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Sm-ch-5
engineering thermodynamics (‫)ریبکریما یتعنص هاگشناد‬
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Chapter 5 Solutions
Engineering and Chemical Thermodynamics 2e
Wyatt Tenhaeff
Milo Koretsky
School of Chemical, Biological, and Environmental Engineering
Oregon State University
milo.koretsky@oregonstate.edu
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5.1
 ∂u 

 - positive. This expression is equal to cv. As T goes up u goes up
 ∂T v
 ∂s 

 - positive. Higher temperature, more energy, more configurations
 ∂T v
 ∂u 
  - zero. The internal energy of an ideal gas depends only on T.
 ∂v T
 ∂s 
  - positive. An adiabatic isothermal expansion is spontaneous.
 ∂v T
2
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5.2
 ∂h 

 - positive. This expression is equal to cP. As T goes up h goes up
 ∂T  P
 ∂s 

 - positive. Higher temperature, more energy, more configurations
 ∂T  P
 ∂h 

 - zero. The enthalpy of an ideal gas depends only on T.
 ∂P T
 ∂s 

 - negative. An adiabatic isothermal expansion is spontaneous.
 ∂P T
 ∂P 

 - infinity. At constant h, T does not change as P changes
 ∂T  h
3
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5.3
 ∂u 

 - positive. This expression is equal to cv. As T goes up u goes up.
 ∂T v
 ∂s 

 - positive. Higher temperature, more energy, more configurations
 ∂T v
 ∂u 
  - positive. As the molecules move further apart, the attractive interactions decrease
 ∂v T
 ∂s 
  - positive. An adiabatic isothermal expansion is spontaneous.
 ∂v T
4
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5.4
 ∂h 

 - positive. This expression is equal to cP. As T goes up h goes up
 ∂T  P
 ∂s 

 - positive. Higher temperature, more energy, more configurations
 ∂T  P
 ∂h 

 - negative. As the molecules move closer together (higher P), the attractive interactions increase.
 ∂P T
 ∂s 

 - negative. An adiabatic isothermal expansion is spontaneous.
 ∂P T
5
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5.5
 ∂h 

 - positive. This expression is equal to cP. As T goes up h goes up
 ∂T  P
 ∂s 

 - positive. Higher temperature, more energy, more configurations
 ∂T  P
 ∂h 

 - close to zero. Liquid properties weakly depend on P.
 ∂P T
 ∂s 

 -- close to zero. Liquid properties weakly depend on P.
 ∂P T
β − positive. Liquids expand upon heating
κ − positive and close to zero
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5.6
c
 ∂g 
 ∂g   ∂h 
 ∂s 
cP − T P =
0

 Zero - 
 =

 −T 
 =
T
 ∂T  P
 ∂T  P  ∂T  P
 ∂T  P
 ∂g   ∂h 
 ∂s 
 ∂g 
   −T  
  positive =
 ∂P T  ∂P T
 ∂P T
 ∂P T
 ∂h 

 - zero. The enthalpy of an ideal gas depends only on T.
 ∂P T
 ∂s 

 - negative. An adiabatic isothermal expansion is spontaneous.
 ∂P T
 ∂g 
  zero holding P constant
 ∂P  P
 ∂v 
  negative – in an isentropic expansion, the volume decreases.
 ∂P  s
7
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5.7
Closest to ideal – methane, high T, low P
Methane at 240 oC and 1 bar
Methane at 180 oC and 1 bar
Water at 240 oC and 1 bar
Water at 180 oC and 1 bar
Methane at 180 oC and 10 bar
Water at 180 oC and 10 bar
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5.8
(b) CCl3H and CO(CH3)2 form a hydrogen bond in a specific orientation giving the mixture more
structure – higher magnitude of entropy departure.
9
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5.9
s = s (T , v) - For pressure explicit equation of state – the choice of T and v leads to derivatives that you can
explicitly differentiate.
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5.10
Volume
Path 1
State 2
(T2,v2)
∆u
State 1
(T1, v1)
Path 4
Temperature
We choose Path 1 since at we want to change the temperature when the gas behaves as an ideal gas;
this is most likely to occur at large volume
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Pressure
5.11
We choose Path 4 since at we want to change the temperature when the gas behaves as an ideal gas;
this is most likely to occur at low pressure.
State 2
(T2,P2)
State 1
(T1, P1)
∆h
Path 4
Temperature
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5.12
(a)
A sketch of the process is provided below
The diagram shows an infinitesimal amount of mass being placed on top of the piston of a
piston-cylinder assembly. The increase in mass causes the gas in the piston to be compressed.
Because the mass increases infinitesimally and the piston is well insulated, the compression is
reversible and adiabatic. For a reversible, adiabatic process the change in entropy is zero.
Therefore, the compression changes the internal energy of the gas at constant entropy as the
pressure increases.
(b)
To determine the sign of the relation, consider an energy balance on the piston. Neglecting
potential and kinetic energy changes, we obtain
∆U = Q + W
Since the process is adiabatic, the energy balance reduces to
∆U = W
As the pressure increases on the piston, the piston compresses. Positive work is done on the
system; hence, the change in internal energy is positive. We have justified the statement
 ∂u 
  >0
 ∂P  s
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5.13
(a)
Following the example given by Equation 5.5 in the text
 ∂u 
 ∂u 
du = 
 dT +   dP
 ∂T  P
 ∂P T
(b)
(c)
14
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5.14
The internal energy can be written as follows
 ∂u 
 ∂u 
du = 
 dT +   dv
 ∂T  v
 ∂v T
Substituting Equations 5.38 and 5.40
  ∂P 

 ∂u 
 ∂u 
 = cv and   = T 

 − P
 ∂T  v
 ∂v T   ∂T  v

into the above expression yields
  ∂P 

du = cv dT + T 
 − P  dv
  ∂T  v

From the ideal gas law, we have
R
 ∂P 
 =

 ∂T  v v
Therefore,

 RT
du = cv dT + 
− P  dv

 v
which upon noting that P =
RT
for an ideal gas, becomes
v
du = cv dT
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5.15
The heat capacity at constant pressure can be defined mathematically as follows
 ∂h 
 ∂u + ∂ (Pv ) 
 ∂u 
 ∂v 
cP = 
 =
 =
 + P

∂T
 ∂T  P 
 ∂T  P
 P  ∂T  v
For an ideal gas:
R
 ∂v 

 =
 ∂T  P P
Therefore,
 ∂u 
cP = 
 +R
 ∂T  v
One mathematical definition of du is
 ∂u 
 ∂u 
du = 
 dT +   dP
 ∂P T
 ∂T  P
 ∂u 
We can now rewrite 
 :
 ∂T  v
 ∂u   ∂T   ∂u   ∂P 
 ∂u 
 = cv
 +  
 
 =

 ∂T  v  ∂T  P  ∂T  v  ∂P T  ∂T  v
For an ideal gas:
 ∂u 
  =0
 ∂P  T
so
 ∂u 
cv = 

 ∂T  P
Substituting this result into our expression for c P gives
c P = cv + R
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5.16
In terms of P, v, and T, the cyclic equation is
 ∂P   ∂T   ∂v 
−1 = 
  
 
 ∂T  v  ∂v  P  ∂P T
For the ideal gas law:
Pv = RT
so the derivatives become:
R
 ∂P 

 =
 ∂T  v v
P
 ∂T 
 =

 ∂v  P R
− RT − v
 ∂v 
  = 2 =
P
 ∂P T
P
Therefore,
 ∂P   ∂T   ∂v 
 R  P  − v 
 = −1
 
