi + S ex = −8.314 × 2 × (0.5 ln 0.5) + 0.045
= 5.984 J mol K
7.24 We start with
ln
fi
1
=
xi P RT
zc
P
h
Vi − Vi IGM dP =
0
1
RT
Eqn. (7.2 - 3a)
z LMMNFGH
P
0
∂V
∂ Ni
IJ
K
−
T , P , N j ≠i
OP
PQ
RT
dP
P
Now
dP =
1
P
RT
P
P
P
d ( PV ) − dV =
dZ − dV = dZ = − dV
V
V
V
V
Z
V
Also, by triple product rule
FG ∂V IJ
H∂ N K
i
⇒ Vi =
⋅
T , P , N j≠ i
FG ∂V IJ
H∂ N K
i
So
FG ∂ P IJ
H ∂V K
P , T , N j≠ i
FG ∂ N IJ
H ∂P K
F ∂ P IJ
= −G
H∂ N K
i
⋅
T ,N j
i
= −1
T ,V , N j≠i
T ,V , N j ≠i
FG ∂V IJ
H ∂ PK
T ,N j
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
ln
z FGH
z FGH
IJ
K
fi
1
∂P
=
−
xi P RT
∂ Ni
T ,V , N
V
=−
=
=
7.25
1
∂P
RT V = ∞ ∂ Ni
1
RT
1
RT
z LMMN
z LMMN
V
V =∞
V
V =∞
IJ
K
j≠ i
FG ∂V IJ
H ∂ PK
z
dP −
T ,N j
LM
N
1 P
P
dZ − dV
P Z
V
z
z
V
NdV +
T ,V , N
j≠ i
OP
Q
Z
1
1
dV −
dZ
V
Z
V =∞
P= 0
Z =1
FG IJ
H K
RT
F ∂ P IJ
− NG
V
H∂ N K
RT
∂P
−N
V
∂ Ni
T ,V , N j≠i
i
T ,V , N j≠i
OPdV − ln Z
Z =1
PQ
OPdV − ln Z
PQ
LM∑ x Λ OP
N
Q
LM ∑ x Λ OP
G
NG
=
= −∑ N lnM
RT
RT
MN N PPQ
G
∂ FG I
⇒
=
G J
RT
∂ N H RT K
LM ∑ x Λ OP
LM Λ ∑ N Λ dN OP
N
= − ln M
−∑N
−
N
dN P
∑ N Λ MM N
MN N PPQ
PQ
N
x ∑x Λ
L
O
G
xΛ
ln γ =
= − ln M∑ x Λ P − ∑
+
RT
N
Q ∑x Λ ∑ ∑x Λ
L
O xΛ
= 1 − lnM ∑ x Λ P − ∑
N
Q ∑x Λ
C
G ex
= − ∑ xi ln
RT
i =1
ex
j
j
ij
j
C
ex
i =1
ex
1
ij
j
i
ex
T , P , N j≠ i
1
j
j
ij
C
i =1
i
j
j
j
i1
ij
2
ij
1
j
1
C
ex
1
j =1
C
j =1
C
j
ij
i =1
C
j
ij
i= 1
i
i
j
C
i1
j
ij
i =1
i
j
ij
j
j
j
ij
i1
j
ij
j
Which is the answer to part b... To obtain the result of part a, we restrict i and j to
the values 1 and 2, and note that Λii = 1 . Thus
ln γ 1 = 1 − ln x1 + x2 Λ12 −
x1
x2 Λ 21
−
x1 + x2 Λ12 x1Λ 21 + x2
but
1−
x1
x + x Λ −x
x2 Λ12
= 1 2 12 1 =
x1 + x2 Λ12
x1 + x2 Λ12
x1 + x2 Λ12
so that
ln γ 1 = − ln x1 + x2 Λ 12 + x2
LM Λ
Nx + x Λ
−
12
1
2
12
Λ 21
x1 Λ 21 + x2
OP
Q
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.26 All the calculations for this problem were done using the program VLMU with the
binary interaction parameter for CO 2 − n C4 equal to 0.13 as given in Table 7.4-1.
The results are only given in graphical form here.
T = 377.6 K
kij = 0.13
80
supercritical
fCO
60
2
40
vapor
20
liquid
vapor
40
60
80
fnC
60
40
4
20
xnC
4
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
T =300 K
kij = 0.13
80
liquid
60
f CO
2
40
40
20
20
liquid
vapor
vapor
80
40
f
20
20
nC 4
liquid
x nC
4
60
40
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.27
a
G ex = ax1 x2 x1 − x2
N G ex =
f = ALM N N − N N OP
aN + N f
N aN + N f Q
L 2 N N − N − 2c N N − N N
= AM
aN + N f
MN a N + N f
a
AN1 N2 N1 − N 2
2
1
2
2
2
1
2
1
∂ N G ex
∂ N1
f
2
2
1
1
N 2 ,T , P
2
2
1
2
2
2
2
1
2
= A 2 x1 x2 −
1
− 2 x12 x2
x22
= Ax2 2 x1 − x2 −
2 x12
+
2
1 2
3
2
2
hOP
PQ
2 x1 x22
a
f
+ 2 x1x2 = Ax2 2 x1 1 − x1 − x2 + 2 x1 x2
a
f
= Ax2 2 x1 x2 − x2 + 2 x1 x2 = Ax22 4 x1 − 1 = Ax22 4 − 4 x2 − 1
a3 − 4x f = RT ln γ
L N − 2 N N − 2c N N − N N h OP
= AM
a N + N f PQ
MN a N + N f
=
∂ NG
∂ N2
ex
Ax22
2
2
1
N1 , T , P
1
1
=A
x12
1
−
2
1
2
2
2
1
2 x1 x2 − 2 x12 x2
+
2
1 2
3
2
2
2 x1 x22
a
f
= Ax1 x1 − 2 x2 − 2 x1 x2 + 2 x22 = Ax1 x1 + 2 x2 x2 − 1 − 2 x1 x2
= Ax1 x1 − 2 x1 x2 − 2 x1 x2 =
a
f
a
f
Ax12
1 − 4 x2 =
a
1 − 4 1 − x1
Ax12
f
=
−3 + 4 x1 = RT ln γ 2
(This is just a check since by symmetry of original equation 1 ↔ 2 gives minus
sign. Therefore 1 ↔ 2 on lnγ must give minus sign!)
Does this expression satisfy the Gibbs-Duhem Equation?
d ln γ 1
d ln γ 1
d ln γ 1
d ln γ 1
x1
+ x2
= 0 or − x1
+ x2
=0
dx1
dx1
dx2
dx1
Ax12
d ln γ 2
A d
A d
=
x12 −3 + 4 x1 =
−3 x12 + 4 x13
dx1
RT dx1
RT dx1
c
c
h
A
−6 x1 + 12 x12
RT
d ln γ 1
A d
A d
=
x22 3 − 4 x2 =
3 x22 − 4 x32
dx2
RT dx2
RT dx2
=
h
a
c
f
c
h
h
A
6 x2 − 12 x22
RT
Gibbs-Duhem Equation
A
A
− x1
6 x2 − 12 x22 + x2
−6 x1 + 12 x12
RT
RT
A
=
−6 x2 x1 + 12 x1 x22 − 6 x2 x1 + 12 x12 x2
RT
A
12 A 2 2
=
x1 x2 −12 x1 x2 + 12 x1 x22 + 12 x12 x2 =
x1 x2 x1 + x2 − 1
RT
RT
12 A 2 2
=
x1 x2 ( 0) = 0
RT
⇒ Satisfies Gibbs-Duhem Equation
=
c
h
c
c
h
h
a
f
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.28 To check the utility of these models we will use the Gibbs-Duhem equation in the
form
x1
FG ∂ lnγ IJ
H ∂x K
+ x2
1
1
T ,P
FG ∂ln γ IJ
H ∂x K
=0
2
1
T ,P
For the model
a
ln γ 1 = Ax22 = A 1 − x1
f
2
ln γ 2 = Bx12
Gives
x1
FG ∂ lnγ IJ
H ∂x K
+ x2
1
1
T ,P
FG ∂ln γ IJ
H ∂x K
1
a
f
= 0 = x1 ⋅ 2 ⋅ A 1 − x1 (− 1) + x2 B ⋅ 2 x1
2
T ,P
= 2 x1 x2 ( B − A) = 0
The only way this equation can be satisfied is if A = B ; if not the Gibbs-Duhem
equation is violated.
For the model lnγ 1 = Ax2n ; lnγ 2 = Ax1n
x1
FG ∂ lnγ IJ
H ∂x K
+ x2
1
1
T ,P
FG ∂ln γ IJ
H ∂x K
a
= x1 ⋅ n ⋅ A 1 − x1
2
1
T ,P
f
n −1
( −1) + x2 ⋅ n ⋅ Ax1n −1 = 0
= nAx1 x2 − x2n − 2 + x1n − 2 = 0
The only way the Gibbs-Duhem equation can be satisfied for all values of x1 and
x2 (with x1 + x2 = 1 ) is if n = 2 in which case the term in brackets is always zero.
For the model lnγ 1 = Ax2n ; lnγ 2 = Bx1n . We have
x1
FG ∂ lnγ IJ
H ∂x K
+ x2
1
1
T ,P
FG ∂ lnγ IJ
H ∂x K
1
a
= x1 ⋅ n ⋅ A 1 − x1
2
T, P
f
n −1
( −1) + x2 ⋅ n ⋅ Bx1n −1 = 0
= nx1 x2 − Ax2n − 2 + Bx1n −2 = 0
For this equation to be satisfied, the term in the brackets must be zero. This can
only be in n = 2 and A = B .
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.29 We will write the Flory-Huggins expression as
a
ex
G
RT
G ex
RT
G ex
RT
f
= χ x1 + mx2 φ1φ2 =
res
= N1 ln
comb
=
res
F
GH
N G ex
RT
∂ G
∂ N1 RT
ex
χx1 x2 m
x1 + mx2
φ1
φ
x1
mx 2
+ N2 ln 2 with φ1 =
, φ2 =
x1
x2
x1 + mx 2
x1 + mx2
χN1 N2 m
N1 + mN 2
=
res
I = χN m − χN N m
JK N + mN a N + mN f
2
1
2
2
res
1
2
res
I = χN m
JK N + mN
1
a
f
a
f
= χ φ2 − φ1φ2 = χφ2 1 − φ1 = χφ22
2
Similarly
F
GH
∂ G ex
∂ N2 RT
G ex
RT
1
= N1 ln
comb
F
GH
∂ G ex
∂ N1 RT
2
f
φ1
φ
+ N2 ln 2
x1
x2
I = ln φ + N
JK x
1
comb
a
χN1 N 2 m
= χm φ1 − φ1φ2 = xm φ12
N1 + mN 2
−
1
1
1
∂
φ
∂
φ
ln 1 + N 2
ln 2
∂ N1 x1
∂ N 1 x2
φ1
1
N
=
=
x1 x1 + mx 2
N1 + mN 2
and
LM
N
∂
φ
N + mN 2
1
N
ln 1 = 1
−
∂ N1 x1
N
N1 + mN 2
N1 + mN 2
a
φ2
m
Nm
=
=
x2
x1 + mx2
N1 + mN 2
LM
N
OP; N ∂
f Q ∂N
ln
1
2
1
φ1
= x1 − φ1
x1
OP
Q
∂
φ
m
1
∂
φ
φ
ln 2 =
−
; N2
ln 2 = x2 − 2
∂ N 1 x2
Nm N1 + mN 2
∂ N1 x2
m
Therefore
I + ∂ FG I
JK ∂ N GH RT JK
φ
φ
= ln + a x − φ f + F x − I + χφ
H
x
mK
ln γ 1 =
F
GH
∂ G ex
∂ N1 RT
1
1
1
ex
1
comb
1
2
res
2
2
2
F
H
I
K
φ
φ
φ
1
= ln 1 + 1 − φ1 − 2 + xφ22 = ln 1 + φ2 1 −
+ χφ22
x1
m
x1
m
Also
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
F
GH
∂ G ex
∂ N2 RT
I = ∂ L N ln φ + N ln φ O
JK ∂ N MN x
x PQ
1
2
1
conf
2
2
1
2
φ
x ∂ φ1
x ∂ φ2
= ln 2 + N1 1
+ N2 2
x2
φ1 ∂ N 2 x1
φ2 ∂ N2 x2
144244
3 1442443
x1 − mφ 1
a
x2 − φ 2
f a
f
φ
ln γ 2 = ln 2 + x1 − mφ1 + x2 − φ2 + χφ12
x2
= ln
φ2
φ
+ φ1 − mφ1 + xφ21 = ln 2 + (1 − m)φ1 + χφ21
x2
x2
= ln
φ2
− ( m − 1)φ1 + χφ12
x2
7.30 (also available as a Mathcad worksheet)
7.30
i
0 , 1 .. 10
0.1 . i
xi
fmai
xi . exp 1.06 . 1
fmi
1
xi
xi . exp 1.06 . xi
. 1.126
2
2
. 0.847
0
x=
0
0 0
0 0
0
0 0.847
1 0.1
1 0.266
1 0.77
2 0.2
2 0.444
2 0.707
3 0.3
3 0.568
3 0.652
4 0.4
fma =
4 0.66
fm =
4 0.602
5 0.5
5 0.734
5 0.552
6 0.6
6 0.8
6 0.496
7 0.7
7 0.867
7 0.427
8 0.8
8 0.94
8 0.334
9 0.9
9 1.024
9 0.2
10 1
10 1.126
10 0
Hma
Hm
1.126 . exp ( 1.06 )
0.847 . exp ( 1.06 )
Hma = 3.25
Hm = 2.445
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7.31 Show AEOS ( P → ∞ ) = C
LM a
Nb
− ∑ xi
mix
mix
start
−P=
OP
Q
ai
bi
z
V
∂A
→ A V − A V → ∞ = − PdV
∂V T
∞
(V → ∞ is convenient since we have ideal gas and ideal gas mixtures)
P-R
RT
a
P=
−
← pure component
V − b V (V + b) + b (V − b)
z
z
V
AV − AV →∞ = −
V
RT
a
+
dV
V − b ∞ V (V + b ) + b(V − b )
∞
= − RT ln(V − b ) − ln ∞ +
z
V
a
dV
2
V
+
2
bV − b 2
∞
14442444
3
Need to integrate
From Problem 4.2 we have that
z
V
∞V
1
2
+ 2bV − b
2
dV =
1
2 2b
F V + d1 − 2 ib I
GH V + d1 + 2 ib JK
ln
⇒ A V − A V →∞ = − RT ln(V − b ) − ln ∞ +
a
2 2b
ln
F V + d1 − 2 ibI
GH V + d1 + 2 ibJK
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
So for pure component:
ka
f
F V + b d1− 2 iI
G
J
2 2 b H V + b d1 + 2 i K
f − ln ∞p + 2 a2b lnFG VV ++ bb dd11−+ 22 iiIJ
H
K
p
ai
Ai V − Ai V → ∞ = − RT ln V − bi − ln ∞ +
i
ln
i
By exact analogy for mixture
ka
A V − A V →∞ = − RT ln V mix − bmix
i
mix
mix
mix
mix
mix
mix
Now when V → ∞ get ideal gas or ideal gas mixture:
AV − A
IG
a
= − RT {ln(V − b ) − ln ∞} +
ka
f
2 2b
F V + bd1− 2 iI
GH V + bd1+ 2 iJK
ln
p
A V − A IGM = − RT ln V mix − bmix − ln ∞
+
a mix
2 2 bmix
ln
FV
GH V
d i IJ
d1 + 2 i K
mix
+ bmix 1 − 2
mix
+ bmix
and by definition Aex = A − AIM (i.e., A = AIM + A ex ) so; have A − AIGM so
d
i
Aex = A − AIGM − AIM + A IGM and
A
IGM
=∑
+ RT ∑ xi ln xi ; AIM = ∑ xi Ai + RT ∑ xi ln xi
xi AiIG
Aex = A − AIGM − ∑ xi Ai − RT ∑ xi ln xi + ∑ xi AiIG + RT ∑ xi ln xi
A
ex
d
= A− A
IGM
i −∑ x dA − A i
i
IG
i
i
this is why we found ∆ A for pure component
z
V
So putting in results for A V − A ∞ = − PdV for pure i and mixture.
∞
iIJ
2 2b
+b
2 iK
R|
F V + b d1− 2 iI U|
a
− ∑ x S − RT lnaV − b f + RT ln ∞ +
ln G
JV
2 2 b H V + b d1 + 2 i K |
|T
W
Need to collect some terms.
L
O
A = − RT MlnaV
− b f − ln ∞ − ∑ x lnaV − b f − ln ∞ P
N
Q
F
I
F
V
+ b d1 − 2 i
V + b d1 −
a
a
+
− lnG
− ∑x
lnG
J
2 2b
2 2 b H V + b d1 +
H V + b d1 + 2 iK
a
f
a mix
Aex = − RT ln V mix − bmix − ln ∞ +
FV
GH V
ln
mix
i
i
i
mix
i
ex
i
mix
mix
mix
mix
mix
mix
i
i
i
i
i
mix
i
i
i
i
mix
mix
i
i
d
d1 +
+ bmix 1 − 2
i
mix
i
i
i
i
Now let P → ∞ which is the same as V i → bi and V mix → bmix
= −RT
+
[ ln
a
a mix
2
2 b mix
1
b mix
ln
− b mix
Fc
GH c
2
2
f
3
ln
∞ − ∑ x i [ ln
h I−∑
h JK
−
2 b mix
+
2 b mix
xi
a
2
bi
ai
2
f
4
− b i − ln ∞]
ln
2 bi
Now 1 and 2 cancel ln 0 − ∑ xi ln 0 = 0
Fc
GH c
hI
h JK
2
−
2 bi
2
+
2 bi
iIJ
2 iK
2
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
and 3 and 4 cancel − ln ∞ + ∑ xi ln ∞ = 0 . i.e., −∞ + 3 ⋅ ∞ = 0
So
Aex
P- R =
P→ ∞
FG
H
IJ
K
FG
H
1 a mix
2− 2
ai
2− 2
⋅
ln
− ∑ xi
ln
2 2 bmix
2+ 2
2 2bi
2+ 2
FG 2 − 2 IJ ⋅ 1 LM a
H 2 + 2 K 2 2 Nb
La
A
= −0.6232 M
Nb
= ln
mix
ex
P- R
P→ ∞
mix
FG 2 − 2 IJ 1 = −0.6232 = C
H2+ 2K 2 2
where ln
mix
*
OP
Q
a O
−∑x P
bQ
mix
− ∑ xi
i
IJ
K
ai
bi
i
i
for P - R
Now for van der Waals ⇒ same process though solution is briefer.
V RT
V a
∂A
−P =
⇒ A V − AV →∞ = −
dV +
dV
∞V −b
∞ 2
∂V T
V
z
Pure component
AV− A
IG
a
z
f
= − RT ln V i − bi + RT ln ∞ −
ai
Vi
and for the mixture
AV− A
IGM
a
f
= − RT ln V M − bM + RT ln ∞ −
d
i
d
aM
VM
i
and Aex ≡ A − A IM = A − AIGM = ∑ xi Ai − A IG ← same as above
z
V
Putting in results for A V − A ∞ = − PdV for pure i and for the mixture.
∞
a
Aex = − RT ln V
RS
T
M
f
− b M − RT ln ∞ −
a
f
aM
VM
− ∑ xi − RT ln V i − bi − RT ln ∞ −
a
= − RT ln V
M
f
a
ai
Vi
f
− b M + ∑ xi ln V i − bi + ln ∞ − ∑ xi ln ∞ −
Take limit P → ∞ , V i → bi ; V M → b M .
First 4 terms cancel!
aM
a
Aex
− ∑ xi i
VDW = − 1
bM
bi
P→ ∞
LM
N
OP
Q
7.32 Starting from
aij
a
b M − M = ∑ ∑ xi x j bij −
≡ Q and
RT
RT
i j
G ex = C*
LM a
Nb
FG
H
M
M
− ∑ xi
i
UV
W
ai
bi
OP
Q
IJ
K
aM
a
+ ∑ xi i
VM
Vi
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
or
ex
aM
G
a
a
= * + ∑ xi i ≡ DRT and b M = M
bM
C
bi
DRT
i
Substituting, we then obtain
aM
a
− M = Q so that
DRT RT
and
bM =
7.33
aM
D
=Q
RT
1− D
aM
D
Q
+Q= Q
+Q=
RT
1− D
1− D
Equation (7.10-11) is easily derived, is generic, and applies to any mixing rule.
This will be used as the starting point. With the Wong-Sandler mixing rule
Note that derivatives must be taken with respect to mole numbers.
Therefore
FG
H
Q = ∑ ∑ xi x j bij −
i
j
FG
H
aij
RT
IJ
K
N 2 Q = ∑ ∑ Ni N j bij −
i
j
needs to be in the form of
aij
RT
IJ
K
Similarly
D = ∑ xi
i
ai
G ex
+ *
needs to be in the form of
bi RT C RT
ND = ∑ Ni
i
ex
ai
NG
+
bi RT C * RT
From this starting point, eqns. (7.10 -12 and 13) are easily derived.
7.34
Starting from eqn. (7.2-13)
ln
fk
1
=
xk P RT
z
V = ZRT / P
V =∞
LM RT F ∂P I
MN V − N GH ∂N JK
k
T ,V , N j≠k
OP
PQdV − ln Z
The Soave-Redlich-Kwong equation of state is
P=
RT
a( T )
NRT
N 2a( T )
−
=
−
V − b V (V + b ) V − Nb V (V + Nb )
with
Nb = ∑ Ni bi and N 2 a = ∑ ∑ Ni N j aij
i
i
j
Now taking the derivative, we obtain
N
FG ∂P IJ
H ∂N K
k
=
T ,V , N j≠k
FG IJ
H K
RT
∂P
−N
V
∂N k
2 ∑ xi a ik
RT
RTbk
abk
i
+
−
+
V − b (V − b ) 2 V (V + b ) V (V + b) 2
=
T ,V , N j≠k
2 ∑ xi aik
RT
RT
RTbk
abk
i
−
−
+
−
V
V − b (V − b) 2 V (V + b) V (V + b )2
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
and then
ln
OPdV − ln Z
z
x P
PQ
LM
R|2∑ x a
ZRT / P
RTb
RT ln
+
+S
ZRT / P − b ZRT / P − b |
b
1 M
=
M
T
RT M
MM− b(ZRTab/ P + b)
N
f
k
k
=
1
RT
V = ZRT / P
V =∞
LM RT − N F ∂P I
MN V GH ∂N JK
k
T ,V , N j≠k
i ik
k
i
−
abk
k
b2
U| ZRT / P OP
V| ln ZRT / P + b P
PP − ln Z
W
PP
Q
Now using B=Pb/RT and A = Pa/(RT) 2 we obtain
fk
Z
Bk
A
ln
= ln
+
+
xk P
Z − B Z − B B
R| 2∑ x A
S| A
T
i
i
ik
U|
V|
W
B
Z
A Bk
− k ln
−
− ln Z
B
Z + B B Z+ B
However, the Soave-Redlich-Kwong equation of state can be rewritten as
follows
RT
a
P=
−
V − b V (V + b )
PV
V
a
Z
A
=Z=
−
=
−
RT
V − b RT (V + b) Z − B Z + B
B
A
−
Z− B Z+ B
Using this expression in the 2nd and 4th terms on the right-hand side of the
fugacity expression yields the desired result
2∑ xi Aik
f
Z
A
B
Z
B
i
ln k = − ln Z − B
+
− k ln
+ k ( Z − 1)
xk P
Z − B B
A
B
Z + B B
Z −1 =
R|
S|
T
U|
V|
W
Note that in this derivation, we have used the following
dx
1
x
= ln
x( x + b) b
x +b
dx
1
x
1
= 2 ln
+
2
x
+
b
b
(
x
+ b)
x( x + b)
b
z
z
z
F
H
F
H
I
K
I
K
dx
= ln( x − b )
( x − b)
7.35 See Mathcad worksheet.
a and b) See Mathcad file 7-35.MCD and figures contained there
c) Clearly
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
a
f
a
Ax2 RT
x1 + mx2
T
H ex = χRT x1 + mx2 φ1φ2 =
f ax +xmx f a x mx
+ mx f
1
1
2
2
1
2
Ax22 x1Rm
=
x1 + mx2
a
ex
f
G
φ
φ
Ax2
= x1 ln 1 + x2 ln 2 +
x1 + mx2 φ1φ2
RT
x1
x2
T
F F Ax
II
I GG ∂H N T a x + mx fφ φ K JJ
K G
∂N
JK
H
ax + mx fφ φ IK IJ
A ∂ L Nx a x + mx f x mx O A ∂ L
N N mN
OP
=
=
M
P
M
J
∂N
T ∂N N
ax + mx f Q T ∂N N a N + mN fa N + N fQ
JK
AL
N mN
N N mN
N N mN
OP
= M
−
−
T N a N + mN fa N + N f a N + mN f a N + N f a N + mN fa N + N f Q
FG
H
ex
1 ∂N G
ln γ 1 =
RT
∂ N1
F ∂F N Ax
GG H T
GH
IJ
K
T , P, N 2
F
H
2
φ
1
= ln 1 + 1 −
φ2 +
x1
m
1
2
1 2
1
T , P, N 2
2
1
2
1 2
2
1
1
1
2
1
2
1
2
2
2
1
1
1
2
2
1
2
T ,P ,N 2
2
1
2
2
2
1
2
1
1
2
2
2
2
1
2
2
1
2
1
2
1
2
f IK IJ
JJ
K
T , P, N 1
A
Ax2φ2
Ax2 φ2
x2 φ2 − x2φ2 φ1 − x1 x2φ2 =
1 − φ1 − x1 =
φ2 − x1
T
T
T
φ
1
Ax2φ2
ln γ 1 = ln 1 + 1 −
φ2 +
φ2 − x1
x1
m
T
Ax2
∂ N
x1 + mx2 φ1φ2
1 ∂ N G ex
φ2
T
ln γ 2 =
= ln + (1 − m)φ1 +
RT
∂ N2 T , P , N
x2
∂ N2
=
F
H
I
K
FG
H
IJ
K
FF
GG H
GH
1
a
F ∂FH N Ax ax + mx fφ φ IK I
GG T
JJ = A ∂ LM Nx ax + mx fx mx OP = A ∂ LM N N m OP
∂N
T ∂N N
ax + mx f Q T ∂N N a N + mN faN + N fQ
GH
JK
OP
AL
2 N mN
N Nm
N Nm
= M
−
−
T N a N + mN fa N + N f a N + mN f a N + N f a N + mN fa N + N f Q
2
1
2
1 2
2
2
2
1
2
1
2
1
2
2
2
2
2
1
1
2
1
T , P , N1
1
2
1
2
1
2
2
2
1
2
2
2
2
1
2
2
1
2
1
2
A
Ax1φ2
Ax2φ2
2 x1φ2 − x1φ22 − x1 x2φ2 =
2 − φ2 − x2 =
φ1 + x1
T
T
T
φ
Ax2 φ2
ln γ 2 = ln 2 + (1 − m )φ2 +
φ1 + x1
x2
T
=
d) and e) See figures in Mathcad worksheet 7-35.mcd.
1
1
2
2
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 7
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8
8.1-1
fiV = fi L ⇒ xi Pi vapγ i = yi P , since the pressure is low enough that fugacity coefficient corrections
will be small.
(a) For the ideal solution, γ i = 1 for all species; 1 = EB, 2 = nH ; yi = xi Pi vap P . Thus,
y1 = 0.4723 × 0.7569 0.4537 = 0.7879
and
y2 = 0.5277 × 0.0773 0.4537 = 0.0899 ;
.
∑ yi = 08778
which indicates that the ideal solution assumption is invalid!
(b) Regular solution behavior:
V 1 = 75 cm3
a
f
δ1 = 8.9 cal cc
a
V 2 = 148 cm 3
12
a
δ2 = 7.4 cal cc
f
12
f
φ1 = x1V 1 x1V 1 + x2 V 2 = 0.4723 × 75 (0.4723 × 75 + 0.5277 × 148) = 0.312
φ2 = 0.688
ln γ 1 =
V 1φ22 δ1 − δ2
= 0133
.
;
γ 1 = 1142
.
.
Similarly
RT
y1 = 1142
.
× 0.7879 = 0.8998 ; y2 = 1.055 × 0.0899 = 0.0948 ; and
closer to unity.
(c) UNIFAC:
using
the
program
UNIFAC
we
(d) First the expression γ i = yi P
Therefore,
∑ yi = 0.9946
have
⇒ y1 = 1173
.
× 0.7879 = 0.9242 ; y2 = 1118
.
× 0.0899 = 01005
.
;
xi P1vap
γ 2 = 1055
.
.
γ 1 = 1173
.
.
∑ yi = 10247
which is much
and
γ 2 = 1118
.
;
which is too high.
and the given vapor-liquid equilibrium data will be used
to compute the species activity coefficients in the given solution:
0.8152 × 0.3197
= 1.211 ; similarly, γ 2 = 1.069
0.2843 × 0.7569
Using eqns. (7.5-10) we obtain β = 0.3055 and α = 0.6747 . Thus, using the van Laar eqn.
γ1 =
ln γ 1 =
0.6747
1 + 2 .2085 x1 x2
2
and ln γ 2 =
0.3055
1 + 0.4528 x2 x1
2
at x1 = 0.4723 , γ 1 = 1.079 , γ 2 = 1144
.
, so that y1 = 08504
.
, y2 = 0.1029 and
Since none of the models yields
.
.
∑ yi = 0953
∑ yi = 10. , none of the solution models is completely correct.
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
Since the regular solution model comes closest to meeting this criteria, it presumably leads to
the best predictions—however, this is merely a hypothesis.
8.1-2
The van Laar equation will be used to fit the data given. Starting from
1 = H2O
γ i = yi P xi Pi vap
2 = FURF
we obtain, at 10 mole % water, γ 1 = 5.826 and γ 2 = 1.266 . Using eqn. (7.5-10) we get
α = 8.5648 and β = 0.7901 . Thus,
ln γ 1 =
8.5648
a1 + 10.841 x x f
2
1
and ln γ 2 =
2
0.7901
a1 + 0.0922 x x f
2
2
1
which we will assume is valid at all temperatures.
At the new temperature we have xiγ i P1vap = yi P or x1γ 1 × 10352
.
= y1 × 1013
.
and
x2γ 2 × 01193
.
= y2 × 1.013 which must be solved together with the activity coefficient expressions
above, and the criteria that x1 + x2 = 1 and y1 + y2 = 1 . Solution procedure I used was to guess a
value of x1 , compute x2 from x2 = 1 − x1 , compute γ 1 and γ 2 from the expression above, yi
from yi = xi γ i Pi vap P for i = 1 and 2, and then check to see if
∑ yi = 1 .
Proceeding this way,
the following results were obtained
calculated measured
Note
x1
0.075
0.20
x2
0.925
0.80
y1
0867
.
0.89
y2
0129
.
0.11
∑ yi = 0.996 which is not quite equal to 1. The discrepancy between the calculated and
experimental results indicates the dangers of using approximate solution models.
8.1-3
The desired result may be proved a number of different ways.
∂P
Simplest proof: We have show that at an azeotropic point
∂x1
FG IJ
H K
= 0 using the triple product rule
T
in the form
FG ∂P IJ F ∂T I F ∂x I
H ∂x K H ∂P K H ∂T K
= −1
1
1 T
x1
P
yields
FG ∂P IJ
H ∂x K
FG ∂T IJ F ∂P I = 0
H ∂x K H ∂T K
R
∂P
∂P
Since there is no reason to believe F I = 0 Sin fact, F I
H ∂T K
H
∂T K
T
1 T
=−
1
x1
P
x1
=
x1
P∆ H vap
RT 2
UV
W
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
FG ∂T IJ
H ∂x K
We then have
1
= 0.
P
Alternate proof: P = x1γ 1 P1vap + x2γ 2 P2vap = RT V for an ideal gas phase. Thus
FG ∂T IJ
H ∂x K
1
=
P
RS
T
LM
N
OP
Q
V
∂ ln γ 1 ∂ ln P1vap
γ 1 P1vap + x1γ 1 P1vap
+
− γ 2 P2vap
R
∂ x1
∂ x1
+ x2γ 2 P2vap
LM ∂ lnγ
N ∂x
∂ ln P2vap
∂ x1
+
2
1
OP UV = 0
QW
P
Now:
i)
FG ∂ ln P IJ
H ∂x K
vap
1
1
= 0 Since the pure component vapor pressure does not depend on the mixture
P
FG ∂T IJ
H ∂x K
composition at fixed P and T since
1
=0
P
ii) Gibbs-Duhem eqn. is
FG IJ
H K
H ex ∂ T
T
∂ x1
→0
c
⇒ γ 1 P1vap
FG ∂P IJ + x FG ∂ ln γ IJ + x FG ∂ lnγ IJ
H ∂x K H ∂x K H ∂x K
L F ∂ lnγ IJ OP = 0
− γ P hM1 + x G
N H ∂x K Q
− V ex
1
P
1
1
1 P
→0
vap
2 2
=0
2
2
P
1
1
P
1
1
P
1
or γ 1 P1vap = γ 2 P2vap . From here on it is the same argument as in the text.
Alternative to proof above: start with P = x1γ 1 P1vap + x2γ 2 P2vap
FG ∂P IJ
H ∂x K
1
RSFG ∂ ln γ IJ
TH ∂ x K
IJ + FG ∂ ln P
K H ∂x
≡ 0 = γ 1 P1vap + x1γ 1 P1vap
1
P
+ x2γ 2 P2vap
RSFG ∂ lnγ
TH ∂x
1
1
2
P
vap
2
P
1
FG ∂ ln P IJ UV − γ P
H ∂x K W
IJ UV
KW
+
1
vap
1
1
vap
2 2
P
P
Now using an argument similar to (i) above, and also using (ii), gives γ 1 P1vap = γ 2 P2vap
8.1-4
In general, we have xiγ i Pi vap = yi P and
know
vap
PET
(a) Ideal
= 02321
.
bar and
solution:
PBvap
∑ xiγ i Pi vap = P .
Also, from the experimental data, we
= 0.2939 bar.
xi Pi vap = yi P
P = xET × 0.2321 + xE = 0.2939 or
xET × 0.2321
and
P
P = 0.2321 − 0.0617 xET . Consequently x − y and P − x
and
diagrams are given on following page.
∑ xi Pivap = P .
Thus
yET =
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
(b)
Regular solution model:
δi
V Li
ethanol
12.5 58.4
benzene
9.2
Ethanol solubility parameters at
89.0
25° C was computed using ∆H vap
ET = 9674
d
vap
vap
L
∆U vap
ET = ∆ H ET − RT and δET = ∆U ET V ET
φ2BV 2ET δET − δ B
cal/mol
i
12
2
= 1.0059φ2B and ln γ B = 15329
.
φ 2ET . These activity coefficient
RT
expressions are used with the general equations above to obtain the solution. The results are
given below.
(c) The program UNIFAC was used to obtain the predictions shown in the figures below.
(d) First we evaluate the activity coefficients at the given data point using γ i = yi P xi Pi vap to
Thus ln γ ET =
obtain γ ET = 1.2244 and γ B = 2.0166 . Next using eqns. (7.5-10) we obtain α = 2.0271 and
β = 1.4993 . {This is to be compared with α = 195
. and β = 116
.
in Table 7.5-1]. Thus we
obtain
RS
2.0271
1
+
13520
.
x a1 − x
T
R
14993
.
= expS
T 1 + 0.7396a1 − x f x
γ ET = exp
ET
and
γB
ET
ET
UV
fW
UV
W
2
2
ET
Using these expressions in the general equations we obtain the results plotted below.
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-5 (a) Using the vapor pressure data (Plotting
vap
PAC
≈ 0.6665 bar and
ln P vap vs 1 T ), I find that at T = 1054
. °C,
PTvap ≈ 08793
.
bar.
a
f
γ AC = 1013
.
0.6665 = 1520
.
Thus
and
γ T = 1.013 0.8793 = 1152
.
at the azeotropic composition of xT = 0.627 . Next, using eqns. (7.510) and treating toluene as species 1, I find α = 1075
.
and β = 1.029 . Thus,
ln γ T =
1075
.
1 + 1045
.
and ln γ AC =
xT 2
xM
1029
.
1 + 0.957
xAC 2
xT
.
These expressions have been used to obtain the results plotted below.
(b) fi L = fi V ⇒ xi γ i Pi vap = yi P . Thus, γ i = xi γ i Pi vap P and ∑ yi = 1 . Procedure I used was, for
each value of xT , to
i) Guess an equilibrium temperature T
ii) Compute yT and yAC , and check to see if
∑ yi = 1
iii) If not, guess a new value of T and repeat the calculation
A simpler procedure is to use Mathcad or another computer algebra program
Results
xT
yT
T (° C)
Experiment
0.25 0.50
0.75
0.25
0.43
0.25
Ideal
0.50
0.75
0.30
0.56
0.80
104.5 100.8 100.6 107.5 105.8 105.8 116.5
0.43
114
111
0.57
0.69
van Laar
0.50 0.75
0.57
0.70
Ideal solution results were obtained in a similar matter, except that all activity coefficients were
set equal to unity.
8.1-5
0 , 1 .. 20
i
0.05 . i
xi
Write van Laar model this way to avoid division by zero.
1.075 . 1
gamt ( i)
exp
1
xi
xi
1.029 . xi
2
1.045 . xi
gamac( i)
2
exp
xi
0.957 . 1
3
xi
2
1.5
gamt ( i )
gamac ( i )
2
ln( gamt ( i ) )
2
1
ln( gamac ( i ) )
0.5
1
0
0.5
x
i
1
0
0
0.5
x
i
1
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
Pi
xi . gamt ( i) . 0.8793
xi . 0.8793
Pid i
1
1
xi . gamac( i) . 0.6665
xi . gamt ( i) .
yi
xi . 0.6665
yid i
xi .
0.8793
Pi
0.8793
Pid i
1
y
P
i
yid
0.5
Pid
i
1
i
i
0
0.8
0.6
0
0.5
x
i
1
8.1-6 (a) We start with eqn. (6.2-12b):
0
∑ xidGi + SdT − VdP = 0 .
a ff
dGi = RTd ln f i = RTd ln xiγ i fi T1 P ,
so that
RT ∑ xi d ln xi + RT ∑ xi d lnγ i + RT∑ xi d ln fi −VdP = 0
However, for the pure fluid fugacity, we have, from eqn. (7.2-8a)
RTd ln fi = d G i = V i dP
Thus
RT ∑ xi d ln xi + RT ∑ xi d lnγ i +
b∑ x V −V gdP = 0
i
i
Also
∑ xiV i − V = ∑ xiV i − ∑ xiV i = −∑ xiV i
= −V ex ⇒ RT ∑ xi d lna xiγ i f − V exdP = 0
ex
Now assuming
i) Ideal gas-phase behavior: xiγ i Pi vap = yi P or xiγ i = yi P Pi vap and
ii) That PV ex / RT << 1 we obtain
∑ xid lnaxiγ i f = ∑ xid lnc yi P
or
h
Pi vap =
PV ex
d ln P
RT
∑ xid ln yi + ∑ xid ln P − ∑ xid ln Pivap = d PV ex
1
And note that at constant temperature
SdT = 0 and
a
0.5
x
i
i
RT d ln P
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
∑ xid ln Pi vap = 0 , since
Now noting that
Pi vap is a function of temperature only and
F PV
∑ xid ln P = d ln Pb∑ xi g = d ln P , yields ∑ xid ln yi = GH
FG
H
FG
H
IJ
K
ex
RT
IJ
K
− 1 d ln P or
IJ
K
x1
x
x
x
PV ex
dy1 + 2 dy2 = 1 − 2 dy1 =
− 1 d ln P ≈ −d ln P
y1
y2
y1 y2
RT
Since
a
y1 + y2 = 1 ,
f
dy2 = −dy1 .
a
Also
f a f
a f
a f
a y − x f dy = d ln P
y a1 − y f dx
dx
y2 = 1 − y ,
x1 1 − x1
x 1 − y1 − y1 1 − x1
x − y1
−
= 1
= 1
y1 1 − y1
y1 1 − y1
y1 1 − y1
1
1
1
1
1
1
and
x2 = 1 − x1 ,
so
1
To obtain the x − y diagram, I used the equation above in a finite difference form. Using
the argument i to denote the ith data point, the equation above becomes
y1(i) − x1(i )
⋅ y1 (i) − y1(i − 1) = ln P(i) − ln P(i − 1)
y1 (i) 1 − y1 (i)
a
f
y1(i ) is unknown, however, P1(i ) , P1(i − 1) , x1 (i) are known. Also y1(1) is either 0 or 1
depending on which end of the data one starts with. In fact, I started at both ends, in two
separate calculations, to check the results. I solved this problem using the equation above
rewritten as
y1(i ) =
B ± B 2 − 4C
2
where
B=
x1 (i) + y1(i − 1) + ∆ ln P
x (i) y (i − 1)
and C = 1 1
1 + ∆ ln P
1 + ∆ ln P
and averaged the results from starting at the x1 = 0 and x1 = 1 ends.
