Chem 420/523 Chemical Thermodynamics
Homework Assignment # 5
1. *Assume O2 gas obeys the virial equation
pVm = RT + Bp
with B = −12.5 cm3 mol−1 at 298.15 K. Calculate the fugacity of oxygen at p = 10.0 MPa and
T = 298.15 K.
Answer
The simplest equation for calculating fugacity, given an equation of state, is
ln
µ ¶
f
p
=
Z
0
p µZ
¶
−1
dp.
p
In this case, from the equation of state and the definition Z = pVm /(RT ), we get
Z=1 +
ln
µ ¶
B
p
RT
Z
f
B
=
p
RT
Bp
=
.
RT
p
dp
0
Therefore, paying special attention to units in the exponential, we get
f=peBp/(RT )
"
−12.5 × 10−6 m3 mol−1 × 106 Pa
=10.0 MPa × exp
8.3145 J K−1 mol−1 × 298.15 K
=9.95 MPa
#
2. Fugacity of liquid water at 298.15 K is approximately 3.17 kPa. Take the ideal enthalpy of vaporization
of water as 43,720 J mol−1 and calculate the fugacity of water at 300 K.
Answer
Using the Clausius-Clapeyron equation, we get
ln
µ
¶
µ
¶
f2
∆vap H ◦ 1
1
=
−
f1
R
T1 T2
µ
¶
1
1
43720 J mol−1
−
=
8.3145 J K−1 mol−1 298.15 300
=0.1088
So, given that f1 = 3.17 kPa, we calculate f2 :
f2 =f1 e0.1088
=3.5344 kPa
1
3. * The second virial coefficient of methane is given by
B/(cm3 mol−1 ) = 460 − 88, 000/T.
(a) Obtain an expression for the fugacity of methane as a function of temperature and pressure.
(b) Calculate ∆Gm for the expansion of methane at T = 175 K from p = 4.0 MPa to p = 0.20 MPa.
‡ − H for methane gas at T = 175 K and p = 4.0 MPa.
(c) Determine Hm
m
Answer
(a) From the virial equation pVm = RT + Bp, we get
µ
¶
88000
pVm = RT + 460 −
p
T¶
µ
p
88000
.
Z = 1 + 460 −
T
RT
Therefore, fugacity is given by
ln
µ ¶
f
p
=
µ
Z
0
p µZ
¶
−1
dp
p
¶Z
p
88000
1
460 −
dp
=
RT µ
T
¶0
p
88000
.
= 460 −
T
RT
So, after accounting for the units in the exponent, we get
f = p × exp
·µ
¶
p (Pa)
88000
460 −
× 10−6 m3 mol−1
T
R(J K−1 mol−1 )T (K)
¸
(b) For an isothermal expansion, we have
∆Gm = RT ln
µ
f2
f1
¶
From the temperature and the two pressures given, we use the expression derived in part (a) to calculate
f1 and f2 at T = 175 K, with B = −42.857 cm3 mol−1 ;
p1 = 4.0 MPa; f1 = 3.556 MPa.
p2 = 0.20 MPa; f2 = 0.199 MPa.
Therefore,
−1
∆Gm = 8.3145 J K
−1
mol
µ
¶
0.199
× 175 K × ln
3.556
= −4195.0 J mol−1
(c) From the expression derived in part (a), we write
b
ln f − ln p = e(a− T )p/RT .
From Eq. (6.37),
µ
∂ ln f
∂T
¶
=
p
2
‡ −H
Hm
m
.
RT 2
(1)
Evaluating the left hand side of Eq. (1), (note that p is held constant) we get
µ
∂ ln f
∂T
¶
p
·
µ
−0= − a−
b
T
¶
¸
b
p
b p
+ 2
e(a− T )p/RT
2
RT
T RT
aT − 2b 2 (a− b )p/RT
=−
p e T
.
RT 3
Therefore,
‡
Hm
− Hm = RT
µ
¶
2
µ
∂ ln f
∂T
¶
p
b
2b
=− a−
p × e(a− T )p/RT .