   =   

 ∂T  v  ∂v  P  ∂P T  v  R  P 
The ideal gas law follows the cyclic rule.
17
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5.17
For a pure species two independent, intensive properties constrains the state of the system. If we
specify these variables, all other properties are fixed. Thus, if we hold T and P constant h cannot
change, i.e.,
 ∂h 
=0
 
 ∂v T , P
18
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5.18
Expansion of the enthalpy term in the numerator results in
 T∂s + v∂P 
 ∂h 

 =

∂T
s
 ∂T  s 
 ∂P 
 ∂h 
∴

 = v
 ∂T  s
 ∂T  s
Using a Maxwell relation
 ∂s 
 ∂h 
 = v 

 ∂v  P
 ∂T  s
 ∂s   ∂T 
 ∂h 
∴

 
 = v
 ∂T  P  ∂v  P
 ∂T  s
We can show that
c
 ∂s 
 = P

 ∂T  P T
(use thermodynamic web)
1  RT
2a 2ab 
 ∂T 

−
+
 = 

 ∂v  P R  v − b v 2 v 3 
(differentiate van der Waals EOS)
Therefore,
 v
vc  RT 2a 2ab 
2a  b  
 ∂h 
 = c P 
−
−
+
1 −  
 = P 

2
3
 ∂T  s RT  v − b v
v 
 v − b vRT  v 
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5.19
 ∂h 
  :
 ∂P T
 ∂s 
 T∂s + v∂P 
 ∂h 
 = T  + v
  =
∂P
 ∂P T
T
 ∂P T 
(
R
 ∂v 
 ∂s 
2
  = −
 = − 1 + B' P + C ' P
P
 ∂T  P
 ∂P T
(
)
)
RT
 ∂h 
1 + B ' P + C ' P 2 + v = −v + v
  =−
∂
P
P
 T
 ∂h 
  =0
 ∂P T
 ∂h 
  :
 ∂P  s
 ∂h   T∂s + v∂P 
  =
 =v
∂P
 ∂P  s 
s
RT
 ∂h 
1 + B' P + C ' P 2
  =
P
 ∂P  s
(
)
 ∂h 

 :
 ∂T  P
 ∂h 

 = cP
 ∂T  P
(Definition of cP)
 ∂h 

 :
 ∂T  s
 ∂P 
 ∂h 
 T∂s + v∂P 

 =
 = v

∂T
s
 ∂T  s
 ∂T  s 
c  ∂T 
c
P
 ∂P 
 ∂s   ∂P 
 = P
   = P
 = −

T  ∂v  P T R 1 + B' P + C ' P 2
 ∂T  s
 ∂T  P  ∂s T
(
20
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)
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(
(
Pv
1
1 + B' P + C ' P 2
 ∂h 
=
=
c
c


P
P
RT 1 + B' P + C ' P 2
 ∂T  s
1 + B' P + C ' P 2
(
)
)
)
 ∂h 
 = cP

 ∂T  s
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5.20
(a)
By definition:
1  ∂v 

v  ∂T  P
β= 
and
1  ∂v 

v  ∂P T
κ =− 
Dividing, we get:
 ∂v 
 
β
 ∂v   ∂P 
 ∂T  P
= −   
=−
κ
 ∂v 
 ∂T  P  ∂v  T
 
 ∂P  T
where derivative inversion was used. Applying the cyclic rule:
 ∂v   ∂P   ∂T 
−1 =      
 ∂T  P  ∂v  T  ∂P  v
Hence,
β  ∂P 
= 
κ  ∂T  v
(b)
If we write T = T(v,P), we get:
 ∂T 
 ∂T 
dT =   dv +   dP
 ∂P  v
 ∂v  P
(1)
From Equations 5.30 and 5.32
ds =
cv
c
 ∂v 
 ∂P 
dT +   dv = P dT −   dP
T
T
 ∂T  P
 ∂T  v
We can solve for dT to get:
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dT =
T  ∂P 
T  ∂v 
  dP
  dv +
c P − cv  ∂T  v
c P − cv  ∂T  P
(2)
For Equations 1 and 2 to be equal, each term on the left hand side must be equal. Hence,
T  ∂P 
 ∂T 
 
  =
 ∂v  P c P − cv  ∂T  v
or
β  ∂v 
 ∂P   ∂v 
c P − cv = T     = T  
κ  ∂T  P
 ∂T  v  ∂T  P
where the result from part a was used. Applying the definition of the thermal expansion
coefficient:
Tvβ
 ∂P   ∂v 
c P − cv = T     =
κ
 ∂T  v  ∂T  P
2
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5.21
We need data for acetone, benzene, and copper. A table of values for the molar volume, thermal
expansion coefficient and isothermal compressibility are taken from Table 4.4:
[ ]
 m3 
v × 10 6 

 mol 
73.33
86.89
7.11
Species
Acetone
Benzene
Copper
[ ]
β × 103 K -1
κ × 1010 Pa -1
1.49
1.24
0.0486
12.7
9.4
0.091
We can calculate the difference in heat capacity use the result from Problem 5.20b:
c P − cv =
vTβ 2
κ
or
[ ])
(
c p − cv =
Species
Acetone
Benzene
Copper

 3 
 73.33 × 10 − 6  m  (293 K ) 1.49 × 10 − 3 K -1 2


 mol  

12.7 × 10
−10
[Pa ]
-1
 J 
c p − cv 
 mol ⋅ K 
37.6
41.6
0.5
 J 
= 37.6 
 mol ⋅ K 
 J 
cp 
 mol ⋅ K 
125.6
135.6
22.6
% difference
30%
31%
2%
We can compare values to that of the heat capacity given in Appendix A2.2. While we often
assume that cP and cv are equal for condensed phases, this may not be the case.
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5.22
We know from Equations 4.32 and 4.33
1  ∂v 

v  ∂T  P
β= 
and
1  ∂v 

v  ∂P T
κ =− 
Maxwell relation:
 ∂P 
 ∂s 

  =
 ∂v T  ∂T  v
Employing the cyclic rule gives
 ∂P   ∂v 
 ∂P 

 = −  

 ∂v T  ∂T  P
 ∂T  v
which can be rewritten as
1  ∂v 


v  ∂T  P
 ∂P 
 ∂s 
  =
 =
 ∂v T  ∂T  v − 1  ∂P 
 
v  ∂v T
Therefore,
β
 ∂s 
  =
 ∂v T κ
Maxwell Relation:
 ∂v 
 ∂s 

  = −
 ∂T  P
 ∂P T
From Equation 4.32:
 ∂v 
 = βv

 ∂T  P
Therefore,
 ∂s 
  = − βv
 ∂P T
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5.23
(a)
An isochor on a Mollier diagram can be represented mathematically as
 ∂h 
 
 ∂s  v
This can be rewritten:
 ∂P 
 ∂h 
 T∂s + v∂P 
  =
 =T + 
∂s
 ∂s  v
 ∂s  v 
v
Employing the appropriate Maxwell relation and cyclic rule results in
 ∂h 
 ∂T   ∂s 
  = T + v
  
 ∂s  v
 ∂s  v  ∂v  T
We know
T
 ∂T 
 ∂P 
 ∂s 
and   = 

 =

 ∂s  v c v
 ∂v  T  ∂T  v
For an ideal gas:
R
 ∂P 
 ∂s 
  =
 =
 ∂v  T  ∂T  v v
Therefore,

R
T R
 ∂h 
= T 1 + 
  =T + v
cv 
cv v
 ∂s  v

(b)
In Part (a), we found
T  ∂P 
 ∂h 
  =T + v 

c v  ∂T  v
 ∂s  v
For a van der Waals gas:
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R
 ∂P 