Once x1 and
γ 1 = y1 P
b.
x1 P1vap
y1 , were known, the activity coefficients were calculated from
and γ 2 = y2 P x2 P2vap . Results are given below.
CCl 4 + n-Heptane System
= exprmntl
x-y data
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
c.
Ethylene bromide + 1-nitropentane System
Azeotrope
= exprmntl
x-y data
8.1-7
A simpler solution using Mathcad is available as a Mathcad worksheet.
(a) At the bubble point we have
yi = xi Pi vap P where P = 5 bar
xET = 0.05 ; xP = 0.10 , xNB = 0.40 and xMP = 0.45 . Procedure used was
i) Guess T,
ii) Compute each yi , and the sum
guessed T is too high; if
recalculate.
Solution:
∑ yi .
∑ yi = 1 , guessed T is correct; if ∑ yi > 1 ,
∑ yi < 1 , guessed T is too low. If ∑ yi ≠ 1 , we correct T and
If
T = 29366
.
K (bubble point)
yET = 0.4167 ,
yP = 0.1730 ,
yNB = 0.1601 and yMP = 0.2502 .
(b) The dew point calculation is similar. Here, yET = 0.05 , yP = 010
. , yNB = 0.40 and yMP = 0.45 .
P = 5 bar, and T and the xi 's are the unknowns.
Thus, here we guess the dew point
temperature, compute each of the xi 's from xi = Pyi Pi
vap
guessed temperature is the dew point temperature; if
∑ xi <1 , guessed T is too high.
and evaluate
∑ xi >1 ,
∑ xi .
If
∑ xi = 1 , the
guessed T is too low; if
Solution (obtained using the computer) T = 314.23 K (dew
point) xET = 0.0039 , xP = 0.0337 , xNB = 0.5215 and xMP = 0.4409 .
(c) The advantage of the Mathcad worksheet for the isothermal flash calculation is that one can use
the initial flash equations directly, rather than having to make the substitutions below.
For the isothermal flash vaporization calculation, we proceed as in Illustration 8.1-3. First, we
calculate the K factors, i.e.
10 + −817.08 303.15 + 4 .402229
KET =
= 10185
.
,
5
and, similarly KP = 2.238 , KNB = 0.546 and KMP = 0.743 . Thus, the equations to be solved
are:
xET + xP + xNB + xMP = 1
(1)
a
f
yET + yP + yNB + yMP = 1 ⇒ 10185
. xET + 2.238 xP + 0.546 xNB + 0.743 xMP = 1
Also,
a
f
xET L 1 − KET + KET = 0.05 ⇒ xET (10185
.
− 9 .185 L) = 0.05
and, similarly
(2)
(3)
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
xP ( 2.238 − 1.238 L ) = 0.10
(4)
xNB (0.546 + 0.454 L ) = 0.40
(5)
xMP (0.743 + 0.257 L ) = 0.45
(6)
Solution procedure I used was to guess L, compute the xi 's from eqns. (3 to 6), and then
ascertain whether eqns. (1) and (2) were satisfied. After a number of iterations, I obtained the
following solution:
L = 0.86667
V = 0.13333
xET = 0.0225
yET = 0.2289
xP = 0.0858
yP = 0.1921
xNB = 0.4258
yNB = 0.2326
xMP = 0.4659
yMP = 0.3464
∑ xi = 1.000
∑ yi = 1.000
(d) For an adiabatic flush vaporization, shown below, the energy balance must also be satisfied
liquid
liquid
X
vapor
pressure reducing valve or device
This is a (two-phase) Joule-Thomson expansion, so that the energy balance yields H in = H out ,
or
L
L
∑ xi H i (T , P, x ) inlet cond itions = L ∑ xi H i (T , P , x) outlet liquid conditions +
V
∑ yi H i cT , P, yh outlet vapor conditi ons
V
This equation must be satisfied, together with the mass balances and phase equilibrium
equations of part c. Thus, we have one new unknown here, the outlet temperature, and an
additional equation from which to find that unknown.
8.1-8 (a) Starting from xiγ i Pi vap = yi P , we obtain γ i = yi P xi Pi vap , and using the data in the problem
a
statement, we can compute each γ i , and then ln γ P γ A
These results, together with G
a
ex
f and G
ex
RT = xP ln γ P + xA lnγ A .
RT xP xA and its inverse are tabulated on the following page.
f
Also, there is a plot of ln γ P γ A vs. xA . This plot indicates that the data appears to be
thermodynamically consistent (i.e.
z
a
f
ln γ P γ A dxA ≈ 0 ), though the points at the composition
extremes ( xA = 0.021 and xA = 0.953 ) look suspect
(b) See Problem 7.22 the plots of G ex RT xP xA and xP xA G ex RT appear on the following
page. The fact that neither is linear indicates that neither the two-constant Margules, nor the van
Laar equation will accurately fit the data. Hence, one will have to use at least a 3-constant
Redlich-Kister expansion for the Gibbs free energy to obtain a good fit of the experimental data
for this system!
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
γP
γA
0.021 0.979
3.3407
13140
.
0.061 0.939
3.6497
10643
.
0134
.
31326
.
11142
.
0.210 0.790 2.5617
11700
.
0.292 0.708 2.2196 1.2187
xA
xP
FG γ IJ
Hγ K
G ex
RT
0.2927
G es
RTxA xP
14.237
xA xP RT
G ex
0.0702
12323
.
01367
.
2.3866
0.4190
1.0337
0.2467
2.1259
0.4704
0.7837
0.3216
1.9385
0.5159
0.5995
0.3729
18037
.
0.5544
0.3000
ln
P
A
0.405
0866
.
0.4198
17421
.
0.5740
0.503 0.497
15132
.
15582
.
− 0.0293 0.4288
1.7153
0.5830
0.611 0.389
13299
.
18214
.
− 0.3145 0.4074
17141
.
0.5834
0.728 0.272
11689
.
2 .3716 − 0.7075 0.3485
1.7600
0.5682
3.1505 − 10948
.
01962
.
1.7235
0.5802
10191
.
4 .5509 − 14964
.
0.0893
1.9937
0.5016
0.869
0.595 18190
.
0.131 10542
.
0.953 0.047
13475
.
0.9331
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
The next step is to fit parameters in the Gibbs free energy models to the experimental data. I
have done this assuming small errors in all the variables (T, P, x and y) and using the maximum
likelihood method. The results of the different models are given below:
Wilson model Λ12 = 347.82
Λ21 = 107523
.
Pmeas
(hPa)
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
Pcalc
(hPa)
1008.46
1012.01
1012.35
1012.89
1012.21
1008.83
1024.86
1007.84
1011.79
1010.94
1012.31
Tmeas
(o C)
49.15
45.76
39.58
36.67
34.35
32.85
33.35
31.97
31.93
32.27
33.89
Tcalc
(o C)
49.17
45.76
39.58
36.67
34.35
32.87
33.31
31.99
31.93
32.28
33.89
x1,meas
x1,calc
y1,meas
y1,calc
.0210
.0610
.1340
.2100
.2920
.4050
.5030
.6110
.7280
.8690
.9530
.0317
.0556
.1273
.1942
.2958
.4290
.4291
.6224
.7287
.8640
.9554
.1080
· 3070
.4750
.5500
.6140
.6640
.6780
.7110
.7390
.8100
.9060
.2353
.3404
.5048
.5761
.6330
.6731
.6725
.7128
.7380
.7957
.8933
1013 hPa = 1013
.
bar = 1013
. × 105 Pa
The sum of squares of weighted residuals for this model is 1298, the mean deviation in y1 is
2.55%, and in P is 0.29%
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
NRTL model
Pmeas
(hPa)
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
Pcalc
(hPa)
1008.14
1011.69
1011.87
1012.35
1012.18
1010.35
1025.99
1007.37
1010.33
1009.50
1011.83
τ12 = 777.95
Tmeas
(o C)
49.15
45.76
39.58
36.67
34.35
32.85
33.35
31.97
31.93
32.27
33.89
Tcalc
(o C)
49.17
45.76
39.58
36.67
34.35
32.86
33.30
31.99
31.94
32.28
33.89
τ21 = 432 .53
x1,meas
x1,calc
y1,meas
y1,calc
.0210
.0610
.1340
.2100
.2920
· 4050
.5030
.6110
.7280
.8690
.9530
.0379
.0650
.1399
.2034
.2939
.4186
.4173
.6208
.7296
.8581
.9505
.1080
.3070
.4750
.5500
.6140
.6640
.6780
·7110
.7390
.8100
.9060
.2400
.3469
.5131
.5843
.6403
.6793
.6784
.7131
.7348
.7896
.8890
Sum of squares of weighted residuals =1547
Mean deviation in y1 is 3.00%; in P = 0.34%
UNIQUAC Model
The parameters are τ12 = 572.61 and τ21 = –72.84
Mean deviation in y1 is 3.0790; in P is 0.34%
Pmeas
(hPa)
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
Pcalc
(hPa)
1008.12
1011.66
1011.82
1012.26
1012.10
1010.40
1026.29
1007.25
1010.16
1009.43
1011.80
Tmeas
(o C)
49.15
45.76
39.58
36.67
34.35
32.85
33.35
31-97
31.93
32.27
33.89
Tcalc
(o C)
49.17
45.76
39.58
36.67
34.35
32.86
33.30
31.99
31.94
32.28
33.89
x1,meas
x1,calc
y1,meas
y1,calc
.0210
.0610
·1340
.2100
·2920
.4050
.5030
·6110
.7280
.8690
.9530
.0381
.0655
.1412
.2048
.2946
.4180
.4177
.6206
.7295
.8579
.9503
.1080
.3070
.4750
.5500
.6140
.6640
.6780
.7110
.7390
.8100
.9060
.2402
.3474
.5141
.5855
.6417
.6807
.6800
.7139
.7351
.7895
.8888
van Laar modelα = 15032
.
β = 18534
.
Mean % P = 0.34 ; y = 318%
.
sum of squares of wt. residuals =1625
Pmeas
(hPa)
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
Pcalc
(hPa)
1008.13
1011.66
1011.78
1012.19
1012.06
1010.66
1025.98
1007.14
1009.92
1009.57
1011.89
Tmeas
(o C)
49.15
45.76
39.58
36.67
34.35
32.85
33.35
31.97
31.93
32.27
33.89
Tcalc
(o C)
49.17
45.76
39.58
36.67
34.35
32.86
33.30
31.99
31.94
32.28
33.89
x1,meas
x1,calc
y1,meas
y1,calc
.0210
.0610
.1340
.2100
.2920
.4050
.5030
.6110
.7280
.8690
.9530
.0380
.0656
.1419
.2057
.2947
.4164
.4121
.6202
.7293
.8585
.9512
.1080
.3070
.4750
.5500
.6140
.6640
.6780
.7110
.7390
.8100
.9060
.2403
.3478
.5154
.5874
.6441
.6837
.6822
.7162
.7362
.7897
.8894
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
2-Constant Margules Model
A = 17737
.
B = 19259
.
Mean % deviation in P = 0.33% and in y = 2.90% , and sum of squares of weighted residuals =1401
Pmeas
(hPa)
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
1013.00
Pcalc
(hPa)
1008.60
1012.14
1012.66
1013.56
1013.81
1004.67
1023.95
1005.18
1014.78
1013.07
1012.36
Tmeas
(o C)
49.15
45.76
39.58
36.67
34.35
32.85
33.35
31.97
31.93
32.27
33.89
Tcalc
(o C)
49.17
45.76
39.58
36.67
34.35
32.88
33.31
32.00
31.92
32.27
33.89
x1,meas
x1,calc
y1,meas
y1,calc
.0210
.0610
.1340
.2100
.2920
.4050
.5030
.6110
.7280
.8690
.9530
.0287
.0502
.1135
.1731
.2764
.4396
.4598
.6465
.7317
.8863
.9578
.1080
.3070
.4750
.5500
.6140
.6640
.6780
.7110
.7390
.8100
.9060
.2337
.3382
.5022
.5736
.6314
.6653
.6670
.6899
.7082
.7932
.8928
So, of the models considered here, the Wilson model provides the best description (of the twoconstant models) for this data set.
8.1-9
Using the program UNIFAC taking T and xP as known, and computing γ P , γ A and P as well as
yP we obtain
xP
T (° C)
γ calc
P
γ calc
A
0.021
49.15
4.1390
1006
.
0.061
45.76
3.7515
P calc =
∑xγP
yPcalc
yexp
P
0.922
0.147
0.108
10053
.
0.983
0.325 0.307
i i i
( bar )
vap
0134
.
39.58
31680
.
10258
.
0.976
0.498
0.475
0.210
36.67
2 .6687 1.0649
0.995
0.583
0.550
0.292
34.35
2 .2487
11302
.
0.993
0.635 0.614
0.405
32.85
18162
.
1.2697
0.993
0.677 0.664
0.503
33.35
15416
.
14540
.
1004
.
0.697
0.678
0.611
3197
.
13247
.
17663
.
1002
.
0.717
0.711
0.728
3193
.
11592
.
2 .3176
1001
.
0.742 0.739
0.869
32.27
10388
.
3.5893
1.005
0.804
0.810
0.953
3389
.
10053
.
4.9909
1.018
0.897
0.906
A
measured
value =1013
.
bar
Solutions to Chemical and Engineering Thermodynamics, 3e
While the prediction is not perfect, it is relatively good.
Section 8.1
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-10
First we should check to see if this problem is soluble (i.e.., well-posed in the sense of the Gibbs
phase rule). The Gibbs phase rule, eqn. (6.9-6) is F = C − M − P + 2 . Thus here we have
F = 3 − 0 − 2 + 2 = 3 degrees of freedom. Since the temperature, and two independent liquidphase mole fractions are specified, the problem is well posed.
For the solution of this problem, the following subscripts will be used: 1 = ethanol ,
2 = benzene and 3 = ethyl acetate . As the first step, compute the pure component vapor
pressures. This is done by using the vapor pressure data in the “Chemical Engineers' Handbook ”,
making plots of ln Pi vap vs 1 T , and then determining ln Pi vap (and hence Pi vap ) at T = 78° C
c1 T = 2.847 × 10 Kh we find
−3
P1vap ≅ 10
. bar; P2vap ≅ 0.9666 bar and P3vap ≅ 1053
.
bar
Next, we need compute the liquid phase activity coefficients. This will be done using the ternary
van Laar eqn. (eqn. (A7.3-2)). [See also Problem 7.8c], together with the entries in Table 7.6-1.
Here one has to be careful about the order in which the species appear in the table. I obtained the
following:
α12 = 1.946
β12 = 1610
.
⇒
α21 = β12 = 1.610
α32 = 115
.
β32 = 0.92
⇒ α23 = β32 = 0.92
α31 = 0.896
β31 = 0.896 ⇒
β21 = α12 = 1946
.
β23 = α32 = 115
.
α13 = β31 = 0.896
β13 = α31 = 0.896
Now from eqn. (A7.3-2)
ln γ 1
{x α e j
=
2
2
β 12 2
12 α 12
e j
eα + α − α j}
x + x e j+ x e j
+ 1730
.
x x r
0.6333
=
= 0.7308
+ x32α13
β 13 2
α 13
1
=
.
x
m13320
2
2
+ 0.896 x32
β
+ x2 x3 α12
12
β 12
2 α 12
β 13
α 13
β 13
3 α 13
12
13
α 12
23 β 12
2
2 3
x1 + 0.8273x2 + x3
2
0.8666
x2 = x3 = 0.4
x1 = 0.2
⇒ γ 1 = 2.0767
To obtain an expression for γ 2 , we interchange subscripts 1 and 2 in eqn. (A7.3-2) to obtain [see
solution to prob. 7.8c]
ln γ 2
{x β e j
=
2
1
α 12 2
12 β 12
+ x32α23
e j
e je jeβ + α − α j}
x +x e j+ x e j
+ 2 .7025x x r
0.5403
=
= 0.4145
β 23 2
α 23
2
=
m2.3521x
2
1
+ 14375
.
x23
+ x1 x3
α 12
1 β 12
x2 + 1.2087x1 + 1.250 x3
α 12
β 12
β 23
α 23
β 23
3 α 23
12
23
β 12
13 α 12
2
1 3
2
x2 = x3 =0 .4
x1 = 0.2
13036
.
⇒ γ 2 = 15136
.
An expression for γ 3 is obtained by interchanging indices 1 and 3 in eqn. (A7.3-2) to obtain [see
solution to Problem 7.8c]
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
ln γ 3
{x β e j
=
α 23 2
2
2 23 β 23
e je jeβ + β − β
x + x e j+ x e j
+ 0.0268x x r
0.1557
=
+ x12 β13
e j
α 13 2
β 13
α 23
2 β 23
3
=
m0.7360x
2
2
+ 0.896 x12
+ x1 x2
α 23
β 23
α 13
β 13
α 13
1 β 13
23
β 23
12 α 23
13
j}
2
1 2
x3 + 0.80 x2 + x1
2
x 2 = x3 = 0.4
x1 =0 .2
0.8464
⇒ ln γ 3 = 01840
.
and γ 3 = 12020
.
With these "preliminaries" taken care of, we can now proceed on to the solution. The equilibrium
equations are
xiγ i Pi vap = yi P and ∑ xiγ i Pi vap = P
Therefore
x1γ 1 P1vap = 0.2 × 2.0767 × 1.000 = 0.41534 bar
x2γ 2 P2vap = 0.4 × 15136
.
× 0.9666 = 0.58522
x3γ 3 P3vap = 0.4 × 12020
.
× 1053
.
= 0.50620
P = 150684
.
bar
and
y1 = 0.2756
y2 = 0.3884
y3 = 0.3360
Note that the vapor composition is only very slightly different than the liquid composition. (This
is because the vapor pressures and activity coefficients of the species are all quite similar).
An alternative is to use the program UNIFAC to estimate the activity coefficients. Using
the program with ethanol 1 − CH 3 , 1 − CH 2 , 1 − OH , benzene (6–ACH) and ethyl acetate
a
a
f
f
1 − CH 3 , 1 − CH 2 , 1 − CH 3COO we obtain, at 20 mole % ethanol, 40 mole % benzene and 40
mole % ethyl acetate at 78°C ) that
γ 1 = 2.2062 ; γ 2 = 11931
.
and γ 3 = 10038
.
The solution is P = 12965
.
bar and
y1 = 0.3181 ; y2 = 0.3558 and y3 = 0.3261
Clearly this result is different from the ternary van Laar prediction. In the absence of experimental
data for this ternary mixture, it is difficult to say which model is better.
8.1-11
For the simpler models, it is possible to show by simple mathematics that the model either does or
does not permit a double azeotrope. For example, the van Laar model is
a
f
G ex 2a12 x1q1 x2 q 2 2 a12
αβx1x2
αβx1 1 − x1
=
×
=
=
RT
x1q1 + x2 q2
2 a12 αx1 + βx2 αx1 + β 1 − x1
a
f
Now for the benzene-hexafluorobenzene system G ex has an interior maximum and an interior
minimum. That is, dGex dx1 is zero twice in the region 0 ≤ x1 ≤ 1 . To see if the van Laar model
permits this we examine
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
FG IJ
a f
a f
H K a a ff
a f
a f
αβa1 − 2 x f
αβx a1 − x f(α − β)
−
=0
αx + βa1 − x f
dx + βa1 − x f
⇒ a x − x faαx + βx f − x x (α − β) = 0
d G ex
αβ 1 − x1
αβx1( −1)
αβx1 1 − x1
=
+
−
dx RT
αx1 + β 1 − x1
αx1 + β 1 − x1
αx1 + β 1 − x1
or
1
1
1
1
αx1 x2 − αx12
(α − β) = 0
1
2
1
2
2
1
1
2
− βx1 x2 +
or
αx12 = βx22 ⇒
1
1 2
βx22
FG IJ
H K
α
x
= 2
β
x1
− αx1 x2 + βx1 x2 = 0
2
or
x2
α
=±
x1
β
Now α and β must be of same sign (otherwise we get the square root of a negative number).
Also, since 0 ≤ x1 ≤ 1 and 0 ≤ x2 ≤ 1 , only positive sign is allowed. Thus x2 x1 = α β when
dGex dx1 = 0 . And only an interior maximum (if α > 0 and β > 0 ) or an interior minimum (if
α < 0 and β < 0 ) can occur, but not both! Therefore, van Laar model can not describe the
observed behavior.
Similarly, obviously the one-constant Margules model Gex = Ax1 x2 can not give both an interior
minimum & maximum, so it can not describe observed behavior.
Instead of continuing this extreme argument, we will look at the results of merely fitting the
experimental data.
Two-constant-Margules model
Pmeas
(hPa)
521.60
525.70
525.68
522.87
518.18
509.89
507.73
503.50
499.74
497.57
497.94
501.55
Pcalc
(hPa)
521.60
518.42
517.53
517.19
516.24
514.09
511.32
507.72
503.06
498.70
496.96
501.55
x1,meas
x1,calc
y1,meas
y1,calc
.0000
.0941
.1849
.2741
.3648
.4538
.5266
.6013
.6894
.7852
.8960
1.0000
.0000
.0940
.1849
.2741
.3648
.4539
.5268
.6015
.6896
.7852
.8960
1.0000
.0000
.0970
.1788
.2567
.3383
.4237
.4982
.5783
.6760
.7824
.8996
1.0000
.0000
.0880
.1777
.2679
.3605
.4522
.5275
.6051
.6970
.7960
.9063
1.0000
- only 1 azeotrope at x1 ≥ 0.9
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
Wilson model
Pmeas
(hPa)
521.60
525.70
525.68
522.87
518.18
509.89
507.73
503.50
499.74
497.57
497.94
501.55
Pcalc
(hPa)
521.60
510.55
500.76
492.21
484.89
479.35
476.28
474.74
475.30
479.24
488.48
501.55
y1,meas
y1,calc
.0000
.0970
.1788
.2567
.3383
.4237
.4982
.5783
.6760
.7824
.8996
1.0000
.0000
.0757
.1556
.2407
.3335
.4300
.5118
.5974
.6982
.8038
.9145
1.0000
Only a single azeotrope predicted to occur.
NRTL model
Pmeas
Pcalc
x1,meas
y1,meas
y1,calc
(hPa)
(hPa)
521.60
521.60
.0000
.0000
.0000
525.70
518.93
.0941
.0970
.0896
525.68
516.49
.1849
.1788
.1773
522.87
514.23
.2741
.2567
.2647
518.18
512.08
.3648
.3383
.3545
509.89
510.11
.4538
.4237
.4436
507.73
508.62
.5266
.4982
.5170
503.50
507.19
.6013
.5783
.5927
499.74
505.66
.6894
.6760
.6825
497.57
504.17
.7852
.7824
.7804
497.94
502.69
.8960
.8996
.8938
501.55
501.55
1.0000
1.0000
1.0000
No azeotrope results form the least squares fitting of parameters
UNIQUAC model
Pmeas
(hPa)
521.60
525.70
525.68
522.87
518.18
509.89
507.73
503 50
499.74
497.57
497.94
501.55
Pcalc
(hPa)
521.60
528.19
526.27
521.58
515.87
510.44
506.61
503.46
500.91
499.57
499.80
501.55
x1,meas
y1,meas
y1,calc
.0000
.0941
.1849
.2741
.3648
.4538
.5266
.6013
.6894
.7852
.8960
1.0000
.0000
.0970
.1788
.2567
.3383
.4237
.4982
.5783
.6760
.7824
.8996
1.0000
.0000
.0953
.1717
.2503
.3368
.4276
.5046
.5852
.6807
.7832
.8978
1.0000
Double azeotrope predicted, as indicated.
←
←
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
Therefore, of the models considered only the UNIQUAC model is capable of producing the
peculiar behavior of G ex for this system.
8.1-12
From Table 8.1-1 we have (assuming an ideal vapor phase)
x1
γ1
γ2
log 10
FG γ IJ
Hγ K
1
2
See figure below.
0.0503 3.4337 1.0247
0.5251
01014
.
31394
.
1.0192
0.4885
01647
.
2.6218 10445
.
0.3997
0.2212 2.2340 10918
.
0.3109
0.3019 19334
.
11332
.
0.2320
0.3476 17879
.
11637
.
0.1865
0.4082 15928
.
12643
.
0.1003
0.4463 15237
.
13068
.
0.0666
0.5031 14284
.
13755
.
0.0164
0.5610 1.3225 1.4984
−0.0542
0.6812
11841
.
1.7837
− 01779
.
0.7597
11285
.
2 .0086
−0.2504
0.8333 10648
.
2 .4539
− 0.3625
0.9180 1.0223
31792
.
−0.4927
Solutions to Chemical and Engineering Thermodynamics, 3e
The two areas I and II appear to be approximately
equal, so the data will be presumed to be
thermodynamically consistent. In fact, from proper
numerical analysis, we find the data to be consistent.
Log
Section 8.1
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-13
P = x1γ 1 P1vap + x2γ 2 P2vap
FG ∂P IJ
H ∂x K
2
∂x
∂γ 1
∂γ
+ γ 2 P2vap + x2 P2vap 2 ; where we have used that 1 = −1 as
∂x2
∂ x2
∂x2
= −γ 1 P1vap + x1 P1vap
T
∂γ 1
= 0 . So
∂ x2
x2 → 0 γ 1 → 1 and
FG ∂P IJ
H ∂x K
2
∂γ 2
∂x2
= − P1vap + γ 2 P2vap + x2 (→ 0)
T , x2 → 0
so that
F I
GH JK
P1vap + ∂P
∂x
2
γ 2 x2 → 0 =
vap
P2
a
f
T , x2 → 0
constant temperature ebulliometer
Now
P = x1γ 1 P1vap + x2γ 2 P2vap
and
FG ∂P IJ
H ∂x K
2
FG ∂γ IJ
H ∂x K
F ∂γ IJ P
+x G
H ∂x K
= 0 = −γ 1 P1vap + x1
P
P
2
+γ 2 P2vap
vap
2
2
2
P
2
as x2 → 0, γ 1 → 1 and
FG ∂P IJ FG ∂T IJ
H ∂T K H ∂x K
F ∂P IJ FG ∂T IJ
+xγ G
H ∂T K H ∂x K
vap
P1vap + x1γ 1
1
1
2
P
vap
2
2
2
T
2
P
∂γ 1
→0
∂ x2
0 = − P1vap +
FG IJ
H K
dP1vap ∂T
dT ∂ x2
a
f
+ γ 2 x2 → 0 P2vap
P
or
a
f
γ 2 x2 → 0 =
8.1-14
P1vap −
FG dP IJ FG ∂T IJ
H dT K H ∂x2 K
vap
1
P, x2 → 0
P2vap
constant pressure ebulliometer
(also available as a Mathcad worksheet)
Clearly many different thermodynamic models can be used. We will use the van Laar model
ln γ 1 =
α
αx
1 + βx 1
2
2
and
ln γ 2 =
β
1+
β x2 2
α x1
which gives lnγ ∞1 = α and lnγ ∞2 = β . Using the data in the problem statement
α = ln (1.6931) = 0.5266 and β = ln(19523
.
) = 0.6690 . Using these parameter values in the activity
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
coefficient equations above, together with Pi = xiγ i Pi vap ; P = P1 + P2 and yi = Pi P gives, at
x1 = 0.2
x2 = 0.2
y1 = 0.4483
y2 = 0.5517 and P = 06482
.
bar
Also
8.1-15
x2 = 0.500
y1 = 0.7036
y2 = 0.2946
P = 0.8431
0.700
0.8118
01882
.
0.9262
0.850
0.8943
01057
.
0.9732
0.900
0.9256
0.0744
0.9861
0.950
0.9604
0.0396
0.9972
0.975
0.9795
0.0205
1.0019
(also available as a Mathcad worksheet)
Using Mathcad I obtained the following results
T = 300 K
KEOH− EAC
T = 400 K
KEOH− EAC
xEOH = 01
.
58340
.
xEOH = 01
.
15.318
0.5
0.4255
0.5
11172
.
0.9
0.03103
0.9
0.0815
Thus the results exhibit strong composition and temperature dependence. For an ideal solution
y
P vap
y x
Pvap
xi Pi vap = yi P ⇒ i = i ⇒ Kij = i i = i vap
xi
P
yj xj
Pj
Thus, for an ideal solution, the relative volatility Kij has no composition dependence, but can be
dependent on temperature (unless, fortuitously, Pi vap and Pjvap have the same temperature
dependence, that is, ∆ H ivap = ∆ H vap
j ). The composition dependence arises from the non-ideal
solution behavior. Since the activity coefficients dependent on temperature, nonideal solution
behavior also contributes to the temperature dependence of the relative volatility.
8.1-16
This system was used for illustration in the first edition. The figures which appear below are from
that source. [I changed to the hexane-triethylamine system since the x and y were too close in the
benzene-ethylene chloride system because the pure component vapor pressures are so close.]
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-17
x1γ 1 P1vap
FfI=yP
H PK
1
a f P = γ P a f Pf = P e a f Pf
a f P γ P a f Pf P e a f P f
P
i a f Pf = P e d a f i a f Pf
=
e d
P
a f Pf P
a f Pf
P
a 24f 3 a f Pf
=
e14
P
a4f P43f
;
12
(a) α21 =
y2 x2 γ 2 P2vap f P
=
y1 x1 γ 1 P1vap f P
vap
2
vap
1
A x12 − x 22
vap
2
vap
1
vap
2 2
vap
1 1
2
1
RT
2
1
− A 1 − 2 x1 RT
vap
2
vap
1
2
Composition
1
T dependenceand T dependence T and P
dependence
(b) Ideal mixture at low pressure
f
A = 0 above and also all
=1
P
F I
H K
α21 =
P2vap (T )
P1vap (T )
1
424
3
T dependence
A
vap Ax12 RT
2
vap Ax22 RT
1
2
1
x12 −
1− x1
2
RT
2
1
2
1
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-18 a) Starting from
L
L
f i = f i we get, for the ideal solution, that xi Pi vap = yi P
Now adding such equations for both components, we get
x1 P1vap + x2 P2vap = P so that
xi Pi vap
;
x1 P1vap + x2 P2vap
yi =
a
f
yi x1 = 0.5 =
In a equimolar mixture
Pi vap
P1 + P2vap
vap
b) For the nonideal mixture
xiγ i Pi vap = yi P
Now adding these equations for both components, we get
x1γ 1 P1vap + x2 γ 2 P2vap = P so that
xi γ i Pi vap
yi =
x1γ 1 P1vap + x2 γ 2 P2vap
;
For the one - constant Margules model
a
f
xi Pi vap exp( a 1 − xi 2 )
yi =
x1 P1vap exp( ax22 ) + x2 P2vap exp( ax12 )
In a equimolar mixture with the
a
0.5 Pi vap exp( a ( 05
.) )
Pi vap
=
2
2
vap
vap
0.5 P1vap exp(a ( 0.5) ) + 0.5 P2vap exp( a (0.5) ) P1 + P2
which is exactly the same result as for the ideal solution. However, these two
different models only give the same vapor-phase composition in an equimolar
mixture. However, even in this case, the pressures for the ideal and oneconstant Margules mixtures are different.
f
2
yi x1 = 0.5 =
8.1-19
G
ex
= Ax1 x2 ⇒ γ1 = exp
FG Ax IJ ; γ
H RT K
2
2
2
FG Ax IJ
H RT K
= exp
2
1
x1γ 1 P1vap = y1 P
Azeotrope x1 = y1 ⇒ γ1 =
P
P1vap
2
γ2
P P vap P vap e Ax1
= vap ⋅ 1 = 1vap =
2
γ1
P
P2
P2
e Ax2
ln
c
h
RT
RT
d a
=e
d
A x12 − x 22
P1vap
A 2
A 2
=
x1 − x22 =
x1 − 1 − x1
vap
P2
RT
RT
a
i RT
f i = RTA cx
2
2
1
− 1 + 2 x1 − x12
h
f
A
P vap
RT
P vap
RT
Pvap
2 x1 − 1 = ln 1vap ⇒ A =
ln 1vap =
ln 2vap
RT
P2
2 x1 − 1 P2
1 − 2 x1 P1
So for an azeotrope to form
(1) If P2vap > P1vap
Azeotrope will form near x1 = 0 if A = RT ln
P2vap
P1vap
Azeotrope will form at x1+ = 05
. if A = ∞ ; at x1− = 05
. if A = −∞ .
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
Azeotrope will form near x1 = 1 if A = − RT ln
FG
H
FG
H
1
RT
P vap
1−
ln 2vap
2
A
P1
Or in general x1 =
IJ IJ
KK
P2vap
P1vap
So we can draw figures of regions in which azeotropes can be expected to form.
RT
P vap
(2) If P1vap > P2vap then A =
ln 1vap
1 − 2 x1 P2
Mirror image of point 1
2 Ax1 1 − x1
RT
(b) T =
; A=
at x1 = 0.5 ; A = 2RT
R
2 x1 1 − x1
a
8.1-20
f
a
f
(also available as a Mathcad worksheet)
8.1-20
T
69
p5
26799
10.422
p6
p5
exp ( p5 )
p5 = 2.721
8.314 . ( 273.15
T)
p6
exp ( p6 )
p6 = 1.024
35200
.
8.314 ( 273.15
T)
p7
exp ( p7 )
p7 = 0.389
29676
10.456
p7
T)
8.314 . ( 273.15
11.431
x5
0.25
x6
P
x5 . p5
x6 . p6
0.45
x7
x7 . p7
( x5 . p5 )
y5
0.3
P = 1.258
y6
( x6 . p6 )
P
y7
( x7 . p7 )
P
P
P = 1.258
Bubble point pressure
Bubble point compositions
y5 = 0.541
y6 = 0.366
y7 = 0.093
Now on to dew point calculation
Initial guesses
z5
0.25
GIVEN
x5 x6
soln
x5
P
z6
1
0.45
x5
z7
x5 . p5 z5. P
x7 1
0.1
x6
0.6
x7
0.3
0.3
x6 . p6 z6. P
x7 . p7 z7. P
FIND( x5 , x6 , x7 , P )
soln 0
x6
soln 1
Dew point pressure
Dew point compositions
x7
soln 2
P
soln 3
P = 0.768
x5 = 0.071
x6 = 0.338
x7 = 0.592
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-21
(also available as a Mathcad worksheet)
8.1-21
Solving for the bubble point pressure
T
69
P
p5( T )
26799
exp 10.422
p6( T )
8.314 . ( 273.15
8.314 . ( 273.15
y6
T)
T)
K5( T , P )
8.314 . ( 273.15
0.33
y7
T)
0.33
z5
p5( T )
P
p6( T ) = 1.024
K6( T , P )
p6( T )
p7( T ) = 0.389
K7( T , P )
p7( T )
35200
exp 11.431
0.33
p5( T ) = 2.721
29676
exp 10.456
p7( T )
y5
1.013
0.25
z6
0.45
z7
P
P
0.3
GIVEN
K5( T , P ) . z5
soln
y5
K6( T , P ) . z6
K7( T , P ) . z7 1
y5 K5( T , P ) . z5
y6 K6( T , P ) . z6
y7 K7( T , P ) . z7
FIND( y5 , y6 , y7 , P )
soln 0
y6
y5 = 0.541
soln 1
y7
soln 2
y6 = 0.366
P
soln 3
y7 = 0.093
P = 1.258
This is the bubble-point pressure solution. Now on to the dew-point pressure problem.
x5
0.33
x6
0.33
x7
Note that xi=yi/Ki
0.33
GIVEN
z5
z6
z7
K5( T , P )
K6( T , P )
K7( T , P )
soln
x5
1
x5
z5
K5( T , P )
x6
z6
K6( T , P )
x7
z7
K7( T , P )
FIND( x5 , x6 , x7 , P )
soln 0
x5 = 0.071
x6
soln 1
x7
x6 = 0.338
soln 2
P
x7 = 0.592
soln 3
P = 0.768
This is the dew-point pressure solution.
So for a mixture of the composition z5=0.25, z6=0.45 and z7=0.30, at a temperature of 69 C, the
mixture will be all liquid at pressures above 1.258 bar, and all vapor at pressures below 0.768 bar.
Vapor-liquid equilibrium will exist at this temperature only between 0.768 and 1.258 bar, so this is
the pressure range we will examine.
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
T
69
L
0.99
P
1.2
K5( T , P ) . x5
GIVEN
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
x5
soln 0
x6
x5 = 0.223
y5
K5( T , P ) . x5
P
soln 1
y6
L
L
K6( T , P ) . x6 y7
K6( T , P ) . x6
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x6
x5 = 0.18
soln 1
x7
x6 = 0.458
K5( T , P ) . x5
y6
y5 = 0.445
soln 2
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x5 = 0.141
K5( T , P ) . x5
y5 = 0.383
L = 0.906
K7( T , P ) . x7
soln
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
V
1
L
V = 0.264
L = 0.736
K7( T , P ) . x7
0.60
K6( T , P ) . x6
x5
L
y7 = 0.128
K5( T , P ) . x5
GIVEN
1
V = 0.094
x7 = 0.362
K6( T , P ) . x6 y7
y6 = 0.427
L
V
0.80
K5( T , P ) ) z5
1.0
x7 ) 0
y7 = 0.104
K5( T , P ) )
y5
x6
K7( T , P ) . x7
x5 . ( L. ( 1
x5
( x5
FIND( x5 , x6 , x7 , L)
soln 3
x7 = 0.32
K5( T , P ) . x5
GIVEN
K7( T , P ) . x7
soln
soln 2
y6 = 0.389
1.1
y5
x7
x6 = 0.456
y5 = 0.507
P
K6( T , P ) . x6
x6
soln 1
x6 = 0.445
y6
x7
soln 2
soln
L
soln 3
x7 = 0.414
K6( T , P ) . x6 y7
y6 = 0.456
K7( T , P ) . x7
K7( T , P ) . x7
y7 = 0.161
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
V
1
V = 0.451
L
L = 0.549
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
P
0.9
L
0.40
K5( T , P ) . x5
K6( T , P ) . x6
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
GIVEN
x5
soln 0
x6
x5 = 0.107
y6
y5 = 0.323
P
0.8
L
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x6
soln 1
x7
x6 = 0.359
K5( T , P ) . x5
y6
y5 = 0.267
0.77
soln 2
K6( T , P ) . x6 y7
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x6
x5 = 0.071
y6
y5 = 0.251
1.25
soln 1
x7
x6 = 0.339
K5( T , P ) . x5
y5
soln 2
K6( T , P ) . x6 y7
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x5 = 0.246
K5( T , P ) . x5
y5 = 0.536
K7( T , P ) . x7
K7( T , P ) . x7
soln
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
V
1
L
V = 0.91
L = 0.09
K7( T , P ) . x7
K7( T , P ) . x7
soln
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
V
1
L
V = 0.995
L = 5.309 10
K7( T , P ) . x7
0.95
K6( T , P ) . x6
x5
L = 0.337
y7 = 0.298
K5( T , P ) . x5
GIVEN
L
V = 0.663
x7 = 0.59
y6 = 0.451
L
1
0.10
K6( T , P ) . x6
x5
V
y7 = 0.274
K5( T , P ) . x5
GIVEN
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
x7 = 0.563
y6 = 0.459
L
x6
y7 = 0.208
K6( T , P ) . x6
y5
L
( x5
0.20
x5 = 0.078
y5
K6( T , P ) . x6 y7
K5( T , P ) . x5
x5
P
soln 2
soln
x7 = 0.481
y6 = 0.469
GIVEN
P
x7
x6 = 0.412
K5( T , P ) . x5
y5
soln 1
K7( T , P ) . x7
x6
soln 1
x6 = 0.451
y6
x7
soln 2
soln
L
soln 3
x7 = 0.303
K6( T , P ) . x6 y7
y6 = 0.369
K7( T , P ) . x7
K7( T , P ) . x7
y7 = 0.094
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
V
1
V = 0.013
L
L = 0.987
3
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
i
0 , 1 .. 9
PP
8.1-22
0.768
0.071
0.338
0.592
0
.77
0.071
0.339
0.590
0.0053
.8
0.078
0.359
0.563
0.090
.9
0.107
0.412
0.481
0.337
1.0
xx5
0.141
xx6
0.445
xx7
0.414
LL
0.549
1.1
0.180
0.458
0.362
0.736
1.2
0.223
0.456
0.320
0.906
1.25
0.246
0.451
0.303
0.987
1.258
0.250
0.450
0.300
1.0
(also available as a Mathcad worksheet)
8.1-22
Solving for the bubble point temperature
T
69
p5( T )
p6( T )
p7( T )
P
1.013
exp 10.422
exp 10.456
exp 11.431
26799
.