T
Substituting the constants (with proper units) and evaluating, we get
µ
¶
2 × 88000
× 10−6 m3 mol−1 (4.0 × 106 Pa)
175
"
#
−42.857 × 10−6 m3 mol−1 × 4.0 × 106 Pa
× exp
8.3145 J K−1 mol−1 × 175 K
‡
− Hm = − 460 −
Hm
= 1940.25 J mol−1 .
4. The partial pressure of Br2 above a (x1 CCl4 + x2 Br2 ) solution is 1.369 kPa. The composition of the
solution is x2 = 0.0250. The vapor pressure of pure bromine at the same temperature is 28.4 kPa. Assume
Raoult’s law standard state for bromine and calculate the activity coefficient of Br2 in the solution.
Answer
The activity coefficient is related to the activity ai and the mole fraction xi as
ai = γR,i xi .
The activity ai is defined, with reference to Raoult’s law standard state, as
ai =
fi
pi
' ∗.
fi∗
pi
Therefore,
1.369
= 0.0482.
28.4
a2
0.0482
γR,2 = =
= 1.928
x2
0.0250
a2 =
5. The following data are for various mixtures of isopropanol (I) in benzene (B) at 25 ◦ C.
xI
pI (torr)
ptot (torr)
0
0
94.4
0.059
12.9
104.5
0.146
22.4
109.0
0.362
27.6
108.4
0.521
30.5
105.8
0.700
36.4
99.8
0.836
39.5
84.0
0.924
42.2
66.4
1.00
44.0
44.0
(a) Construct a pressure-composition diagram showing the variation of pI , pB , and ptot with xI .
(b) Does this solution exhibit positive or negative deviation from Raoult’s Law?
3
(c) From a pressure composition diagram, determine the activities aI and aB and the activity coefficients γI
and γB at xI = 0.20, 0.50, and 0.80. (Assume Raoult’s law standard state.)
Answer
(a) The pressure-composition diagram is shown below. The partial pressure of benzene is calculated at each
composition as pB = ptot − pI .
Ptot
110
Piso
100
Pbenz
Pressure/Torr
90
80
70
60
50
40
30
20
10
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Mole fraction of Isopropanol
(b) It is clear that the solution exhibits positive deviation from Raoult’s law since at each point, pI > xI p∗I ,
and pB > xB p∗B .
(c) The activities are defined with respect to Raoult’s law standard state as ai = fi /fi∗ ' pi /p∗i and the
activity coefficients are defined as γR,i = ai /xi . The pressures for the table below are read from the
graph at the given mole fractions.
xI
0.20
0.50
0.80
pI (torr)
25.0
30.0
39.0
aI
0.568
0.682
0.886
γR,I
2.84
1.36
1.11
pB (torr)
84.5
76.0
50.0
aB
0.895
0.805
0.530
γR,B
1.12
1.61
2.65
6. A certain substance exists in two solid forms A and B and also in liquid and gas states. Construct a p-T
diagram for this system given the following triple point data and additional observations.
T /K
200
300
400
p/kPa
100
300
400
Phases in equilibrium
A, B, gas
A, B, liquid
B, liquid, gas
Below 200 K, the gas is in equilibrium with A and B does not exist. Above 400 K, the liquid is in
equilibrium with gas and neither solid form exists. Above 300 K, the equilibrium between solids A and
B does not exist. Above 300 K and 400 kPa, A is in equilibrium with liquid up to moderately higher
pressures, and solid A is denser than the liquid. The experiments did not go high enough in pressure or
temperature to establish the critical point of the gas. (In other words, limit the upper limits of your p-T
diagram to about 500 K.)
4
Answer
The qualitative phase diagram is shown below. The line representing the boundary between A(s) and
liquid has a positive slope because the solid is denser than the liquid.
Triple points and two-phase equilibria
450
Liquid
p (torr)
350
A(s)
250
Gas
B(s)
150
50
100
150
200
250
300
T (K)
5
350
400
450
500