 =
 ∂T  v v − b
Therefore,
RT  v 
 ∂h 
  =T +


cv  v − b 
 ∂s  v
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5.24
(a)
The cyclic rule can be employed to give
 ∂T 
 ∂T   ∂s 
  
 = −

 ∂P  s
 ∂s  P  ∂P T
Substitution of Equations 5.19 and 5.29 yields
T  ∂v 
 ∂T 


 =

 ∂P  s c P  ∂T  P
For an ideal gas:
R
 ∂v 

 =
 ∂T  P P
Therefore,
RT 1
v
 ∂T 
=
 =

P cP cP
 ∂P  s
(b)
Separation of variables provides
∂T
R ∂P
=
T
cP P
Integration provides
R
T 
 P c
ln 2  = ln 2  P
 T1 
 P1 
which can be rewritten as
R
 P2  cP
T2
= 
T1  P1 
The ideal gas law is now employed
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R
P2 v 2  P2  cP
= 
P1v1  P1 
 R
 R
1− 
1− 
cP 
c

P2
v 2 = P1 P  v1
where
1−
R c P − R cv 1
=
=
=
cP
cP
cP k
If we raise both sides of the equation by a power of k, we find
P2 v 2k = P1v1k
∴ Pv k = const.
(c)
In Part (a), we found
T  ∂v 
 ∂T 
 =



 ∂P  s c P  ∂T  P
Using the derivative inversion rule, we find for the van der Waals equation
Rv 3 (v − b )
 ∂v 

 =
 ∂T  P RTv 3 − 2a(v − b )2
Therefore,
1
RTv 3 (v − b )
 ∂T 
 =

 ∂P  s c P RTv 3 − 2a (v − b )2
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5.25
The development of Equation 5.39 is analogous to the development of Equation E5.3D. We
want to know how the heat capacity changes with pressure, so consider
 ∂c P 


 ∂P  T
which can be rewritten as
 ∂  ∂h  
 ∂  ∂h  
 ∂c P 
 = 

  =   
 ∂P T  ∂P  ∂T  P  T  ∂T  ∂P T  P
 ∂h 
Consider the   term:
 ∂P T
 ∂h 
 T∂s + v∂P 
 ∂v 
 ∂s 
 = T   + v = −T 
 +v
  =
∂P
T
 ∂P T 
 ∂T  P
 ∂P T
 ∂c 
Substitution of this expression back into the equation for  P  results in
 ∂P T
 ∂   ∂v 

 ∂c P 

 =   − T 
 + v 
 ∂P T  ∂T   ∂T  P
 P
 ∂ 2v 
∂T  ∂v 
 ∂c P 

 +  ∂v 
T
−
=
−




2


∂T  ∂T  P
 ∂P T
 ∂T  P  ∂T  P
 ∂ 2v 
 ∂c P 
 = −T  2 



 ∂P T
 ∂T  P
Therefore,
c Preal
∫ dc P =
c Pideal
 ∂ 2v  

T
−
∫   ∂T 2   dP

P 
Pideal 
Preal 
and
ideal
c Preal = c P
−
 ∂ 2v  

T
∫   ∂T 2   dP
P 
Pideal  
Preal 
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5.26
In order to solve this problem we need to relate the change in entropy from 10 to 12 bar to the
change in molar volume (for which we have complete data). First, we can rewrite the change in
entropy as
12 bar
∆s = s 2 − s1 =
 ∂s 
∫  ∂P T dP
10 bar
Applying a Maxwell relation, we can relate the above equation to the change in molar volume:
12 bar
12 bar
s 2 = s1 +
 ∂v 
 ∂s 
∫  ∂P T dP = s1 + ∫ −  ∂T  P dP
10 bar
10 bar
As 10 bar:
 m3 
 ∆v 
 ∂v 
−4
=
×
5
.
60
10
≅






 ∂T  P  ∆T  P
 kg ⋅ K 
At 12 bar:
 m3 
 ∆v 
 ∂v 
−4
=
×
≅
4
.
80
10






 ∂T  P  ∆T  P
 kg ⋅ K 
 ∂v 
To integrate the above entropy equation, we need an expression that relates 
 to pressure.
 ∂T  P
Thus, we will fit a line to the data. We obtain

 3 
 m3  
 ∂v 
−10
−4 m


  P + 9.6 × 10 
 =  − 4.0 × 10

 ∂T  P 
 kg ⋅ K 
 kg ⋅ K ⋅ Pa  
Now integrate the equation to find the entropy:
∫ [(4.0 ×10
1.2×10 6 Pa
s 2 = s1 +
1.0×10 6 Pa
−10
)P − 9.6 ×10− 4 ]dP =5.4960 − 0.104  kgkJ⋅ K  = 5.392  kgkJ⋅ K 
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5.27
(a)
From the Maxwell relation:
∂s ∂ P
  =

 ∂ v T  ∂ T v
The Redlich-Kwong equation gives
RT
a
P
=
−
v −b
T v (v + b)
For the Redlich-Kwong parameters:
 J ⋅ m3 ⋅ K1/2 
0.42748 R 2Tc2.5
14.24 
=
a =

2
Pc
 mol

3
0.08664 RTc
−5  m 
=
= 2.11×10 
b

Pc
 mol 
At 250 oC and 800 kPa
 m3 
v v = 0.00531 

 mol 
Taking the derivative:
R
1
a
 ∂P 
+
 =

3/2
 ∂T v v − b 2 T v ( v + b )
Pa
 ∂P 

 = 1,592
K
 ∂T v
(b)
6.9246 − 7.1816
kPa
Pa
∂s
=
= 1.595
= 1.595
 
K
K
 ∂ v T 0.23268 − 0.39383
Remarkably close!
(c)
For simplicity choose where the specific volume is the same: 1.2 MPa, 500 oC:
Pa
 ∂P  400, 000
=
= 1, 600


250
K
 ∂T v
Again this is close – could do better by interpolating.
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5.28
(a)
c= c
real
P
ideal
P
  ∂ 2v  
− ∫ T  2   dP
  ∂T P 
RT
+b
P
R
 ∂v 

 =
 ∂T  P P
=
v
 ∂ 2v 
 2
 ∂T P
real
ideal
c=
c=
35 J / ( mol K )
P
P
(b)
We will neglect kinetic and potential energy effects, and
consider the system to be at steady-state. Since this is an
adiabatic throttling process, we can immediately state that:
P
T1,P1
P1=10
∆h =
0
  R   RT

dh Cp dT +  −T   + 
dP Cp dT + bdP
=
+ b =

 P  P
Δh1
  ∂v 

 ∂h   ∂h 
=
dh 
 + =
 cp dT +  −T 
 + v  dP
 ∂T  P  ∂P T
  ∂T  P

T2,P2
P2=1
Δh2
T1=300°C
T2=?
First, we need to find the enthalpy change of each leg of the hypothetical path. Since we are
changing only one variable at a time, we can break up Equation 7 and integrate each leg of the
path separately.
∆h=
1
P2
∫ bdP= b ( P − P )
2
1
P1
=
∆h2
T2
dT c (T − T )
∫c =
ideal
P
ideal
P
2
1
T1
Combining these, we can solve for the unknown T2:
∆h = ∆h1 + ∆h2 = b ( P2 − P1 ) + c Pideal (T2 − T1 ) = 0
T2= T1 −
b ( P2 − P1 )
c Pideal
3


5 J
5 J 
−4 m
×
5
10

 1×10 3 − 10 × 10 3 
mole  
m
m 
=
T2 573K − 
J
35
mole ⋅ K
The final temperature of the gas will be T2 = 586K, or T2 = 313°C.
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T
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(c) For the system described above, find the Joule-Thompson coefficient, µ JT.
From the definition of the Joule-Thompson coefficient,
 ∂T 