8.314 ( 273.15
T)
29676
.
8.314 ( 273.15
T)
35200
.
8.314 ( 273.15
T)
p5( T ) = 2.721
K5( T , P )
p5( T )
P
p6( T ) = 1.024
K6( T , P )
p6( T )
p7( T ) = 0.389
K7( T , P )
p7( T )
P
P
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
y5
0.33
y6
0.33
y7
0.33
z5
0.25
z6
0.45
z7
0.3
GIVEN
K5( T , P ) . z5
K7( T , P ) . z7 1
y5 K5( T , P ) . z5
y6 K6( T , P ) . z6
y7 K7( T , P ) . z7
FIND( y5 , y6 , y7 , T )
soln
y5
K6( T , P ) . z6
soln 0
y6
soln 1
y5 = 0.548
y7
soln 2
y6 = 0.363
T
soln 3
y7 = 0.088
T = 61.788
This is the bubble-point temperature solution. Now on to the dew-point temperature problem.
x5
0.33
x6
0.33
x7
Note that xi=yi/Ki
0.33
GIVEN
z5
z6
z7
K5( T , P )
K6( T , P )
K7( T , P )
x5
soln 2
T
z5
x6
K5( T , P )
z6
K6( T , P )
x7
z7
K7( T , P )
FIND( x5 , x6 , x7 , T )
soln
x5
1
soln 0
x6
soln 1
x5 = 0.074
x7
x6 = 0.346
soln 3
x7 = 0.579
T = 77.436
This is the dew-point pressure solution.
So for a mixture of the composition z5=0.25, z6=0.45 and z7=0.30, the mixture will be all liquid
at temperatures below 61.79 C, and all vapor at temperatures above 77.44 C. Vapor-liquid
equilibrium will exist only between 61.79 and 77.44 C, so this is the temperature range we
will examine.
T
62
L
0.99
P
1.013
K5( T , P ) . x5
GIVEN
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
x5
soln 0
x6
x5 = 0.246
y5
K5( T , P ) . x5
65
soln 1
x6 = 0.451
y5 = 0.543
T
K6( T , P ) . x6
y6
0.80
soln 2
soln
L
soln 3
x7 = 0.303
K6( T , P ) . x6 y7
y6 = 0.367
L
x7
K7( T , P ) . x7
K7( T , P ) . x7
y7 = 0.09
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
V
1
V = 0.012
L
L = 0.988
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
K5 ( T , P ) . x5
GIVEN
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
x5
soln 0
x6
x5 = 0.198
y6
y5 = 0.476
68
L
soln 2
K6( T , P ) . x6 y7
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x6
x5 = 0.157
soln 1
x7
x6 = 0.453
K5( T , P ) . x5
y6
y5 = 0.411
soln 2
K6( T , P ) . x6 y7
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x6
x5 = 0.124
y5
soln 1
x7
x6 = 0.432
K5( T , P ) . x5
y5 = 0.352
74
y6
soln 2
L = 0.813
K7( T , P ) . x7
K7( T , P ) . x7
soln
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
V
1
L
V = 0.365
L = 0.635
K7( T , P ) . x7
K7( T , P ) . x7
soln
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
V
1
L
V = 0.551
L = 0.449
K7( T , P ) . x7
y7 = 0.183
0.20
K5( T , P ) . x5
K6( T , P ) . x6
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
GIVEN
L
V = 0.187
x7 = 0.443
K6( T , P ) . x6 y7
y6 = 0.464
L
1
0.40
K6( T , P ) . x6
x5
V
y7 = 0.144
K5( T , P ) . x5
GIVEN
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
x7 = 0.389
y6 = 0.444
L
L
x6
0.60
x5 . ( L. ( 1
71
soln
( x5
y7 = 0.114
K6( T , P ) . x6
y5
K7 ( T , P ) . x7
x7 = 0.343
K5( T , P ) . x5
x5
T
x7
y6 = 0.41
GIVEN
T
soln 1
x6 = 0.459
K5( T , P ) . x5
y5
T
K6 ( T , P ) . x6
K7( T , P ) . x7
soln
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
x5
soln
x6
0
x5 = 0.098
y5 = 0.301
T
76
L
K6( T , P ) . x6 y7
y6
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x6
x5 = 0.083
soln 1
x7
x6 = 0.369
K5( T , P ) . x5
y6
y5 = 0.27
soln 2
L = 0.251
K7( T , P ) . x7
K7( T , P ) . x7
soln
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
V
1
L
V = 0.891
L = 0.109
K7( T , P ) . x7
y7 = 0.27
K5( T , P ) . x5
K6( T , P ) . x6
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x6
x5 = 0.077
soln 1
x7
x6 = 0.353
K5( T , P ) . x5
y5 = 0.256
78
L
0.10
GIVEN
y5
1
V = 0.749
x7 = 0.548
K6( T , P ) . x6 y7
y6 = 0.46
L
x5
V
0.10
x5 . ( L. ( 1
77
soln 3
y7 = 0.232
K6( T , P ) . x6
y5
L
x7 = 0.504
K5( T , P ) . x5
x5
T
soln 2
y6 = 0.467
GIVEN
T
x7
x6 = 0.398
K5( T , P ) . x5
y5
soln 1
y6
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
soln 3
V
1
L
V = 0.966
L = 0.034
K7( T , P ) . x7
y7 = 0.291
0.05
K5( T , P ) . x5
K6( T , P ) . x6
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
GIVEN
soln
x7 = 0.57
K6( T , P ) . x6 y7
y6 = 0.453
L
soln 2
K7( T , P ) . x7
K7( T , P ) . x7
( x5
x6
x7 ) 0
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
x7 . ( L. ( 1
x5
K7( T , P ) )
soln 0
x6
x5 = 0.071
y5 = 0.242
T
61
L
soln 3
x7 = 0.592
y6 = 0.445
V
1
L
V = 1.045
L = 0.045
K7( T , P ) . x7
y7 = 0.313
0.9
x5 . ( L. ( 1
K5( T , P ) )
K5( T , P ) ) z5
x6 . ( L. ( 1
K6( T , P ) )
K6( T , P ) ) z6
x7 . ( L. ( 1
K7( T , P ) )
K7( T , P ) ) z7
soln 0
x5 = 0.264
K5( T , P ) . x5
y5 = 0.566
L
K6( T , P ) . x6 y7
y6
FIND( x5 , x6 , x7 , L)
soln
soln 2
K6( T , P ) . x6
x5
i
x7
K5( T , P ) . x5
GIVEN
y5
soln 1
x6 = 0.337
K5( T , P ) . x5
y5
K7( T , P ) ) z7
x6
soln 1
x7
x6 = 0.446
y6
K7( T , P ) . x7
soln
soln 2
soln 3
x7 = 0.29
K6( T , P ) . x6 y7
y6 = 0.351
L
( x5
x6
x7 ) 0
FIND( x5 , x6 , x7 , L)
V
1
L
V = 0.047
L = 1.047
K7( T , P ) . x7
y7 = 0.083
0 , 1 .. 9
TT
61.788
0.25
0.45
0.3
1.0
62
0.246
0.451
0.303
0.988
65
0.198
0.459
0.343
0.813
68
0.157
0.453
0.389
0.635
71
xx5
0.124
xx6
0.432
xx7
0.443
LL
0.449
74
0.098
0.398
0.504
0.251
76
0.083
0.369
0.548
0.109
77
0.077
0.353
0.570
0.034
77.436
0.074
0.346
0.570
0.0
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-23
(also available as a Mathcad worksheet)
8.1-23a
i
Pi
0 .. 10
T
xi
380
Pvap1
3120.29
exp 9.3225
T
Pvap2
T
630 .
gam1
x=
exp
57.57
. 1
xi
323.15
2
8.314 . T
T
335 . 1
323.15
xi . gam1 . Pvap1
1
. x
i
2
8.314 . T
x . gam2 . Pvap2
i
0
0
0 0.329
1 0.1
1 0.368
2 0.2
2 0.403
3 0.3
3 0.433
P=
T
323.15
0 0
4 0.4
T
335 . 1
323.15
exp
630 .
gam2
63.63
3341.62
exp 9.2508
T
Pi
0.1 . i
4 0.461
5 0.5
5 0.485
6 0.6
6 0.508
7 0.7
7 0.529
8 0.8
8 0.548
9 0.9
9 0.566
10 1
10 0.583
P in bar
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
T
380
Pvap1
exp 9.3225
3120.29
T
T
630 .
gaml i
exp
y1i
T
T
. 1
2
xi
323.15
8.314 . T
630 .
gam2 i
3341.62
exp 9.2508
63.63
335 . 1
323.15
Pvap2
T
335 . 1
323.15
exp
T
323.15
. x
i
2
8.314 . T
xi . gaml i . Pvap1
y2i
Pi
0.582581
1
y1i
0.6
0.5
P
i
0.4
0.328551 0.3
0
0
0.2
0.4
0.6
x
i
0.8
1
1
57.57
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-23b generalized at fixed pressure
0 .. 10
i
TTi
0.1 . i
xi
T
373.
DT
5
DT > 0.00001
while
Pvap1
3120.29
exp 9.3225
T
Pvap2
63.63
3341.62
exp 9.2508
T
T
630 .
gam1
T
10 . ln
T
T
TTi
T
335 . 1
323.15
exp
xi
2
323.15
. x
i
2
8.314 . T
xi . gam1 . Pvap1
DT
. 1
323.15
8.314 . T
630 .
P
T
335 . 1
323.15
exp
gam2
57.57
xi . gam2 . Pvap2
1
380
P . 750
DT
T
TTi
Have to recalculate vapor pressures, activity coefficients and vapor phase mole fractions since
these variables are only defined within the subprogram.
Pvap1i
exp 9.3225
Pvap2i
exp 9.2508
TTi
i
TTi
323.15
exp
63.63
3341.62
TTi
630 .
gam1
3120.29
57.57
335 . 1
TTi
. 1
xi
323.15
.
8.314 TT
i
630 .
gam2
i
exp
TTi
335 . 1
323.15
TTi
323.15
8.314 . TTi
. x
i
2
2
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
P
i
xi . gam1 i . Pvap1i
xi . gam1 i . Pvap1i
y1i
Pi
P=
gam1 =
1
xi . gam2 i . Pvap2i
1
y2i
xi . gam2 i . Pvap2i
Pi
0
0 0.507
0
0 0
0 0
0
0 394.064
1 0.507
1 0.1
1 0.193
1 390.316
2 0.507
2 0.2
2 0.339
2 387.383
3 0.507
3 0.3
3 0.456
3 385.011
4 0.507
4 0.4
4 0.554
x=
0
y1 =
TT =
4 383.041
5 0.507
5 0.5
5 0.639
5 381.368
6 0.507
6 0.6
6 0.717
6 379.919
7 0.507
7 0.7
7 0.789
7 378.646
8 0.507
8 0.8
8 0.859
8 377.513
9 0.507
9 0.9
9 0.929
9 376.498
10 0.507
10 1
10 1
10 375.584
0
0 0.507
0
0 1.293
0
0 1
0
0 0.886
1 1.23
1 1.003
1 0.795
1 0.453
2 1.177
2 1.01
2 0.73
2 0.414
3 1.133
3 1.023
3 0.679
3 0.385
4 1.096
4 1.041
4 0.64
gam2 =
Pvap1 =
Pvap2 =
4 0.362
5 1.065
5 1.065
5 0.608
5 0.343
6 1.041
6 1.095
6 0.581
6 0.328
7 1.023
7 1.132
7 0.558
7 0.315
8 1.01
8 1.175
8 0.539
8 0.303
9 1.003
9 1.227
9 0.522
9 0.293
10 1
10 1.286
10 0.507
10 0.285
400
1
390
TT
i
y1
i
380
0.5
370
0
0.5
x
i
1
0
0
0.5
x
i
1
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-24
(also available as a Mathcad worksheet)
8.1-24
NB
Benzene activity in benzene - polyisobutylene (40,000) mixtures
1
NPIB
40000
VB
88.26
VIB
131.9
R
χ
8.314
104
mb
0.8331
mpib
1
mb
WtB
mb
WtB
78
xb
xpib
WtB
WtPIB
78
40000
PhiB = 0.4264
m
lnGB
ln
PhiB
1
1 .
PhiP
PhiP
3
xb . VB
WtPIB
xb . VB
xpib . NPIB . VIB
PhiP
PhiP = 0.5736
1
1
WtB
PhiB
xb = 0.9977
χ . PhiP
2
γB
exp ( lnGB )
γ B = 1.0529
χ . PhiB
2
γP
exp ( lnG2 )
γP= 0
m
(1
m) . PhiB
xpib
ab
lnG2 = 238.9603
lnGB = 0.0516
xb . γ B
ab = 1.0505
activity of benzene
Partial pressure of benzene =
mb
PhiB
WtB = 0.4545
m = 574.7878
NB . VB
xb
lnG2
xb
xpib = 2.3352 10
NPIB . VIB
ln
1
mpib
1.0
0.5543
mpib
1
0.0606 . ab = 0.0637
bar
mb
WtB
mb
mpib
WtB = 0.3566
WtPIB
xb . VB
xpib . NPIB . VIB
PhiP
1
WtB
WtB
78
xb
xpib
WtB
WtPIB
78
40000
xb = 0.9965
PhiB = 0.3309
1
xb
PhiB
xpib = 3.5056 10
3
xb . VB
PhiP = 0.6691
1
PhiB
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
lnGB
ln
PhiB
1
1 .
PhiP
m
χ . PhiP
2
γB
exp ( lnGB )
γ B = 1.0133
(1
m) . PhiB
χ . PhiB
2
γP
exp ( lnG2 )
γP= 0
ab
xb . γ B
xb
lnG2
PhiP
ln
xpib
lnGB = 0.0132
lnG2 = 184.5044
mb
0.291
mpib
1
activity of benzene
0.0606 . ab = 0.0612
Partial pressure of benzene =
ab = 1.0097
bar
mb
WtB
mb
mpib
WtB = 0.2254
WtPIB
xb . VB
xpib . NPIB . VIB
PhiP
1
WtB
WtB
78
xb
xpib
WtB
WtPIB
78
40000
xb = 0.9933
lnGB
ln
ln
xb
PhiB
PhiB = 0.2061
PhiB
PhiP
xb . VB
3
xpib = 6.6564 10
1 .
PhiP
m
χ . PhiP
2
γB
exp ( lnGB )
γ B = 0.8608
(1
m) . PhiB
χ . PhiB
2
γP
exp ( lnG2 )
γP= 0
ab
xb . γ B
xpib
lnGB = 0.1499
lnG2 = 113.4422
10
B
ln ( 40 ) A
B
273.15
con
A
ln ( 60 ) A
7.6
B
273.15
15.4
find( A , B)
A = 18.6885
con 0
B
con 1
3
B = 4.2111 10
B
exp A
Pvap
ab = 0.8551
3000
given
273.15
750
10
Pvap = 0.0606
PhiB
activity of benzene
Partial pressure of benzene = 0.0606 . ab = 0.0518
bar
Calculation of pure component vapor pressure of benzene
Data from Perry's
Pvap = 40 mm Hg at T = 7.6 C and 60 mm Hg at 15.4 C
A
1
PhiP = 0.7939
1
xb
lnG2
1
bar
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8.1-25
(also available as a Mathcad worksheet)
8.1-25
NC
Cyclohexane activity in cyclohexane - polyisobutylene (40,000) mixtures
1
NPIB
40000
84.16
VC
104
mc
1.318
mpib
1
mc
84.16
xpib
WtC
WtPIB
84.16
40000
PhiC = 0.5716
lnGC
PhiC
ln
ln
1 .
PhiP
1
m) . PhiC
(1
xpib
mpib
xc
PhiC
3
xc . VC
WtC = 0.5686
WtPIB
xc . VC
xpib . NPIB . VIB
PhiP
PhiP = 0.4284
0.475
1
1
WtC
PhiC
xc = 0.9984
2
γC
exp ( lnGC )
γ C = 0.9578
χ . PhiC
2
γP
exp ( lnG2 )
γP= 0
xc . γ C
ac
lnG2 = 262.067
lnGC = 0.0431
ac = 0.9563
activity of cyclohexane
Partial pressure of cyclohexane = 0.1303 . ac = 0.1246
mc
χ
8.314
χ . PhiP
m
PhiP
R
m = 469.5731
xc
lnG2
1
xpib = 1.5938 10
NPIB . VIB
NC . VC
m
131.9
mc
WtC
WtC
xc
VIB
.779
0.434
mpib
1
bar
mc
WtC
mc
WtC = 0.3026
mpib
WtPIB
1
WtC
WtC
84.16
xc
WtC
WtPIB
84.16
40000
xc = 0.9952
lnGC
xpib
ln
PhiC = 0.3052
PhiC
ln
PhiP
xpib
lnGC = 0.2593
xc
PhiC
xpib = 4.8245 10
xc . VC
xpib . NPIB . VIB
xc . VC
3
PhiP
1
PhiC
PhiP = 0.6948
1
1 .
PhiP
m
χ . PhiP
2
γC
exp ( lnGC )
γ C = 0.7716
(1
m) . PhiC
χ . PhiC
2
γP
exp ( lnG2 )
γP= 0
lnG2 = 137.998
ac
xc . γ C
xc
lnG2
1
ac = 0.7679
activity of cyclohexane
Partial pressure of cyclohexane = 0.1303 . ac = 0.1001
bar
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
mc
0.147
mpib
1
mc
WtC
mc
mpib
WtC = 0.1282
WtPIB
xc . VC
xpib . NPIB . VIB
PhiP
1
WtC
WtC
84.16
xc
PhiC
WtPIB
84.16
40000
xc = 0.4174
lnGC
ln
xpib
ln
PhiC
xc
PhiC = 0.1556
PhiC
PhiP
xc . VC
xpib = 0.5826
1 .
PhiP
m
χ . PhiP
2
γC
exp ( lnGC )
γ C = 1.2146
(1
m) . PhiC
χ . PhiC
2
γP
exp ( lnG2 )
γP= 0
lnG2 = 72.5182
ac
xc . γ C
xpib
lnGC = 0.1944
PhiC
ac = 0.507
activity of benzene
0.1303 . ac = 0.0661
Partial pressure of cyclohexane =
1
PhiP = 0.8444
1
xc
lnG2
1
bar
So while not perfect, the value of the Flory parameter chosen, 0.475, gives a reasonably good
description of the cyclohexane-polyisobutylene system.
Calculation of pure component vapor pressure of cyclohexane
Data from Perry's
Pvap = 60 mm Hg at T = 14.7 C and 100 mm Hg at 25.5 C
A
10
B
3000
given
ln ( 60 ) A
B
273.15
con
A
A = 18.2201
con 0
273.15
25.5
B
con 1
3
B = 4.0661 10
B
273.15
Pvap
25
Pvap = 0.1303
750
bar
(also available as a Mathcad worksheet)
8.1-26
NP
14.7
find( A , B)
exp A
8.1-26
B
ln ( 100 ) A
Pentane activity in pentane - polyisobutylene (40,000) mixtures
1
NPIB
40000
104
VP
72.15
.630
VIB
131.9
R
8.314
χ
0.85
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
mp
1.405
mpib
1
mp
WtP
mp
WtP
84.16
xp
xpib
WtP
WtPIB
84.16
40000
PhiP = 0.6012
lnGP
1 .
PhiPib
1
xp
lnG2
ln
PhiP
3
WtPIB
xp . VP
xpib . NPIB . VIB
xp . VP
PhiPib = 0.3988
PhiPib
1
WtP
1
PhiP
xp = 0.9985
m = 442.9714
PhiP
ln
xp
xpib = 1.4953 10
NPIB . VIB
NP . VP
m
1
WtP = 0.5842
mpib
m
PhiPib
(1
m) . PhiP
xpib
lnGP = 0.0258
χ . PhiPib
2
χ . PhiP
2
γP
γ Pib
γ Pib = 0
exp ( lnG2 )
xp . γ P
aP
lnG2 = 259.8176
γ P = 1.0261
exp ( lnGP )
aP = 1.0246
activity of n-pentane
0.3778 . aP = 0.3871
Partial pressure of n-pentane =
mp
0.269
mpib
1
bar
mp
WtP
mp
WtP = 0.212
mpib
WtPIB
1
WtP
WtP
84.16
xp
WtP
WtPIB
84.16
40000
xp = 0.9922
lnGP
xpib
ln
PhiP = 0.224
PhiP
1
xp
lnG2
ln
PhiPib
xpib
lnGP = 0.2023
(1
1 .
PhiPib
m
m) . PhiP
1
xp
PhiP
xpib = 7.7609 10
χ . PhiPib
χ . PhiP
2
lnG2 = 94.3441
2
xp . VP
3
xp . VP
xpib . NPIB . VIB
PhiPib
1
PhiP = 0.224
γP
γ Pib
ap
exp ( lnGP )
exp ( lnG2 )
xp . γ P
γ P = 0.8169
γ Pib = 0
ap = 0.8105
activity of n-pentane
Partial pressure of n-pentane =
0.3778 . ap = 0.3062
bar
PhiP
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
mp
0.0294
mpib
1
mp
WtP
mp
WtP = 0.0286
mpib
WtPIB
1
WtP
WtP
84.16
xp
PhiP
WtPIB
84.16
40000
xp = 0.1264
lnGP
ln
xpib
ln
xp
PhiP = 0.0355
PhiP
1
xp
lnG2
1
PhiP
PhiPib
xpib = 0.8736
1 .
PhiPib
m
m) . PhiP
(1
xpib
lnGP = 0.4822
xp . VP
xp . VP
xpib . NPIB . VIB
χ . PhiPib
2
γP
χ . PhiP
2
ap
PhiP
exp ( lnGP )
exp ( lnG2 )
xp . γ P
γ P = 1.6197
γ Pib = 1.7316 10
7
ap = 0.2047
activity of n-pentane
0.3778 . ap = 0.0773
Partial pressure of pentane =
1
PhiPib = 0.9645
γ Pib
lnG2 = 15.5691
PhiPib
bar
So while not perfect, the value of the Flory parameter chosen, 0.85, gives a reasonably good description
of the pentane-polyisobutylene system.
Calculation of pure component vapor pressure of pentane
Data from Perry's
Pvap = 200 mm Hg at T = 1.9 C and 400 mm Hg at 18.5 C
A
10
B
3000
given
ln ( 200 ) A
B
273.15
con
A
1.9
B
273.15
find( A , B)
A = 17.4764
con 0
B
con 1
3
B = 3.3496 10
B
exp A
Pvap
ln ( 400 ) A
273.15
750
10
Pvap = 0.3778
bar
18.5
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u u u Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
8.4-1 (a) From eqn. (8.4-2) we have, at equilibrium, that
d i
d i for all species i
xiI γ iI x I = xiII γ iII x II
and from regular solution theory we have
a
RT lnγ i = V i φ 2j δ 1 − δ 2
G
and
ex
a
f
2
= x1V 1 + x2V 2 φ1φ2 δ1 − δ 2
=
f
2
a
x1 x2V 1V 2
δ1 − δ2
x1V 1 + x2V 2
f.
2
The
critical
solution
temperature is found from
FG ∂ G IJ
H ∂x K
2
2
1
=0=
T, P
FG ∂ G IJ
H ∂x K
2
ex
2
1
FG ∂ G IJ
H ∂x K
2
+
IM
2
1
T, P
T, P
where
a
G IM = x1 G 1( T , P) + x2 G 2 (T , P ) + RT x1 ln x1 + x2 ln x2
f
By taking derivatives, we find that
FG ∂ G IJ
H ∂x K
2
IM
2
1
=
T ,P
FG
H
RT
∂ 2 G ex
and
x1 x2
∂x12
IJ
K
=−
T, P
a
2V 21 V 22 δ1 − δ2
ax V
1
1 + x2 V 2
f
f
2
3
Thus, setting
FG ∂ G IJ
H ∂x K
2
IM
2
1
=
T ,P
FG ∂ G IJ
H ∂x K
2
ex
2
1
=0
T ,P
we obtain
RTC =
a
2 x1x2 V 12V 22 δ1 − δ 2
ax V
1 1+
x2V 2
f
3
f
2
= 2φ1φ2
a
f
f
V 1V 2 δ1 − δ 2
x1V 1 + x2 V 2
a
(b) To find the upper consolute temperature, that is, TC ,max , we use
2
(*)
FG ∂T IJ
H ∂x K
C
1
= 0 or
P
FG ∂T IJ = 2V V aδ − δ f RS x − x − 3x x aV − V f UV = 0
H ∂x K ax V + x V f T
ax V + x V f W
clearly the term k p must be zero at the upper consolute temperature. Thus, we obtain
R
2
1
C
1
P
2
2
2
1
2
1 2
3
1
x1 =
1
2
V1
2
1
1 1
2
V1 −V2
±
1
2
2
2
V 12 + V 22 − V 1V 2
aV
1
−V 2
f
Only the negative solution is realistic. Thus, the composition at the upper consolute point is
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
cV
x1 = 1 − x2 =
2
1
h
+ V 2 − V 1V 2 − V 1
2
aV
2
f
−V 1
Note that this composition depends only on the molar volumes of species, and not their
solubility parameters! Note also, that as V 2 → 1 , x1 → 0.5 , as should be expected.
Substituting this result into eqn. (*) above yields
RTC =
8.4-2
For G
ex
a
aV
f {aV + V fdV + V − V V i − V
− V f {dV + V − V V i − aV + V f}
2V 12V 22 δ1 − δ2
2
2
2
1
= Ax1 x2 we have, from Gi
RT lnγ 2 =
Ax12
2
1
1
2
2
1
2
2
ex
12
2
2
1
2
− V 22
}
3
12
1
2
1
2
1
F ∂dN G
= RT ln γ = G
H ∂N
i
ex
i
2
iIJ
K
. That RT lnγ 1 = Ax22 and
T , P , N j≠ i
.
I
II
(a) The equilibrium curve is one for which G i = G i or
xiIγ Ii = xiIIγ iII for both species (i.e.,
i = 1, 2 ). Thus, the equilibrium phase envelope is the solution to the equations
R| Ac1 − x h U|
R − x h U|
S| RT V| = x exp|S| Ac1RT
V|
T
W
T
W
c1− x h expR|S| AxRT U|V| = c1 − x h expR|S| AxRT U|V|
T W
T W
I 2
i
xiI exp
and
I2
i
I
i
II 2
i
II
i
II2
i
II
i
and we have two equations for the two unknowns: xiI , xiII
FG ∂ G IJ
H ∂x K
2
(b) Limit of stability criterion is
2
1
= 0 , where
T, P
a
f
G = x1 G 1 + x2 G 2 + RT x1 ln x1 + x2 ln x2 + Ax1 x2
144444424444443 123
G
setting
FG ∂ G IJ
H ∂x K
2
2
1
IM
= 0 yields x1 x2 =
T, P
solving for x1 yields x1 =
G
a
ex
f
RT
= x1 1 − x1 = x1 − x12
2A
1 1
2 RT
1 1
2 RT
1 1
2 RT
±
1−
or xiI = −
1−
and xiII = +
1−
2 2
A
2 2
A
2 2
A
8.4-3 (a) Regular solution theory suppose that Sex = 0 , or, since G ex = Hex − TS ex , that G ex = H ex .
This is the case for the C6 H 6 - CCl4 system, but not for the C6 H 6 - CS2 system. Therefore,
regular solution theory is not applicable to the C6 H 6 - CS2 system. To test the HildebrandScatchard model we use
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
G ex =
x1V 12 x2V 22
2
aδ − δ f
2
2
x1V 1 + x2 V 2
1
2
at x1 = 0.5 we obtain
G ex =
(0.5) × 89 × (0.5) × 61
2
. cal mol = 48.45 J mol
× (10 − 9.2) = 1158
0.5(89 + 61)
compared with ≈ 105 J mol experimentally. Thus, we concluded that while the CCl4 - C6 H 6
system has Sex = 0 and thus may satisfy the regular solution model, it is not well represent by
Scatchard-Hildebrand regular solution theory.
(b) Since G ex is a symmetric function of composition for the CS 2 - C6 H 6 system, we will represent
the composition dependence of G ex by the one-constant Margules expression Gex = Ax1 x2 with
a
f
A = 1160 J mol , so that G ex = x1 = 05
. = 290 J mol , as is observed experimentally. For the
one-constant Margules eqn., by eqn. (8.4-14),
A
1160
TUC =
=
≈ 70 K
2 R 2 × 8.314
Thus, if a liquid phase(s) were to exist at very low temperatures, it would exist as two phases
below 70 K, and a single stable phase above 70 K. However, since 70 K is well below the
melting points of either of the pure components, and, presumably, the eutectic point as well (see
section 8.7), no liquid-liquid phase separation will be observed. [Note: we can improve our
estimation of the upper consolute temperature by taking into account the temperature
∂ GT
H
dependence of the excess Gibbs free energy. In particular, from
= − 2 we obtain
∂T
T
bg
G ex
T
G ex
T
−
T2
=−
T1
z
T2
T1
FG
H
H ex
1
ex 1
dT ≈ H
−
2
T
T2 T1
IJ
K
where we have assumed, for simplicity, that H ex is temperature independent. At x1 = 0.5 ,
H ex ≈ 525 J mol . Thus
a
f LMN 298290.15 + 525RST T1 − 2981.15UVWOPQ
G ex x1 = 0.5, T2 = T2
2
and, at T2 = 80 K ,
a
f
G ex x1 = 05
. , 80 K ≅ 462 J mol , and A(T = 80 K) ≅ 1848 J mol
implying an upper consolute temperature of
A
TUC =
= 111 K
2R
Guess again TUC = T2 = 105 K
a
f
G ex x1 = 05
. , 105 K = 1056
. K , TUC = 106.4 K (which is close to guess)
Since this temperature is still well below the melting points of the species, our conclusion does
not change, there is no phase separation. We do, however, see the importance of accounting for
the temperature dependence of the excess Gibbs free energy (and activity coefficients)!]
(c) For vapor-liquid equilibrium we have fi L = fi V ⇒ xi γ i Pi vap = yi P at an azeotropic point
xi = yi , so that γ i Pi vap = P for our system
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
A
1160
=
= 0.4365
RT 8.314 × 319.65
also
vap
γ C 6 H 6 PCvap
= P = γ CS 2 PCS
6H 6
2
or
vap
ln γ C 6 H 6 + ln PCvap
= ln γ CS2 + ln PCS
6H6
2
or
2
0.4365xCS
+ ln
2
PCvap
6H 6
c
= 0.4365 1 − xCS 2
vap
PCS
2
h
2
This equation has, as its only solution,
xCS 2 ≅ 182
.
which is not between 0 and 1. Thus, we conclude that no azeotrope is formed. [Note: we could
get a better estimate of G ex at the temperature of interest by talking into account the
temperature dependence of G ex as was done in part b. Then we find that
a
f
G ex T = 465
. ° C, x1 = 05
. = 273 J mol
Thus, the solution becomes even less non-ideal, and an azeotrope will not be formed.]
8.4-4 (a) Starting from
xiIγ Ii
=
xiI
L Ac1 − x h OP
exp M
MN RT PQ = x γ
I 2
i
II II
i i
=
xiII
L Ac1 − x
exp M
MN RT
h OP
PQ
II 2
i
which we can solve for A
e j
A
=
RT c1 − x h − c1 − x h
ln
xiII
xiI
I 2
i
II 2
i
c
h
Using the data for benzene xBI = 0.48 and xBII = 0.94 yields
A
= 2.52 , while using the data
RT
A
= 3.31 . Since the two values of A are different, we conclude
RT
that the one-constant Margules equation is not consistent with the experimental data.
(b) Regular solution theory gives
for perfluoro-n-heptane yields
LV cφ h aδ − δ f OP
= exp M
RT
MN
PQ = x γ
which, solving for aδ − δ f yields
xiIγ Ii
i
xiI
I 2
j
2
P
B
II II
i i
=
xiII exp
LMV cφ h aδ − δ f OP
RT
MN
PQ
i
II 2
j
2
P
B
2
P
B
aδ − δ f = V φ −e φj ; where φ = x V x +V x V
{c h c h }
RT ln
2
P
B
i
I 2
j
xiII
j
xiI
II 2
j
j
P
P
j
B
B
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
From Table 7.6-1 V B = 89 cc mol ; V P = 0.226 m 3 mol from the problem statement. Thus
φ IB = 0.2666 , φ IIB = 08605
.
, φ IP = 07334
.
and φ IIP = 01395
.
a
f
Using the benzene data, we obtain δ P − δB 2 = 108
. cal mol or δP = δ B ± 3.3 = 12.5 or 5.9.
Note that we can not choose between these two values solely on the basis of the data here.
Activity coefficient data on perfluoro-n-heptane in other fluids would be needed to fix δ P . The
value 5.9 is, however, quite close to the value of 6.0 given in Table 7.6-1.
Doing a similar calculation to the one above, but not using the perfluoro-n-heptane data
yields
aδ
f
− δ B 2 = 7 .75 or δP = δ B ± 2 .8 = 12.0 or 6.4
Thus, regular solution theory is also not completely consistent with the experimental data!
8.4-5
P
The condition for material stability is that d 2G > 0 for all variations at constant T and P. Here
this implies
FG ∂ G IJ
H ∂x K
2
2
1
> 0 or equivalently
T, P
FG ∂ ∆G IJ
H ∂x K
2
mix
2
1
T ,P
> 0.
Looking at the curve in the problem statement, we see that at points B and C this derivative is
zero, and between points B and C it is negative. This implies phase instability or phase separation,
with points B and C being the limits of stability.
The condition of phase equilibrium is
G1I = G1II and G2I = G2II
Now from Chapter 6 we have
FG ∂∆V IJ
H ∂x K
T, P
FG ∂∆ H IJ
H ∂x K
T ,P
FG ∂∆ G IJ
H ∂x K
T ,P
∆V mix − x1
and
∆H mix − x1
∆G mix − x1
= H 2 − H2
mix
1
similarly we have
= V2 −V2
mix
1
= G2 − G 2
mix
1
(1)
(2)
Therefore, the equilibrium conditions can be written as
a
f = ∆G d T , P, x i − x ∂ a ∆G f
∂x
(3a)
f = ∆ G dT, P, x i − x ∂a∆ G f
∂x
(3b)
d
i
∂ ∆ G mix
d
i
∂ ∆ G mix
∆G mix T , P, x I − x2I
and
∆G mix T , P, x II − x1I
∂x2I
a
∂x1I
II
mix
II
mix
Subtracting the second of these equations from the first yields
II
2
II
1
mix
II
2
mix
II
1
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
x1I
a
∂ ∆ G mix
∂ x1I
f − x ∂a∆ G f = x ∂a∆ G f − x ∂a∆ G f
I
2
II
1
mix
∂x2I
mix
∂ x1I
II
2
mix
∂ x2II
and using dx2 = −dx1 gives
cx
I
1
+ x2I
or, since x1i + x2i = 1
h ∂a∆∂Gx f = cx
mix
I
1
a
∂ ∆G mix
∂x1I
II
1
+ x2II
h ∂a∆∂Gx f
mix
II
1
f = ∂a∆G f
mix
(4)
∂ x1II
Thus, at the equilibrium state not only is Gi I = Gi II , but eqn. (4) is satisfied also. Using this last
result in eqn. (3a) gives, at equilibrium
i FGH ∂∆∂Gx IJK
d
∆G mix T , P, x I − xiI
1
d
i FGH ∂∆∂Gx IJK
= ∆G mix T , P, xII − xiII
mix
T ,P
mix
1
T ,P
Now eqn. (4) implies that the slope of the ∆G mix curve must be the same at the two points at
which the phases are in equilibrium. Further, from eqns. (1 and 2) we have that since Gi I = Gi II ,
the two lines have the same intercept. Since the two tangent lines have the same slope and the
same intercept, the lines must be identical, i.e., the equilibrium points are on a common tangent
line.
8.4-6
At the bubble point (assuming an ideal vapor phase)
∑ xiγ i Pivap = ∑ yi P = P .
So xB γ B PBvap + xW γ W PWvap = P
(a) (0.04)γ B (0.427 ) + ( 0.96)(1)(0.784 ) = 1.013 bar ⇒ γ B = 15.244
so yB =
15.244 × 0.04 × 0.427
0.784
= 0.257 and yW = 0.96 × 1 ×
= 0.743
1.013
1.013
(b) At
γ IIW =
(c) Since
equilibrium
I I
xW
γW
xIIW
xiIγ Ii
=
xBIγ IB = xBIIγ IIB ⇒ γ IIB =
xIBγ IB 0.04 × 15244
.
=
II
xB
0.4
or
γ IIB = 1524
.
;
also
0.96 × 1
= 1.60
0.6
= xiIIγ iII and
∑ xiIIγ IIi Pi vap = ∑ xiIγ Ii Pi vap = P .
The second liquid phase will also
have a bubble point pressure of 1.013 bar.
8.4-7
Though the overall composition is 50 mole % isobutane, in fact there are really two phases ... one
liquid of composition 11.8 mole % isobutane and the other liquid of composition 92.5 mole %
isobutane. Since, at liquid-liquid equilibrium xiIγ Ii = xiIIγ iII and
Pbubble = ∑ xiIγ Ii Pi vap = ∑ xiIIγ iII Pi vap ,
we need to calculate the bubble point pressure for one phase since the other phase will have the
same bubble point pressure. Also, since the pressures are not expected to be very high, we will
assume an ideal vapor phase. So the equations we will use are
xiγ i Pi vap = yi P and
∑ xiγ i Pivap = ∑ yi P = P
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
We will use the 92.5 mole % liquid for the calculations.
a
ax
f
= 0.925 f = ?
γ isob xisob = 0.925 = 1019
.
from the problem statement
γ furf
isob
However, from the liquid-liquid equilibrium condition we have
a
f
I
II
xfurf
γ Ifurf = xfurf
γ IIfurf ⇒ γ Ifurf xisob = 0.925 =
xIIfurfγ IIfurf 0.882 × 1033
.
=
= 12.148
I
xfurf
0.075
Therefore
P = ∑ xi γ i P = 0.925 × 1019
.
× 4 .909 + 12148
.
× 0.075 × 4.93 × 10−3 = 4.632 bar
yisob = 0.925 × 1019
.
×
4.909
= 0.999 ; yfurf = 0.001
4.632
8.4-8 (a) At LLE xiIγ Ii = xiIIγ iII . Here this implies
RS A c1 − x h UV = x expRS A c1 − x h UV
T RT
W
T RT
W
and
c1− x h expRST RTA cx h UVW = c1 − x h expRST RTA cx h UVW
These equations are symmetric with respect to the interchange on the subscripts 1 and 2 (that is
replacing x by x = 1 − x and x = a1 − x f by x yields exactly the same set of equations).
I 2
1
x1I exp
I 2
1
I
1
1
II 2
1
II
1
2
1
II 2
1
II
1
2
1
1
This suggests that the equilibrium is symmetric. Of course, that is exactly what we would
expect with the one-constant Margules expression. Therefore, we have
xHI = 0.0902
xIIH = 0.9098
and
= 0.9098
I
xEtOH
xIIEtOH = 0.0902
(b) Here, as in the previous two problems, the bubble point pressure can be computed from either
liquid phase since xiIγ Ii = xiIIγ iII . Assuming an ideal vapor phase, we have
xiγ i Pi vap = yi P and P = ∑ xi γ i Pi vap
where the activity coefficients and vapor pressure are given in the problem statement. The
solution (putting all the equations into Mathcad) is
P = 19657
.
bar, yH = 0.5795 , yEtOH = 0.4205
8.4-9
The condition for liquid-liquid equilibrium is xiIγ Ii = xiIIγ iII using the one-constant Margules eqn.
c
we have RT ln γ Ii = A 1 − xiI
a
f
h
2
. Now
a
f
0.097γ 1 x1 = 0.097 = 0.903γ 1 x1 = 0.903 ⇒ ln
=
a
a
f
f
γ 1 x1 = 0.097
0.903
= ln
γ 1 x1 = 0.903
0.097
A
A
(1 − 0.097 ) 2 − (1 − 0.903) 2 ⇒
= 2 .768
RT
RT
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
so that
d a fi
= expd 2.768 a1 − x f i
γ 1 = exp 2.768 1 − x1
γ2
2
2
2
Now to compute the pressure in the one phase liquid region we use
P = x1γ1 P1vap + x2γ 2 P2vap
d
a
= x1 exp 2.768 1 − x1
xacet
P( bar )
0
0.00313
0.02
0.0507
0.04
0.0887
0.06
0.1186
0.08
0.1420
0.09
01517
.
0.097
Two
01577
.
phase
region
0.903
01577
.
0.92
0.1590
0.94
01607
.
0.96
0.1626
0.98
01647
.
10
.
0.167
8.4-10
f i 0.167 + a1 − x f expc2.768 x h3.13× 10
2
1
2
1
−3
Since we have the concentrations of the coexisting equilibrium liquid phase we can determine two
binary parameters. Also, since we are interested in two different temperatures (LLE at 20°C and
VLE at 734
. ° C ) we want an activity coefficient model with some built in temperature dependence
(otherwise, we will get LLE with the same compositions at all temperatures.) Consequently, I will
use the two constant Margules equation
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
a
f
aα + B x f
⇒ RT ln γ 1 = x22 α1 + B1 x2
RT ln γ 2 =
x12
2
2 1
where αi = A + 3(−1)i +1 B ; Bi = 4(−1)i B . These equations are to be used with x1Iγ I1 = x1IIγ 1II ;
x2Iγ I2 = x2IIγ II2 ; x1I = 00850
.
; x2I = 1 − x1I ; x1II = 0.6363 ; x2II = 1 − x2I . Putting all this into Mathcad, I
find
A = 48334
. J mol
B = − 19802
. J mol
Now using T = 734
. ° C , the same constants as above, I find
xMEK
P( bar)
yMEK
0
0.3603
0
.1
0.7241 0.540
.2
0.8313
0.617
.3
0.8601 0.637
.4
0.8696
0.646
.5
0.8776
0.656
.6
0.8867
0.677
.7
0.8931 0.714
.8
0.8903
0.775
.9
0.8718
0.867
1.0
0.8337
1.000
Azeotrope is predicted to occur at xMEK ≈ 0.7287 and P = 08935
.
bar
Note: LLE does not occur at this higher temperature. If it did the calculated P − x diagram would have
both an interior maximum and minimum as a function of temperature, and the predicted x-y diagram would
be like a sideways S, with the x-y line crossing the x = y line twice.
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
8.4-11
The Wilson model is
LM∑ x Λ OP
N
Q
L cλ − λ h OP
expM −
N RT Q
G ex = − RT ∑ xi ln
with
V jL
Λij =
Vi L
j
j
ij
ij
ji
which, for a binary mixture, reduces to
k a
f a
+ RT a x ln x + x
G ex = −RT x1 ln x1 + x2 Λ12 + x2 ln x1Λ21 + x2
and
G IM = x1 G 1 + x2 G 2
so
1
FG
H
1
2
ln x2
fp
f
IJ
K
x1
x2
+ x2 ln
x1 + x2 Λ12
x1Λ 21 + x2
G = x1 G 1 + x2 G 2 + RT x1 ln
Now we look at the derivative of G
FG ∂G IJ
H ∂x K
1
= G 1 − G 2 + RT ln
T, P
x1
x1 + x2 Λ12
FG
H
x1 + x2 Λ12 ∂
x1
x1
∂x1 x1 + x2 Λ12
+ RTx1
− RT ln
x2
x1Λ 21 + x2
= G 1 − G 2 + RT ln
a
+ RT x1 + x2 Λ12
+ RTx2
FG
H
IJ
K
x1Λ 21 + x2 ∂
x2
x2
∂x1 x1Λ21 + x2
x1
x1 + x2 Λ12
fRST x + 1x Λ − axx a+1 −x ΛΛ ff UVW
1
12
2
1
− RT ln
IJ
K
2
12
1
2
12
x2
x1Λ 21 + x2
a
+ RT x1Λ21 + x2
fRST x Λ −1+ x − a xx ΛaΛ + −x1ff UVW
2
21
2
1
= G 1 − G 2 + RT ln
21
2
1
21
2
x1
x1 + x2 Λ12
x
x + x Λ − x + x Λ p − RT ln
k
a
f
xΛ +x
RT
+
ax Λ + x f k− x Λ − x − x Λ + x p
+
RT
x1 + x2Λ 12
2
1
2
21
1
12
21
2
2
21
2
2
= G 1 − G 2 + RT ln
− RT ln
1
1
1
1
12
x2
x1Λ 21 + x2
x1
RT Λ12
+
x1 + x2 Λ12 x1 + x2Λ 12
−
RT Λ21
x1Λ 21 + x2
21
2
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
Then
FG ∂ G IJ
H ∂x K
2
2
1
FG
IJ
H
K a
f = a1 − Λ f
xΛ +x ∂ F
− RT
G x IJ + RT Λ aΛ − 1f
x
∂x H x Λ + x K a x Λ + x f
RT
=
kax + x Λ f − x a1 − Λ fp
x ax + x Λ f
RT
+
kax Λ + x f + x aΛ − 1fp
x ax Λ + x f
= RT
T, P
x1 + x2 Λ12 ∂
x1
RTΛ12
−
x1
∂x1 x1 + x2 Λ12
x1 + x2 Λ 12
1
21
2
2
21
2
2
1
1
1
1
1
2
1
2
12
1
1
21
21
2
12
12
1
2
2
21
2
12
2
21
21
2
2
21
2
So
FG ∂ G IJ
H ∂x K
2
2
1
= RT
T, P
LM Λ
N x ax + x Λ f
2
12
1
1
2
2
+
12
a
OP = 0
+x f Q
Λ221
x2 x1Λ 21
2
2
at upper critical solution temperature
Now 0 ≤ x1 ≤ 1 and 0 ≤ x2 ≤ 1 ; in particular, neither x1 nor x2 is negative. Also, clearly
FG Λ IJ
Hx +x Λ K
FG Λ IJ > 0
HxΛ +x K
F ∂ G IJ = 0 is if T = 0 K . Thus, the upper
So, for the Wilson model, the only way for G
H ∂x K
12
1
2
2
2
> 0 and
12
21
1
21
2
2
2
1
T, P
consolute temperature for the Wilson model is T = 0 K , and there is no liquid-liquid equilibrium.
8.4-12 (a) Clearly from
RT ln γ i = 8163
. x 2j we have
G
ex
= x1 RT ln γ 1 + x2 RT ln γ 2 = x18.163 x22 + x2 8.163 x12
a
f
kJ
mol
Therefore, the upper consolute temperature for this model is
A
8163
Tuc =
=
= 490.9 K
2 R 2 × 8.314
First do the LLE calculation. At LLE xiIγ Ii = xiIIγ iII . Here this implies
= x1 x2 8.163 x1 + x2 = x1 x2 8163
.
RS A c1 − x h UV = x expRS A c1 − x h UV
T RT
W
T RT
W
c1− x h expRST RTA cx h UVW = c1 − x h expRST RTA cx h UVW
x1I exp
and
I
1
I 2
1
I 2
1
II 2
1
II
1
II
1
II 2
1
where A=8163. The results using the MATHCAD worksheet with this problem number are
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
T (K)
xHI
xHII
250
275
300
325
350
375
400
425
450
475
485
0.023
0.035
0.050
0.0688
0.0921
0.1206
0.1558
0.2001
0.2584
0.3461
0.4053
0.977
0.965
0.950
0.9312
0.9079
0.8794
0.8442
0.7999
0.7416
0.6539
0.5949
F
GH
Now do VLE calculation. Basis is xi exp −
xH
0
0.01
0.02
0.025
0.03
0.04
0.05
0.06
0.07
0.075
0.08
0.09
0.10
0.125
0.15
0.175
0.825
0.85
0.875
0.9
0.91
0.92
0.925
0.93
0.94
0.95
0.96
0.97
0.975
0.98
0.99
1.00
IP
RT JK
Ax 2j
i
vap
= yi P and that
F
H
∑ xi exp G −
i
IP
RT JK
Ax2j
i
vap
The results using the MATHCAD worksheet with this problem number are
P=0.1013 bar
P=1.013 bar
P=10.13 bar
yH
T K
yH
T K
yH
0
300.18
0
351.56
0
0.4276
290.08
0.6305
282.65
0.3185
342.29
0.1578
0.7359
277.28
0.7959
273.22
0.4892
335.70
0.2642
0.5867
331.07
0.6197
329.31
0.3376
0.6457
327.84
0.6827
325.60
0.5818
327.24
0.6110
328.39
0.6540
330.06
0.7503
0.7898
274.67
275.60
0.7188
332.50
0.8409
0.9083
1.0
276.75
278.22
280.10
0.8216
0.8973
1.0
336.15
338.66
341.86
=P
T K
424.17
418.62
414.45
411.36
0.3888
0.4250
0.4505
0.4683
0.4568
0.4718
0.4934
0.5249
409.09
407.45
406.29
405.49
405.41
406.44
407.95
410.17
0.5717
413.43
0.6444
418.30
0.7661
425.85
0.885
1.0
432.58
438.59
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
550
500
LLE
Temperature, oC
450
Vapor
400
P = 10.13 bar
Liquid
350
Vapor
P = 1.013 bar
300
Vapor
P = 0.1013 bar
250
200
0.0
Liquid
0.2
0.4
0.6
Mole fraction of n-hexane
Liquid
0.8
1.0
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
360
Vapor
Temperature, oC
340
VLE
Region
VLE Region
VLLE Line
320
L2
L1
300
LLE Region
280
260
240
0.0
0.2
0.4
0.6
0.8
1.0
Mole fraction of n-hexane
8.4-13(a) Bubble point pressure. Assume vapor phase is ideal. Then
xiγ i Pi vap = yi P
Solvent
LM
N
x1 exp ln
F
H
OP
Q
I
K
LMF
NH
I
K
LMF
NH
I
K
OP
Q
φ1
1
1
+ 1−
φ2 + xφ22 P1vap = y1 P ⇒ φ1 exp 1 −
φ2 + xφ22 P1vap = y1 P
x1
m
m
Polymer
LM
N
x2 exp ln
OP
Q
OP
Q
φ2
1
− (m − 1)φ1 + xφ12 P2vap = y2 P ⇒ φ2 exp 1 −
φ2 + xφ22 P2vap = y2 P
x2
m
but for the polymer
P2vap ~ 0 ⇒ y2 ~ 0, y1 ~ 1
Therefore, from eqn. (1)
LMF
NH
P = Pbubble = φ1 exp 1 −
point
I
K
OP
Q
1
φ2 + xφ22 P1vap
m
(1)
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
(b)
Liquid-liquid equilibrium
LM
N
x1I exp ln
F
H
I
K
c
φI1
1 I
+ 1−
φ2 + x φI2
m
x1I
h OPQ P
2
vap
1
LM φ + F1 − 1 I φ + x cφ h OP P
N x H mK
Q
1I
L
F
O
LF 1 I
O
⇒ φ exp 1 −
NMH mK φ + xcφ h QP = φ expNMH 1 − mK φ + xcφ h QP
= x1II exp ln
II
1
II
1
I
1
I
2
Similarly
c h
φI2 exp (1 − m)φ1I + x φI1
2
II
2
II 2
2
I 2
2
II
1
vap
1
II 2
2
II
2
c h
= φII2 exp (1 − m)φ1II + x φ1II
2
Data needed
(a) bubble point
volume of solvent
volume of polymer
solvent-polymer χ parameter
vapor pressure of solvent
(b) liquid-liquid equilibrium
volume of solvent
volume of polymer
solvent-polymer χ parameter
8.4-14
Consider the condition for liquid-liquid phase equilibrium of a solute assuming that undissolved
solute is also present
I
d
i
II
d
i
f i T , P, x I = f i T , P, xII = fi L ( T , P)
or
d
i
d
i
xiIγ T , P, xI fi L ( T , P) = xiIIγ T , P , x II f i L ( T , P) = fi L ( T , P)
and then
d
i
d
i
xiIγ T , P, xI = xiIIγ T , P, x II = 1
Therefore, if the activity of a species in solution is ever greater than unity, a separate phase pure
(or very concentrated) in that species will form and reduce the activity of the species in the other
phases to unity.
8.4-15 (also available as a MATHCAD worksheet)
8.4-15
V1
1.6 . 10
5
1.5 . 10
5
V2
m
V2
1895
chi( T )
V1
phi11( x11)
phi12( x12)
x11
0.01
x11. V1
x11. V1
(1
x11) . V2
x12. V1
x12. V1
(1
x12
x12) . V2
0.99
x21
1
T
phi21( x11)
1
phi11( x11)
phi22( x12)
1
phi12( x12)
x11
x22
1
x12
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
T
ln
600
Given
phi11( x11)
phi11( x11) ) . 1
( phi12( x12)
phi12( x12)
ln
1
chi( T ) . phi21( x11)
2
phi22( x12)
2
0
2
phi12( x12)
2
0
m
phi21( x11)
phi12( x12) ) . ( 1
( phi11( x11)
m)
chi( T ) . phi11( x11)
phi22( x12)
find( x11, x12)
v
T
700
0.94392
Given
phi11( x11)
ln
0.05979
v=
1
phi11( x11) ) . 1
( phi12( x12)
phi12( x12)
2
phi22( x12)
2
0
2
phi12( x12)
2
0
2
0
2
0
m
phi21( x11)
ln
chi( T ) . phi21( x11)
phi12( x12) ) . ( 1
( phi11( x11)
m)
chi( T ) . phi11( x11)
1
chi( T ) . phi21( x11)
phi22( x12)
find( x11, x12)
v
T
ln
800
v=
0.11360
0.89956
Given
phi11( x11)
phi11( x11) ) . 1
( phi12( x12)
phi12( x12)
ln
2
phi22( x12)
2
phi12( x12)
m
phi21( x11)
phi12( x12) ) . ( 1
( phi11( x11)
m)
chi( T ) . phi11( x11)
phi22( x12)
v
T
ln
find( x11, x12)
825
phi11( x11)
v=
0.21265
0.83659
Given
( phi12( x12)
phi11( x11) ) . 1
phi12( x12)
ln
phi21( x11)
find( x11, x12)
chi( T ) . phi21( x11)
2
phi22( x12)
2
0
2
phi12( x12)
2
0
m
( phi11( x11)
phi12( x12) ) . ( 1
phi22( x12)
v
1
v=
0.25958
0.81687
m)
chi( T ) . phi11( x11)
Section 8.4
Solutions to Chemical and Engineering Thermodynamics, 3e
T
ln
835
Given
phi11( x11)
( phi12( x12)
phi11( x11) ) . 1
phi12( x12)
ln
phi21( x11)
1
chi( T ) . phi21( x11)
2
phi22( x12)
2
0
2
phi12( x12)
2
0
m
( phi11( x11)
phi12( x12) ) . ( 1
chi( T ) . phi11( x11)
m)
phi22( x12)
find( x11, x12)
v
T
ln
843
phi11( x11)
v=
0.28957
0.80795
Given
( phi12( x12)
phi11( x11) ) . 1
phi12( x12)
ln
phi21( x11)
1
chi( T ) . phi21( x11)
2
phi22( x12)
2
0
2
phi12( x12)
2
0
m
( phi11( x11)
phi12( x12) ) . ( 1
m)
chi( T ) . phi11( x11)
phi22( x12)
v
find( x11, x12)
v=
0.34590
0.79783
The temperature 843 K is the highest at which a nontrivial solution is obtained. At higher
temperature on the trivial solution of both phases being equal is obtained. Thus, the FH
model predicts that LLE will occur up to this temperature. That is, 916.2 is the UCST for
SAN and PMMA. Of course, at this high temperature the polymers are likely to
decompose.
8.4-16
There are many different algorithms that could be used. One is a sequential one of first testing
for LLE, if LLE does not occur then test for VLE. If LLE occurs, one must also test for VLLE,
etc. In all the possibilities that must be tested for are only a liquid phase as stable phase, only a
vapor as the stable phase, VLE with a liquid rich in component 1 as the equilibrium phases,
VLE with a liquid rich in component 2 as the equilibrium phases, or VLLE.
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VXEVWDQFHRI 0: LHILOWHUSDSHU Section 8.9
Solutions to Chemical and Engineering Thermodynamics, 3e
8.9-1 (also available as a Mathcad worksheet)
8.9-1
kow
5.5
10
0.4 . 0.05 . 1.3 . kow
ksedw
ksw
3
ksedw = 8.221922 10
0.0001
250
xs
xs = 7.2 10
1000
3
ksw = 3.794733 10
equilibrium
solubility
9
0.4 . 0.02 . 1.5 . kow
1
gam
8
gam = 1.388889 10
xs
18
7
2 . 10 . 1.013 . gam
H
H
0.2164 .
kaw
H = 28.138889
kaw = 0.020423
298.15
cw
0.3 . 10
cb
0.05 . kow. cw
cb = 4.743416
ca
kaw. cw
ca = 6.127039 10
cs
ksw. cw
cs
9
g/g water or
cs
1.5 . 10
1.3 . 10
6
8.9-2
g/m^3 of water
g/m^3 of air
6
g/m^3 of soil
7
csed = 2.466577
csed
3
g/m^3 water or 4743 ppb by weight
cs = 7.589466 10
ksedw. cw
csed
0.3 . 10
cw
cs = 1.13842
6
csed
bar/mol frac
g/g soil or 0.7589 ppm by wt.
g/m^3 of sediment
csed = 1.897367 10
6
g/g sediment or 1.897 ppm by wt.
(also available as a Mathcad worksheet)
8.9-2
i
0 .. 3
LKow0
S0
Si
Ci
5.52
LKow1
27
S1
Si . 10
9
140
solubility in g/liter
LKow
i
0.05 . Si . 10
5.16
LKow2
S2
7000
3.66
LKow3
S3
3.31
40000
Section 8.9
Solutions to Chemical and Engineering Thermodynamics, 3e
4.47 10
C=
4
1.012 10
1.6 10
Fish concentration in g/liter
3
4.083 10
CPi
3
3
0.447
Ci . 1000
CP =
Fish concentration in ppm
1.012
1.6
4.083
8.9-3 (also available as a Mathcad worksheet)
Water = 4 m3 ;
fish = 200 cm3 = 2 × 102 cm3 × 10−6 m3 cm3 = 2 × 10− 4 m3
Soil = 3 m3
air = 10 − 4 − 3 − 0.0002 = 29998
.
m3
(a) Benzene
vapor pressure (25°C) = 0127
.
bar
solubility in water (25°C) = 0.0405 mol %
1
xBsatγ ∞B ~ 1 ⇒ γ ∞ =
= 2.469 × 103
0.000405
H B = γ ∞ Pi vap = 0127
.
× 2 .469 × 10 3 = 313.6 bar mol fraction
0.2164
= 0.2276
298.15
= 2.13; K0W = 135; KBW = 0.05 × 135 = 6.75;
KAW = 313.6 ×
log 10 K0W
KSW = 0.4 × 0.02 × 135 = 1.08
By a mass balance
10 × 10−3 g = 4 m3H 2 O CB,H 2 O + 3 m 3soilCB,soil + 2 .9998 m3 air × CB,air
+2 × 10−4 m 3fish × CB,fish
c
h
c
h
= 4CB,H 2 O g m3 + 3KSW CB,H 2O g 106 g soil 15
.
+2.9998 m3 air × KAW CB,H 2O
c
+2 × 10−4 m 3f KBW CB,H 2 O g m3 B
−3
10 × 10
h
g
c
h
= CB,H 2 O 4 + 3 × 108
. × 15
. + 2.9998 × 0.2276 + 2.4 × 10 −4 × 6.75
CB,H 2 O = 1.048 × 10
−3
g m = 1.048 ppb ; in water 1 g m =1 ppm
3
CB,soil = 1.08 × 1.048 × 10 −3 =
3
1132
.
× 10−3 g
106 g soil
= 1698
.
× 10−3 g m3
CB,air = 0.2276 × 1048
.
× 10− 3 = 0.239 × 10−3 g m3
CB,fish = 6.75 × 1048
.
× 10 −3 = 7.074 × 10 −3 g m 3
Section 8.9
Solutions to Chemical and Engineering Thermodynamics, 3e
(b) DDT KAW, DDT = 9.5 × 10 −4
K0W, DDT = 1549
.
× 10 6
Proceeding the same way
CDDT,H 2 O = 1.793 × 10 −7 g m3 = 17.93 ppt
CDDT,air = 3.332 × 10− 3 g m 3
CDDT,fish = 13883
.
× 10−3 g m3 = 1388
. ppb
8.9-4 (also available as a Mathcad worksheet).
8.9-4
Kow
224
3.
Pvap
10
2
bar
S
440
mg/liter
MW
157.5
750
0.440
157.5
x
x = 5.029 10
1000
5
mole fraction of ClNO2benzene
18
Gam
1
4
Gam = 1.989 10
x
H
Pvap . Gam
H = 0.795
bar/mole fr
Mass balance: 100 kg = CW*7*10^6 + CA*6*10^9 + CS*4.5*10^4 + Csed*2.1*10^4
Equilibrium relations: CA=Kaw*CW
CS=Ksw*CW
Csed=Ksedw*Cw
Ksw
0.02 . 0.4 . Kow. 1.5
Ksw = 2.688
Cw
3
7 . 10
9
6 . 10 . Kaw
H
298.15
Kaw = 5.773 10
4
4
4.5 . 10 . Ksw
4
2.1 . 10 . Ksedw
g/m^3 or 9.339 ppb by wt
Kaw. Cw
Ca = 5.392 10
Cs
Ksw . Cw
Cs = 0.025
Ksedw. Cw
0.2164 .
Kaw
( 100 . 1000 )
Ca
Csed
0.05 . 0.4 . Kow. 1.3
Ksedw
Ksedw = 5.824
6
Cw = 9.339 10
Kaw=0.2164*H/298.15
Ksw=0.02*0.4*Kow*1.5 (g/m^3)/(g/m^3)
Ksedw=0.05*0.4*Kow*1.3
6
Csed = 0.054
g/m^3
g/m^3 or
Cs
Cs .
1000
Cs = 16.736
1.5
g/m^3 or
Csed
Csed .
1000
1.3
Csed = 41.84
ppb by wt
ppb by wt
Section 8.10
Solutions to Chemical and Engineering Thermodynamics, 3e
8.10-1 This is a simple algebraic exercise, so the details will not be given.
8.10-2 The starting point for the liquidus line is
c h expLM ∆ H bT g RS1 − T UVOP
γ cx h
N RT T T WQ
=
γ cx h
L ∆ H bT g RS1 − T UVOP − γ cx h expLM ∆ H bT
exp M
γ cx h
N RT T T WQ γ cx h N RT
γ L2 x1L
1−
x1L
L
1
S
1
L
1
S
1
S
2
fus
1
fus
2
m ,2
S
1
m ,2
L
2
S
2
m,1
m ,1
L
1
S
1
fus
2
m, 2
g RS1 − T
T T
m ,2
UVOP
WQ
and for the solidus line is
c h expLM− ∆ H bT g RS1− T UVOP
cx h N RT T T WQ
=
γ cx h
L ∆ H bT g RS1 − T UVOP − γ c x h expLM− ∆ H bT g RS1 − T UVOP
exp M−
γ cx h
N RT T T WQ γ c x h N RT T T WQ
1−
x1S
S
1
L
1
S
1
L
1
γ S2 x1S
γ 2L
fus
1
fus
2
m ,2
L
1
m ,2
m ,1
m,1
S
2
L
2
S
1
L
1
fus
2
m ,2
m ,2
a) Regular solution model for the liquid, the solid phase is ideal
the liquidus line is
F Ωc1 − x h I L ∆ H bT g R T UO
1 − expG
GH RT JJK expMN RT ST1 − T VWPQ
=
F Ωc1 − x h I L ∆ H bT g R T UO F Ωcx h I L ∆ H bT g R T UO
expG
GH RT JJK expMN RT ST1 − T VWPQ − expGGH RT JJK expMN RT ST1 − T VWPQ
L 2
1
x1L
fus
2
m, 2
m ,2
L 2
1
fus
1
L 2
1
m ,1
fus
2
m, 2
m,1
m, 2
and for the solidus line is
F Ωcx h I L ∆ H bT g R T UO
1 − expG −
GH RT JJK expMN− RT ST1 − T VWPQ
h IJ expLM− ∆ H bT g RS1 − T UVOP − expFG − Ωcx h IJ expLM− ∆ H bT g RS1− T UVOP
JK N RT T T WQ GH RT JK N RT T T WQ
L 2
1
x1S =
F Ωc1 −
GG RT
H
exp −
fus
2
m ,2
m ,2
2
x1L
fus
1
L 2
1
m ,1
fus
2
m ,2
m ,1
m ,2
So these nonlinear equations must be solved simultaneously for the liquidus and solidus lines,
together with the equations for the second component, and that the sum of the mole fractions in
each phase must be unity.
b) Regular solution model for the solid, and the liquid phase is ideal
the liquidus line is
F Ωc1− x h I L ∆ H bT g R T UO
GG RT JJ expM RT ST1 − T VWP
Q
H
K N
=
F Ωc1 − x h I L ∆ H bT g R T UO F Ωcx h I L ∆ H bT g R T
expG −
GH RT JJK expMN RT ST1 − T VWPQ − expGGH − RT JJK expMN RT ST1 − T
S 2
1
1 − exp −
x1L
fus
2
m ,2
m ,2
S 2
1
fus
1
S 2
1
m ,1
m ,1
fus
2
m ,2
m ,2
UVOP
WQ
Section 8.10
Solutions to Chemical and Engineering Thermodynamics, 3e
and for the solidus line is
F Ωcx h I L ∆ H bT g R T UO
1 − expG
GH RT JJK expMN − RT ST1 − T VWPQ
=
F Ωc1 − x h I L ∆ H bT g R T UO F Ωcx h I L ∆ H bT g R T
expG
GH RT JJK expMN− RT ST1 − T VWPQ − expGGH RT JJK expMN− RT ST1− T
S 2
1
x1S
fus
2
m, 2
m, 2
S 2
1
fus
1
S 2
1
m ,1
fus
2
m ,2
m ,1
m ,2
UVOP
WQ
So these nonlinear equations must also be solved simultaneously for the liquidus and solidus lines, together
with the equations for the second component, and that the sum of the mole fractions in each phase must be
unity.
c) Regular solution model for the liquid and the solid phases, but with different values of Ω.
The liquidus line is
F Ω c1 − x h − Ω c1 − x h I L ∆ H bT g R T UO
GG
JJ expM RT ST1 − T VWP
RT
Q
H
K N
h − Ω c1− x h IJ expLM ∆ H bT g RS1 − T UVOP − expFG Ω c x h − Ω cx h IJ expLM ∆ H bT g RS1 − T UVOP
JK N RT T T WQ GH
JK N RT T T WQ
RT
RT
x1L =
F Ω c1−
expG
GH
L
L 2
1
L
1 − exp
S
S 2
1
fus
2
m, 2
m, 2
2
x1L
S 2
1
S
fus
1
m ,1
L 2
1
L
S 2
1
S
fus
2
m ,2
m ,1
m ,2
and for the solidus line is
F Ω cx h − Ω cx h I L ∆ H bT g R T UO
GG
JJ expM− RT ST1 − T VWP
RT
Q
H
K N
=
F Ω c1 − x h − Ω c1 − x h I L ∆ H bT g R T UO F Ω cx h − Ω cx h I L ∆ H bT g R T
expG
JJ expM− RT ST1 − T VWP − expGG
JJ expM− RT ST1 − T
GH
RT
RT
Q H
K N
K N
1 − exp
x1S
S 2
1
S
L
L 2
1
fus
2
m ,2
S
S 2
1
m ,2
S 2
1
S
L 2
1
L
fus
1
m,1
L
L 2
1
fus
2
m ,1
m ,2
Again, we have a set of nonlinear equations that must be solved simultaneously for the liquidus and solidus
lines, together with the equations for the second component, and that the sum of the mole fractions in each
phase must be unity.
8.10-3
(also available as a Mathcad worksheet).
Problem 8.10-3
x11
File: 8-10-3.MCD
0.1
x12
0.9
T
225
TUC
Given
x11. exp
5000 . ( 1
x11)
8.314 . T
5000 . ( x11)
8.314 . T
(1
x11) . exp
v
find( x11, x12)
2
x12. exp
2
(1
v=
5000 . ( 1
x12) . exp
0.111
0.889
x12)
2
8.314 . T
5000 . ( x12)
8.314 . T
2
m ,2
5000
2 . 8.314
TUC = 300.698
UVOP
WQ
Section 8.10
Solutions to Chemical and Engineering Thermodynamics, 3e
T
250
x11. exp
Given
5000 . ( 1
5000 . ( x11)
8.314 . T
x11) . exp
v
find( x11, x12)
5000 . ( 1
find( x11, x12)
285
2
5000 . ( x12)
8.314 . T
2
0.169
0.831
2
x12. exp
2
(1
v=
5000 . ( 1
x12)
2
8.314 . T
x12) . exp
5000 . ( x12)
8.314 . T
2
0.256
0.744
Given
5000 . ( 1
x11)
2
8.314 . T
5000 . ( x11)
8.314 . T
(1
x11) . exp
v
find( x11, x12)
290
x11. exp
x11)
5000 . ( x11)
8.314 . T
v
T
x12)
8.314 . T
x12) . exp
(1
8.314 . T
x11) . exp
x11. exp
2
5000 . ( 1
Given
(1
T
x12. exp
v=
275
x11. exp
2
8.314 . T
(1
T
x11)
x12. exp
2
(1
v=
5000 . ( 1
x12)
2
8.314 . T
x12) . exp
5000 . ( x12)
8.314 . T
2
0.306
0.694
Given
5000 . ( 1
x11)
8.314 . T
5000 . ( x11)
8.314 . T
(1
x11) . exp
v
find( x11, x12)
2
x12. exp
2
(1
v=
5000 . ( 1
x12) . exp
0.339
0.661
x12)
2
8.314 . T
5000 . ( x12)
8.314 . T
2
Section 8.10
Solutions to Chemical and Engineering Thermodynamics, 3e
T
298
x11. exp
Given
5000 . ( 1
5000 . ( x11)
8.314 . T
x11) . exp
v
find( x11, x12)
5000 . ( 1
x12. exp
2
x12)
2
8.314 . T
x12) . exp
(1
v=
300
x11. exp
2
8.314 . T
(1
T
x11)
5000 . ( x12)
8.314 . T
2
0.418
0.582
Given
5000 . ( 1
x11)
2
x12. exp
8.314 . T
5000 . ( x11)
8.314 . T
(1
x11) . exp
v
find( x11, x12)
2
(1
v=
5000 . ( 1
x12)
2
8.314 . T
5000 . ( x12)
8.314 . T
x12) . exp
2
0.458
0.542
8.10-4 (also available as a Mathcad worksheet)
Solid-liquid phase diagram:
Given: Ω S = 10,000 , Tm,1 = 800 K , Tm, 2 = 600 K ; ∆H 1f = 6200 J mol ; ∆H f2 = 4900 J mol .
liquid phase ideal:
→ solidus line
RS
T
1 − γ S2 exp −
x1S =
RS− ∆ H F 1− T I UV − γ
T RT GH T JK W
f
1
γ 1s exp
m1
F
GH
∆ H f2
T
1−
RT
Tm2
s
2
I UV
JK W
R ∆ H F1 − T I UV
expS
T RT GH T JK W
f
2
(1)
m2
liquidus line
RS F I UV
T GH JK W
FG 1− T IJ UV − 1 expRS ∆ H FG1 − T IJ UV
H T K W γ T RT H T K W
1−
x1L =
RS
T
1
∆ H 1f
exp
γ S1
RT
∆ H f2
1
T
exp
1−
RT
Tm 2
γ S2
f
2
m1
S
2
(2)
m2
Start with Equation (1); pick x1S , find T so that (1) is satisfied. Use T, x1S in (2).
Section 8.10
Solutions to Chemical and Engineering Thermodynamics, 3e
LM Ω cx h OP
N RT Q
L Ω cx h OP
= exp M
N RT Q
S
γ S1 = exp
γ S2
S
S 2
2
S 2
1
To find solid-solid equilibrium, use
S,II S,II
f1S,I = f1S,II → x1S,Iγ 1S,I = x1S,IIγ 1S,II and x2S,I ⋅ γ S,I
2 = x2 γ 2
Ω = 10000
x
TS
TL
0.0001
599.95
599.73
T
x1I
x1II
0.05
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.95
0.9999
573.39
543.92
480.02
418.14
378.88
393.54
458.60
545.01
635.78
722.65
762.83
799.93
476.34
387.62
350.42
366.34
376.07
372.61
351.32
290.99
300
350
400
450
500
550
600
575
601.4
0.021
0.041
0.070
0.111
0.169
0.256
0.458
0.322
0.979
0.959
0.930
0.889
0.831
0.744
0.542
0.678
0.975
0.925
781.83
743.08
698.52
472.85
10000
= 601.40
2 × 8.314
Tuc =
x1Iγ I1 = x1IIγ 1II
c1− x hγ = c1 − x hγ
I
1
I
2
II
1
II
2
595.06
799.59
Liquid – Liquid Eq.
0.5
Section 8.10
Solutions to Chemical and Engineering Thermodynamics, 3e
L
L
S1LE
S1
S2LE
S2
S1S2E
8.10-5
F
H
Ω
RT
G ex = x1 x2 Ω 1 − x1x2
I
K
so that
IJ
Ω
f K
a
f ZRT
FG ∂N G IJ = LM N − N N OPΩ − LM 2 N N − 3N N OP Ω
H ∂N K
N N + N aN + N f Q N a N + N f a N + N f Q ZRT
Ω
Ω
= a x − x x fΩ − 2 x x − 3 x x
= x a1 − x fΩ − x x 2 − 3 x
ZRT
ZRT
F ∂N G IJ = x Ω − x x 2 − 3x Ω
RT ln γ = G
H ∂N K
ZRT
ex
NG =
FG
H a
N1 N2
N1 N2
Ω 1−
N1 + N 2
N1 + N2
2
Ω
N1 N 2
N12 N22
=
Ω−
ZRT
N1 + N2
N1 + N 2
ex
2
1
2
2
2
1
2
T , P, N2
1
1
2
1
2
1
2
3
2
2
3
2
1
2
4
2
1
2
2
2
2
1 2
1 2
2
2 2
1 2
2
ex
1
2
1 2
2
2
2
1
2
1 2
1
T , P , N2
1
and by symmetry
FG ∂N G IJ
H ∂N K
ex
RT ln γ 2 =
2
= x12Ω − x2 x12 2 − 3x2
T , P, N1
Ω2
ZRT
8.10-6 (also available as a Mathcad worksheet).
8.10-6
Treat as freezing point depression problem and use Eqn. 8.5-12.
i
0 , 1 .. 9
xi
1
0.1 . i
1
Section 8.10
Solutions to Chemical and Engineering Thermodynamics, 3e
T
1410
0
T6
T1
1261
∆ Cp ( T )
∆H
1385
T7
26.606
1350
1242
T8
Tm
T3
1215
3
2.469 . 10 . T
23.932
50626
1316
T9
1290
T5
1090
R
8.314
1410
∆ Cp ( x) d x
Tm
i
T4
5 2
4.142 . 10 . T
Ti
1 .
R. T
Term2 i
T2
Term3 i
Ti
1.
R
∆ Cp ( x)
dx
x
Tm
Term1 i
∆H .
1
8.314 . T
i
γ
i
1.
xi
Ti
Tm
Term2 i
Term3 i
exp Term1 i
γ
i
1
1.029
1.041
1.074
1.158
1.341
1.592
2.005
2.774
3.809
4
γi
2
0
0
0.5
x
i
1
1278
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
8.11-1
Clearly, it is only water that condenses out at the dew point, since O 2 and N 2
are far above their critical temperatures. Thus, at the dew point
PHvap
= yH 2O P = PH 2 O = partial pressure of H 2 O in air.
2O
[In writing this expression, all fugacity coefficients have been assumed equal to
unity.]
From the data in Problem 5.12 we have, at the dew point,
54328
.
vap
ln PH 2 O = 26.3026 −
= 7.8086
27315
. + 256
.
and
PHvap
= (dew point) = PH 2 O = 24618
. Pa
2O
at the air conditions PHvap
( T = 25.6° C) = 33549
.
2O
⇒ relative humidity =
8.11-2
( T = 20.6° C)
PH 2 O = PHvap
2O
PHvap
( T = 25.6° C)
2O
× 100% = 73.38%
Equilibrium condition for V-L-S equilibrium: fHV2 O = f HL2O = f HS2 O where
fHV2 O = yH 2O Patm
f HL2 O = vapor pressure of liquid water
f HS2 O = vapor (sublimation) pressure of ice.
[Here, again, we have neglected all fugacity coefficient departures from unity.]
Now, in fact, we know that at normal pressures the liquid is the stable phase above
0°C and the solid at temperatures below 0°C. Thus, liquid droplets will stable at
saturation conditions above 0°C, and water (ice) crystals will be stable at
saturation conditions below 0°C.
At –25°C (248.15 K) and P = 1 2 bar, we have, for equilibrium with the liquid
L
yeq,
H2O =
fHL2 O
Patm
=
Pvap ( water)
= 1.644 × 10− 3
0.5 bar
[from equating f HV2 O = fHL2 O ]
For equilibrium with the solid
eq, S
yH 2 O =
P vap(ice)
= 1268
.