 ∂P h
µ JT ≡ 
(1)
Since the Joule-Thompson coefficient is defined at constant enthalpy, dh = 0. Starting from
Equation 7, we find an expression for the above derivative.
dh= cPideal dT + bdP= 0
 ∂T 
cPideal 
0
 +b =
 ∂P  h
b
 ∂T 
µ JT =   = − ideal
cP
 ∂P h
The Joule-Thompson coefficient for the gas is negative, with a value of µ JT = -1.43x10-6
m3·K/J.
Does this make sense from a physical perspective? Intuitively, we expect gases to cool as they
expand (e.g. the formation of mist in a soda bottle when the cap is removed). However, our gas
was hotter after the expansion and pressure drop. Certain gases, such as helium, have very low
inversion temperatures, and have negative Joule-Thompson coefficients at normal laboratory
conditions.
(d) Since this process is adiabatic:
∆suniv =
∆ssys
cP
 ∂s 
 ∂s 
 ∂v 
dssys =

 dT +   dP =dT − 
 dP
T
 ∂T  P
 ∂P T
 ∂T  P
=
dssys
cP
R
dT − dP
T
P
T
P
J
∆suniv = ∆ssys = ∆s1 + ∆s2 = cP ln 2 − R ln 2 = 19.9
mol K
T1
P1
It is irreversible
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5.29
For the van der Waals EOS:
=
P
RT
a
− 2
v −b v
Solving for T:
a  v − b 

T  P + 2 
=

v  R 

For methane: Tc = 190.6 K and Pc = 46 bar, so
 J ⋅ m3 
27 ( RTc )
=
a =
0.23 
2 
64 Pc
 mol 
2
and
3
RTc
−5  m 
=
b = 4.31×10 

8 Pc
 mol 
so
T1 = 300 K and T2 = 365 K
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Picking T and v as independent properties:
0.05 m3
∆ua
Volume
State 1
(T1, v1)
∆ub
State 2
(T2, v2)
0.003 m3
300 K
365 K
Temperature
  ∂P 

 ∂u 
 ∂u 
du =
cv dT + T 

 dT +   dv =
 − P  dv
 ∂T v
 ∂v T
  ∂T v

Differentiating and substituting
a
c

du = R  P − 1 dT + 2 dv
v
R

c

dua R  P − 1 dT
=
R

Using data for methane from Appendix
365K
−6 2
A: ∆ua R ∫ ( 0.702 + 9.081×10−3 T − 2.164 × 10=
=
T )dT 1,970
300 K
dub =
a
dv
v2
3×10−3
a
J
∆ub =
−
=
−72
v 0.05
mol
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J
mol
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w =∆u =1,900
J
mol
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5.30
(a) First find a and b using
a
= 37.9 bar and
= 1.39
and
= 425.2 K:
b
= 1.17 x 10-4
Next Re-arrange the van der Waals EOS to solve for initial temperature (Tinitial)
P = Pinitial = 10 bar
v = vinitial = 3 x 10-3
So…Tinitial = Tfinal = 400.32 K = 400. K
Use the original van der Waals EOS to solve for Pfinal using the newly calculated Tfinal
(b)
vfinal = 0.05
6.61 x 104 Pa
(c)
To find work use
The problem statement indicated that PE is constant throughout the expansion process
therefore PE = Pfinal The integral can be modified as below:
wirrev = -3.11 x 103
(d)
To find q, you must first find ∆u
0
with
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Energy Balance: ∆u = qirrev + wirrev
q = ∆u – w = 436 + 3.11 x 103 = 3540
(e)
Last, to find the entropy change of the universe
∆suniv = ∆ssurr + ∆ssys
qsurr = -qsys
∆ssurr = -11.9
To solve for the change in entropy use the thermodynamic web:
and
 ∂s 
 ∂s 
 ∂s 
dssys = 
 dT +   dv =   dv
 ∂T v
 ∂v T
 ∂v T
 ∂P 
 ∂s 

  =
 ∂v T  ∂T  v
For the van der Waals EOS
R
 ∂P 
 =

 ∂T  v (v − b )
Now we can combine everything and calculate the change in entropy
v final
=
∆ssys
R
 v final − b 

J

dv R ln  =
 23.7 
∫ (=
v − b)
v
b
−
 mol ⋅ K 


vinitial
initial
Alternative for
Energy Balance for a reversible system: ∆u = qrev + wrev
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because internal energy is path independent
where P is given by the van der Waals EOS
wrev = -9050
qrev = ∆u - wrev = 9.48
Calculating entropy change of the system we get
Finally,
∆suniv = -11.9 + 23.7 = 11.8
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5.31
(a) P1 is low, so we can use an ideal gas for state 1:
=
T1
PV
1 1
=
300.7 K
n1 R
(b) It is an adiabatic reversible process so ∆s = 0
(c)
0.05 m3/mol
∆s1
Volume
State 1
(T1, v1)
∆s2
State 2
(T2, v2)
0.001 m3/mol
300 K
365 K
Temperature
For ethane: Tc = 305.4 K and Pc = 48.74 bar, so
 J ⋅ m3 
27 ( RTc )
=
a =
0.558 
2 
64 Pc
 mol 
2
and
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=
b
 m3 
RTc
= 6.51×10−5 

8 Pc
 mol 
so
cv
cv
R
 ∂P 
ds =
dT + 
dv
 dv =dT +
T
T
v −b
 ∂T v
Integrating
365K
∆s1 =
R ∫
( 0.131 + 0.019225T − 5.561×10 T )dT
−6
2
T
300 K
= 0.131ln
T2
5.561×10−6
+ 0.019225 × (T2 − 300.7 ) −
× (T22 − 300.7 2 )
300.7
2
and
∆s2 =
∫
J
R
dv =
−13.4
mol K
v −b
Implicit solution for T2:
T2 = 532 K and P2 = 41.7 bar
(d)
0.05 m3/mol
∆u1
Volume
State 1
(T1, v1)
∆u2
State 2
(T2, v2)
0.001 m3/mol
300 K
365 K
Temperature
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  ∂P 

 ∂u 
 ∂u 
du =
cv dT + T 

 dT +   dv =
 − P  dv
 ∂T v
 ∂v T
  ∂T v

Differentiating and substituting
a
c

du = R  P − 1 dT + 2 dv
v
R

c

du1 R  P − 1 dT
=
R

Using data for methane from Appendix
504.1K
A: ∆ua R
=
T )dT 11, 700
∫ ( 0.131 + 0.019225T − 5.561×10=
mol
−6
2
J
300 K
du2 =
a
dv
v2
1×10−3
a
J
∆ub =
−
=
−550
v 0.05
mol
w =∆u =11, 200
J
mol
W = nw = 22,400 J
(e)
Q=0
(f)
Larger – part of the compression is aided by the attractive force as CH4 gets closer
together. With no intermolecular interactions, this additional energy needs to be added.
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5.32
A schematic of the process follows:
We also know the ideal gas heat capacity from Table A.2.1:
cP
= 1.213 + 28.785 × 10 − 3 T − 8.824 ×10 − 6 T 2
R
Since this process is isentropic (∆s=0), we can construct a path such that the sum of ∆s is zero.
(a) T, v as independent variables
Choosing T and v as the independent variables, (and changing T under ideal gas conditions), we
get:
or in mathematical terms:
 ∂s 
 ∂s 
ds = 
 dT +   dv = 0
 ∂v T
 ∂T  v
However, From Equation 5.30:
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c
 ∂ P
dv
ds = v dT +
 ∂T  v
T
To get ∆s1
P=
RT
a
− 2
v−b v
 ∂ P
R
=
 ∂T  v v − b
so
and
v2
∆s1 = ∫ ds = ∫
v1
v2
 v − b
R
 ∂P 
dv = ∫
dv = R ln 2
 v1 − b 
 ∂T  v
v−b
v1
or, using the ideal gas law, we can put ∆s1 in terms of T2:
 RT 2 − b 
P

∆s1 = R ln  2
 v1 − b 
For step 2
T2
T2
T1
623.15 K
c
∆s 2 = ∫ v dT = R
T
∫
0.213 + 28.785 × 10 − 3 T − 8.824 × 10 − 6 T 2
dT
T
Now add both steps
∆s = ∆s1 + ∆s 2 = 0

 RT2
 P − b
8.824 × 10 − 6 2
 T2 
−3
2


T2 − (623.15 K )2
+ 0.213 ln
= ln
 + 28.785 × 10 (T2 − 623.15 K ) −
2
 v1 − b 
 623.15 




[
Substitute
T1 = 623.15 K
[
v1 = 600 cm 3 /mol
P2 = 1 atm
]
 cm 3 ⋅ atm 
R = 82.06 

 mol ⋅ K 
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]
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and solve for T2:
T2 = 448.3 [K ]
(b) T, P as independent variables
Choosing T and P as the independent variables, (and changing T under ideal gas conditions), we
get:
Mathematically, the entropy is defined as follows
 ∂s 
 ∂s 
ds = 
 dT +   dP = 0
 ∂P T
 ∂T  P
Using the appropriate relationships, the expression can be rewritten as
c
 ∂v 
ds = P dT − 
 dP = 0
T
 ∂T  P
For the van der Waals equation
 R 
 (v − b ) 


 ∂v 
 =

 ∂T  P 
RT
2a 
+ 
−
2
v 3 
 (v − b )
Therefore,
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T2
 R 
 (v − b ) 


P2
c
∆s = ∫ P dT − ∫
dP = 0
T


2
RT
a
T1
P1 −
+ 

 (v − b )2 v 3 
We can’t integrate the second term of the expression as it is, so we need to rewrite dP in terms
of the other variables. For the van der Waals equation at constant temperature:
 2a
RT 
dP =  −
 dv
3
(v − b )2 
 v
Substituting this into the entropy expression, we get
2
1.213 + 28.785 ×10 − 3 T − 8.824 ×10 − 6 T 2
R
∆s = ∫
dv = 0
dT − ∫
(v − b )
T
T2
v
T1
v1
Upon substituting
T1 = 623.15 K
v1 = 600 cm 3
v2 =
RT2
(gas acts ideally at 1 atm)
P2
 cm 3 
b = 91 

 mol 
 cm 3 ⋅ atm 
R = 82.06 

 mol ⋅ K 
we obtain one equation for one unknown. Solving, we get
T2 = 448.3 K
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5.33
(a)
Attractive forces dominate. If we examine the expression for z, we see that at any absolute
temperature and pressure, z < 1. The intermolecular attractions cause the molar volume to
deviate negatively from ideality and are stronger than the repulsive interactions.
(b)
Energy balance:
h2 − h1 = q
Alternative 1: path through ideal gas state
Because the gas is not ideal under these conditions, we have to create a hypothetical path that
connects the initial and final states through three steps. One hypothetical path is shown below:
Choosing T and P as the independent properties:
 ∂h 
 ∂h 
dh = 
 dT +   dP
 ∂P T
 ∂T  P
or using Equation 5.37
  ∂v 

dh = cP dT + − T 
 + v  dP
  ∂T  P 
The given EOS can be rewritten as
1

v = R + aT 1 / 2 
P

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Taking the derivative gives:
R
 ∂v 
− 0.5

 = + 0.5aRT
 ∂T  P P
so
(
)
dh = cP dT + 0.5aRT 0.5 dP
For step 1
∫ (0.5aRT1
0
0.5
∆h1 =
50 bar
)dP = −0.5aRT10.5 P = 252  molJ 
For step 2
∫ (3.58 + 3.02 ×10 T − 0.875T
500 K
∆h2 = R
−3
− 0.5
300 K
)dT = 7961  molJ 
For step 3:
∫ (0.5aRT2 )dP = 0.5aRT2 P = −323  mol 
50 bar
∆h3 =
0.5
 J 
0 .5
0
Finally summing up the three terms, we get,
 J 
q = ∆h1 + ∆h2 + ∆h3 = 7888 
 mol 
Alternative 2: real heat capacity
For a real gas
∆h = c Preal
From Equation 5.39:
 ∂ 2v 
T
∫  ∂T 2  dP
ideak
P

P
P real
ideal
c Preal = c P
−
For the given EOS
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1

v = R + aT 1 / 2 
P

Therefore,
 ∂ 2v 

 = −0.25aRT −1.5
 ∂T 2 

P
and
(
[ ])
P =50 bar
 ∂ 2v 
 dP =

T
− 0.25aRT −0.5 dP = 0.875 K 1/2 RT −0.5
∫ideak  ∂T 2 
∫
P

P
P ideak =0 bar
P real
real
We can combine this result with the expression for c Preal and find the enthalpy change.
∫ (3.58 + 3.02 ×10 T − 0.875T
500 K
∆h = R
−3
− 0.5
)dT
300 K
 J 
q = ∆h = 7888 
 mol 
The answers is equivalent to that calculated in alternative 1
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5.34
(a)
Calculate the temperature of the gas using the van der Waals equation. The van der Waals
equation is given by:
P=
RT
a
− 2
v −b v
First, we need to find the molar volume and pressure of state 1.
( [ ])
 m3 
V
Al 0.1 m 2 (0.4 [m])
v1 = 1 =
=
= 0.00016 

n
n
250 [mol]
 mol 
P1 =
mg
+ Patm =
A
(10000 [kg ]) 9.81  m2  

0.1 m 2
[ ]
s  
+ 1.01325 ×10 5 [Pa ] = 1.08 ×10 6 [Pa ]
Substituting these equations into the van der Waals equation above gives
 J ⋅ m3 

 J 
0
.5


 8.314 
T1
 mol 
 mol ⋅ K  

6
1.08 × 10 [Pa ] =
−
2
3

 m3 
3 


−5 m
m
0.00016 
  0.00016 
 − 4 × 10 


mol
 mol  
 mol 
 

T1 = 297.5 K
Since the process is isothermal, the following path can be used to calculate internal energy:
Thus, we can write the change in internal energy as:
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 ∂u 
 ∂u 
 ∂u 
du = 
 dT +   dv =   dv
 ∂v T
 ∂v T
 ∂T  v
Using Equation 5.34
v2
  ∂P 

∆u = ∫ T 
 − P  dv
 ∂T  v

v1 
For the van der Waals EOS:
P=
RT
a
− 2
v −b v
so
R
 ∂P 
 =

 ∂T  v v − b
Therefore,
v2
∆u = ∫
a
2
v v
dv
1
We can assume the gas in state 2 is an ideal gas since the final pressure is atmospheric.
Therefore, we calculate v 2 ,
v2 =
 m3 
RT2
= 0.0244 