× 10−3
0.5 bar
[from equating f HV2 O = fHS2 O ]
Thus, if the relative humidity (with respect to equilibrium with the liquid) is only
.
× 10
c1268
−3
h
1.644 × 10 −3 × 100% ~ 77.1% the ice crystals will be stable. At
higher relative humidities it is possible to have water vapor in equilibrium with
liquid droplets in a metastable state, at lower relative humidities the ice crystals
will sublime.
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
8.11-3
We will assume, since Hydrochloric acid is a strong acid, that the HCl molecule
will be completely ionized at all concentrations.
Let y = wt% HCl ; 100 − y = wt% of H 2 O .
y
36.5
100 − y
mol H 2 O 100 grams solution =
18.0
2y
mol H + , Cl− ions 100 grams solution =
365
.
(100 − y ) 18.0
2.0278(100 − y )
mole fraction of water =
=
(100 − y ) 18.0 + 2 y 36.5 2.0278(100 − y) + 2 y
mol HCl 100 grams solution =
Mole fractions for each solution are given in the table below.
Next, we use the partial pressure (vapor-liquid equilibrium) data. For water, we
have
fWL = fWV ⇒ xWγ W PWvap = PW ,
where, again, we have neglected all fugacity coefficient corrections. Using, from
Problem 5.12, that for
ln PWvap = 263026
.
− 5432.8 T in Pa;
for bar
ln PWvap = 14.7898 − 54328
. T and from the problem statement that
F
H
ln PW = 2.3026 A −
we obtain
B
T
I
K
a
f
ln xW γ W − ( 2.3026 A − 1.47898) +
1
(54328
. − 2 .3026B)
T
From which we obtain the following results:
y
10
20
30
40
xW
0.9012
0.8022
0.7029
0.6033
a
f
ln xW γ W
–0.19172
–0.53917
–1.16719
–2.12638
xWγ W
0.82554
0.5832
0.31124
0.11927
γ W = xWγ W xW
0.9160
0.7270
0.4428
0.1977
(0.8707)
(0.6472)
(0.3770)
(0.1585)
Note that the activity coefficient for water is significantly less than unity.
[Numbers in parentheses are γ W calculated assuming HCl not ionized.]
8.11-4
Equilibrium between water in air and water in aqueous solution requires that
f HL2 O = fHV2 O . Neglecting fugacity coefficient corrections, we have
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
fHL2 O = xH 2 O γ H 2 O PHvap
= 4 .24 xH 2 Oγ H 2 O
2O
Aentry in table for
bar
0 wt% Na 2 CO 3
fHV2 O
f
= yH 2O P( ) = yH 2 O P = PH 2 O
P
=1
⇒ γ H2O =
PH 2 O
4 .24 xH 2 O
, for PH 2 O in bar
Let W = wt% Na 2 CO 3 ; MWNa 2CO 3 = 106 g mol
W
106
3W assuming Na 2 CO 3 is
moles ions 100 g solution =
106 completely ionized
moles Na2 CO 3 100 g solution =
FG
H
IJ
K
100 − W
18
(100 − W ) 18
100 − W
mole fraction water =
=
(100 − W ) 18 + 3W 106 100 − 0.4906W
moles H 2 O 100 g solution =
Thus
γ H2O =
PH 2 O (100 − 0.4906W )
4.24(100 − W )
Values of γ H 2 O calculated from above equation are listed below:
W
PH 2 O (kPa )
0
4.24
γ H2O
1.00 1.007 1.010 1.015 1.021 1.023 1.011
xH 2 O
1.0
0.974 0.946 0.918 0.887 0.855 0.821
γ H 2 O (assuming
1.0
0.990 0.974 0.959 0.944 0.924 0.891
5
4.16
10
4.05
15
3.95
20
3.84
25
3.71
Na 2 CO3 did not ionize)
8.11-5
xair = mole fraction of air in water (liquid)
xair =
Pair
H
=
Henry’s
law
constant
01333
.
kPa × 103 Pa kPa
≅ 0.3 × 10 −7
4.3 × 104 bar × 105 Pa bar
⇒ xH 2 O = 1 (Do not have to consider air trapped in water)
At equilibrium
fHL2 O = f HS 2O = f HV2 O ;
⇒ xH 2O PHvap
= PHsub
= yH 2 O P = PHvap
since xH 2 O = 1
2O
2O
2O
30
3.52
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
[Note: Because of the low pressures involved, we have neglected f P terms and
Poynting corrections.]
For comparison, in an air-free measurement, we have
PHvap
= PHsub
= PTP
2O
2O
where PTP is the true triple point temperature.
Since PHvap
= PHsub
has to be satisfied in both cases, we would obtain the same
2O
2O
triple point temperature in both the air-free experiment, and the measurement with
air.
In the air-free experiment we measure P = PH 2 O and get the triple point pressure.
In the experiment with air we measure P = PH 2O + Pair and, mistakenly, assume
this is the triple point pressure; actually PH 2 O is the triple point pressure. The
error, ∆P , is equal to the partial pressure of air; here 0.1333 kPa. Thus, we have
∆P
01333
.
× 100 =
× 100 = 218%
.
0.6113
PHTP2 O
% error =
[Note: From the Steam Tables, triple point pressure is 0.6113 kPa.]
8.11-6
S ex = −
=−
FG ∂G IJ
H ∂T K
ex
=−
P, x
∂
∂T
mx G
ex
1 1
P, x
k
∂
∂T
RT x1 ln γ 1 + x2 ln γ 2
P, x
= − R ∑ xi ln γ i − RT ∑ xi
+ x2 G2ex
r
p
FG ∂ln γ IJ
H ∂T K
i
P, x
LM ∂dG Ti OP = − RT ∂ x lnγ
∑
∂T
MN ∂ T PQ
F ∂ln γ IJ
= − RT ∑ x G
H ∂P K
F ∂G IJ = ∂ ∑ x RT ln γ = RT ∑ x FG ∂ln γ IJ
V =G
H ∂P K
H ∂P K ∂P
F ∂ ln γ IJ − PRT ∑ x FG ∂ lnγ IJ
U = H − PV = − RT ∑ x G
H ∂T K
H ∂P K
L ∂ lnγ I + F ∂ lnγ I OP
= − RT ∑ x MF
NH ∂ln T K H ∂ ln P K Q
F ∂H IJ = −2 RT ∑ x FG ∂ lnγ IJ − RT ∑ x FG ∂ ln γ IJ
C =G
H ∂T K
H ∂T K
H ∂T K
ex
H ex = − T 2
2
i
2
i
P, x
P, x
i
i
P ,x
ex
ex
i
ex
ex
2
i
i
i
T, x
ex
ex
P
i
P, x
T, x
i
P, x
i
i
i
P, x
i
T ,x
i
i
i
i
i
T ,x
T ,x
ex
i
2
i
2
P, x
i
i
2
P, x
8.11-7 (a) Since tartaric acid is a weak acid, we will assume it is not ionized. Letting
z = grams of tartaric acid per 100 grams of water, we obtain
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
xT =
mole fraction
of tartaric acid
=
z 150
; xW = 1 − xT .
z 150 + 100 18
xWγ W PWvap = PW = 1013
.
bar .
At the boiling point we have
γ W =1013
.
xW PWvap
Thus,
. The results of the computations appear below.
z
xT
0.0945
0.1752
0.2461
87
177
272
xW
0.9055
0.8248
0.7539
γW
0.9445
0.8744
0.8105
Thus, solution is not ideal.
(b) Now we use Eqn. (8.7-2)
a
f
ln γ W xW = −
a f LM T − T OP − ∆C LM1 − T
N TT Q R N T
∆ H fus Tm
R
m
f
P
m f
From Problem 5.26
CP (liquid ) = 4.22 J g ° C
CP (solid ) = 2.1 J g ° C
m
+ ln
f
Tm
Tf
OP
Q
⇒ ∆CP = 3816
. J mol ° C
Also, from the Chemical Engineers Handbook ∆H fus = 6008.2 J mol . Let
y = Tm Tf , we obtain
LM T − 1OP − 3816
.
. L T
a f 8.314− 60082
1−
× 27316
. NT
Q 8.314 MN T
lna x γ f = −2.6457 y − 1 + 4.590 1 − y + ln y
ln xW γ W =
m
m
f
f
+ ln
Tm
Tf
OP
Q
W W
Procedure is to use the data tabulated above to obtain the product xWγ W , and
then compute y by trial and error. Below are the results of solving the
equation above, and also of neglecting the ∆CP term, i.e., solving
a
f
ln xW γ W = −2 .6457 y − 1
T f also
T f also
z
xw
87
0.9055
∆ T with ∆ CP term
-15.95° C
∆T without ∆CP term
−15.25° C
177
0.8248
−33.76
−30.03
272
0.7539
–49.99
–42.88
8.11-8 (a) Imagine the separation process to occur continuously, as below:
1 mole of mixture
Separator
x1 moles of pure species 1
x2 moles of pure species 2
Q W
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
Mass balance: 0 = 1 − x1 − x2
Energy balance: 0 = H mix − x1 H 1 − x2 H 2 + Q + W
Entropy balance: 0 = S mix − x1 S 1 − x2 S 2 +
Q
+S gen
T
0
Subtracting the product of temperature and the entropy balance from the
energy balance yields
a
f a
f a
f
− x G − x G p = −∆G
= − RT ∑ x ln x − G
0 = H mix − TS mix − x1 H1 − TS1 − x2 H 2 − TS 2 + W
or
k
W = − G mix
1
2
1
2
mix
i = 1, 2
i
i
Thus
W = − RT x1 ln x1 + x2 ln x2 − Ax1 x2
and
Q = −W H mix − x1 H 1 − x2 H 2 = −W − H ex
However,
d
∂ G ex T
i
∂T
=−
P , xi
⇒ H ex = − T 2
∂
∂T
FG IJ
H K
ex
ex
H
∂ G
⇒ H ex = − T 2
2
T
∂T T
RS Ax x UV = Ax x
T T W
1 2
1 2
Therefore
Q = −W − H ex = + RT
= G ex ⇒ S ex = 0
∑ xi ln xi + G ex − G ex
i =1, 2
or
k
Q = RT x1 ln x1 + x2 ln x2
a
p
f
(b) W = 0 ⇒ RT x1 ln x1 + x2 ln x2 = − Ax1 x2 or
T W =0 =
− Ax1x2
.
R x1 ln x1 + x2 ln x2
Upper consolute temperature: T UC =
For an equimolar mixture
T W =0 =
A
2R
a fa f
a fa f
−A 1 2 1 2
A
T UC
=
=
R 1 2 ln 1 2 × 2 4 R ln 2 2 ln 2
and, for this case
Q = − R ln 2 ⋅
T UC
1
A
= − RT UC = −
2 ln 2
2
4
(c) If A is a function of temperature, then
∂ Ax1 x2
Ax x
∂A
ex
H = −T 2
= T 2 12 2 − Tx1x2
∂T
T
T
∂T
P, x
= G ex
I
K
F ∂ A IJ
− x x TG
H∂T K
FG IJ
H K
F
H
1 2
P ,x
P, x
ex
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
a
f
W = − RT x1 ln x1 + x2 ln x2 − Ax1 x2
as before but
a
f
Q = RT x1 ln x1 + x2 ln x2 + x1 x2 T
and, for W = 0 , we have, as before
T=
but T UC is no longer equal to
8.11-9
d
i
FG ∂ A IJ
H ∂T K
P, x
A 4R
ln 2
A
.
2R
d
i
i
d
At equilibrium Gi I T , P, x I = Gi II T , P, xII ; i =1, 2 and along the equilibrium
curve
d
i
dGi I T , P , xI = dGi II T , P, xII ; i =1, 2
Treating T, P and one mole fraction as the independent variables in this binary
system, we obtain
FG ∂ G IJ dx U|
H ∂ x K |V i = 1, 2
F ∂ G IJ dy |
dP + G
H ∂ y K |W
i
dGi I = −S iI dT + Vi IdP +
I
1
1
dGi II = − S iII dT + Vi II
i
II
1
1
Equating dGi I and dGi II ,
−S iI dT + Vi I dP +
FG ∂ G IJ
H ∂x K
i
I
1
dx1 = − Si IIdT + Vi IIdP +
T, P
FG ∂ G IJ
H ∂y K
i
II
1
dy1 ; i =1, 2
T, P
Now multiplying by yi , summing and rearranging gives
2
c
h
IJ
K
c
h
−∑ yi SiI − S iII dT + ∑ yi Vi I − Vi II dP
i =1
= ∑ yi
⇒
FG ∂G
H ∂y
i
II
1 T, P
= 0 by the GibbsDuhem equation
dy1 − ∑ yi
FG ∂ G IJ
H ∂x K
i
I
1
dx1
T, P
F ∂ G IJ
K
∑ yi cVi I − Vi II hdP = ∑ yi cSiI − SiII hdT − ∑ yi GH ∂ xi
I
dx1
1 T, P
Since x’s and y ’s are mix ed,
the Gibbs- Duhem equation
does not apply
c
h
Since Gi I = Gi II , Hi I − TSi I = Hi II − TSiII or SiI − S iII = H iI − Hi II T .
For vapor-liquid equilibrium, with phase I = liquid, and phase II = vapor, we have,
at low and moderate pressures, that
Vi II >> Vi I , and Vi II ≈
RT
P
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
Therefore
FG IJ
H K
2
2
RT
1 2
∂ Gi
−∑ yi
dP = ∑ yi HiI − HiII dT − ∑ yi
T i =1
∂ x1
142P
1i =4
44
3
144
42444
3 i =1
c
− RTd ln P
8.11-10
(a)
dx1
T ,P
∆ H vap for y1 moles
component 1 and
y2 =1 − y1 moles of
component 2 from solution.
⇒ − RTd ln P = −
FG ∂ ln P IJ
H ∂T K
h
=
x
FG IJ
H K
2
1
∂ Gi
∆ H vapdT − ∑ yi
T
∂ x1
i =1
dx1
T ,P
∆ H vap
RT 2
Start from the Gibbs-Duhem equation for each phase
− S K dT + V K dP + ∑ xiK dGi K = 0
where K designates the phase, and is equal to I, II or III here. The criterion
for equilibrium is
Gi I = Gi II = Gi III = Gi (no need to designate phase on Gi )
Along the equilibrium coexistence line
dGi I = dGi II = dGi III = dGi
Also, the pressures are equal in each phase as are the temperatures. Thus, we
have the three equations
dP
dG
+ x1I 1 + x2I
dT
dT
dP
dG1
V II
+ x1II
+ x2II
dT
dT
dG
III dP
V
+ x1III 1 + x2III
dT
dT
V
I
dG2
I
=S
dT
dG2
= S II
dT
dG2
III
=S
dT
However,
d
G = H K − T S K = ∑ xiK Gi ⇒ T S K = H K − ∑ xiK Gi = H K − Gi
Using this result gives
VK
FG
H
IJ
K
dP
dGi 1
1
+ ∑ xiK
+ Gi = ∑ xiK HiK ; K = 1, 2, 3
dT
dT T
T
i
or, in matrix form
i
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
LM V
MMV
NV
I
II
III
OP dP LM x
PP dT + MM x
Q Nx
I
1
II
1
III
1
OPF dG 1 I LM x
PPGH dT + T G JK + MM x
Q
Nx
1
I
2
II
2
III
2
1
OP F dG G I 1 LM ∑ x H
PP GH dT + T JK = T MM ∑ x H
Q
N∑ x H
2
I
i
II
i
III
i
2
I
i
II
i
III
i
OP
PP
Q
Thus, we have 3 algebraic equations for the three unknowns
dG1 G1
dG 2 G2
dP
,
+
and
+
dT
T
dT
T
dT
dP
Using Cramer’s rule and solving for
gives
dT
F I FG
H K H
IJ
K
FG
H
IJ
K
1
∑ xiI HiI x1I x2I
T
1
∑ xiII HiII x1II x2II
T
1
∑ xiIII HiIII x1III xIII2
dP
T
=
dT
VI
x1I
x2I
V II
x1II
xII2
V III
x1III
x2III
This type of relationship was first derived by Gibbs.
(b) The Gibbs Phase Rule is F = C − P − M + 2
i) For liquid-liquid miscibility (only one liquid phase)
P = 2 (vapor, liquid), C = 2 and M = 0
F = 2 − 2 − 0 + 2 = 2 degrees of freedom.
Thus if, at fixed temperature, the liquid phase mole fraction is varied, the total
pressure will change.
ii) liquid-liquid immiscibility (two liquid phases)
P = 3 (vapor + 2 liquids)
Thus
F = 2 − 3 − 0 + 2 = 1 degree of freedom
Consequently, at fixed temperature the two phase compositions and the
pressure are fixed. Varying the average mole fraction would change the mass
distribution between the two phases, but would not change the composition of
either phase or the total pressure. That is, when two liquid phases and a vapor
phase exist in a binary mixture, the equilibrium pressure depends only on
temperature and not on average composition.
8.11-11 Types of equilibrium that could occur are:
i) solid-liquid
ii) liquid-vapor
iii) solid-liquid-vapor
iv) solid-vapor
We will assume
i) Ideal solutions ⇒ fi L = xi Pi vap
ii) Ideal gas phase ⇒ fi V = yi P
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
iii) That oxygen and nitrogen are immiscible in the solid phase.
fi S = fi S = Pi sub .
Thus
For vapor-liquid equilibrium, we have fi V = fi L ⇒ xi Pi vap = yi P
For solid-liquid equilibrium we have fi S = fi L ⇒ Pi sub = xi Pi vap
Calculation of solid-liquid equilibrium:
1) Assume N 2 is the solid phase in equilibrium with the liquid
Choose T, use data in the problem statement to calculate PHsub
and PHvap
, and
2
2
2)
xN 2 = PNsub
PNvap
2
2
Repeat calculation for other values of T
Repeat calculation assuming O 2 is the solid phase, and calculating
3)
4)
xO 2 = POsub
POvap
2
2
5)
At each composition, determine which solid freezes out by determining which
results in the highest melting temperature.
[In this calculation, the tabulated vapor pressure and sublimation pressure data
were plotted as ln P vs 1 T , and this graph was used for interpolation.] Some
results are shown below:
T(K)
PNsub
(T ) PNvap
(T )
2
2
= xN 2
T(K)
POsub
POvap
2
2
= xO 2
34
38
42
46
50
54
58
62
63.2
0.126/0.4467
1.806/4.7867
15.427/32.2533
89.827/154.80
390.93/572.93
1356.8/1733.7
3934.7/4471.2
9873.3/10136.5
0.282
0.377
0.478
0.580
0.682
0.783
0.880
0.974
1.0
45.46
47.62
50.0
52.6
54.35
3.8933/5.5733
10.560/13.733
29.867/34.000
78.667/84.000
0.699
0.769
0.878
0.937
1.0
Since the sublimation and vapor pressures below the normal melting point are so
far below the total system pressure of 1 atm (1.013 bar), we do not have to
consider either solid-vapor or solid-liquid-vapor equilibrium.
For the calculation of vapor-liquid equilibrium we use xi Pi vap = yi P and
.
bar .
∑ xi Pivap = 1013
Thus
c
h
c
P = xN2 PNvap
+ 1 − xN2 POvap
⇒ P − POvap
= xN2 PNvap
− POvap
2
2
2
2
2
h
Therefore
xN2 =
P − POvap
2
PNvap
− POvap
2
2
and
yN2 =
xN2 PNvap
2
P
and the procedure is to choose T, calculate PNvap
and POvap
, and then xN 2 and
2
2
yN 2 . The results are given below:
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
T
P
vap
N2
75.5
1.013
77.5
1.333
80
1.849
82.5
2.467
85
3.284
87.5
4.200
0.2187 0.3147 0.4267 0.5867 0.7600
POvap
2
90.1
K
bar
1.013
bar
xN 2
1.0
0.713
0.455
0.288
0.158
0.074
0
yN 2
1.0
0.938
0.830
0.700
0.512
0.305
0
Below is the vapor-liquid-solid phase diagram for O 2 and N 2 determined by experiment
(B.F. Dodge and A. K. Dunbar, J. Amer. Chem., Soc. 49, 501 (1927); B. F. Dodge, Chem. & Met.
Eng. 35, 622 (1928); M. Ruhemann and B. Ruhemann, “Low Temperature Physics”, Cambridge
Univ. Press, London, 1937, p. 100; and R.B. Scott, “Crogenic Engineering”, Van Nostrand,
Princeton, 1959, p. 286). The main difference between this figure and our calculations is that O 2
and N 2 actually form mixed solids on freezing, which we presumed would not occur.
[I am grateful to my former colleague at the University of Delaware, Prof. K. Bischoff for
bringing these data to our attention.]
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
8.11-12 Note: Error in Problem statement in 1st printing vapor pressure of isobutane is
490.9 kPa not 4.909 kPa.
Based on Illustration 8.4-2 this is a problem in vapor-liquid-liquid (3 phase)
equilibrium. Also, from Problem 8.9-10, we have that the coexistence pressure is
constant over the whole range of average (or total) mole fractions for which two
liquid phases exist. From Illustration 8.4-2, one liquid phase is present for
xisobutane = x1 ≤ 0.1128 and x1 ≥ 0.9284 . For overall mole fractions in the range
0.1128 ≤ x1 ≤ 0.9284 , two liquid phases exist. To compute the V-L-L coexistence
pressure in the one-liquid phase region, we use (neglecting fugacity coefficient
corrections)
x1γ 1 P1vap + x2 γ 2 P2vap = P
where
P1vap = 490.9 kPa ,
and
P2vap = 0493
.
kPa , and γ 1 and γ 2 (or x1γ 1 and x2γ 2 ) are given in Table in
Illustration 8.4-2. Also, the van Laar constants are given there, so γ 1 and γ 2 can
be computed at other compositions. Results are given below
x1
0
0.025
0.05
0.075
0.10
0.1128
↓
0.9284
0.95
0.975
1
x1γ 1 P1vap
+
0
0.3068 × 490.9
05491
.
× 490.9
0.7384 × 490.9
08843
.
× 490.9
0.945 × 4909
.
+
+
+
+
+
0.9582 × 490.9
0.9771 × 490.9
+
+
↓
490.9
P
x2γ 2 P2vap
1 × 0493
.
0.493 kPa
0.9764 × 0.493
151.1
0.9555 × 0493
.
270.0
0.9371 × 0.493
362.9
0.9231 × 0.493
434.6
0.914 × 0.493
464.4
↓
0.7325 × 0493
.
0.4318 × 0493
.
↓
464.4
470.7
479.9
490.9
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
8.11-13 The vapor pressure of water is 1.013 bar at 100°C. To compute the vapor
pressure of acetone, we fit the data in the problem statement to
ln Pvap = A T + B
and
find
A = −36189
. ,
B = 109930
.
and
vap
PAC
(100° C) = 3650
.
bar .
(a) To compute activity coefficients, we will use the van Laar model with
α = 2.05 , β = 150
. , as given in Table 7.5-1. Thus
ln γ AC =
α
aa
ff
2
ff
2
1 + αxAC β 1 − xAC
=
⇒ γ AC = 1.050 at xAC = 0.8
2.05
a a
1 + 1.3667 xAC 1 − xAC
ff
2
and
ln γ W =
β
aa
1 + βxW α 1 − xW
⇒ γ W = 2.921 at xW = 0.2
At vapor-liquid equilibrium
=
1.50
a a
1 + 0.7317 xW 1 − xW
ff
2
vap
xWγ W PWvap + xACγ AC PAC
=P
Substituting the results for γ i , Pi vap above yields
P = 3658
.
bar
Thus, for all pressure above 3.658 bar only a liquid of composition xW = 0.2 ,
xAC = 0.8 will be present.
(b) This calculation is more difficult, since we can not calculate the dew point
“pressure” (at fixed temperature) until the liquid phase composition and
activity coefficients are known. Thus the problem involves a trial-and-error
solution of the equations
vap
xWγ W PWvap = yW P ; xACγ AC PAC
= yAC P , and xAC + xW = 1
where xW , xAC and P are the unknowns (The γ i can be calculated from the
xi using the van Laar equations). By repeated guesses, I find
P = 3601
.
bar; xW = 0.295 and xAC = 0.705
Thus, for all pressures below 3.601 bar, only the vapor (of composition
yW = 0.20 , yAC = 0.80 ) is present.
Note: One should check the conditions of both parts (a) and (b) to the above
problem for the possible occurrence of two coexisting liquid phases.
8.11-14 Using the program VLMU with kCO 2 − n C6 = 011
. (Table 7.4-1) results in no
solution at 140 bar and 75°C. However, trying the bubble point and dew point
pressure programs we obtain the following results (at T = 34815
. K)
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
xCO 2
yCO 2
Program doesn’t
converge at higher
0.001
0.05
0.1
0.3
0.4
0.5
0.6
0.7
0.73
0.75
0.78
0.80
dew point
P, bar
1.21
1.28
1.35
1.76
2.07
2.51
3.17
4.31
4.83
5.25
6.03
6.70
CO 2 concentrations
0.82
7.56
0.1
0.3
0.5
0.7
0.72
0.74
0.76
0.78
0.80
bubble point
P, bar
12.96
37.93
64.15
89.14
91.36
93.49
95.52
97.43
99.17
yCO 2
0.9
0.92
0.93
0.94
0.945
0.947
0.949
0.9495
dew point
P, bar
15.06
20.25
24.64
32.02
38.56
42.62
49.39
52.75
Program doesn’t
converge at higher
CO 2 concentrations
Since the program doesn’t converge at higher concentrations of CO 2 along either
the bubble point or dew point curves, we have to make an estimate of the CO 2
concentration based on the data above. There are two possibilities: (1) The CO 2
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
saturation of the liquid at 140 bar is in the retrograde region at somewhere between
xCO 2 of 0.8 and 0.95 [Note, simple equation of state programs, such as VLMU
typically do not converge in the retrograde region, and more sophisticated
algorithms and numerical methods must be used]; (2) at 140 bar only the vapor
exists, that is, all the hexane vaporizes.
An alternative is to use the activity coefficient approach. We do this here using
regular solution theory and corresponding states for the fugacity coefficients. The
starting point is the equilibrium condition
xiγ i fi L = yi fi V
which, for hexane translates to
xH γ H
FfI
H PK
PHvap exp
sat, H
R|V cP − P
S| RT
T
L
H
vap
H
hU|V = y PF f I
W| H P K
H
H
while for CO 2 , which is considerable above its critical point, we have
xCO 2 γ CO 2
FG f
H
L
( P = 1013
.
bar )
PC
IJ
K
PC , CO 2
CO 2
|RV
expS
T|
L
CO 2
( P − 1.013 bar )
|UV = y PF f I
H PK
W|
CO 2
RT
CO 2
Also, fitting vapor pressure data for n-hexane in “The Chemical Engineers
Handbook” we find PHvap (T = 75° C) = 1226
.
bar . Also T = 75° C = 34815
. K
Next, we have
n-hexane
CO 2
n-hexane
CO 2
Thus
TC ( K)
PC ( bar )
507.4
304.2
29.69
73.76
a
δ cal cc
7.3
6.0
f
a
←Table 7.6-1→
←Table 8.3-1→
a f Pf c f
Tr
Pr
0.686
1.144
4.715
1.898
sat
L
PC
h
0.732
132
55
cf
V
−6
P
h
~0.08←Fig. 5.4-1
0.65←Fig. 8.3-1
~1.15
R| 55 × 10 m mol × (140 − 1013
bar U
|Vγ
ST| 348.15 K × 8.314 × 10 cbar m. h ()mol
K) W
|
xCO 2 × 115
. × 73.76 × exp
f
V L cc mol
12
3
−5
3
CO 2
= yCO 2 × 0.650 × 140
⇒ xCO 2 γ CO 2 = 0.8238 yCO 2
and
RS132 × 10 × (140 − 1226
. )U
T 348.15 × 8.314 × 10 VWγ
xH × 0.732 × 1226
.
× exp
−6
⇒ xH γ H = 6.628 yH
−5
H
= yH × 0.08 × 140
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
a
f
The ideal mixture γ i = 1 solution is
yCO 2 = 0.9696
xCO 2 = 07988
.
(Easily solved using
yH = 0.0304
xH = 0.2012
Mathcad)
To obtain a more accurate solution, regular solution theory will now be used to
compute the γ i ’s
ln γ H =
c
V LHφ2CO 2 δ H − δ CO 2
2
RT
and
ln γ CO 2 =
h
c
V LCO φ2H δ H − δCO 2
2
h
2
RT
using the ideal solution as a first guess and iterating, or using Mathcad and solving
directly, I obtain the following
yCO 2 = 0.9620
xCO 2 = 07747
.
Qualitatively in agreement
yH = 0.0380
xH = 0.2253
with the P-R e.o.s. results.
Note the enormous solubility of carbon dioxide in hexane and, indeed, in reservoir
crude! That is why carbon dioxide has been used in enhanced oil recovery (crude
oil swells so more is recovered, and viscosity drops so the trapped oil in the earth
matrix flows more easily.)
8.11-15 Possibilities:
1 liquid phase
1 vapor phase
2 phases vapor + liquid
2 phases liquid + liquid
3 phases liquid + liquid + vapor
We will assume that only one liquid phase exists and show that this assumption is
correct.
From the data in the problem statement
vap
vap
PEAC
= 09475
.
bar and PEOH
= 08879
.
bar
The bubble point pressure of an equimolar mixture is
P = ∑ xi γ i Pi vap = 0.5 exp 0.896 × 0.52 (0.9475 + 0.8879) = 1148
.
bar
c
h
Since the applied pressure is 1.8 bar, no vapor is in equilibrium with an equimolar
mixture at the specified temperature of 75°C. Now we have to check to see
whether one or two liquids are present at equilibrium. To determine this we start
with
A 2
RT ln γ i = Ax2j ⇒ ln γ i −
x j = A′ x2j
RT
with A′ = 0.896 given in the problem statement. Therefore A = A′RT = 0.896 RT .
Now from eqn. 8.4-14 we have that the upper consolute temperature (the highest
temperature at which two liquids exist) for the one-constant Margules equation is
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
A 0.896 RT
=
= 0.448 T = 155.97 K
2R
2R
TUC =
This temperature is so much below the system temperature of 348.15 K, that the
single liquid phase is the stable phase. Therefore, the equimolar mixture at 75°C
and 1.8 bar is stable, and the only phase present.
8.11-16 For an azeotropic mixture, from eqn. (8.1-3)
γ i = P / Pi vap
a
γ ax
f 00.2747
= 11167
.
.246
0.2747
= 0.51f =
= 11258
.
0.244
γ C xC = 0.51 =
B
C
Since the two activity coefficients are so close, and the azeotrope occurs near
0.5 mole fraction, so I will use the one-constant Margules equation.
a
f
G ex xC = 0.51
= xC ln γ C + xB ln γ B = 0.51 ln(11167)
.
+ 0.49 ln(11258
.
) = 0.11435
RT
= AxC xB = A(0.51)(0.49) = 0.11435
A = 0.4576 therefore Gex =8.314×(273.15+40) ×0.4576 = 1191xCxB
RT ln γ C = 1191xB2 or γ C = exp 0.4576x2B
b
g
RT ln γ = 1191x or γ = expb0.4576x g
Pa x f = x expc0.4576a1 − x f h0.246 + (1 − x ) expb0.4576x g0.244
2
C
B
2
C
B
2
C
C
C
Also
yC ( xC ) =
c
a
xC exp 0.4576 1 − xC
a f
2
C
B
f h0.246
2
P xC
1
0.28
P
y
i
0.5
P
i
0.26
i
0
0.24
0
0.5
x
i
1
b) The LLE upper consolute temperature is
A 0.4576 × 31315
. × 8.314
TUC =
=
= 71.65 K
2R
2 × 8.314
which is much below the freezing point of each compound.
0
0.5
x ,y
i i
1
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
c) Freezing point depression
fus
∆ H Tm,solvent 1
1
ln xSγ S = −
−
R
T Tm,solvent
g FG
H
b
a f
g a f
1
1
R
=
−
ln xSγ S
T Tm, solvent ∆ H fus Tm,solvent
b
Tm,solvent
RTm,solvent
T=
1−
∆H
fus
bT
IJ
K
or
or
g lnax γ f
S S
m, solvent
For cyclohexane freezing
27315
. + 6.6
279.75
=
8.314 × ( 27315
. + 6.6)
1
0
.
8844
ln xBγ B
−
1−
ln xBγ B
2630
For benzene freezing
T=
T=
a f
a f
27315
. + 5.53
278.68
=
8.314 × ( 27315
. + 5.53)
1 − 0.2378 ln xCγ C
1−
ln xCγ C
9953
a f
a f
400
Tc
Tb
i
200
k
0
0
0.5
x ,x
i k
1
8.11-17
c
a)
h
C1 Do
∇ µo + RT ln x1γ 1 = CD∇ x1
RT
1 ∂ ln γ 1
x1 Do ∇ ln x1 + ∇ ln γ 1 = D∇ x1 = x1 Do
+
x1
∂x1
j1 = −
LM
N
D = Do 1 +
∂ ln γ 1
∂ ln x1
OP
Q
LM
N
RT ln γ 1 = A(1 − x1 ) 2
b)
∂ ln γ 1
∂ ln γ1
∂ ln γ 1
= x1
= − x1
= − x1
∂ ln x1
∂x1
∂ x2
LM
N
D = Do 1 +
c)
OP
Q
FH
∂ ln γ1
2A
= Do 1 −
x1 x2
∂ ln x1
RT
FH A x IK
RT
= −2 Ax x
∂
IK
1) Infinite dilution x1 → 0 and D → Do
OP
Q
2
2
∂ x2
1 2
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
2) At LLE critical point TUC =
LM
N
D = Do 1 −
OP
Q
A
at x1 = x2 = 0.5
2R
2 × 2 × R × TUC
=0
R × TUC
3) Negative deviations from Raoult’s law ⇒ A is negative
F
H
D = Do 1 −
I
K
2A
x1x2 with A negative, D > Do
RT
8.11-18 (also available as a Mathcad worksheet).
8.11-18
γ Pinf
γ Hinf
16
6.34
PvapP
20.277
PvapH
28.022
Using the van Laar equation
α
i
γ Pi
ln ( γ Pinf)
β
1 , 2 .. 99
ln ( γ Hinf)
0.01 . i
xi
α
exp
1
xi
α
β 1
γ Hi
2
β
exp
1
xi
β 1
α
xi
2
Pi
xi . PvapP. γ Pi
1
xi . PvapH. γ Hi
xi
50
40
P
yPi
i
30
20
0
0.920522
yP
i
0.097206
0.5
x
i
1
1
0.5
0
0
0.01
0.5
x
i
1
0.99
xi . PvapP. γ Pi
Pi
Section 8.11
Solutions to Chemical and Engineering Thermodynamics, 3e
So this system exhibits either azeotropy or LLE. Test for LLE
x11
0.01
x12
0.90
Given
α
x11. exp
1
(1
α
x11
β 1
y
1
x11
β
x11) . exp
1
α
x12. exp
2
β 1
α
(1
x11
α
β 1
2
x12
x12
β
x12) . exp
2
1
x11
β 1
α
x12
2
x12
find( x11, x12)
x11
y0
x11 = 0.113
x21
1
x11
x12
y1
x12 = 0.684
x22
1
x12
x12 = 0.684
So there is LLE
x21 = 0.887
x22 = 0.316
9
Note that many of the problems in this chapter can be solved relatively easily with two
programs. The first is CHEMEQ which makes the calculation of the chemical equilibrium
constant at any temperature very easy. The second is an equation solving program, such
as Mathcad, for solving the nonlinear algebraic equation(s) which result. It is advisable
that students know how to use both. [I have used Mathcad for many of the problem
solutions reported here.]
9.1
From Equation 9.1-18
ln Ka (T = 25° C) =
− ∆ G °rxn (25° C)
−17,740
=
RT
8.314 × 29815
.
⇒ ln Ka = − 7.1566 and Ka (T = 25° C) = 7.7967 × 10 −4
Next using Eqn. 9.1-23b with ∆a = 16.736 J mol K ; ∆b = ∆c = ∆ d = ∆e = 0 gives
[Note: Error in problem statement of first printing.
∆CP = 16.736 J mol K not
kJ/mol K]
ln
Ka ( T = 500 K )
∆a
500
=
ln
Ka ( T = 298.15 K )
R
298.15
1
o
− ∆ H rxn (298 .15) + ∆ a × 298.15 500 − 1 − 298.15−1
R
16.736
500
=
ln
8.314
298.15
1
1
1
+
−55,480 + 16.736 × 298.15
−
8.314
500 298.15
= 10407
.
+ 8.2228 = 9.2635
ln Ka (T = 500 K ) = ln Ka (T = 298 .15) + 9.2635 = −7 .1566 + 9.2635 = 2 .107
+
F
H
⇒ Ka (T = 500 K) = 8.225
I
K
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Mass Balance Table
Species
In
Out
yi
yi ( X = 0.9436)
IPOH
1
1− X
(1 − X ) (1 + X )
0.0290
Prop
0
X
X (1 + X )
0.4855
H2
0
X
X (1 + X )
Now Ka =
0.4855
→ calculated after X was
found below
1+ X
Total
a H2 a Prop
aIPOH
=
a
a
f
X 2 P = 1013
.
bar 1 bar
X2
=
.
(1 + X )(1 − X )
1− X 2
f
b
a
g
f
X = Ka (1 − X ) ; 1+ Ka X = Ka or X = Ka 1 + Ka ;
2
2
2
Ka = 8.225 ⇒ X = 0.9436 ⇒ 94.36% of alcohol is converted.
9.2
Reaction: CaC2 O 4 = CaCO3 + CO
Ka =
T (° C)
375
Pdiss = PCO ( kPa )
Ka = PCO 100
ln Ka
a CaCO 3 aCO
a CoC2 O 4
388
403
= aCO = PCO
410
416
418
1.09
4.00
17.86
33.33
78.25
91.18
0.0109
0.0400
0.1786
0.3333
0.7825
0.9181
–4.5282 –32.315 –1.7356 –1.1112 –0.2581 –0.1052
T(K)
648.15
661.15
676.15
683.15
689.15
691.15
24.401
− RT ln Ka = ∆G °rxn
17.763
9.757
6.3113
1.4788
0.6351
kJ / mol CaC2O 4
reacted
Now
b
o
d ∆G rxn
RT
dT
g = − d ln K
dT
a
=−
o
∆H rxn
RT 2
(1)
o
o
∆H rxn
− ∆G rxn
(2)
T
Plot the data for ln Ka vs. T. It falls on a reasonably straight line of slope ~ 0.106.
o
and ∆S rxn
=
d ln Ka
o
~ 0.103 . Thus, ∆Hrxn
≅ 0.103 RT 2 , which follows from Eqn. (1). Once
dT
o
∆Horxn is evaluated, Eqn. (2) can be used to get ∆Srxn
. The results are given
i.e.,
below:
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
T (° C)
375°C
403°C
418°C
∆Horxn
358.7
390.4
409.9
kJ/mol CaC2 O 4 reacted
o
∆Srxn
0.5159
0.5630
0.5984
kJ/mol K CaC2 O 4 reacted
0
ln Ka
2
4
6
640
650
660
670
680
690
700
Temperature (K)
9.3
Reactions: C + CO 2 ( g ) = 2CO( g )
(1)
2CO( g ) = 2C + O 2 ( g )
(2)
Mole balance table
Species
Initial
yi
Final
C
CO 2
—
1
1 − X1
CO
0
2 X1 − 2 X 2
O2
0
X2
a1− X f a1+ X − X f
2 a X − X f a1 + X − X f
X a1 + X − X f
1
1
1
2
2
1
2
1
2
2
1 + X1 − X 2
yi P
= yi since P = 1 bar
1 bar
(a) From the program CHEMEQ we find that Ka,1( T = 2000 K ) = 39050
ai =
at T = 2000 K Ka ,2 = 2.445 × 10 −19 ⇒ Ka ,2 ~ 0 , X 2 ~ 0 and Ka ,1 = 39050
Ka ,1 =
a
f
a f
f a fa f
2
Ka ,1
aCO
4 X1 − X 2
4 X1
=
≅
⇒ X 12 =
=1
a CaCO 2
1 − X1 1 + X1 − X 2
1 − X 1 1 + X1
4 + Ka ,1
a
2
fa
2
(as one would expect with such a large equilibrium constant)
(b) At 1000 K using CHEMEQ
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Ka ,1( T = 1000 K ) = 1835
.