P2
 mol 
and
 J ⋅ m3 
0
.
5


0.0244
 mol 
 J 
dv = 3104.5 
∆u = ∫
 mol 
v2
0.00016
or
∆U = 776.1 [kJ ]
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(b)
From the definition of entropy:
∆suniv = ∆s sys + ∆s surr
First, let’s solve for ∆s sys using the thermodynamic web.
 ∂s 
 ∂s 
ds sys = 
 dT +   dv
 ∂v T
 ∂T  v
Since the process is isothermal,
 ∂s 
ds sys =   dv
 ∂v T
v2
 ∂P 
∴ ∆s sys = ∫ 
 dv
∂T  v

v
1
Again, for the van der Waals equation,
R
 ∂P 
 =

 ∂T  v v − b
Substitution of this expression into the equation for entropy yields
v2
∆s sys = ∫
v1
R
dv
v−b
 J 
8.314 
 mol ⋅ K  dv = 44.17  J 
∆s sys = ∫
 mol ⋅ K 
 3
0.00016 v − 4 × 10 − 5 m


 mol 
J
∆S sys = 11042.5  
K
0.0244
The change in entropy of the surroundings will be calculated as follows
∆s surr =
Qsurr
Tsurr
where
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Qsurr = −Q
(Q is the heat transfer for the system)
Application of the first law provides
Q = ∆U − W
We know the change in internal energy from part a, so let’s calculate W using
v2
W = −n ∫ Pdv
v1
Since the external pressure is constant,

 m 
m 
W = −(250 [mol])(1.01325 × 10 5 [Pa ]) 0.0244 

 − 0.00016 

mol
mol 
3


W = −614030 [J ]
3

Now calculate heat transfer.
Q = 776100 [J ] − (− )614030 [J ] = 1.39 × 10 6 [J ]
Therefore,
∆S surr =
− 1.39 × 10 6 [J ]
J
= −4672  
297.5 [K ]
K 
and the entropy change of the universe is:
J
J
J
∆S univ = 11042.5   − 4672   = 6370.5  
K 
K 
K 
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
 
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5.35
First, calculate the initial and final pressure of the system.
(
20000 [kg ])(9.81 [m/s 2 ])
Pi = 10 × 10 [Pa ] +
= 4.92 × 10 6 [Pa ]
2
[ ]
(30000 [kg ])(9.81 [m/s ]) = 6.89 × 10 [Pa ]
P = 10 × 10 [Pa ] +
0.05 [m ]
5
0.05 m
2
5
f
6
2
To find the final temperature, we can perform an energy balance. Since the system is wellinsulated, all of the work done by adding the third block is converted into internal energy. The
energy balance is
∆u = w
To find the work, we need the initial and final molar volumes, which we can obtain from the
given EOS:
[
vi = 8.37 ×10 −4 m 3 /mol
vf =
]
8.314T f
(
)

25 
6.89 × 10 6 1 +


 Tf 
[
+ 3.2 × 10 - 5 m 3 /mol
]
Now, calculate the work






8.314T f
w = − Pf v f − vi = − 6.89 ×10 6 Pa 
+ 3.2 ×10 -5 − 8.37 ×10 −4 
 25 


6

6
.
89
10
1+
×


 Tf 






We also need to find an expression for the change in internal energy with only one variable: Tf.
To find the change in internal energy, we can create a hypothetical path shown below:
(
)
(
)
(
)
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For step 1, we calculate the change in internal energy as follows
v = RT / Plow
∫
∆u1 =
vi
∆u1 =
 ∂u 
  dv =
 ∂v T
v = RT / Plow
∫
vi
  ∂P 

 − P dv
T 
  ∂T  v

v = RT / Plow 
2
2 
 aRTi  dv = aRTi ln RTi / Plow − b 

 (T + a )2  (v − b ) (T + a )2 
vi − b

i

 i
∫
vi
Similarly, for step 3:
vf
v
f
  ∂P 

 ∂u 
−
P
dv
T
=
∆u3 =





∫  ∂v T
∫   ∂T  v dv
v = RT / Plow
v = RT / Plow
 aRT 2  dv


aRT f 2
vf −b
f



=
ln
∫  T + a 2  (v − b ) T + a 2  RT f / Plow − b 


f
v = RT / Plow  f

vf
∆u3 =
(
)
(
)
Insert the expression for the final molar volume into the equation for ∆u3 :


8.314T f


∆u3 =
ln

2 
6
Tf + a
 6.89 × 10 1 + 25 / T f RT f / Plow − b 
(
aRT f 2
)
(
)(
)(
)
Since the pressure is low (molar volume is big) during the second step, we can use the ideal heat
capacity to calculate the change in internal energy.
∆u 2 =
Tf
Tf
Ti = 500 K
Ti = 500 K
∫ cv dT =
(
∫ (20 − R + 0.05T )dT
)
(
∆u 2 = 11.686 T f − 500 + 0.025 T f2 − 500 2
)
If we set the sum of the three steps in the internal energy calculation equal to the work and
choose an arbitrary value for Plow, 100 Pa for example, we obtain one equation with one
unknown:
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(
)
 RT / P − b 
 + 11.686 T f − 500 + 0.025 T f2 − 500 2 +
ln i low
2
(Ti + a )  vi − b 
(
aRTi 2
)


8.314T f

=
ln

2 
6
Tf + a
 6.89 × 10 1 + 25 / T f RT f / Plow − b 
(
aRT f 2
)
(
)(
)(
)






8.314T f
− 6.89 × 10 6 Pa 
+ 3.2 × 10 -5 − 8.37 × 10 −4 



25 
6

 6.89 × 10 1 +



 Tf 


(
)
(
)
Solving for Tf we get
T f = 536.2 K
The piston-cylinder assembly is well-insulated, so
∆suniv = ∆s sys
Since the gas in the cylinder is not ideal, we must construct a hypothetical path, such as one
shown below, to calculate the change in entropy during this process.
For steps 1 and 3
Plow
P
low
  ∂v  
 ∂s 
∆s1 = ∫   dP = ∫ − 
  dP
∂T  P 
∂P T



P
P
i
i
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Pf
P
low
low
f
  ∂v  
 ∂s 
∆s3 = ∫   dP = ∫ − 
  dP
∂
∂P T
T


P 
P
P 
We can differentiate the given EOS as required:
Plow 
∆s1 = ∫ 
Pi 
− RTi (2a + Ti )  Plow 
− RTi (2a + Ti ) 

dP =
ln

P
(a + Ti )2 P 
(a + Ti )2
i


Pf 
∆s3 = ∫ 

Plow 
(
)
(
)
− RT f 2a + T f
− RT f 2a + T f 
 Pf 
 dP =

ln

P
a +Tf 2
a + T f 2 P 
low


(
)
(
)
For step 2
Tf
Tf
Tf
i
i
i
c

 20
 ∂s 
∆s 2 = ∫ 
 dT = ∫ P dT = ∫  + 0.05  dT
T

T
 ∂T  P
T
T
T
Tf 
 + 0.05 T f − Ti
∆s2 = 20 ln

T
 i 
(
)
Sum all of the steps to obtain the change in entropy for the entire process
∆suniv = ∆s sys = ∆s1 + ∆s 2 + ∆s3
(
)
RT f 2a + T f
 Pf 
Tf 
− RTi (2a + Ti )  Plow 


 + 0.05 T f − Ti −
 + 20 ln
ln
ln
P 
T 
2
(a + Ti )2
a
T
+
i
low
 Pi 




f
Arbitrarily choose Plow (try 100 Pa), substitute numerical values, and evaluate:
(
∆suniv =
)
 J 
∆suniv = 0.388 
 mol ⋅ K 

J
 J 
 = 0.766  
∆S univ = (2 mol) 0.388 

K 
 mol ⋅ K  

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)
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5.36
A schematic of the process is given by:
(a)
The following equation was developed in Chapter 5:
v   2  
∂ P
dv
cvreal = cvideal + ∫ T 
 ∂T 2  

v 
v ideal  
For the van der Waals EOS
 ∂2P 

=0
 ∂T 2 


Therefore,
cvreal = cvideal
From Appendix A.2:

3100 
 J 
 − R = 22.0 
cvreal = R 3.376 + 5.57 × 10 − 4 (500 ) −
 mol ⋅ K 
500 2 

(b)
As the diaphragm ruptures, the total internal energy of the system remains constant. Because the
volume available to the molecules increases, the average distance between molecules also
increases. Due to the increase in intermolecular distances, the potential energies increase. Since
the total internal energy does not change, the kinetic energy must compensate by decreasing.
Therefore, the temperature, which is a manifestation of molecular kinetic energy, decreases.
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(c)
Because the heat capacity is ideal under these circumstances we can create a two-step
hypothetical path to connect the initial and final states. One hypothetical path is shown below:
For the first section of the path, we have
Tf
Tf
∆u1 = ∫
∫ cv
cvreal dT =
ideal
Ti
Ti
(
Tf
∆u1 = R
(
dT
)

3100 
−4
2.376 + 5.57 ×10 T − 2  dT
T 
500 K 
∫
i
)
∆u1 = 2.32 ×10 − 3 T f2 + (19.75)T f +
25773.4
− 10507.4
Tf
For the second step, we can use the following equation
vf
 ∂u 
∆u 2 = ∫   dv
 ∂v T
v
i
If we apply Equation 5.34, we can rewrite the above equation as
vf
  ∂P 

∆u 2 = ∫ T 
 − P  dv
 ∂T  v

vi 
For the van der Waals EOS, P =
RT
a
− 2 ,
v −b v
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R
 ∂P 
 =

 ∂T  v v − b
Therefore,
v f = 0.002
v f = 0.002
a
 J 
 RT

∆u1 =
∫  v − b − P dv = ∫ v 2 dv = 73.7  mol 
v = 0.001
v = 0.001
i
i
Now set the sum of the two internal energies equal to zero and solve for Tf:
(
)
∆u1 + ∆u 2 = 2.32 × 10 − 3 T f2 + 19.75T f +
25773.4
− 10507.4 + 73.7 = 0
Tf
T f = 497 K
(d)
Since the system is well-insulated
∆suniv = ∆s sys
To solve for the change in entropy use the following development:
 ∂s 
 ∂s 
ds sys = 
 dT +   dv
 ∂v T
 ∂T  v
Using the thermodynamic web, the following relationships can be proven
cv
 ∂s 
 =

 ∂T  v T
 ∂P 
 ∂s 

  =
 ∂v T  ∂T  v
For the van der Waals EOS
R
 ∂P 
 =

 ∂T  v (v − b )
Now we can combine everything and calculate the change in entropy
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497 K
∆s sys =
∫
500 K
cv
dT +
T
0.002
R
∫ (v − b ) dv
0.001
(
)
0.002

497 K  2.376
dv
3100 
dT
+
∆s sys = R  ∫ 
+ 5.57 × 10 − 4 −

∫ v − 3.95 × 10 − 5 
500 K  T
T3 
0.001

 J 
∆suniv = ∆s sys = 5.80 

 mol ⋅ K 
(
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5.37
A schematic of the process is given by:
Energy balance:
∆u = 0
Because the gas is not ideal under these conditions, we have to create a hypothetical path that
connects the initial and final states through three steps. One hypothetical path is shown below:
For the first section of the path, we have
v =∞
∆u1 =
 ∂u 
∫  ∂v T dv
vi
If we apply Equation 5.34, we can rewrite the above equation as
v =∞
∆u1 =
  ∂P 

∫ T  ∂T v − P  dv
vi
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For the van der Waals EOS
R
a
 ∂P 
+

 =
2
 ∂T  v v − b T v 2
Therefore,
v =∞
∆u1 =
v =∞

 RT
2a
a
 J 
∫ −4  v − b + Tv 2 − P  dv = ∫ −4 T v 2 dv = 1120  mol 
i
v = 2.5×10
v = 2.5×10
i
i
Similarly for step 3:
v f = 5×10 −4
∆u3 =
∫
v =∞

 RT
a
 v − b + 2 − P  dv =
Tv


v f = 5×10 −4
∫
v =∞
2a
Tf v
2
dv =
− 168000  J 
 mol ⋅ K 
Tf
For step 2, the molar volume is infinite, so we can use the ideal heat capacity given in the
problem statement to calculate the change in internal energy:
∆u 2 =
(
3
R T f − 300 K
2
)
If we set sum of the changes in internal energy for each step, we obtain one equation for one
unknown:
(
)
− 168000
 J  3
 J 
+  8.314 
∆u1 + ∆u 2 + ∆u3 = 1120 
=0
 T f − 300 K +


Tf
 mol  2 
 mol ⋅ K  
Solve for Tf:
T f = 261.6 K
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5.38
A schematic of the process is shown below:
(a)
Consider the tank as the system. Since kinetic and potential energy effects are negligible, the
open system, unsteady-state energy balance (Equation 2.22) is
 dU 
 = ∑ nin hin − ∑ n out hout + Q + W s

 dt  sys in
out
The process is adiabatic and no shaft work is done. Furthermore, there is one inlet stream and no
outlet stream. The energy balance reduces to
 dU 
 = nin hin

 dt  sys
Integration must now be performed
U2
t
U1
0
∫ dU = ∫ n h dt
in in
t
n2 u 2 − n1u1 = hin ∫ n in dt = nin hin = (n2 − n1 )hin
0
Since the tank is initially a vacuum, n1=0, and the relation reduces to:
u 2 = hin
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As is typical for problems involving the thermodynamic web, this problem can be solved in
several possible ways. To illustrate we present two alternatives below:
Alternative 1: path through ideal gas state
Substituting the definition of enthalpy:
u 2 = u in + Pin vin
or
u2 (at 3 MPa, T ) − uin (at 3 MPa, 500 K ) = Pin vin
(1)
From the equation of state:
[
)]

 J 
 J 
(2)
Pin vin = RT (1 + B ' P ) =  8.314 
(552 K ) 1 − 2.8 × 10 −8 3 × 10 6 Pa = 3,800 

 mol 
 mol ⋅ K  

(
The change in internal energy can be found from the following path:
For steps 1 and 3, we need to determine how the internal energy changes with pressure at
constant temperature: From the fundamental property relation and the appropriate Maxwell
relation:
 ∂v 
 ∂v 
 ∂v 
 ∂s 
 ∂u 
 − P 
  = T   − P  = −T 
 ∂P  T
 ∂T  P
 ∂P  T
 ∂P  T
 ∂P  T
From the equation of state
RT
 ∂u 
(1 + B'P ) − P − RT2  = − B' RT
  =−
P
 P 
 ∂P T
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So for step 1:
 ∂u 
∆u1 = ∫   dP = − ∫ B ' RTdP =B ' RT Pin = −349 [J/mol]
∂P  T
Pin 
Pin
0
0
(3)
and for step 3:
 ∂u 
∆u 3 = ∫   dP = − ∫ B ' RTdP = − B ' RT P2 = 0.7T
∂P  T
0
0
P2
P2
(4)
For step 2
T
T
 ∂h   ∂Pv  
 ∂u 
∆u 2 = ∫ 
 dT = ∫ [c P − R ]dT
 −
 dT = ∫ 
∂T  P  ∂T  P 
∂T  P
500 
500 
500
T
or
∆u 2 = R
∫ [0.131 + 19.225 ×10 T − 5.561×10 T ]dT
T2
−3
−6
2
T1 =500 K
Substituting Equations 2, 3, 4, and 5 into 1 and solving for T gives:
T2 = 552 K
Alternative 2: real heat capacity
Starting with:
u 2 = hin
The above equation is equivalent to
h2 − P2 v2 = hin
∴ h2 − hin = P2 v2
To calculate the enthalpy difference, we can use the real heat capacity
P   2  
∂ v
ideal
dP
− ∫ T 
c Preal = c P
2 