Ka , 2 (T = 1000 K) = 3.984 × 10−23
X1 =
1835
.
= 0.561
4 + 1835
.
Thus, the composition of the gas leaving the graphite bed is
Species
CO 2
2000 K
CO
O2
9.4
RS
T
Ka = exp −
=
1000 K
0.283
2.594 × 10−5
1.0
0
UV
W
0.717
0
RS
T
UV
W
o
∆Grxn
−2866
= exp
= 0.3147
RT
8.314 × 298.15
RS
TA
UV
W
a diamond
fi (T , P )
Vi
; where ai =
= exp
( P − 1 bar )
agraphite
fi (T , P = 1 bar )
RT
k
n
p
s
R|d
S|
T
Poynting correction
terms assu med
incompressible solid
i
V dim − V gr ( P − 1 bar )
adiamond exp V dim ( P − 1 bar) RT
=
= exp
agraphite
RT
exp V gr ( P − 1 bar) RT
U|
V|
W
where
V diam =
12 g mol
= 3.4188 cc mol = 3.4188 × 10 −6 m 3 mol
3.51 g cc
V graph =
12 g mol
= 5.3333 cc mol = 5.3333 × 10−6 m3 mol
2.25 g cc
o
( 3.4188 − 5.3333)( P − 1) cc - bar − ∆ G rxn −2866
=
=
J mol
RT
RT
R
2866 J mol
⇒ P − 1 bar =
= 14970
. J cc = 14970 bar
19145
.
cc mol
ln 0.3147 =
or P = 14971 bar .
Thus for
P < 14971 bar ;
adiam
> Ka and graphite is stable phase
agraph
for
P > 14971 bar ;
adiam
< Ka and diamond is stable phase
agraph
⇒ Need a hydraulic press capable of exerting 14971 bar to convert pencil leads to
diamonds. (Also, should consider a higher temperature!)
9.5
For convenience, write reaction as N 2 + O 2 = 2NO 2
Species balance table
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Initial
Final
yi
O2
1
(1 − X ) 4.762
N2
0.79 0.21 = 3.762
NO
0
1− X
3762
. −X
2X
Species
(3.762 − X ) 4.762
2 X 4.762
4.762
yi P
1 bar
( 2 X )2
=
which has the solution
(1 − X )(3.762 − X )
Since P = 1013
.
bar ; a i =
2
yNO
yN 2 yO 2
Ka =
X=
−2.381 + 19073
.
+ 15048
.
Ka
4 Ka − 1
a f
Using the program CHEMEQ and its data base the following results are obtained
(which agree with Figure 9.1-2)
T (° C)
1500
0.9795 ×10
X
0.00954
T (° C)
0.2154 × 10
0.0438
T (° C)
Ka
X
01924
.
×10
2000
−3
05861
.
× 10
01455
.
× 10−2
0.0231
2200
−2
X
1800
−3
0.0133
2100
Ka
9.6
1600
−4
Ka
0.0361
2400
−2
0.3077 × 10
0.0520
2500
−2
05718
.
×10
0.007487
0.0701
0.0796
2600
2800
2900
3000
0.009539
0.01450
0.01732
0.02028
0.0893
0.1086
0.118
0.1269
(a) From Appendix IV
∆H of
∆G of
Na 2SO 4 ⋅ 10H 2 O
–4322.5
–3642.3 kJ/mol
Na 2SO 4
–1382.8
–1265.2
H2O
–241.8
–228.6
⇒ ∆H rxn(T = 25° C) = (− 43225
. ) − ( −13828
. ) − 10(− 228.6 ) = −521.7 kJ mol
∆G rxn(T = 25° C) = −911
. kJ mol
⇒ ln Ka ( T = 25° C) =
+91100
,
= +36.751
(8.314 × 29815
. )
Ka ( T = 25° C) = 9 .139 × 1015
Now
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Ka =
a Na 2SO 4 ⋅10H 2 O
a Na2 SO 4 ⋅ a 10
H2O
F 1 bar I
=G
H P JK
10
= 0.9139 × 1015
H 2O
⇒ PH 2 O ( 25° C) = 2.503 × 10 −2 bar = 0.0253 bar
(b) At 15°C. Since 15°C is near 25°C we will correct Ka for temperature using only
o
the ∆H rxn
term, i.e.
ln
F
H
I
K
o
Ka ( T = 15° C)
∆H rxn
1
1
=−
−
= 7.3438
Ka (T = 25° C)
R
288 .15 29815
.
Ka ( T = 15° C) = Ka (T = 25° C) exp( 7.304 ) = 1358
.
× 1019
PH2O (15° C) = 1.221 × 10− 2 bar = 0.01221 bar
Experimental data (Baxter and Lansing, J.A.C.S. 42, 419 (1920))
PH 2 O (0° C) = 0.003693 bar
PH 2 O (15° C) = 0.01228 bar
PH 2 O ( 25° C) = 0.0256 bar
9.7
(also available as a Mathcad worksheet)
aCS 2
yCS2
Ka =
=
. Since a C = 1 (solid), and
aC aS2
yS2
P = 1 bar = standard state
pressure.
Species balance table:
Species
Initial
Final
C
S2
—
1
1− X
CS 2
0
X
—
Ka = X (1 − X )
yi
—
(1− X ) ⇒ or
X = Ka (1 + Ka )
X
1
Using CHEMEQ I find Ka (750 ° C) = 8.478 and Ka (1000° C) = 6.607 . Therefore
X (750 ° C) = 0.894
and
X (1000° C) = 0.869 .
X = yCS 2
is the percentage
equilibrium conversion of sulfur.
9.8
a f
(a) Ba NO 3
2
solution
∑
m
1
1
( 2) 2 CBa + (1)2 CNO3 + (1)2 CAg + (1) 2 CCl
zi2Ci =
2
2
1
=
4 CBa + CNO3 + CAg + CCl
2
I=
m
r
but CAg = CCl = CAgCl ; CBa = CBa aNO 3 f 2 ; CNO 3 = 2CBa a NO 3 f2
r
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
I=
CBaa NO 3 f2
n
s
1
4 CBaa NO 3 f2 + 2 CBa a NO3 f2 + 2 CAgCl = 3CBaa NO 3 f 2 + CAgCl
2
I
CAgCl
I
ln Ks , where
2
Ks = CAgCl
2111
. × 10−4
7.064 × 10−4
44.02 × 10−4
560
. × 10−4
(mol/liter)
a f
(b) La NO 3
2
01309
.
×10−4 6.4639 × 10−4 2.542 × 10−2
01339
.
×10−4 213259
.
× 10−4 4.618 × 10−2
01450
.
× 10−4 132.21 × 10−4 11498
.
×10−2
01467
.
×10−4 168147
. ×10−4 12.967 × 10−2
–22.4873
–22.4420
–22.2827
–22.2594
(mol/liter)
solution
Using similar analysis to that above yields I = 6CLa aNO 3 f 3 + CAgCl .
CLa aNO 3 f 2
I
CAgCl
I
ln Ks , where
2
Ks = CAgCl
1438
.
× 10−4
5780
.
× 10−4
166
. × 10−4
2807
. × 10−4
(mol/liter)
01317
.
×10−4 87597
.
× 10−4
01367
.
×10−4 348167
.
× 10−4
01432
.
×10−4 99.7432 ×10−4
01477
.
×10−4 168.568 ×10−4
2.9597 × 10−2
59006
.
× 10−2
9.9872 × 10−2
12.983 × 10−2
–22.4751
–22.4006
–22.3077
–22.2458
(mol/liter)
I
( mol / liter)
1
2
Except for the single point or high ionic strength (AgCl in KNO 3 ), all the data
fall on a straight line.
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.9
For BaSO4 = Ba+ + + SO4 − − , we have
I=
1
2
∑z M
2
i
i
≅
o
t
1 2
2 CBa ++ + 2 2 CSO −− = 4 CBaSO 4 = 4 S
4
2
where S = solubility of BaSO 4 in moles/liter
also Ks = CBa + + CSO
−−
4
= S 2 . Note we have neglected the difference between M
and C. Thus
T (° C)
α(Table 7.6)
5
10
15
20
25
S
Ks = S 2
−6
156
. ×10
1.140
1..149
1.158
1.167
1.178
16.7
18.3
19.8
21.6
Ka =
Ka (1 molal)ν + +ν −
Ks °
(1 molal)ν + +ν −
Ks =
and
γ ν±+ +ν −
7.9 × 10−3
2.43 × 10
2.789
3.349
3.920
4.666
(mol/liter)
Now Ks =
I =2 S
−10
8.173
8.556
8.900
9.295
amol liter f amol literf
2
Ks °
ν + +ν −
γ±
.
12
Ks ° = Ka (1 molal)ν + +ν −
or
and
ln Ks = ln Ks °− lnν±+ + ν −
e
= ln Ks °+ νBa ++ + νSO
−−
4
jz
z
Ba ++ SO 4 −−
α I
= ln Ks °+ 2 ⋅ (2 × 2)α I = ln Ks °+ 16α I
d
i
d
i
or Ks ° = Ks exp −16α S = S 2 exp −16α S .
Note that Ka is the equilibrium constant for the reaction
a
f
a
BaSO4 (s) = Ba+ + aq, ideal 1 molal + SO4− − aq, ideal 1 molal
∆Grxn
= − R ln Ka . Thus we have
T
∆Grxn
T (° C)
Ks °
ln Ks °
= − R ln Ka
T
5
184.654
2.261 × 10− 10 –22.2100
10
2.587
–22.0754
183.535
15
3.094
–21.8964
182.047
20
3.607
–21.7430
180.771
25
4.275
–21.5731
179.359
J
mol lit 2
mol K
f
and
a
f
F
H
I
K
∆H rxn
∆Grxn
20490 51,362
21234 51,968
21991 52,457
22760 52,993
23543 53,476
(J/mol) (J/mol)
∆Srxn
–110.99
–108.54
–105.73
–103.13
–100.40
(J/mol)
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
∆Grxn T is essentially a linear function of T as can be seen by plotting the data.
Also, from the plot we have
a
f a
f
∂ ∆Grxn T
∆ ∆Grxn T
~
~ −0.26485 J mol K 2
∂T
∆T
However
a f
a
f
∂G T
H
∂ ∆Grxn T
= − 2 ⇒ ∆H rxn = −T 2
= T 2 ( 0.26485)
∂T
∂
T
T
Finally
∆S =
∆H − ∆G
∆H rxn − ∆Grxn
⇒ ∆S rxn =
T
T
Both ∆H rxn and ∆Srxn are given in the table on previous page.
9.10 (also available as a Mathcad worksheet)
(a) Using the program CHEMEQ, the following results are obtained
rxn 1
C3 H 8 + 3H 2 O = 3CO + 7H 2
rxn 2
C3 H 8 + 6H 2 O = 3CO2 + 10H 2
Ka
rxn
Ka
∆Horxn
o
∆Grxn
∆Horxn
o
∆Grxn
1000 K
1100 K
1000 K
1000 K
1100 K
12
14
1
537.260
–213.030
538.140
01343
.
× 10
0.4806 × 10
2
0.3332 × 1012 0.3851× 1014 432.380 –220.590 436.440
All energies in kJ/mol of C3H8
3
3
a H7 2 a CO
a10
H 2 aCO 2
(b) Ka,1 =
and
K
=
where, since P = 1 bar
a,2
aC 3 H 8 a H3 2 O
aC 3H 8 a H6 2O
yi P
= yi .
1 bar
Species balance table
ai =
Species
In
yi
Out
a1− X + X f ∑
a
f
C3 H 8
1
1 − X1 − X 2
H2O
10
10 − 3 X1 − 6 X 2
CO
0
3 X1
3 X1 ∑
CO 2
0
3 X2
3 X2 ∑
H2
0
7 X1 + 10 X 2
1
a
∑ = 11 + 6 X 1 + X 2
2
10 − 3 X1 − 6 X 2 ∑
a7 X + 10X f ∑
1
f
2
Thus
Ka,1 =
a
27 X 13 7 X 1 + 10 X 2
f
7
a1 − X − X fa10 − 3X − 6 X f a11+ 6 X + 6 X f
3
1
2
1
2
1
2
6
1100 K
–288.11
–286.09
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Ka,2 =
a
27 X 23 7 X 1 + 10 X 2
f
10
a1 − X − X fa10 − 3X − 6 X f a11+ 6 X + 6 X f
6
1
2
1
2
1
6
2
Also
Ka,1
Ka ,2
a
X 13 10 − 3 X 1 − 6 X 2
=
a
X 23 7 X1 + 10 X 2
f
f
3
3
In view of the very large numerical values of the equilibrium constants, we
expect X1 + X 2 ~ 1 . Using this approximation we get
Ka,1 =
a f
a1 − X − X fa7 − 3X f (17)
7
27 X 13 7 + 3 X 2
3
1
2
6
; Ka,2 =
2
a f
a1 − X − X fa7 − 3X f (17)
27 X 23 7 + 3 X 2
10
1
2
6
2
6
and
Ka,1
Ka ,2
a
f ⇒ FG K IJ
=
X a7 + 3X f
HK K
X 13 7 − 3 X 2
3
2
3
13
a ,1
3
2
a ,2
=
a
a
f a fa
f
a
X1 7 − 3 X2
1 − X 2 7 − 3X 2
=
X2 7 + 3 X2
X2 7 + 3 X2
f
f
Note that this last equation is a simple quadratic equation for X 2 given Ka,1
and Ka ,2 . Also, then X1 = 1 − X 2 .
(c) at 1000 K: X1 = 0.527 ; X 2 = 0.473
at 1100 K: X1 = 0.603 ; X 2 = 0.397
Thus we obtain
Species
yi (1000 K )
yi (1100 K )
C3 H 8
0
0
H2O
0.328
0.342
CO
CO 2
0.093
0.083
0.106
0.070
H2
0.495
0.482
9.11 (also available as a Mathcad worksheet)
1
Reaction: SO 2 + O 2 = SO 3
2
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Species balance table
yi
Species
Initial
Final
SO2
1
1− X
1
2
1 0.79
×
= 188
.
2 0.21
1
(1 − X )
2
0
X
O2
N2
SO3
1.88
3.38 −
Since P = 1013
.
bar ; a i =
Ka =
1− X
3.38 − 0.5 X
1 2 (1 − X )
3.38 − 0.5 X
1.88
3.38 − 0.5 X
X
3.38 − 0.5 X
1
X
2
yi P
= 1013
. yi
1 bar
a SO 3
aSO 2 a 1O 22
=
ySO 3
ySO 2 y1O 22
=
X ( 3.38 − 0.5 X )1 2
1.0131 2 (1 − X )3 2 ( 05
. )1 2
The chemical equilibrium constant for this problem was calculated using the
program CHEMEQ and then the problem was solved using Mathcad. The results
appear below
T (° C)
Ka
X
ySO 2
yO 2
ySO 3
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1016
.
× 1014
08625
.
× 109
01012
.
× 107
01265
.
× 105
05903
.
× 103
0.6188 × 102
01104
.
× 102
0.2847 ×101
0.9566
0.3951
0.1863
0.09969
0.05862
0.03722
0.02518
~1
~1
0.9998
0.9967
0.9750
0.8935
0.7090
0.4569
0.2467
0.1252
0.0655
0.0366
0.0220
0.0141
0.0096
~0
~0
~0
~0
6.148 × 10 −5
0.001139
0.008635
0.0363
0.0962
0.1723
0.2313
0.2637
0.2792
0.2866
0.2903
0.2923
0.2934
3.074 × 10 −5
0.000569
0.0004317
0.0182
0.0481
0.0862
0.1157
0.1319
0.1396
0.1433
0.1451
0.1462
0.1467
0.3472
0.3472
0.3472
0.3459
0.3371
0.3046
0.2343
0.1450
0.0757
0.0377
0.0196
0.0109
0.00652
0.00418
0.00285
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
1
1
X
i
YSO2
i
0.5
YO2
i
YSO3
i
0
0
0
0
500
1000
T
1500
3
1.4 .10
i
9.12 Reaction: C2 H 4 + Cl2 = C2 H 4 Cl2
However, C2 H 4 Cl2 is a liquid at 50°C — 2 phase reaction!
Some physical property data
Species
Normal boiling point
P vap (50° C)
TC
PC bar
C2 H 4 Cl2
83.47°C
C2 H 4
–88.63°C
0.288 bar
122.5 bar‡
9.2°C
50.36
Cl2
–34.6°C
14.87 bar
‡ Since ethylene is above its critical temperature, its “liquid-phase” vapor-pressure
will have to be estimated if we are to do the vapor-liquid equilibrium calculation.
However, since we need only a moderate extrapolation (from T = 9.2° C to
T = 50° C ), we will do an extrapolation of the vapor-pressure data, and not use
Shair’s correlation. Using vapor-pressure equation in the Handbook of Chemistry
and Physics, we find Pvap (50° C) ~ 1225
. bar for ethylene.
Note: To be consistent, all vapor pressure data for this problem have been taken
from the Handbook of Chemistry and Physics. The data differs, in many cases by
±20% from The Chemical Engineers’ Handbook. I believe the latter may be more
accurate.
Species balance table
Species
Initial
Final
C2 H 4
1
1− X
= yC 2 H 4 V + xC 2 H 4 L
Cl2
1
1− X
C2 H 4 Cl2
0
X
= yCl 2V + xCl 2 L
= yC 2 H 4 Cl 2V + xC2 H 4 Cl 2 L
yi P
= yi for species whose standard state is a vapor.
1 bar
Note: An obvious first guess is that no C2 H 4 Cl2 is present in the vapor phase,
Also, since P = 1 bar , ai =
and no Cl2 or C2 H 4 is present in the liquid phase. Since the standard states of
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
C2 H 4 and Cl2 are as pure vapors and C2 H 4 Cl2 is as a liquid, this would result in
an equilibrium relation of the form
Ka =
a C 2 H 4 Cl 2
aC 2 H 4 a Cl 2
=
1
=4
1 2 ⋅1 2
which clearly can not be true! Therefore, to obtain the correct solution to this
problem we must consider the possibility that all species may be present in all
phases!
In the table above, L and V are the total number of moles in the liquid and vapor
phases, respectively. L + V = 1
Phase equilibrium: f1V = fi L ⇒ yi P = xi γ i Pi vap
Chemical equilibrium: (standard state: C2 H 4 and Cl2 = vapor; C2 H 4 Cl2 = pure
liquid)
Ka =
xC 2 H 4Cl 2 γ C 2 H 4Cl 2
yC2 H 4 yCl 2
Now using data in Appendices II and IV, and Eqn. 9.1-23b we obtain
Ka (T = 50° C) ~ 11
. × 1023 a huge number.
c
h
⇒ yC 2 H 4 Cl 2 yCl 2 × 11
. × 1023 = xC2 H 4 Cl 2 γ C2 H 4 Cl 2
⇒ Reaction goes, essentially, to completion.
Vapor-liquid equations
yC 2H4 Cl2 = xC2 H4Cl 2 γ C2 H4Cl 2 (0.284 ); yC2 H4 = xC 2H4 γ C2 H4 (121); yCl 2 = xCl 2 γ Cl 2 (14.68)
Now going back to chemical equilibrium relation
xC 2 H 4 γ C 2 H 4 xCl 2 γ Cl 2 × (14.68)(121) 11
. × 10 23 = xC 2 H 4 Cl 2 γ C2 H 4 Cl 2
c
hc
h
c
h
Since, for this system, we expect all the activity coefficients to be of reasonable size
(less than, say, 10), it is clear that the only solution is xC2 H 4 Cl 2 ~ 1 ,
xC2 H 4 = xCl 2 = 0 [actually, these latter mole fractions will be of the order 10−13 ]
Plugging these values back into the vapor-liquid equilibrium equation, we find
∑ yi < 1 . ⇒ No vapor phase!
Thus, the solution to this problem is that there is no vapor phase present at
equilibrium, only a liquid phase. The reaction goes to completion in the liquid
phase, so that
xC2 H 4 Cl 2 = 1 , xC2 H 4 = 0 , xCl 2 = 0 .
9.13
Gi = G i + RT ln γ i xi = G i + RT ln γ i÷ xi÷ where
xi = apparent mole fraction =
xi÷ = actual mole fraction =
N i°
∑ N j°
Ni
∑N j
In the model, γ ÷i = 1 , since the ternary mixture is ideal.
⇒ γi =
xi÷
Ni
N + NB°
N
NA + NB°
=
× A
= i ×
xi
N A + N B + N B2
Ni °
N i ° N A + N B + N B2
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
NA + NB°
N A + N B + N B2
⇒ γA =
γB =
and
NB
N A + NB°
=
N B ° N A + N B + N B2
Now consider the chemical equilibrium:
Initial
Number moles
Final
number moles
Mole fraction
xi
A
NA
NA
NA
N A + N °B − N B 2
B
N °B
N B° −2 X
B2
—
X
N B° − 2 N B 2
N A + N °B − N B 2
N B2
N A + N °B − N B 2
N A + N °B − X
= N A + N B° − N B2
Ka =
a B2
=
a 2B
c
cN
N B2 N A + N °B − N B 2
°
B
− 2 N B2
h
h
2
Solving this equation gives
N B2
cN
=
h cN
°
B
+ NA k −
°
B
+ NA k
h
2
c h
2
− 4 K A N °B
k
2
where k = 4 Ka + 1 , and
2 kN B 2
NA +
N °B
c
kx2B
and δ = (2 k − 1) xa + kxB +
Also we obtain
N B2 =
cN
NA + NB
c
= xa + kx B − kxB2 + 2 kxa x B + xa2
h
12
= 2 kxa + 2 kxB − δ = 2 k − δ
+ 2 kxa xB + xa2
h
12
h 2k − δ ; N = N − c N + N h 2k − δ
2k
k
δ
N
+N °
2k
+ N = a N + N °fF I
H 2k K N + N + N = δ
A
+ N B°
B
B2
A
A
and
B
°
B
A
°
B
A
B
B
B2
Thus
γA =
γB =
LM
N
c
N A + N °B
2k
=
and
N A + N B + N B2
δ
N B N A + N B°
N °B
= 1−
cN
A
h
+ N B + N B2
OP
Q
h
( 2 k − δ) 2 k
2
=
x Bk
δ
x Bδ
{c x
2
A
+ 2 kxA xB + kx2B
h
12
− xA
}
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
9.14 (also available as a Mathcad worksheet)
1
O 2 = CH 3CHO + H 2 O
2
C2 H 5OH = CH 3 CHO + H 2
Reactions: C2 H 5OH +
rxn 1
rxn 2
Species balance table:
Species
C2 H 5OH
In
1
Out
1 − X1 − X 2
O2
0.75
0.75 − 0.5 X 1
H2O
0.79
× 0.75 = 2.8214
0.21
0
CH 3 CHO
0
X1 + X 2
H2
0
X2
N2
Σ
2.8214
X1
4 .571 +
1
X1 + X2
2
Using the program CHEMEQ we obtain
Ka,1 = 7.228 × 1013
aerobic reaction
Ka, 2 = 6.643
is greatly favored!
With these values of the equilibrium constant we obtain X1 ~ 1 and X 2 ≅ 0 [all ethyl alcohol used
up in first reaction].
Therefore,
yC 3 H 5OH ~ 0 ,
yO 2 ~ 0.049 ,
yN 2 ~ 0556
. ,
yCH 3CHO ~ 0197
.
and yH 2 ~ 0 .
9.15 (also available as a Mathcad worksheet)
C3 H 8 = C3H 6 + H 2
rxn 1
C3 H 8 = C2 H 4 + CH 4
Species balance table
rxn 2
Species
Initial
Final
yi
C3 H 8
1
1 − X1 − X 2
1 − X1 − X2
1 + X1 + X2
C3 H 6
0
X1
H2
0
X1
C2 H 4
0
X2
X2
1 + X1 + X2
CH 4
0
X2
X2
1 + X1 + X2
1 + X1 + X 2
Σ
In general, we have a i =
yi P
, thus
1 bar
X1
1 + X1 + X2
X1
1 + X1 + X2
yH 2 O ~ 0197
. ,
Solutions to Chemical and Engineering Thermodynamics, 3e
Ka,1 =
a C3 H 6 a H 2
=
aC 3H 8
Chapter 9
yC 3 H 6 yH 2
F P I= P
H 1 barK 1 bar a1 − X
yC 3 H 8
X12
1 − X 2 1 + X1 + X 2
fa
f
Similarly
Ka,2 =
Also
Ka ,2
Ka ,1
=
a C2 H4 a CH 4
=
aC 3H8
P
X 22
1 bar 1 − X1 − X 2 1 + X1 + X 2
a
X
X 22
, so define α = 2 =
2
X1
X1
Ka,1 =
fa
Ka , 2
Ka ,1
f
. Thus
a P 1 barf X
aP 1 barf X
=
a1 − (1 + α) X fa1 + (1 + α) X f 1 − (1+ α) X
2
1
1
2
1
2 2
1
1
(a) Constant pressure case: P 1 bar = 1
⇒ X1 =
Ka ,1
1+
d
Ka1,12
+
2
Ka1, 22
i
Ka ,2
and X 2 =
1+
d
Ka1 ,12
12 2
+ Ka ,2
i
Results are given in table below.
(b) Constant volume case: Assume gas is ideal
PV = NRT ⇒ Pf =
N f Tf
Ni Ti
LMa
N
= 1 + X1 + X 2
f 298T.15OPQ bar
or
1 + (1 + α) X1 T
29815
.
1 + (1 + α) X 1 T 298.15 X12
X 12T
⇒ Ka ,1 =
=
1 + (1 + α) X1 1 − (1 + α) X 1
298.15 1 − (1 + α) X 1
Pf =
a
⇒ X1 =
298.15 1 2
Ka,1
T
⇒ X2 =
298.15 1 2
Ka ,2
T
RS
T
RS
T
f
2
4T
− K1a,21 + Ka1,22
298 .15
2
4T
− Ka1,12 + Ka1,22
29815
.
Ka1,12 + Ka1,22 +
Ka1,21 + Ka1,22 +
UV
W
UV
W
Results are given in table below.
T(K)
Ka,1
1000
2.907
1200
38.88
1400
246.0
1500
512.6
1600
972.4
1800
2809
2000
6511
Ka,2
534.3
2581
7754
11950
17350
31900
50870
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
Part a
yC 3 H 8
0
0
yC 3H 6 = yH 2
0.034
yC 2 H 4 = yCH 4
0.465
0.055
0
0.076
0
0.086
0
0.096
0
0.114
0
0.132
0.445
0.424
0.414
0.404
0.386
0.368
Part b
yC 3 H 8
0.003
0.001
0.000
0.000
0.000
0.000
0.000
yC 3H 6 = yH 2
0.034
0.055
0.076
0.086
0.096
0.114
0.132
yC 2 H 4 = yCH 4
0.464
0.445
0.424
0.414
0.404
0.386
0.368
P bar
6.69
8.045 9.389 10.061 10.732
12.074
13.416
As usual, all equilibrium constant were computed using the program CHEMEQ and Mathcad.
9.16 (also available as a Mathcad worksheet)
Reaction: N 2 + O 2 = 2NO
Species balance table
Since P = 1 bar , ai =
Species
Initial
Final
N2
1
1− X
O2
1
1− X
NO
Σ
0
2X
yi
1− X
2
1− X
2
X
2
yi P
= yi . Thus
1 bar
a NO
4X2
=
2
a N2 aO2 (1 − X )
Ka =
or
K1a 2 =
Ka
2X
⇒X =
1− X
2 + Ka
(1)
Now, the energy balance for the adiabatic reactor, Eqn. (9.7-10b) is
0 = ∑ Ni
a f
Tout
in
z
CP,i dT + ∆ H rxn Tout X
a f
Tin
or
Tout
X =−
zm
CP, N 2 + CP, O 2 dT
r
Tin
∆ H rxn Tout
a f
(2)
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
Using the program CHEMEQ and the data in Appendix II, together with eqns. (1 and 2), the
following results are obtained.
T(K)
Ka
∆H rxn
2800
2820
2840
2860
2880
2900
2920
2940
2960
2980
3000
0.008002
0.008406
0.008822
0.009251
0.009691
0.01014
0.01060
0.01108
0.01156
0.01206
0.01256
(kJ/mol)
162.26
161.38
160.48
159.54
158.58
157.58
156.55
155.49
154.39
153.26
152.10
X eq
X energy
0.0428
0.0438
0.0449
0.0459
0.0469
0.0955
0.0865
0.0773
0.0681
0.0587
0.0493
0.0397
0.0300
0.0201
0.0101
0.0
0.0479
0.0490
0.0500
0.0510
0.0521
0.0531
The solution is T ~ 29025
.
. K and X = 00482
.
so that yNO = 0.0482 , yN 2 = yO 2 = 04759
.
9.17 Suppose we start with 1 mole of hydrogen and z moles of nitrogen. The species balance table is:
Species
Initial
Final
H2
1
1− 3X
N2
z
z−X
NH 3
0
2X
yi
1 − 3X
1+ z −2X
z− X
1+ z −2X
2X
1+ z −2X
∑ = 1+ z −2 X
and
Ka =
cP
1 bar
NH 3
cP
N2
⇒ Ka
hc
2
h
h
1 bar PH 2 1 bar
F PI
H 1 barK
2
=
3
=
2
yNH
3
yN 2 yH3 2
F 1 bar I
H PK
2
4 X 2 (1 + z − 2 X ) 2
( z − X )(1 − 3 X )3
(1)
Note: We are assuming P is low enough that no f P corrections are needed!
Now we want to know how X changes with z, so we will look at the derivative dX dz at constant T
and P. Starting from Eqn. (1) we obtain
Solutions to Chemical and Engineering Thermodynamics, 3e
0=
−
⇒
Chapter 9
2
8 X (1 + z − 2 X ) dX 8 X 2 (1 + z − 2 X )
dX
1− 2
+
3
3
dz
( z − X )(1 − 3 X ) dz ( z − X )(1 − 3 X )
F
H
4 X 2 (1 + z − 2 X ) 2
12 X 2(1 + z − 2 X ) 2
dX
1
−
−
dz
( z − X )2 (1 − 3 X )3
( z − X )(1 − 3 X )4
F
H
I
K
I
K
F −3 dX I
H dz K
1 ( z − X ) − 2 (1 + z − 2 X )
dX
=
2 X − 4 (1 + z − 2 X ) + 1 ( z − X ) + 9 (1 − 3 X )
dz
a f
where X must be equal to, or smaller than, the smallest of 1/3 and z.
z− X
⇒ z = 1 . Let z = 1 + δ where δmay be either + or –.
For yN 2 = 0.5 =
1+ z − 2X
dX
1 (1 + δ − X ) − 2 (2 − 2 X + δ)
=
dz
2 X − 4 (2 − 2 X + δ) + 1 ((1 − X ) + δ) + 9 (1 − 3 X )
a f
Since X ≤ 1 3 , the denominator is always positive, so we need only look at the numerator to
determine the sign of dX dz .
Num =
1
1
−
1+δ− X 1− X + δ 2
a f
Now do Taylor series expansions in
Num =
δ
δ
and
.
1− X
2(1 − X )
δ
δ
1
1
−δ
1−
−
1−
=
1− X
1− X
1− X
2(1 − X )
2 (1 − X )2
F
H
FG
H
I
K
IJ
K
Thus, the sign of dX dz is the same as the sign of ( −δ) . If δ > 0 , i.e., more N 2 is added than is
. , dX dz < 0 , so that NH 3 decomposes, and N 2 is produced.
required for yN 2 = 05
. , the addition of N 2 causes
If, however, δ < 0 , i.e., less N 2 is added than is needed for yN 2 = 05
more NH 3 to be formed.
Note: If, instead of N 2 addition at constant pressure, the nitrogen was added at constant total
volume, so that partial pressure of either species were unaffected, and the partial pressure of N 2
increased, then, from
Ka =
cP
NH 3
cP
H2
1 bar
hc
2
h
h
1 bar PH 2 1 bar
3
it is clear that the reaction would always go in the direction of increased ammonia production. This
is an important distinction between reactions at constant volume and at constant pressure in this case.
9.18 (also available as a Mathcad worksheet)
Reactions: C4 H10 = C4 H 8 + H 2
C4 H 8 = C 4 H 6 + H 2
rxn 1
rxn 2
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
Species balance table
Species
Initial
Final
C4 H 10
1
1 − X1
C4 H 8
0
X1 − X 2
C4 H 6
0
X2
H2
0
X1 + X 2
yi
1 − X1
1 + X1 + X2
X1 − X 2
1 + X1 + X2
X2
X
1 + 1 + X2
X1 + X 2
1 + X1 + X2
1 + X1 + X 2
Σ
yi P
= yi . Thus
1 bar
Since pressure =1 bar , a i =
Ka,1 =
a X − X fa X + X f
a1 − X fa1 + X + X f
− X − X X h= X − X
yC 4 H 8 yH 2
yC 4 H 10
⇒ Ka, 1 1 + X 2
c
=
1
2
1
1
2
1
1
2
1
2
2
1
2
2
2
(1)
and
Ka,2 =
yC 4 H 6 yH 2
yC 4 H 8
⇒ Ka,2 X1 +
c
X aX + X f
a X − X fa1 + X + X f
− X − X h= X X + X
=
2
1
X 12
2
1
2
2
2
2
1
1
2
2
2
2
(2)
Using CHEMEQ (for equilibrium constants) and Mathcad (for solution) I obtain
T
900
1000
Ka,1
Ka,2
0.9731 0.1191
5.814 0.5575
X1
X2
yC 4 H 10
yC 4 H 8
yC 4 H 6
yH 2
0.724
0.951
0.147
0.464
0.147
0.020
0.308
0.202
0.079
0.192
0.466
0.586
9.19 This problem can be solved graphically, as shown here, or analytically as seen in the Mathcad
worksheet.
1
O 2 = H 2O
2
Using the program CHEMEQ, the equilibrium constant can be computed at each temperature. The
yP
.
bar ai = i = yi × 1013
results are given on the next page. Also, since P = 1013
.
.
1 bar
Reaction: H 2 +
Solutions to Chemical and Engineering Thermodynamics, 3e
(a)
Chapter 9
Stoichiometric amount of pure oxygen
Species
In
Out
H2
1
1− X
O2
0.5
1
(1 − X )
2
H2O
0
X
Σ
⇒ Ka =
1
(3 − X )
2
aH2O
=
a H 2 aO1 2 (1.013)1 2
2
=
yi
2(1 − X )
3− X
1− X
3− X
2X
3− X
( 3 − X )1 2
2X
3− X
⋅
⋅
3 − X 2 (1 − X ) (1 − X )1 2 (1013
. )1 2
X (3 − X )
1
( − X )3 2 (1.013)1 2
12
or
Ka (1 − X )3 2 (1013
. )1 2 − X (3 − X )1 2 = 0
This will be solved using Mathcad.
From the energy balance we obtain the following
T(K)
X Eng (Part a)
X Eng (Part b)
X Eng (Part c)
1000
1200
1300
1400
1500
1600
1800
2000
2200
2400
2600
2800
3000
3200
3400
3600
0.129
0.167
0.194
0.226
0.207
0.279
0.246
0.286
0.328
0.370
0.412
0.455
0.498
0.542
0.585
0.629
0.674
0.333
0.388
0.443
0.500
0.557
0.615
0.674
0.733
0.792
0.852
0.911
0.509
0.658
0.734
0.811
0.899
0.968
1.127
and, from Eqn. (9.7-10b), we get
Ted
C
X = − ∑ Ni
i =1
a f
in
z
CP , i dT
Tin
∆ H rxnaT
ed
f
Solutions to Chemical and Engineering Thermodynamics, 3e
C
where
∑ a Ni f inCP , i = CP , H
i =1
2
+
Chapter 9
1
CP ,O 2 and Tin = 29815
. K.
2
(b) 100% excess oxygen
Species
In
Out
H2
1
1− X
O2
1
H2O
0
1−
yi
1− X
2 − 0.5 X
1 − 0.5 X
2 − 0.5 X
X
2 − 0.5 X
1
X
2
X
2−
1
X
2
⇒ Ka (1 − X )(1 − 05
. X )1 2 − X (2 − 05
. X )1 2 = 0 and, for energy balance
C
∑ a Ni fin CP ,i = CP ,N
i =1
2
+ CP, O 2
(c) 100% excess oxygen in air
Species
In
Out
H2
1
1− X
O2
1
1 − 05
. X
N2
0.79
× 1 = 3.762
0.21
3.762
H2O
0
X
yi
1− X
5.762 − 0.5 X
1 − 0.5 X
5.762 − 0.5 X
3.762
5.762 − 0.5 X
X
5.762 − 0.5 X
5762
.
− 05
. X
⇒ Ka (1 − X )(1 − 05
. X )1 2 − X (5762
.
− 05
. X )1 2 = 0
and
C
∑ a Ni f inCP , i = CP , H
i =1
2
+ CP ,O 2 + 3.762 CP, N 2
From the intersections of the equilibrium and energy balance curves, we obtain the following
solutions [curves on following page] (or directly by solving the equations using MATHCAD)
(a) Tad = 3535 K
yH 2 = 0291
.
X = 0.659
yO 2 = 0146
.
.
yH 2 O = 0563
(b) Tad = 3343 K
X = 0835
.
yH 2 = 0104
.
yO 2 = 0.368
.
yH 2 O = 0528
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
(c) Tad = 1646 K
yH 2 ~ 0
X ≅ 10
.
yO 2 = 0.095
.
yH 2 O = 0190
yN 2 = 0715
.
9.20
Using the data in the problem statement, Tables 2.4 and A6.1, I find
Ka,1( 750 K ) =
a CaOSiO2 aCO 2
a CaCO3 aSiO2
= 148.1 = a CO 2 =
yCO 2 P
1 bar
,
(1)
since the activity of all the solids are unity.
Ka,2 (750 K ) =
3
aSiO
a
a
2 Fe3 O 4 CO
a 3FeO⋅SiO2 aCO 2
= 0.0277 =
aCO
y
= CO ,
a CO 2
yCO 2
(2)
and
Ka,3 ( 750 K ) =
3
a Fe3 O 4 aSiO
2
3
a FeOSiO
a1 2
2 O
= 0.8973 × 10 =
2
⇒
yO 2 P
1 bar
14
1
a 1O 2
2
F 1 bar I
=G
H y P JK
12
O2
.
= 1242
× 10−28
From eqn. (2) we have
yCO
y
~ 0.0277 , while from spectroscopic observations CO ≅ 10 −4
yCO 2
yCO 2
(3)
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
Also, from eqn. (1), PCO 2 ~ 148 bar , while from the probe, the total atmospheric pressure is only
between 75 and 105 bar. Finally, from Eqn. (3), we conclude there is no O2 in the atmosphere,
compared to a trace from spectroscopic observations.
Conclusions? Somewhat ambiguous!
Calculations and data are not in quantitative agreement, but are certainly in qualitative agreement.
Consider the uncertainty in all the measurements, the atmospheric model is undoubtedly a
reasonable one, and can not be rejected.
9.21 (a) The condition for chemical equilibrium is
F −∆G I = a
GH RT JK a a
; vapor phase
(1)
F − ∆G I = a
GH RT JK a a
; liquid phase
(2)
KaV = exp
V
rxn
V
EB
V V
S H
or
KaL = exp
L
rxn
L
EB
L L
S H
where ∆G irxn is the standard state Gibbs free energy change on reaction in phase i.
The phase equilibrium requirements are
L
V
fEB
= fEB
, fSL = fSV and fHL = fHV
(3)
From problem statement ∆G Vrxn = −830
. kJ mol , and KiV = 3482
.
×1014 . This implies that the
reaction will go, essentially, to completion in the gas phase. Now G = H − TS , and for most
liquids neither ∆H vap or T∆ S vap is more than several kJ/mol. Also, since, for hydrogen, the
vapor is the stable phase, G LH 2 > G VH 2 . Therefore, it seems likely that ∆G Lrxn will be of about
the same size and sign as ∆G Vrxn . Consequently, the liquid phase chemical equilibrium constant
will also be large, and the hydrogenation reaction will essentially go to completion in the liquid
phase.
⇒ mole fraction of styrene will be very small in both phases. The problem then reduces to
determining the solubility of the excess hydrogen in the liquid ethyl benzene, and determining
the amount of ethyl benzene in the vapor. Thus, the equations to be solved are
L
V
L
fEB
= fEB
⇒ xEBγ EB fEB
= yEB P
and
fHL = f HV ⇒ xHγ H f HL = yH P
Here we have assumed that the vapor phase is ideal.
As a first guess, we will assume that very little hydrogen is dissolved in the liquid phase. Thus,
γ EB = 1 , φEB = 1 , and, using regular solution theory
V H δ EB − δH
RT
L
ln γ H =
a
f
2
. cc mol × (8.8 − 3.25) cal cc × 4.184 J / cal
31
= 0.1612
8.314 J mol K × 298.15 K
2
=
⇒ γ H = 1175
.
Next we have to estimate the fugacity of hydrogen in the liquid phase. An obvious way to
proceed is to use Shair’s correlation, in Sec. 8.5. However, hydrogen was not used in
developing this correlation, and Prausnitz warns against its use for light gases such as hydrogen
and helium; since experimental data are not available, we have little choice but to use this
Solutions to Chemical and Engineering Thermodynamics, 3e
correlation.
Chapter 9
Note, however, that for hydrogen, TC = 332
. K and PC = 12.97 bar , so that
Tr = T TC = 8.98 , which is off the scale of Fig. 8.3-1. If we extrapolate this correlation to
Tr = 8.98 (a very serious assumption), then we obtain f
c
L
PC
h
1.013 bar
~ 4 , and
f L (1013
.
bar, 25° C) = 4 × 1297
. bar = 5188
. bar
[Note the the Poynting pressure correction of this result to 3 bar total pressure is negligible.]
As a first guess, we will assume that the gas-phase is essentially pure hydrogen. Therefore,
y P
3 bar
= 0.049 , and xEB = 1 − xH = 0.951 .
xH = H L =
γ H f H 1.175 × 51.88 bar
Using the vapor pressure data for ethyl benzene, plotted in the form of ln P vap vs 1 T , we find
vap
= 1273
.
kPa at 25°C.
that PEB
vap
0.951 × 1 × 1.273
xEBγ EB PEB
=
≅ 0.004
3 × 100 kPa
P
and yH = 0.996 [Since the gas phase is almost pure hydrogen, as assumed, there is no need to
yEB =
iterate to a solution].
xH = 0.049
yH = 0.996
at T = 25° C
⇒ xEB = 0.951 and yEB = 0.004
and P = 3 bar
xS ≅ 0
yS ≅ 0.0
An alternative calculation is to use the Peng-Robinson equation of state. The critical properties
of hydrogen are given in Table 4.6-1. The values for ethylbenzene are TC = 617.2 K, PC = 36
bar, ω=0.302, and TB=409.3 K. There is no binary interaction parameters for hydrogen with
other components in Table 7.4-1, so we will assume that its value is zero. Using the isothermal
flash calculation in the program VLMU we obtain the following results
xH = 0.0018
yH = 0.9952
at T = 25° C
xEB = 0.9982 and yEB = 0.0048
and P = 3 bar
xS ≅ 0
yS ≅ 0.0
This may be a more accurate calculation than using regular solution theory which required an
extrapolation of the Prausnitz-Shair correlation.. However, the result is based on the assumption
that k ij = 0. It would be better to have some experimental data to get a better estimate of this
parameter.
(b) At 150°C and 3 bar.
Using the data in
the Problem statement, Appendices II and IV we find
Ka (T = 150° C) = 31
. × 10 . So again we can presume that all the styrene in the vapor and liquid
8
phases is converted to ethyl benzene.
As a first approximation (iteration), we will assume that the liquid phase is essentially pure ethyl
benzene. Thus we obtain
31
. ( 3.25 − 8.8)2
γ H = exp
= 112
.
1987
.
× 423.15
Here again, we find, extrapolating Fig. 8.3-1, that
3
fHL ~ 5188
. bar ⇒ xH =
= 0.052
5188
. × 112
.
and xEB = 1 − 0.052 = 0.948 .
RS
T
UV
W
vap
Now, however, PEB
~ 1303
.
bar
1.303 × 0.948
= 0.412
3
Now using these values for another iteration, we obtain
⇒ yEB =
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
xH = 0.029
yH = 0.586
T = 150° C
xEB = 0.971 yEB = 0.414 3 bar
xST ~ 0
yST = 0.0
Again using the Peng-Robinson equation of state, the program VLMU and the assumption that
kij = 0, we obtain the following
xH = 0.0016
yH = 0.5034
T = 150° C
xEB = 0.9984
yEB = 0.4966 3 bar
xST ~ 0
yST = 0.0
In both parts a and b we see that the results of the equation of state calculation are in qualitative,
but not quantitative agreement with the Prausnitz-Shair correlation. However, the latter predicts
much higher solubilities of hydrogen in the liquid phase. The equation of state calculation is
much easier to do (given the availability of the program VLMU). If some experimental data
were available for hydrogen solubility in ethylbenzene (or other aromatics), the value of k ij
could be adjusted to reproduce that data. Then we would have more confidence in the equation
of state predictions for the problem here. If such experimental data were available, it is not clear
how one would adjust the Prausnitz-Shair correlation to match that such data.
9.22 (a) Energy balance on a fixed mass of gas
dU
= Q& − P
dt
but CV =
Q&
∂T
∂t
dV
0
dt
FG ∂U IJ
FG IJ H ∂ T K
H K
=
+
.
W
0
∂U
⇒ Q& =
∂t
FG IJ = FG ∂U IJ FG ∂T IJ
H K H ∂T K H ∂t K
V
V
V
.
V
V
Now U = ∑ N iUi = ∑ Ni U i , since we will assume the gas is ideal at the temperatures and
T
z
pressures encountered here. Also N i = Ni ,0 + νi X and U i ( T ) = U i TR + CV dT where TR is
a f
TR
some convenient reference temperature.
FG ∂U IJ = FG ∂ IJ ∑ N U
H ∂T K H ∂ T K
F ∂ X IJ + ∑ N C
= ∑ν U G
H ∂T K
CV ,eff =
i
V
i
i
V
i V ,i
i
V
FG ∂ N IJ U + ∑ N FG ∂U IJ
H ∂T K
H ∂T K
F ∂ X IJ + ∑ N C (T)
= ∆U (T )G
H ∂T K
=∑
i
i
i
i
V
V
i V ,i
rxn
V
where ∆U rxn = ∑ ν i U i = internal energy change on reaction.
Species balance table
Species
In
Out
N2O4
1
1− X
NO 2
0
2X
Σ
1+ X
yi
1− X
1+ X
2X
1+ X
Solutions to Chemical and Engineering Thermodynamics, 3e
Ka =
2
aNO
2
cy
=
cy
Chapter 9
1 bar
NO 2 P
a N 2O 4
N 2O4 P
2
h
1 bar
h
By the ideal gas law PV = NRT ⇒
=
2
yNO
P
2
yN 2O 4 (1 bar )
=
a
f
4 X 2 P 1 bar
(1 − X )(1 + X )
P
P
= 0 where the subscript
NT N 0T0
0
denotes the initial
conditions
⇒ P = P0
.
bar(1 + X )T
FG N IJ FG T IJ = 1013
H N KHT K
300
0
⇒ Ka =
or
0
4 X (1 + X )T ⋅ 1013
.
4 ⋅ 1013
.
⋅ X 2T
=
(1 − X ) T0
(1 − X )(1 + X )T0
2
X2
T0 Ka
=
=α
1 − X 4 ⋅ 1.013 ⋅ T
X + αX − α = 0 ⇒ X =
F dX I
H dT K
but
=
V
X 2 − α (1 − X ) = 0
RS
T
α
2
1+
LM RS 1 + 4 − 1UVOP = LM X −
N T α WQ N
d α
dT 2
UV
W
4
−1
α
(1)
OP d lnα
1 + a 4 αf Q dT
1
d ln α d ln Ka 1 ∆U rxn
=
− =
. Also
dT
dT
T
RT 2
∆U rxn = ∑ ν i U i = ∑ ν i Hi − RT = ∆ H rxn − RT ∑ ν i
a
f
(2)
∑ νi = 1 ⇒ ∆U rxn = ∆ H rxn − RT
2
⇒ CV ,eff = (1 − X )CV , N 2 O 4 + XCV , NO 2 +
a∆U f LM X −
RT
N
rxn
2
OP
1 + a 4 αf Q
1
(3)
First two terms give the composition (mole fraction) – weighted heat capacity of the individual
components; the last term is the enhancement of the heat capacity due to the chemical reaction.
This term has one ∆U rxn dependence since that amount of energy is absorbed as the reaction
equilibrium shifts, and a second ∆U rxn dependence, since this determines the extent of a shift in
the equilibrium with temperature.
(b) Using CHEMEQ and the data in Appendix II (for CP* ), ln Ka was determined at each
temperature along with CV for both NO 2 and N 2 O 4 . Then, from Eqns. (1) and (3) X and
CV ,eff as well. These are tabulated and plotted below.
T (K)
300
350
400
450
500
550
600
700
α
X
0.044275 0.1891
1.0016
0.6177
10.0613 0.9163
58.883
0.9835
236.25
0.9958
721.36
0.9986
1796.25 0.9994
7220.4 0.99986
P (bar)
CV ,NO 2
CV , N 2 O 4
CV ,eff
1.205
1.912
2.588
3.014
3.370
3.712
4.051
4.727
37.11
38.94
40.65
42.22
43.67
45.00
46.23
48.36
78.83
84.47
89.55
94.06
98.01
101.39
104.20
108.12
J/mol K
410.5
546.9
195.9
69.11
49.23
46.49
46.71
48.45
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
9.23 This is a very difficult problem. I used the NASA chemical equilibrium program, referenced in Sec.
9.4, in the solution of this problem. I will describe here how this problem could be solved without
this program.
First we need to identify the independent chemical reactions among the components. Starting from
2C + 2H = HCCH
2O = O 2
and first using
C + O = CO
2H = HCCH − 2C
2H + O = H 2 O
to eliminate H, and then
1
O = O2
2
to eliminate O
C + 2O = CO2
2H = H 2
yields
1
O 2 = CO
2
1
HCCH + O 2 = H 2 O + 2C
2
C + O 2 = CO 2
C+
HCCH = H 2 + 2C
From Fig. 9.1-2 we have that Ka for the reaction C + 1 2 O 2 = CO is very large over the whole
a f
temperature range (i.e., K = Oc10 h at 1000 K and Oc10 h at 3000 K). Since O is present in
excess, this implies that their will be no solid carbon present. Thus, we will eliminate C using the
reaction equation C = CO − a1 2 fO ⇒
10
a
6
2
2
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
3
1) HCCH + O 2 = H 2 O + 2CO
2
1
2) CO + O 2 = CO 2
2
3) HCCH + O 2 = H 2 + 2CO
The three equations above form a set of independent reactions that can be used for the description of
this reaction system. In fact, since eqns. (1) and (3) are both expected to go to completion, I used the
following reaction sequence for the description of this system:
5
O 2 = H 2 O + 2CO 2
2
1
CO + O 2 = CO 2
2
1
H2 + O2 = H 2O
2
HCCH +
rxn 1
rxn 2
rxn 3
The first step in the numerical solution of this problem is the calculation of the equilibrium constants
and heats of reaction for the reactions above. Using the program CHEMEQ I obtained:
T
∆Hrxn,1
3000
–1310
3100
–1320
3200
–1332
159
. × 10
2.90 × 10
3300
–1345
3400
–1361
kJ
582
. × 10
127
. × 10
∆H rxn,2
256.2
251.2
245.5
238.8
221.0
Ka,2
0.3246
0.4507
0.6092
0.8027
1.033
∆H rxn,3
261.9
264.9
266.9
269.9
273.5
Ka,3
0.0476
0.0668
0.0922
0.1252
0.1675
3500
–1378
3600
–1398
3700
–1419
3800
–1444
3900
–1470
Ka,1
T
∆Hrxn,1
Ka,1
16
15
14
14
2.97 × 10
13
kJ
kJ
kJ
7.45 × 1012 198
. × 1012 554
. × 1011 1633
. × 1011 5006
.
×1010
∆H rxn,2
222.2
212.1
200.6
187.6
173.0
Ka,2
1.299
1.598
1.926
2.274
2.633
∆H rxn,3
277.6
282.2
287.5
293.5
300.2
Ka,3
0.2213
0.2891
0.3738
0.4793
0.6098
kJ
kJ
Clearly, with such a large value of the equilibrium constant, reaction (1) must go essentially to
completion. I will assume it does. Thus, the reaction stoichiometry is
5
HCCH + O 2 = H 2 O + 2CO2
2
1
CO 2 = CO + O 2
2
1
H2O = H2 + O2
2
U|
| reactions are the
= ?V
|| inverses of
= ? reactions 2 and 3 above.
|W
X1 = 1 Note: These
X2
X3
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
1
15
4
After rxn 1
goes to completion
0
5
4
0
5 1
1
+ X 2 + X3
4 2
2
CO 2
0
2
2 − X2
H2O
0
1
1 − X3
H2
0
0
X3
CO
0
0
X2
Species Initially
HCCH
O2
Equilibrium
∑=
⇒ Ka,2
b g
−1
=
=
17 1
+
X2 + X3
4 2
a
a CO a O1 2
=
2
a CO 2
yi
F
H
0
5 1
1
+ X2 + X3
4 2
2
2 − X2
∑
1 − X3
∑
X3
∑
X2
∑
I
K
∑
f
yCO yO1 2
2
yCO 2
X2 5 4 + 1 2 X2 + 1 2 X3
12
a f a f a f
a2 − X fa17 4f + a1 2f X + a1 2f X
2
2
(1)
12
3
(where for simplicity I have assumed that the standard state and atmospheric pressures were the
same) and
−1
bK g
a,3
12
=
aH 2 aO 2
aH2O
12
=
yH 2 yO 2
yH 2 O
=
X3 5 4 + 1 2 X2 + 1 2 X3 1 2
a f a f a f
a1 − X fa17 4f + a1 2fX + a1 2f X
3
2
12
(2)
3
where the equilibrium constants Ka,2 and Ka,3 are the ones whose numerical values are given in the
table above.
Instead of solving these nonlinear algebraic equations, I used the NASA Gibbs free energy
minimization program to find the equilibrium mole fractions. Since this package uses a different set
of thermodynamic data, the computed mole fractions do not agree with eqns. (1 and 2) and the table
of equilibrium constants given above. The results are:
T (K)
3000
3200
3400
3600
3800
4000
yCO
0.1530 0.2126 0.2605 0.2940 0.3154 0.3284
yCO 2
0.2777 0.2017 0.1395 0.0945 0.0642 0.0442
yH 2
0.0153 0.0254 0.0384 0.0539 0.0705 0.0869
yH 2 O
0.2001 0.1818 0.1616 0.1404 0.1193 0.0995
yO 2
0.3539 0.3785 0.4000 0.4173 0.4307 0.4410
X2
1.3017 1.5126 1.6609 1.7614
X3
0.1920 0.2771 0.3710 0.4658
∆H rxn( kJ)
–1006
TAD
z
Tin
CP dT ( kJ)
616.3
691.2
775.7
–987.8 –997.0
871.9
981.1
–1025
1106
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
In the table above, the mole fractions were computed using the NASA program, X 2 and X 3 were
then computed from the mole fractions using
yH 2 = X 3 ∑ and yCO = X 2 ∑
where ∑ =
17 1
+
X 2 + X 3 . ∆H rxn and the integral
4 2
a
f
TAD
z
CP dT were computed using Eqn. (9.1-19a)
Tin
and the ∆H rxn data in Table A6.1 and CP data in Table 2.4. Also,
∆ H rxn(T) = ∆H rxn,1 (T ) + X 2 ∆ H rxn,2 (T ) + X 3∆ H rxn, 3 (T)
and
TAD
z FHC
P, HCCH
Tin
I
K
15
+ CP ,O 2 dT =
4
TAD
z
CP dT .
Tin
Finally, from eqn. (9.7-10a) we have, at the adiabatic reaction (flame) temperature that
C
0 = ∑ Ni
i =1
a f
in
TAD
M
Tin
j =1
z
CP ,i dT + ∑ ∆Hrxn, j TAD X j
a f
or, in the notation here
TAD
z
CP dT + ∆H rxn TAD = 0
a f
Tin
TAD
Plotting up the results in the previous table, i.e.
z
CP dT vs T and ∆H rxn vs T, leads to the solution
Tin
TAD = 3830 K
X 2 = 1.680
X 3 = 0.387
yCO 2 = 0.0606,
yCO = 0.3180,
yH 2O = 0.1160
yH 2 = 0.0732
and
yO 2 = 0.4322.
Comment: The solution above considered only O 2 , H 2 O , CO 2 , CO, H 2 and HCCH as possible
reaction species. At the high temperatures involved here, other reactions and other species are
possible. This is obvious in the results below. The dashed lines result from the chemical
equilibrium program of NASA with only the species mentioned above as allowed species, and the
solid lines result from the Chemical Equilibrium Program with all species allowed.
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
Note how different the two solutions are!
The actual (unrestricted) solution has O and H (not ions, but atoms) as important reaction products,
but these species have not even been considered in the adiabatic reaction temperature calculation.
The conclusion is that restricting over consideration to just the species in the problem statement is
unjustified!
9.24 (also available as a Mathcad worksheet)
C 6 H6
C 2 H4
C 6 H6
C 2 H4
C 6 H5C2 H5
Using the program CHEMEQ we obtain that at 600 K, Ka = 345.0 and ∆Hrxn = −103.94 kJ/mol
for the reaction
C6 H 6 ( g ) + C2 H 4 ( g ) = C6 H 5C2 H 5 ( g ) .
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
To find the extent of reaction we use
Species
Initial
Final
C6 H 6
1
1− X
C2 H 4
1
1− X
C6 H 5 C3H 5
0
X
yi
1− X
2− X
1− X
2− X
X
2− X
ai
F 1− X IF P I
H 2 − X K H 1 bar K
F 1− X IF P I
H 2 − X K H 1 bar K
F X IF P I
H 2 − X K H 1 bar K
2− X
Ka = 345.0 =
a C6H5C 2H5
X ( 2 − X ) P 1 bar
2
(1 − X ) (2 − X ) P 1 bar
a
=
a C6 H6 ⋅ aC 2H4
a
X (2 − X )
(1 − X ) 2 P 1 bar
Total of moles
2−X
Now P = Pinitial ×
= 1.013 bar ×
=
Initial # of moles
2
=
a
f
f
2
f
⇒ Ka = 345.0 =
2 ⋅ X ⋅ (2 − X )
1.013 ⋅ (1 − X ) (2 − X )
2
=
2⋅ X
1013
.
⋅ (1 − X ) 2
.
bar .
which has the solution X = 0.927 and P = 05434
Heat which must be removed to keep reactor isothermal is
0.927 × 103,940 = 96,352 J ( removed) .
1
1
H 2 ( g) + I 2 (g ) and I 2 (g ) = I 2 ( s) .
2
2
Using the data in the Chemical Engineer’s Handbook we have
9.25 The two “reactions” are HI(g ) =
°
∆Grxn,1
= 1.95 kcal = 8.159 kJ
°
∆G rxn,2
= −4.63 kcal = -19.37 kJ
°
rxn,1
12 12
H 2 I2
RS − ∆G UV = 3.72 × 10 = a a
a
T RT W
R− ∆G UV = 2478.3 = a (s)
= exp S
a (g )
T RT W
Ka ,1 = exp
−2
HI
Ka ,2
°
rxn,2
I2
I2
Solid precipitation of one tiny crystal is just like a dew-point problem, that is, at the pressure at
which the first bit of solid appears the vapor composition is unchanged. Therefore, the first step is to
compute the vapor composition due to reaction 1 only.
Species
Initial
Final
yi
HI
1
1− X
1− X
ai
(1 − X ) P
1 atm
1 2 XP
1 atm
1 2 XP
1 atm
a f
a f
1
1
X
X
2
2
1
1
0
I2
X
X
2
2
1
Σ
Note: standard state pressure in the Chemical Engineers Handbook is 1 atm.
H2
0
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
12 2
−2
⇒ 3.72 × 10
=
a1 2f X aP 1 atmf
(1 − X )a P 1 atmf
=
X
= Ka ,1
2(1 − X )
or
x=
2 Ka , 1
2 Ka , 1 + 1
= 0.069248 (independent of pressure!)
Thus at all pressures (low enough that nonideal vapor phase corrections can be ignored) we have
.
yH 2 = yI2 = 003462
; yHI = 0.93075
which is the composition of the vapor when the first precipitation of solid I2 occurs.
Now consider the second reaction:
aI 2( s) = 1
a I 2 ( g ) = yI 2
⇒ Ka , 2 = 2478.3 =
P=
yI 2
a
F PI
H 1 atmK
1
1 atm
⇒P=
P 1 atm
Ka , 2 yI 2
f
1 atm
= 0.01166 atm = 0.01181 bar.
0.03462 × 2478.3
Thus, if P > 001181
.
bar precipitation of solid I 2 will occur.
9.26 (a) 2 NaHCO 3 ( s) = NaCO3 (s) + CO 2 ( g ) + H 2 O ( g )
Ka =
a NaCO 3 aCO 2 a H 2O
a NaHCO2
but
3
⇒ Ka = a H 2 O a CO 2 =
but Pi = yi P =
NH 2 O = NCO 2
a NaHCO 3 = 1
solids
a NaCO 3 = 1
UV
W
PH 2 O PCO 2
⋅
1 bar 1 bar
Ni
P where N = total moles in gas phase, Ni = moles of i in gas phase and
N
1
⇒ PH 2 O = PCO 2 = P .
2
Therefore
Ka =
2
LM a1 2fP OP
N 1 bar Q
and
LM a1 2f × 0.826 kPa OP = 1706
N 100 kPa Q . × 10
a1 2f × 166.97 kPa
K (110° C) = L
MN 100 kPa OPQ = 0.697
Ka (30° C) =
2
−5
2
a
ln Ka (30° C) = −10.979; ln Ka (110 ° C) = −0.3610
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
Now
ln
∆H rxn 1 1
Ka T2
= ln Ka T2 − ln Ka T1 = −
−
Ka T1
R
T2 T1
a f
af
a f
FG
H
a f
IJ
K
(1)
This assumes that ∆H rxn is independent of T, the only assumption we can make with the limited
data in the problem statement
⇒ ∆H rxn = 128.2 kJ
(b) Going back to Eqn. (1) above we have
1
1
1
∆H rxn 1
−
= 1.706 × 10− 5 exp −15420 −
R
T 30315
T 303.15
.
15420
15420
ln Ka = −10.9788 −
+ 50.8659 = 39.8871 −
T
T
g FH
F
H
Ka (T ) = Ka T = 30o C exp −
b
II
KK
. bar ⇒ PH 2 O = 10
. bar
(c) PCO 2 = 10
⇒ Ka (T ) =
LM a1 2f × 2.0OP
N 1 Q
2
F
H
F
H
II
KK
P = 20
. bar
=1
⇒ T = 386.6 K = 113.45° C for PCO 2 = 1 bar.
9.27 Reaction C + 2H 2 = CH 4
Using the program CHEMEQ we have Ka (T = 1000 K) = 0.09838
Species
Initial (gas)
Final (gas)
yi
H2
1
1− 2X
CH 4
0
X
1−2 X
1− X
X
1− X
C
Σ
Ka =
aCH 4
a Ca H2 2
1− X
=
a CH 4
a 2H 2
=
XP
(1 − X )2 1 bar
⋅
(1 − X )1 bar (1 − 2 X )2
P
F
H
I
K
2
X (1 − X ) 1 bar X (1 − X )
=
(1 − 2 X )2 P
(1 − 2 X ) 2
.
and X = 09231
.
.
The solutions to this equation are X = 00769
.
solution is the correct one.
With such a small value of the equilibrium constant, the X = 00769
yH 2 = 0917
.
and yCH 4 = 0.083
This implies
=
01004
.
This is (probably within experimental and calculational error) essentially the same as the equilibrium
composition. Therefore, the reaction process is thermodynamically limited, not mass transfer
limited. Consequently increasing the equilibration time by slowing the hydrogen flow will have no
effect on the process.
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.28 (also available as a Mathcad worksheet)
Reactions
C + H 2 O = CO + H 2
C + 2H 2 O = CO2 + 2H 2
CO 2 + C = 2CO
CO + H 2 O = CO2 + H 2
For simplicity, let’s write reactions 1, 2 and 4 in reverse
CO + H 2 = C + H 2 O
CO 2 + 2H 2 = C + 2H 2 O
CO 2 + C = 2CO
CO2 + H 2 = CO + H 2 O
Now need to identify the independent chemical reactions.
Start by writing
C + O = CO
Eliminate
2H + O = H 2 O
reaction 4
C + 2O = CO 2
since no
2H = H 2
Now use O =
C + O = CO
⇒ H2 + O = H 2O
C + 2O = CO 2
H present
1
1
CO 2 − C to eliminate O since no atomic oxygen present.
2
2
1
1
CO 2 = C + CO
2
2
1
1
H 2 + CO 2 = C + H 2 O
2
2
C+
or
UV This is one set of
= C + 2H O W independent reactions.
C + CO 2 = 2CO
2H 2 + CO 2
2
Add these two
2H 2 + C + 2CO 2 = 2CO + C + 2H 2 O ⇒ H 2 + CO2 = CO + H 2 O
We will use
H 2 + CO 2 = CO + H 2 O
UV as the
W
2H 2 + CO 2 = C + 2H 2 O independent reactions.
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Species
Initial
Final
yi
CO 2
1
1 − X1 − X 2
H2
1
1 − X1 − 2 X 2
CO
0
X1
1 − X1 − X 2
∑
1 − X1 − 2 X 2
∑
X1
∑
H2O
0
X1 + 2 X 2
C
0
X2
(not in gas phase)
∑ = 2 − X2
Ka,2 =
a CO a H 2 O
=
aCO 2 a H 2
aa = 1fa
C
2
H 2O
=
aH2 2 aCO 2
LM1 − X − X OP P
N ∑ Q 1 bar
LM1 − X − 2 X OP P
N ∑ Q 1 bar
LM X OP P
N ∑ Q 1 bar
LM X + 2 X OP P
N ∑ Q 1 bar
1
a
2
1
1
f
X1 X1 + 2 X 2
1 − X1 − X2 1 − X1 − 2 X2
a
2
1
X1 + 2 X2
∑
0
a f
Ka,1 =
ai
fa
2
f
a X + 2 X f a2 − X f 1 bar
a1 − X − 2 X f a1 − X − X f P
2
1
2
2
2
1
2
1
2
Using the program CHEMEQ, I find the following
T(K)
Ka,1
Ka,2
−1
0.3665 ×10
600
700
800
900
1000
758.6
48.43
5.950
1.137
0.2974
0.1110
0.2493
0.4596
0.7387
(a) No carbon deposits X 2 = 0
Ka,1 =
X12
a1 − X f
2
and Ka,2 =
1
X12 ⋅ 2 1 bar
3
P
1 − X1
a
f
Solving these equations, I find
T(K)
600
700
800
900
1000
P( bar )
1151
. × 10
6111
. × 10
If the pressure for a given temperature is above the pressure calculated, carbon
will deposit.
−4
−3
0.126
1.357
9.237
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Exactly 30% of carbon is deposited X 2 = 0.3
(b)
Ka,1 =
9.29
a
f
a
f
2
X1 X1 + 0.6
x1 + 0.6 (1.7 )
and Ka,2 =
0.7 − X 1 0.4 − X1
0.7 − X 1 0.4 − X1
a
fa
f
a
fa
f
2
1 bar
P
T(K)
600
700
800
900
1000
X1
0.0157
0.0410
0.0750
0.1104
0.1427
P( bar )
0.0084
0.170
1.972
15.266
85.419
The reaction the engineer is concerned about is
Ti(s) + SiO2 (s) = TiO2 (s) + Si(s)
This is equivalent to the first reaction in the problem statement minus the second
reaction. Therefore
kJ
kJ
J
∆G orxn = −674 − ( −644)
= −30
= −30,000
mol
mol
mol
and the equilibrium constant for this re action is
F
GH
Ka = exp −
o
I
JK
F
H
I
K
∆ G rxn
30000
= exp
= 17.02
R⋅T
8.314 ⋅ 1273
Consequently, as the engineer fears, the titanium purity will be effected by high
temperature contact with silicon dioxide.
9.30
From Eqns. (9.1-8 and 9.9-6) we have that
o
nFE o = RT ln Ka = −∆ Grxn
Consequently, by measuring the zero-current cell potential we obtain the
standard state Gibbs free energy change on reaction (if all the ions are in their
standard states). Now if we continue further and measure how the zero-current
standard state cell potential varies as a function of temperature, we have
FG ∂E IJ
H ∂T K
o
nF
=−
P
FG ∂∆ G IJ
H ∂T K
o
rxn
o
≡ ∆S rxn
P
Consequently by knowing the zero - current, standard state cell potential
o
and its temperature derivative we can calculate ∆H rxn
from
o
o
o
o
o
o
∆Grxn
= ∆H rxn
− T∆S rxn
or ∆H rxn
= ∆Grxn
+ T ∆S rxn
Similarly starting fromnFE = −∆Grxn and the measured zero-current potentials,
we can calculate the enthalpy and entropy changes for the reaction when the
ions are not in their standard states.
9.31
The chemical reaction is
CH3 -CHOH-CH3 = CH3 -CO-CH3 + H2
Assuming we start with pure acetone, the mass balance table with all species as
vapors (given the high temperature and low pressure) is
Species
i-prop
acetone
hydrogen
Σ
in
1
0
0
out
1-X
X
X
1+X
y
(1-X)/(1+X)
X/(1+X)
X/(1+X)
a
(1-X)×95.9/(1+X) )×100
X×95.9/(1+X) )×100
X×95.9/(1+X) )×100
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
But X = .564, so a I-P =0.2673, and a ace = a H =0.3458. Therefore
Ka =
F
GH
o
a acea H 0.3458 × 0.3458
∆G rxn
=
= 0.4474 = exp −
ai− P
0.2673
RT
−0.8043 = −
9.32
I
JK
∆ G orxn
J
; so ∆G orxn = 8.314 × 452 .2 × 0.8043 = 30238
.
RT
mol
The reactions are
C6 H6 + H2 = 1,3-cyclohexadiene
C6 H6 + 2H2 = cyclohexene
C6 H6 + 3H2 = cyclohexane
The Gibbs free energy of formation data needed to solve this problem
kJ
kJ
o
o
∆G f ( benzene) = 124.5
∆G f ( cyclohexene) = 106.9
mol
mol
kJ
kJ
o
o
∆G f ( cyclohexane) = 26.9
∆ G f (1,3 − cyclohexdiene) = 178.97
mol
mol
The Gibbs free energy of formation for 1,3-cyclohexadiene is not available in
Appendix IV, Perry’s The Chemical Engineer’s Handbook or the Handbook of
Chemistry and Physics. The value was found using data on the WWW site
http://webbook.nist.gov/chemistry. This Web site contains the National Institute
of Standards and Technology (NIST) chemistry data book. The values found on
this Web site are
kJ
∆H of (1,3 − cyclohexdiene, 298.15 K) = 7141
.
mol
J
S (1,3 − cyclohexdiene, 298.15 K) = 197 .3
mol ⋅ K
J
S ( C,graphite, 298.15 K) = 588
.
mol ⋅ K
J
S ( H 2 , 298.15 K) = 130.68
mol ⋅ K
Note that these entropies are with respect to the entropy equal to zero for the
pure component and 0 K. Also, the entropy change of reaction at 0 K is zero for
all reactions. Therefore
∆S of (1,3 − cyclohexdiene, 298.15 K) =
S (1,3 − cyclohexdiene, 298.15 K) − 6 ⋅ S (C, 298.15 K) − 4 ⋅ S ( H 2 , 298.15 K)
J
mol ⋅ K
∆G of (1,3 − cyclohexdiene, 298.15 K) = ∆ H of − T ∆ S of = 71410 − 29815
. ⋅ ( −360.75)
= 197 .3 − 6 ⋅ 588
. − 4 ⋅ 130.68 = −360.75
= 178967
J
kJ
= 178.97
mol
mol
The mass balance table assuming all the organics are present only in the liquid
phase, and that the hydrogen is present in great excess to keep its partial
pressure fixed at 1 bar. Also, since all the organics are so similar, we will
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
assume they form an ideal liquid mixture, and that there is no hydrogen in the
liquid phase. Mass balance table for liquid phase:
Species
In
Out
x
activity
Benzene
1,3-cyclohex
cyclohexene
cyclohexane
Total
1
0
0
0
1-X1 -X2 -X3
X1
X2
X3
1
1-X1 -X2 -X3
X1
X2
X3
1-X1 -X2 -X3
X1
X2
X3
The equilibrium relations are
a1,3 − cyc
X1
(178970 − 124500)
Ka,1 =
=
= exp −
a benz aH 2 1 − X1 − X 2 − X 3
8.314 × 29815
.
F
H
I
K
= exp( −21973
. ) = 2 .866 × 10 −10
a cychene
X2
(106900 − 124500)
Ka,2 =
=
= exp −
abenz a H2 2 1 − X 1 − X 2 − X 3
8.314 × 298.15
F
H
I
K
= exp(7 .1002) = 1212.2
Ka,3 =
a cychane
=
a benz aH3 2
F
H
( 26900 − 124500 )
X3
= exp −
1 − X1 − X 2 − X 3
8.314 × 298.15
I
K
= exp( 39.374) = 1.2587 × 1017
By examining the values of the equilibrium constants, or more directly by taking
ratios of these equations, we see that X3 is about equal to unity. Then by taking
the ratio of the first of these equations to the third, we have
X1 2 .866 × 10 −10
=
= 2 .27 × 10−27 ≈ X 1
X 3 12587
.
× 1017
and by taking the ratio of the second of these equations to the third
X2
12122
.
=
= 9.63 × 10 −15 ≈ X 2
X 3 1.2587 × 1017
This suggests that X3 ~ 1, X2 is of the order or 10-17, and X1 is of order
10-27. Thus the benzene will react to form essentially all cyclohexane.
9.33
The process is
.
N1 , IN
5 × 10-3 mol/kg
.
N2 , IN
5 × 10 mol/kg
5 × 10-3 mol/kg
.
.
N1 , OUT
5 × 10-6 mol/kg N2
, OUT
-5
Assume all other component concentrations are unchanged since the glucose
concentration is so low. The mass balance is
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
N& 1, in + N& 2 , in = N& 1,out + N& 2,out
& is the rate of glucose transported, then
If δ
&
N&
= N&
+δ
b g
b N& g
b g
= b N& g
G 1,out
G 1,in
G 2,out
G 2,in
&
−δ
The energy balance is
0 = ∑ N& i H i
+ ∑ N& i H i
c
h c
1,in
h − c∑ N& H h
i
i
2,in
1,out
−
c∑ N& H h
i
i
2,out
+ Q& + W&
and the entropy balance is
Q& &
+ S gen
1,in
2,in
1,out
2,out
T
The kidney operates reversibly, and minimum work implies S&gen = 0 .
0=
c∑ N& S h
+
i
i
c∑ N& S h
i
−
i
c∑ N& S h
i
−
i
c∑ N& S h
i
+
i
Subtracting T times the entropy balance form the energy balance gives
0 = ∑ N& i G i
+ ∑ N& i G i
− ∑ N& i G i
− ∑ N& i G i
+ W&
c
h c
h
1,in
or
W& =
2,in
c
h
c
1,out
h
2,out
c∑ N& G h + c∑ N& G h − c∑ N& G h − c∑ N& G h
= c N& G h
+ c N& G h
− c N& G h − c N& G h
& cG h
& cG h
= eb N& g + δ
+ e b N& g − δ
− c N& G h − c N&
j
j
i
i
G
G
G
G
1,out
G
G 1,in
i
i
1,out
G
2,out
i
i
2,out
G
1,in
G
G 2,in
1,out
i
1,in
G
2,out
G
i
2,in
2,in
G
G
1,in
h
G G G 2,in
Now since the concentrations are very low, and have not changed significantly,
GG
= GG
and G G
= GG
c h
c h
1,in
c h
1,out
Therefore
2,in
c h
2,out
c h c h
& GG − G G
W& = δ
1
2
(since, from the previous equation, we can eliminate the subscripts in and out).
Then
fG
W&
1
&δ = G G 1 − G G 2 = RT ln f
c h c h
Fc h I
GH c h JK
G 2
Now assuming ideal solutions (or that the activity coefficients of glucose in blood
and urine are the same)
fG
x
CG 1
W&
1
= RT ln G 1 = RT ln
&δ = RT ln f
x
C
Fc h I
GH c h JK
G 2
FG a f IJ
Ha f K
G 2
FG a f IJ
Ha f K
G 2
where we have assume that both blood and urine, being mostly water, have about
the same molar concentration. Therefore
FG a f IJ = RT lnFG 5 × 10 IJ = RT ln(100)
Ha f K
H 5 × 10 K
W&
CG
&δ = RT ln C
G
= 8.314
−3
1
−5
2
J
J
kJ
× 310.1 K × ln( 100) = 11873
= 11873
.
mol K
mol
mol
Note that body temperature is 98.6o F = 310.1 K
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.34 (also available as a Mathcad worksheet).
9.34
Given:
T
298.15 . K
M1
0.0001 .
mole
M2
liter
0.01 .
mole
liter
1
α
1.178 .
2
mole
R
8.31451 .
liter
joule
K . mole
Using eqn 9.9-10 and the Debye-Huckel limiting law of eqn 7.11-15:
ln ( γ )
α . 4 . 4 . M CuSO4
γ M CuSO4
∆ G 2moles
α . 8 . M CuSO4
exp
2 . R. T . ln
M1
M2
2 . R. T . ln
γ M1
γ M2
The change is Gibb's Free Energy calculated above is for two moles of electrons (n=2). The
number of moles of electrons in this problem is calculated below:
n
2.
0.01 . mole
0.0001 . mole
2
∆G
Wmax
n = 9.9 10
3
mole
n .
∆ G 2moles
2
∆G
Wmax = 92.204
(for a process at constant temperature and pressure)
joule
9.35 (also available as a Mathcad worksheet).
From the Steam Tables
Pvap = 12.349 kPa
From CHEMEQ
Ka = 171.2
HNH3 (from problem statement) = 384.5 kPa/mole fraction
The solution at various pressures is
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
P (kPa)
0.10
0.25
0.50
1.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
18.40
18.50
19.00
20.00
25.00
30.00
35.00
40.00
50.00
60.00
70.00
80.00
90.00
100.00
104.50
X
0.0899
0.2017
0.3467
0.5478
0.7868
1.0322
1.1670
1.2559
1.3204
1.3701
1.4099
1.4429
1.4707
1.4757
1.4797
1.4996
1.5339
1.6437
1.7063
1.7488
1.7806
1.8267
1.8602
1.8869
1.9093
1.9291
1.9472
1.9449
f
0
0
0
0
0
0
0
0
0
0
0
0
0
2.02×10-05
0.0114
0.0680
0.1589
0.3989
0.5068
0.5713
0.6162
0.6792
0.7261
0.7661
0.8030
0.8388
0.8745
0.8908
2.5
X and f
2.0
1.5
X, molar extent of reaction
f, fraction liquid
1.0
0.5
0.0
0
20
40
60
Pressure, kPa
80
100
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.36 (also available as a Mathcad worksheet).
9.36
Rxn: C8H18 + 25/2*O2 = 9*H2O + 8*CO2
Given:
0.
∆ Hf N2
joule
∆ Hf O2
mole
393500 .
∆ Hf CO2
joule
Cp N2( T )
Cp H2O( T )
joule
22.243 .
28.883 .
mole . K
joule
∆ Hf C8H18
mole
285800 .
joule
5 2
3.499 . 10 . T .
mole . K
2
joule
mole
joule
9 3
7.464 . 10 . T .
mole . K
3
5 2
0.808 . 10 . T .
mole . K
joule
mole . K
mole . K
5 2
1.055 . 10 . T .
2
joule
mole . K
4
9 3
2.871 . 10 . T .
3
joule
2
0.192 . 10 . T .
joule
joule
2
mole . K
255100 .
mole
joule
2
5.977 . 10 . T .
2
0.157 . 10 . T .
mole . K
32.218 .
joule
∆ Hf H2O
mole
Cp CO2( T )
0.
joule
mole . K
4
joule
mole . K
3
9 3
3.593 . 10 . T .
joule
mole . K
Mass Balance Table:
Species
C8H18
O2
N2
CO2
H2O
Total
Nin C8H18
Nout CO2
In
1
25/2
(25/2)*(0.71/0.21) = 42.26
0
0
55.76
1 . mole
8 . mole
Nin O2
25 .
Nout H2O
mole
2
Out
0
0
42.26
8
9
59.26
N N2
42.26 . mole
9 . mole
Energy Balance at Steady State:
k
N i. H i
0
i= 1
Q
Pd V W
(where Hi is the partial molar enthalpy of species i)
4
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Assuming no heat flow, no change in volume of the engine, and an ideal gas mixture yields:
k
N i. H i
W
(where is Hi is the molar enthalpy of species i)
i= 1
At 150 C:
423.15 . K
∆ Hf N2
H N2
298.15 . K
Cp N2( T ) d T
423.15 . K
∆ Hf H2O
H H2O
298.15 . K
Cp H2O( T ) d T
423.15 . K
∆ Hf CO2
H CO2
W
298.15 . K
Nout CO2 . H CO2
6
W = 5.233 10
Cp CO2( T ) d T
Nout H2O . H H2O
N N2 . H N2
Nin C8H18 . ∆ Hf C8H18
∆ Hf N2
Nin O2 . ∆ Hf O2
joule
This work obtained is per mole of n-octane.
9.37 (also available as a Mathcad worksheet).
9.37
Given:
T1
298.15 . K
24300 .
∆ G C3H8
P1
joule
mole
∆ H C3H8
104700
. joule
mole
5
10 . Pa
T2
650 . K
50500 .
∆ G CH4
joule
mole
∆ H CH4
74500
. joule
mole
R
6
10 . Pa
P2
68500 .
∆ G C2H4
8.31451 .
joule
mole
∆ H C2H4
52500
. joule
mole
Mass Balance Table:
Species
In
Out
C3H8
CH4
C2H4
Total
1
0
0
1-X
X
X
1+X
joule
K . mole
y
(1-X)/(1+X)
X/(1+X)
X/(1+X)
(at 298.15 K)
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Calculation of mole fractions and activities:
yC3H8 ( X )
1
X
1
X
1
P
yC3H8( X ) .
aC3H8( X , P )
X
yCH4( X )
10 Pa
aC2H4( X , P )
1
X
P
yCH4( X ) .
aCH4( X , P )
5.
X
yC2H4( X )
X
5.
10 Pa
P
yC2H4( X ) .
5.
10 Pa
∆ G rxn
∆ G C2H4
∆ G CH4
∆ G C3H8
4
1
∆ G rxn = 4.23 10 mole
joule
∆ H rxn
∆ H C2H4
∆ H CH4
∆ H C3H8
4
1
∆ H rxn = 8.27 10 mole
joule
Ka 298.15
exp
∆ G rxn
Part (a):
Given
Ka 298.15 = 3.885 10
R. T1
X
Ka 298.15
10
4
(initial guess)
aC2H4( X , P1 ) . aCH4( X , P1 )
aC3H8( X , P1 )
yC2H4( Xa ) = 1.971 10
yC3H8( Xa ) = 1
8
4
Xa
Xa = 1.971 10
Find( X )
yCH4( Xa ) = 1.971 10
4
Part (b):
From equation 9.1-22b:
Ka 650
X
.5
Given
Ka 298.15 . exp
∆ H rxn
R
. 1
T2
1
Ka 650 = 2.704
T1
(initial guess)
Ka 650
yC3H8( Xb ) = 0.079
aC2H4( X , P1 ) . aCH4( X , P1 )
aC3H8( X , P1 )
yC2H4( Xb ) = 0.461
Xb
Find( X )
yCH4( Xb ) = 0.461
Xb = 0.854
4
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Part (c):
X
(initial guess)
.5
Given
aC2H4( X , P2 ) . aCH4( X , P2 )
Ka 650
Xc
aC3H8( X , P2 )
yC3H8( Xc ) = 0.369
yC2H4( Xc ) = 0.316
Find( X )
Xc = 0.461
yCH4( Xc ) = 0.316
9.38 (also available as a Mathcad worksheet).
9.38
R
8.31451 .
joule
K . mole
G rxn
a A xA, γ A
x A .γ A
a C xA, γ C
1
Ka
joule
x A .γ C
T
298.15 . K
V
4 . liter
mole
aB
G rxn
exp
2400 .
1
a D( P )
P
5
10 . Pa
Ka = 2.633
R. T
Part (a):
xA
(initial guess)
0.5
Given
xC
Ka
1
5
a D 0.5 . 10 . Pa . a C x A , 1
a B. a A x A , 1
xA
Find x A
xA
x A = 0.16
x C = 0.84
Part (b):
Recognizing that the partial molar Gibb's excess is in the form of the one constant Margules
expression yields:
2
γ A exp 0.3 . x C
2
γ C exp 0.3 . x A
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Given
xC
Ka
1
5
2
a D 0.5 . 10 . Pa . a C x A , exp 0.3 . x A
a B. a A x A , exp 0.3 . 1
xA
xA
2
Find x A
xA
x A = 0.132
x C = 0.868
Part (c):
Mass Balance Table:
Species
A
B
C
D
In
1
2
0
0
Out
1-X
2-X
X
X
Assuming that D is an ideal gas:
P=(n*R*T)/V
X
P( X )
X .R .T
V
(initial guess)
0.5
aD
Given
Ka
X . R. T . mole .
a C( X , 1 )
V
a B. a A ( 1 X , 1 )
X
Find( X )
X = 0.425
NA
xA
1
1
X
NB
2
X
X
N A = 0.575
xC
N B = 1.575
x A = 0.575
9.39
NC
Without dissociation:
Amount_Adsorbed_Without K1 . a H2
X
ND
PD
X
X . R. T . mole
V
N C = 0.425
N D = 0.425
x C = 0.425
5
P D = 2.633 10 Pa
(also available as a Mathcad worksheet).
9.39
X
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Assuming the activity of molecular hydrogen gas is equal to the pressure of hydrogen gas:
Amount_Adsorbed_Without K1 . P H2
With dissociation:
Amount_Adsorbed_With K1 . a H2
1.
2
K3 . a H
Using the equilibrium constant for the reaction H2 = 2H, the activity of atomic hydrogen can be
solved for in terms of the activity of molecular hydrogen:
2
Given
K2
aH
Find a H
a H2
Using the positive root for the activity of hydrogen yields:
Amount_Adsorbed_With K1 . a H2
1.
2
K3 . K2 . a H2
Assuming the activity of molecular hydrogen gas is equal to the pressure of hydrogen gas:
Amount_Adsorbed_With K1 . P H2
1.
2
K3 . K2 . P H2
If the amount adsorbed varies linearly with the partial pressure of molecular hydrogen then no
dissociation is occurring. If the amount adsorbed varies as the square root of the partial pressure,
then dissociation is occurring.
9.40 (also available as a Mathcad worksheet).
9.40
Since this is a combustion reaction, the reaction can be assumed to go to completion.
Rxn:
C4H10
13 .
2
O2
0.79 . 13 .
N2 4 . CO2
0.21 2
5 . H2O
0.79 . 13 .
N2
0.21 2
Given:
∆ Gf C4H10
16600
. joule
mole
(The values for the Gibbs free energy of formation are given
at one bar. The difference in Gibbs free energy between
one bar and one atmosphere will be ignored because it is
insignificant in this calculation.)
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
∆ Gf O2
0.
∆ Gf N2
0.
joule
mole
joule
mole
1
394400 .
∆ Gf H2O
237100 .
gm
4 . 12.001 .
MW C4H10
NI
∆ Gf CO2
0.71 . 13
2
0.21 2
T
298.15 . K
R
8.31451 .
mole
joule
mole
joule
K . mole
gm
10 . 1.0079 .
mole
13
joule
mole
NF
4
5
0.71 . 13
(Total number of moles)
0.21 2
13
GI
∆ Gf C4H10
13 .
2
0.79 . 13 .
∆ Gf O2
0.21 2
∆ Gf N2
1
13 .
NI
2
R. T . ln
ln
0.79 . 13
2
0.79 . 13 .
NI
0.21 2
ln
0.21 2
NI
0.79 . 13
GF
4 . ∆ Gf CO2
4
G I = 6.067 10
6
G F = 2.82 10
mole
mole
From eqn 9.8-5:
W
0.79 . 13 .
5 . ∆ Gf H2O
1
1
0.21 2
GF
GI
8
W = 9.515 10
9.41 (also available as a Mathcad worksheet).
9.41
Process #1:
Ka 1
N2(gas) = N2(metal)
aN2 metal H N2 . xN2 metal
aN2 gas
xN2 metal
Ka 1 . P N2
H N2
P N2
5 . ln
NF
N C4H10
joule
4
R. T . 4 . ln
joule
W molar
N C4H10 . W molar
∆ Gf N2
R. T . N F
joule
20000 . gm
MW C4H10
NI
5
0.79 . 13 .
NF
0.21 2
ln
0.21 2
NF
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
wt%nitrogen A . P N2
Process #2:
2 . Ka 1
H N2
aN metal
H N . xN metal
aN2 gas
P N2
xN metal
A
N2(gas) =2*N(metal)
2
Ka 2
where
2
Ka 2 . P N2
HN
wt%nitrogen B. P N2
where
Ka 2
B
HN
The empirical expression given in the problem is supported by process #2.
9.42 (also available as a Mathcad worksheet).
Rxn: CH4 + 2*O2 = CO2 + 2*H2O
R
8.31451
∆ Hf H2O
241800
∆ Hf CO2
393500
∆ Hf O2
0
∆ Hf N2
0
∆ Hf CH4
74500
∆ Gf H2O
228600
∆ Gf CO2
394400
∆ Gf O2
0
∆ Gf N2
0
∆ Gf CH4
50500
Heat capacity:
Cp CH4( T )
Cp O2( T )
19.875
28.167
2
5.021 . 10 . T
2
0.630 . 10 . T
5 2
1.268 . 10 . T
5 2
0.075 . 10 . T
9 3
11.004 . 10 . T
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Cp N2 ( T )
2
0.623 . 10 . T
27.318
5 2
0.095 . 10 . T
Cp H2O( T )
29.163
2
1.449 . 10 . T
5 2
0.202 . 10 . T
Cp CO2( T )
75.464
4
1.872 . 10 . T
661.42
T
Defining z as the methane to air ratio:
Mass Balance Table:
Species
CH4
O2
N2
CO2
H2O
Total
∆ Hrxn 25
In
9.524*z
2
7.524
0
0
9.524*(z+1)
∆ Hf CO2
2 . ∆ Hf H2O
∆ Hf CH4
Out
(9.524*z)-X
2-2*X
7.524
X
2*X
9.524*(z+1)
2 . ∆ Hf O2
5
∆ Hrxn 25 = 8.026 10
If z<0.105, then methane is the limiting reactant and X=9.524*z
If z>0.105, then oxygen is the limiting reactant and X=1
X( z)
if( z< 0.105 , 9.524 . z, 1 )
Cp out1( z, T )
( 9.524 . z
Cp out2( z, T )
7.524 . Cp N2( T )
Cp out ( z, T )
X( z) ) . Cp CH4( T )
Cp out1( z, T )
(2
X( z) . Cp CO2( T )
2 . X( z) ) . Cp O2( T )
2 . X( z) . Cp H2O( T )
Cp out2( z, T )
Using equation 9.7-10b:
Tout
2500
(initial guess)
Tout
T ad( z)
Cp out ( z, T ) d T
root
298.15
z
0.01 , 0.02 .. 1
∆ Hrxn 25 . X( z) , Tout
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
2500
2000
T ad( z ) 1500
1000
500
0
0.1
0.2
0.3
0.4
0.5
z
0.6
0.7
0.8
0.9
The solution is approximate because the range for the heat capacity of methane used is only
valid between 273 K and 1500 K.
9.43 (also available as a Mathcad worksheet).
9.43:
From eqn 3.3-4:
W Q 1.
T1
dW dQ 1 .
T2
W
T2
(for a Carnot cycle)
T1
T1
T2
T1
T1
T2
T1
dQ1
T1
dQ 1 Cp out . dT 1
T ad( z)
W Carnot ( z )
T1
298.15
T1
298.15
. Cp
out z, T 1 d T 1
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
W Carnot ( z )
Wobtained Carnot ( z )
(Dividing by 9.524 gives the work per mole of air)
9.524
The work obtained by the Carnot cycle is plotted versus z at the end of Problem 9.44
9.44
(also available as a Mathcad worksheet).
9.44
Nout N2
7.524
Nout CO2( z)
X( z)
2 . X( z)
Nout H2O( z)
9.524 . z
Nin CH4( z)
T
Nout CH4( z)
Nin O2
2
9.524 . z
(from the mass balance table)
X( z)
2 . X( z)
N( z)
Gin N2( z)
∆ Gf N2
R. T . ln
Gin O2( z)
∆ Gf O2
R. T . ln
Nout O2( z)
Nin N2
2
9.524 . ( z
7.524
298.15
Partial Molar Gibb's Free Energy:
Gout N2( z)
∆ Gf N2
R. T . ln
Gout O2( z)
∆ Gf O2
R. T . ln
Nout N2
N( z)
Nout O2( z)
N( z)
Gout H2O( z)
∆ Gf H2O
R. T . ln
Gout CO2( z)
∆ Gf CO2
R. T . ln
Gout CH4( z)
∆ Gf CH4
R. T . ln
Gin CH4( z)
∆ Gf CH4
Nout H2O( z)
N( z)
Nout CO2( z)
N( z)
Nout CH4( z)
N( z)
R. T . ln
Nin CH4( z)
N( z)
Nin N2
N( z)
Nin O2
N( z)
From eqn 9.7-16b:
Out1( z )
Nout N2 . Gout N2 ( z )
Nout O2 ( z ) . Gout O2 ( z )
Nout CH4 ( z ) . Gout CH4 ( z )
1)
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Out2( z )
Nout H2O ( z ) . Gout H2O ( z )
Nin N2 . Gin N2 ( z )
In ( z )
W FuelCell ( z )
Out1( z )
Wobtained FuelCell ( z )
Nout CO2 ( z ) . Gout CO2 ( z )
Nin O2 . Gin O2 ( z )
Out2( z )
Nin CH4 ( z ) . Gin CH4 ( z )
In ( z )
W FuelCell ( z )
9.524
The work obtained has units of joules per mole of air.
9.45
(also available as a Mathcad worksheet).
9.45
Given:
108700 .
∆ G AgCl
∆ G TlCl
R
joule
mole
186020 .
8.31451 .
joule
mole
joule
K . mole
∆ G Ag
77110 .
∆ G Tl
32450 .
T
joule
mole
298.15 . K
joule
mole
∆ G Cl
131170 .
joule
mole
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Part (a):
∆ Grxn AgCl
K AgCl
exp
∆ G Ag
∆ G Cl
4
1
∆ Grxn AgCl = 5.464 10 mole
∆ G AgCl
∆ Grxn AgCl
K AgCl = 2.676 10
R. T
joule
10
The solubility product given in illustration 9.3-2 is 1.607E-10. This experimental value is of the
same order of magnitude as the theoretical value calculated above.
Part (b):
∆ Grxn TlCl
K TlCl
exp
∆ G Tl
∆ G Cl
∆ G TlCl
∆ Grxn TlCl =
∆ Grxn TlCl
joule
K TlCl =
R. T
The solubility product given in illustration 9.3-2 is 1.116E-2. This experimental value is two orders
of magnitude greater than the theoretical value calculated above.
9.46
a) From Table 9.1-4, we have that
kJ
kJ
∆G of , Ag + = 77.11
and ∆G o
.
−− = − 741991
f
,
SO
4
mol
mol
Also, from Perry' s Handbook,
kJ
mol
Now consider the reaction
Ag 2 SO4 = 2Ag+ + SO4 —
The chemical equilibrium relation for this reaction is
(2 × 77110 − 741991 − ( −641210 )
53439
Ka = exp −
= exp −
8.314 × 298.15
8.314 × 298.15
∆G of , Ag 2SO 4 = 614 .21
F
H
= exp( −215580
.
) = 4.3388 × 10−10 =
I
K
a 2Ag + aSO −−
4
a AgSO 4
F
H
= Kso
Now from eqn. (9.2-7), assuming the simple Debye-Hückel equation
I
K
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
a
f
ln Ks = ln Kso + ν+ + ν− z + z− α
= ln Kso + (2 + 1) 1 × 2 α
= ln Kso + 3 × 2α
But M SO −− =
4
1
∑ zi2 Mi
2
e
1
MAg + + 4 MSO −−
4
2
e
1
M Ag + + 4 M SO −−
4
2
j
j
1
M + so that
2 Ag
e
j
2
M Ag + MSO −−
3
4
ln Ks =
+ 6α
M Ag + = ln
= ln M Ag +
2
(1 molal)3
This has the solution
mol
mol
MAg + = 8.224 × 10−4
and MSO −− = 4.112 × 10−4
4
liter
liter
Therefore,
ln Kso
F
H
Ks = 8.224 × 10−4
mol
liter
I
K
2
× 4.112 × 10−4
e
j
3
F I
H K
mol
mol
= 2.781 × 10 −10
liter
liter
3
b) Note that there is a error in the problem statement of the first printing of the
text. The solution should be 0.5 M CuSO4 and saturated with AgSO4 . The The
half-cell reactions are
Ag + + e → Ag( s) for which E o = +0.80 volts
Cu(s) → Cu ++ + e for which E o = −0.34 volts
Therefore for the reaction
2Ag + + Cu(s) → 2Ag (s) + Cu ++
E o = +0.80 − 0.34 = 0.46 volts
Next we have from eqn. (9.9-7) that
c h
c h
e j
e j
F c M h aγ f I
= 0.46 − 0.0257G ln
GH e M j + ln aγ f JJK
F (0.5)
aγ f IJ
= 0.46 − 0.0257G ln
+ ln
GH e M j aγ f JK
a ++
a ++
RT
RT
E =E −
ln Cu 2 = E o −
ln Cu
2F
F
aAg +
aAg +
o
0 .5
Cu ++
Ag +
0.5
c
a f h
e a f j
MCu ++ γ ± Cu + +
RT
=E −
ln
F
M Ag + γ ± Ag +
o
0.5
0.5
± Cu ++
± Cu ++
0 .5
0.5
± Cu ++
Ag +
± Cu ++
To proceed further, we have to compute the solubility of AgSO4 in the 0.5M
CuSO4 solution. For this and all the calculations that follow, we will use the fact
that since the CuSO4 concentration is so much higher than that of AgSO4 , we will
neglect the contribution of AgSO4 to the total solution ionic strength. Also,
because of the high ionic strength of the solution, we will use Eqn. (7.11-18) to
compute the mean ionic activity coefficient, as follows:
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
ln γ ± = −
1178
.
z+ z−
where I =
I
1+ I
+ 0.3 I
c
h
1 2
1
2 × 0.5 + 2 2 × 0.5 = (2 + 2 ) = 2
2
2
Therefore,
a f
lnaγ f
ln γ ±
Ag 2SO 4
± Ag 2SO 4
=−
=−
1178
.
1×2 2
1+ 2
1178
.
2 ×2 2
1+ 2
+ 0.3 × 2 = −0.7801
+ 0.3 × 2 = −2.1602
So now we have
ln Ks = ln Kso + ν+ + ν− ( −0.7801)
a
=
ln Kso
f
+ (2 + 1)( −0.7801)
= −215580
.
− 2.3403 = −23.8983
so
eM j M
=
2
−11
Ks = 4.1793 × 10
Ag +
e M j 0.5
=
2
SO −−
4
3
(1 molal)
Ag +
(1 molal) 3
and
MAg + =
4.1793 × 10−11
= 91425
.
× 10 −6 molal
0.5
F (0.5) + 0.5 × (−2.1602) − 1 × (−0.7801)I
GH c9.1425 × 10 h
JK
0 .5
E = 0.46 − 0.0257 ln
−6
= 0.46 − 0.0257(11256
.
− 10801
.
+ 0.7801) = 0.46 − 0.2816 = 0.178 volt
Since this is positive, it is the potential that is produced by the cell (rather than
must be applied) for metallic silver to form.
9.47 (also available as a Mathcad worksheet).
9.47
R
Given (Ka and Hrxn were calculated on CHEMEQ):
Ka
345.0
Hrxn
103940 .
joule
P
5
10 . Pa
T
600 . K
8.31451 .
Ni
mole
Mass Balance Table:
Species
In
Out
y
C6H6
C2H4
C6H5C2H5
Total
1
1
0
1-X
1-X
X
2-X
(1-X)/(2-X)
(1-X)/(2-X)
X/(2-X)
joule
K . mole
2 . mole
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
aC6H6( X )
1
X
2
X
aC2H4( X )
KaX( X )
aC6H5C2H5( X )
aC6H6( X ) . aC2H4( X )
X
(initial guess)
0.9
Given
KaX( X ) Ka
X = 0.946
X
XX
1
X
2
X
X
aC6H5C2H5( X )
2
X
Find( X )
X . mole
Assuming the contents of the reactor behave as an ideal gas:
Ni . R. T
Vi
Nf
(2
X ) . mole
P
Vf
Nf . R. T
P
Nf = 1.054 mole
An energy balance on the reactor yields:
Q
Vf
XX . Hrxn
Pd V
Vi
5
Q = 1.031 10
joule (Heat must be removed because Q is negative)
9.48 (also available as a Mathcad worksheet).
9.48
Part (a):
Using equation 9.1-20b:
ln ( Ka )
∆ Grxn
R. T
57.33
R. T
d
dT
0.17677
∆ Hrxn
R
R. T
57.33 ∆ Hrxn
R. T
2
R. T
2
∆ Hrxn 57.33 .
kJ
mole
2
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Part (b):
∆ G rxn ( T )
Ka ( T )
57330 .
exp
joule
176.77 .
mole
joule
mole . K
.T
8.31451 .
R
joule
K . mole
∆ G rxn ( T )
R. T
Mass Balance Table:
Species
N2O4
NO2
Total
In
1
0
1
Out
1-X
2*X
1+X
Calculation of mole fractions and activities:
y N2O4( X )
a N2O4( X , P )
KaX( X , P )
X
0.7
Given
Xb 0.1
1
X
1
X
y NO2( X )
y N2O4( X ) .
a NO2( X , P )
P
a NO2( X , P )
5
10 . Pa
2.X
1
X
y NO2( X ) .
P
5
10 . Pa
2
a N2O4( X , P )
(initial guess)
Ka ( T ) KaX( X , P )
4
X 323.15 . K , 10 . Pa
X( T , P )
Xb 1
Find( X )
5
X 323.15 . K , 10 . Pa
Xb 10
6
X 323.15 . K , 10 . Pa
y NO2 Xb 0.1 = 0.91
y NO2 Xb 1 = 0.605
y NO2 Xb 10 = 0.261
y N2O4 Xb 0.1 = 0.09
y N2O4 Xb 1 = 0.395
y N2O4 Xb 10 = 0.739
Part (c):
Xc 0.1
4
X 473.15 . K , 10 . Pa
y NO2 Xc 0.1 = 1
y N2O4 Xc 0.1 = 1.246 10
Xc 1
5
X 473.15 . K , 10 . Pa
y NO2 Xc 1 = 0.999
4
y N2O4 Xc 1 = 1.243 10
Xc 10
6
X 473.15 . K , 10 . Pa
y NO2 Xc 10 = 0.988
3
y N2O4 Xc 10 = 0.012
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.49 (also available as a Mathcad worksheet).
9.49
Given:
58620 .
∆ H rxn
Ta
joule
∆ S rxn
mole
298.15 . K
Tb
373.15 . K
138.2 .
joule
R
K . mole
4
10 . Pa
P1
P2
8.31451 .
5
10 . Pa
joule
K . mole
P3
6
10 . Pa
Mass Balance Table:
Species
In
Out
y
M
D
Total
2
0
2-2*X
X
2-X
(2-2*X)/(2-X)
X/(2-X)
Activities, Equilibrium Constant, and Equilibrium Expression:
2.X .
P
X
5
10 . Pa
a M( X , P )
2
∆ G rxn ( T )
∆ H rxn
KaX( X , P )
2
T . ∆ S rxn
0.999
Given
Ka ( T )
X .
P
2 X 105 . Pa
exp
∆ G rxn ( T )
R. T
a D( X , P )
a M( X , P )
X
a D( X , P )
2
(initial guess for solver)
KaX( X , P ) Ka ( T )
Part (a):
DegreeOfDimerization( Ta , P1 ) = 0.953
DegreeOfDimerization( Ta , P2 ) = 0.985
DegreeOfDimerization( Ta , P3 ) = 0.995
X 1
DegreeOfDimerization( T , P )
Find( X )
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Part (b):
DegreeOfDimerization ( Tb , P1 ) = 0.547
DegreeOfDimerization( Tb , P2 ) = 0.842
DegreeOfDimerization( Tb , P3 ) = 0.949
Part (c):
Nomenclature
N = initial number of moles before dimerization
Nm = number of moles of monomer after dimerization = N-2X
Nd = number of moles of dimer = X
Total # of moles = N-X
yM=(N-2X)/(N-X)
yD=X/(N-X)
P = (Nm + Nd)*R*T/V
P ( Nm Nd ) .
( R. T ) ( Nm Nd ) . ( N . R. T ) ( N X ) . ( N . R. T )
V
N
V
N
V
( X.( N
yd
Ka
P
2
ym .
X) )
1 bar
1 bar
P
4 . Ka .
X
N
N.
2
2
which has the solution
P
2
2 X) .
(N
or
1 bar
1
P
4 . Ka .
X 1.
1
N 2
1
1
(N
X)
X
1
N
1
0.5 .
0.5
N
4 . Ka .
P
1
0.5 . 1
1
4 . Ka .
1 bar
P
1
0.5 . 1
4 . Ka .
P
1
1
1 bar
1 bar
and
P
4 . Ka .
. .
. ( N R T)
V
P
1
1 bar
where
Ka exp
( ∆ Hrxn
T∆ Srxn )
.
RT
1 bar
Note that the EOS goes to the ideal gas limit as Ka goes to zero, and 1/2 the ideal gas limit when
Ka goes to infinity (all dimer). Also, the equation of state will have an other than linear dependence on
temperature due to the temperature dependence of Ka.
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.50 (also available as a Mathcad worksheet).
9.50
Given (All Units are SI):
T1
267
T2
255
2 . 10
6
P1 CH4
P2 CH4
Activities:
a hydrate
1
a H2O
1
a CH4 P CH4
1.5 . 10
6
R
8.31451
P CH4
5
10
Part (a):
a hydrate
Ka P CH4
5.75 .
a H2O
∆ G rxn T , P CH4
R. T . ln Ka P CH4
a CH4 P CH4
3
∆ G rxn T1, P1 CH4 = 6.65 10
3
∆ G rxn T2, P2 CH4 = 5.742 10
Part (b):
∆ H rxn
3
∆ S rxn
10
∆ G rxn T1, P1 CH4
Given
ans
(initial guesses)
10
Find ∆ H rxn , ∆ S rxn
4
∆ H rxn = 1.357 10
T1. ∆ S rxn
∆ H rxn
∆ H rxn
ans 0
∆ G rxn T2, P2 CH4
∆ S rxn
ans 1
∆ S rxn = 75.737
Part (c):
Ka 273
P273 CH4
Given
Ka P1 CH4 . exp
6
10
R
1
1
273
T1
.
Ka 273 = 0.044
(initial guess)
Ka P273 CH4
6
P273 CH4 = 2.288 10
∆ H rxn
Ka 273
P273 CH4
Find P273 CH4
∆ H rxn
T2. ∆ S rxn
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.51 The reactions are
iA1 ⇔ Aij
jB1 ⇔ B ji
iA1 + jB1 ⇔ Ai B j
and
The overall mass balance on species A
NA =
initial number of
=
moles of A1
∑ iN + ∑ ∑ iN
Ai
i
i
Ai Bj
=
j
F
∑ GH iN
+i
Ai
i
I
∑ N JK
Ai Bj
j
Overall mass balance on species B
NB =
initial number of
moles of B1
=
∑ jN + ∑ ∑ jN
Bj
j
j
F
I
∑ jGH N + ∑ N JK
=
Ai Bj
Bj
i
Ai Bj
j
i
At equilibrium
c
h
G Bi − iG B1 = 0 ; GAi B j − iGA1 + jGB1 = 0 ;
G Ai − iG A1 = 0 ;
and
G total = ∑ N Ai G Ai + ∑ N B j G B j + ∑ ∑ N Ai B j G Ai B j
i
j
i
j
At equilibrium dGtotal = 0 with respect to each extent of reaction.
dG total
∑G
A i dN A i
+ ∑ G B i dN
+ ∑N
A id G A i
+ ∑N
=
Bi
Bi d G B i
+ ∑ ∑ G A iB j dN
+ ∑∑N
A iB j
A i B j d G A iB j
=0
0 by the GibbsDuhem equation
Also, using the equilibrium equations
b
g
dGtotal = 0 = ∑ iG A1 dN Ai + ∑ iG B1dN Bi + ∑ ∑ iG A1 + jGB 1 dN Ai B j
= 0 = G A1 ∑ idN Ai + GB 1 ∑ idN Bi + ∑ iG A 1 ∑ dN Ai B j + ∑ jGB 1 ∑ dN Ai B j
= GA1
LM F idN
N∑ GH
i
Ai
i
+ i ∑ dN Ai B j
j
= G A 1dN A + GB 1dN B
FG ∂ G IJ
H ∂N K
total
A
N BT , P
I OP + G LM e jdN
JK Q N∑
≡ G A = G Ai . Also
B1
j
FG ∂ G IJ
H ∂N K
j
j
+ j ∑ dN Ai B j
total
B
NA , T, P
jOPQ
≡ GB = GBi
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
Therefore G A 1 = G A and G B1 = GB
Also, we have that, by definition
∂ Gi
∂P
FG IJ
H K
(1)
= RT
T
FG ∂ln f IJ
H ∂P K
i
(2)
T
Thus, integrating between any two states, we have
a f a f
Gi T , P2 − Gi T , P1 = RT ln
a f
a f
fi T , P2
fi T , P1
(3)
Now using Eqn. (3) with Eqns. (1), and recognizing that Eqns. (1) must be
satisfied at all T and P implies that
f A (T, P ) = f A1 (T, P ) and f B (T, P) = f B1 (T, P)
Alternatively we could integrate Eqn. (2) between P = 0 and the pressure P and
note that Eqn. (1) must be satisfied at all T and P. This implies that
f A1 ( P)
f A ( P)
=
f A1 ( P = 0)
f A ( P = 0)
but
as
P→0
only
A1
will
be
present
(LeChatelier’s
principle)
⇒ f A (P = 0) = f A 1( P = 0) so that f A 1( P) = f A ( P) .
9.52
A1 + A1 ⇔ A2
A1 + A2 ⇔ A3
M
A1 + An ⇔ An +1
etc.
N0 =
Total moles
of A1 initially
= NT
∑ i NN
i
= NT
T
⇒
N0
=
NT
= N1 + 2 N2 + 3 N3 + L + nAn + L =
∑ ix where N
i
T
∑ iN
i
= total number of moles in system
∑ ix
i
Now b = ∑ xi bi = ∑ ixi b1 = b1 ∑ ixi and
a=
which implies that
∑∑ x x
i j
aii a jj =
∑ ∑ ix jx a
i
j 1
d∑ ix i
= a1
2
i
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
(a)
FG N IJ
HN K
a = a1
0
T
2
and b = b1
FG N IJ
HN K
0
T
Also
K j +1 = K =
a j +1
a j a1
=
a f
φ φ x x a P 1 bar f
φj +1 x j +1 P 1 bar
2
j 1 j 1
=
φ j +1 x j +1 (1 bar )
φ j φ1 x j x1 P
Then
φ jφ1 P
φ j +1
=
x j +1
x j x1 K
or
φ j +1
φ jφ1 P
=
x j x1 K
x j +1
For the moment we will assume that
φ jφ1 P
φ j+ 1
=
x j +1
x j x1 K
=α
is independent of the index j and then show that this is indeed the case. Then
x12α K = x2
a
f a
f
a f
x1 x2α K = x3 = x1α K x2 = x1α K x12α K = x13 α K
2
Similarly
a f
x4 = x14 α K
M
a f
3
j− 1
x j = αK
x1j
M
etc.
Then NT = N 0 ∑ ixi = N0 ∑ i( Kα) i −1 x1i .
∑ x j = 1 = ∑ ( Kα) j−1 x1j
Also
=1
Now from the properties of geometric sums
∞
1
∑ θi = 1 − θ
i =0
we have
∞
∞
∞
j =1
j =1
j =0
∑ ( Kα) j−1 x1j = x1∑ aKαx1f j−1 = x1 ∑a Kαx1f j = 1 − K1αx
x
1
=1
(*)
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
so that
x1 = 1 − Kαx1 or x1 (1 + Kα) = 1 ; Kα =
1
−1
x1
Also
F
I
∑ iθi− 1 = dθ GH ∑θi JK = dθ FH 1 − θIK = (1 − θ) 2 = ∑ iθi −1
∞
d
i =0
∞
d
1
∞
1
i= 0
i =1
so that
∞
∞
∞
i =1
i =1
i= 0
∑ i( Kα)i− 1x1i = x1∑ia Kαx1fi −1 = x1∑ iaKαx1fi −1
=
x1
a1 − Kαx f
1
2
FG
H
N
1
x1
= T =
N0
x1 1 − Kαx1
from Eqn. (*) above
FG IJ
H K
IJ
K
FG IJ
H K
N
1
N0
N0
⇒ T =
and b = b1
= x1b1 ; a = a1
N0
x1
NT
NT
2
=
1 2
1
x1
2
= x12 a1
For the van der Waals equation of state we have
P=
RT
a
NRT
N 2a
− 2 =
−
V −b V
V − Nb
V
and
ln φi =
2∑ x j aij
Bi
− ln (Z − B) −
Z−B
RTV
which here becomes
ln φi =
2 ∑ x j ija1
iB1
iB1
2 iai
− ln(Z − B) −
=
− ln( Z − B) −
Z−B
RTV
Z−B
RTV
ln φi =
so that
iBi
2ia1 N0
− ln( Z − B ) −
Z− B
RTV NT
∑ jx j
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
ln
φ j φ1
φ j +1
= ln φ j + ln φ1 − ln φj +1
= ( j + 1 − ( j + 1))
B1
+ (− 1 − 1 + 1) ln( Z − B)
z−B
2a1 N0
( j + 1 − ( j + 1))
RTV NT
−
= − ln( Z − B )
φ j φ1
φ j +1
(b)
or
Pφ jφ1
φ j +1
=
=
1
RT P
=
Z − B V −b
RT
RT
NT RT
=
=
=α
V − b V − b1 N0 N T
NT V − N 0b1
a
f
which is independent of the index j as was assumed. Now that we know that
αis independent of the index j. We can use
x1 (1 + αK) = 1 and α =
FG
H
x1 1 +
IJ
K
NT RT
RT
RT
=
=
NT V − N0b1 V − N 0 N T b V − x1b
a
f
a
f
RTK
= 1 ⇒ x1 V − x1b1 + RTK = V − x1b1
V − x1b1
a
f
− x12b1 + x1 V + RTK + b1 − V = 0
a
f
b x − x1 V + RTK + b1 + V = 0
2
1 1
aV + RTK + b f ± aV + RTK + b f − 4b V
2
(c)
x1 =
1
1
1
(**)
2 b1
Also
P=
RT
a
RT
x2 a
− 2 =
− 1 21
V −b V
V − x1b V
(***)
Equations (**) and (***) are the set which forms the equation of state for the
associating van der Waals fluid. Notice that to solve for V we need x1 which
depends on V ; therefore, the equation is no longer cubic.
Note that if the fluid is non-associating, then K = 0 in this limit
x1 =
aV + b f ± dV
2
1
− 4b1V + b12
2b1
i = aV + b f − aV − b f = 1
so that
P=
RT
a
− 1
V − b1 V 2
which is the usual van der Waals equation.
1
1
2 b1
Chapter 9
Solutions to Chemical and Engineering Thermodynamics, 3e
9.53 The description of HF containing systems is described in the article “Collection of
Phase Equilibrium Data for Separation Technology” by William Schotte in Ind.
Eng. Chem. Process Des. Dev. (1980), 19, 432–439. By a careful examination of
the density and other data, he proposed that HF associates in the vapor phase
according to the reactions
2HF ⇔ ( HF) 2
6HF ⇔ ( HF) 6
8HF ⇔ ( HF)8
and, over the temperature range of 195 to 240 K, the equilibrium constants are
LM
OP
N
Q
211009
.
= exp L
MN T − 69.7292OPQ
252245
.
OP
= exp L
.
MN T − 834689
Q
K2 =
f2
6429.4
= exp
− 241456
.
f12
T
K6 =
f6
f16
K8 =
f8
f18
where Kn has units of (atm)n −1 , and fi is the fugacity of species i. Next, Schotte
used a (questionable) argument by Tamir and Wisniak [Chem. Eng. Sci (1978), 33,
651] that the fugacity coefficients φ i = fi xi P of the monomers, dimers,
hexamers and octamers are all aproximately equal, and can be calculated from the
fugacity of pure HF. This alleviated the need to specify the molecular parameters
for the association complexes. Using this model, Schotte obtained very good
agreement for the association factor (density) of pure HF and an HF-Freon mixture
using a simple equation of state such as Peng-Robinson (which must be solved
iteratively since chemical equilibrium is superimposed on the phase equilibrium
calculation).
An alternative, instead of using the Tamir-Wisniak assumption of equality of
fugacity coefficients is to use the model in Problem 9.52
a j = j2 a1 and bj = jb1
and then treat HF as a chemical reaction system with HF, ( HF)2 , ( HF)6 and
( HF)8 . Similarly, HF + non-associating component would be treated as a five
component system: non-associating component + HF, ( HF)2 , ( HF)6 and ( HF)8 .
In each of these cases the compositions of the HF components change as the
equilibrium changed. The problem with this proposal is that the a parameters for
the association complexes become unrealistically large.
For example,
a 8 = 64 × a1 . Consequently, no completely theoretically-correct model for the HF
associating system exists, though the models now in use are probably satisfactory
for engineering calculations.