∂
T
P 
Pideal  
For the truncated virial equation,
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 ∂ 2v 
 =0

 ∂T 2 
P

Therefore,
ideal
c Preal = c P
Now, we can calculate the change in enthalpy and equate it to the flow work term.
T2
∫ cP
ideal
dT = P2 v2
T1 = 500 K
∫ [1.131 + 19.225 ×10 T − 5.561×10 T ]dT = P v = RT (1 + B' P )
T2
R
−3
−6
2
2 2
2
2
T1 =500 K
Integrate and solve for T2:
T2 = 552 K
(b)
In order to solve the problem, we will need to find the final pressure. To do so, first we need to
calculate the molar volume. Using the information from Part (a) and the truncated virial
equation to do this
v=
RT
(1 + B' P ) =
P

 J 
 8.314 
(552 K )
 mol ⋅ K  

6
3 × 10 Pa
[1 − 2.8 ×10 (3 ×10 Pa )]
−8
6
 m3 
v = 0.0014 

 mol 
This quantity will not change as the tank cools, so now we can calculate the final pressure.

 m3  
P2  0.0014 


mol

 

= 1 − 2.8 ×10 −8 P2

 J 
(293 K )
 8.314 
 mol ⋅ K  

(
)
Solve for P2 :
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P2 = 1.66 × 10 6 Pa
The entropy change of the universe can be expressed as follows:
∆S univ = ∆S sys + ∆S surr
To solve for the change in entropy of the system start with the following relationship:
 ∂s 
 ∂s 
ds sys = 
 dT +   dP
 ∂T  P
 ∂P T
Alternative 1: path through ideal gas state
Using the proper relationships, the above equation can be rewritten as
ds sys =
cP
 ∂v 
dT + 
 dP
T
 ∂T  P
We can then use the following solution path:
Choosing a value of 1 Pa for Plow, for step 1:
R
 ∂v 
'
∆s1 = ∫ 
 dP = ∫ − 1 + B P dP
∂T  P
P
1.66 MPa
1.66 MPa 
1 Pa
(
1 Pa
)
For step 3,
R
 ∂v 
'
∆s1 = ∫ 
 dP = ∫ − 1 + B P dP
P
∂T  P
1 Pa 
1 Pa
3 MPa
3 MPa
(
)
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For step 2:
cP

1.131
dT = R ∫ 
+ 19.225 × 10 −3 − 5.561× 10 −6 T  dT
∫
T
T

552 K 
552 K
293 K
293 K
∆s 2 =
Adding together steps 1, 2 and 3:
 J 
∆s sys = −46.9 
 mol ⋅ K 
______________________________________________________________________________
Alternative 2: real heat capacity
Using the proper relationships, the above equation can be rewritten as
c real
 ∂v 
ds sys = P dT + 
 dP
T
 ∂T  P
For the truncated virial equation
 ∂v 
1


 = R + B ' 
 ∂T  P
P

Now, substitute the proper values into the expression for entropy and integrate:
1.66×10 6 Pa
293 K

1.131

1
∆s sys = R ∫ 
+ 19.225 × 10 − 3 − 5.561 × 10 − 6 T  dT + R
 + B' dP
∫

 T

P
552 K
3×10 ^ Pa
 J 
∆s sys = −46.9 
 mol ⋅ K 
______________________________________________________________________________
In order to calculate the change in entropy of the surroundings, first perform an energy balance.
∆u = q
Rewrite the above equation as follows
∆h − ∆(Pv ) = q
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Since the real heat capacity is equal to ideal heat capacity and the molar volume does not change,
we obtain the following equation
Tf
∫ cP
ideal
(
)
dT − v P f − Pi = q
Ti
T f = 293K
R
∫
Ti = 552 K
m  
[1.131 + 19.225 ×10 T − 5.561×10 T ]dT −  0.0014  mol
 (1.66 × 10 Pa - 3 × 10 Pa ) = q

−3
−6
3
2


6

 J 
q = −15845 
 mol 
Therefore,
 J 
q surr = 15845 
 mol 
and
 J 
∆s surr = 54.08 
 mol ⋅ K 
Before combining the two entropies to obtain the entropy change of the universe, find the
number of moles in the tank.
[ ]
0.05 m 3
= 75.7 mol
n=
 m3 
0.0014 

 mol 
Now, calculate the entropy change of the universe.

 J 
 J 
+ −46.9 
∆S univ = (75.7 mol) 54.08 
 

 mol ⋅ K  
 mol ⋅ K 

J
∆S univ = 544  
K 
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Pressure
5.39
(a) For an isentropic process, ∆s = 0. First, pick path
State 1
(T1, P1)
∆s
∆s1
∆s2
State 2
(T2,P2)
Temperature
cP
cP
R
 ∂v 
ds =
dT − 
 dP =dT − dv
T
T
v
 ∂T  P
∆s1 =− R ln
∆s2 =
P2
P1
T
T1
( cv + R ) ln 2
T
P
∆s = 0 = ( cv + R ) ln 2 − R ln 2
T1
P1
Solve for T2:
T2 = 307.8 K
(b)
 ∂T 

 ∂P  s
µ =
 ∂T   ∂s   ∂P 
−1 =
 
  
 ∂P  s  ∂T  P  ∂s T
cP
 ∂s 

 =
 ∂T  P T
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P
 ∂P   ∂T 
  = 
 = −
R
 ∂s T  ∂v  P
RT
 ∂T 
=
µ =

 ∂P  s PcP
Pressure
(c)
State 1
(T1, P1)
∆h
∆h1
∆h2
State 2
(T2,P2)
Temperature
  ∂v 

 ∂h 
 ∂h 
=
= cP dT +  −T 
dh 
 dT +   dP
 + v  dP
 ∂T  P
 ∂P T
  ∂T  P

=
dh cP dT + bdP
∆h = ∆h1 + ∆h2 = b ( P2 − P1 ) + cP (T2 − T1 ) = 9200
J
mol
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5.40
We write s = s ( P, v ) as:
 ∂s 
∂s
ds 
=
 dP +   dv
 ∂ P v
 ∂ v P
Substituting in Equation (5.40) gives:
 ∂s

∂s
∂s
 ∂s 
∂s
= 
ds
)   β vdT −   +   κ v  dP
 dP +   ( β vdT − κ vdP=
 ∂ P v
 ∂ v P
 ∂ v P
 ∂ P  v  ∂ v  P 
However, from Equation (5.41), we have
=
ds
cP
dT − β vdP
T
The first terms of the right hand sides must be equal:
cP
∂s
  βv =
T
 ∂ v P
So
cP
∂s
  =
 ∂ v  P β vT
Since the second terms of the right hand sides of Equations the last two equations must be equal:
 ∂ s   ∂ s 

βv
 +   κ v =

∂
P
∂
v




v
P


or
c Pκ
 ∂s 

= β v −
βT
 ∂ P v
Substituting back into Equation (E5.5A) gives:

c Pκ 
cP
ds =
 β v − β T  dP + β vT dv


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5.41
We can write h = h (T , v ) as:
∂h
∂h
dh 
=
 dT + 
 dv
 ∂ T v
 ∂ v T
Substituting in Equation (5.40) gives:
 ∂ h   ∂ h 

∂h
∂h
∂h
dh = 
 dT + 
 ( β vdT − κ vdP ) = 
 +
 β v  dT − 
 κ vdP
 ∂ T v
 ∂ v T
 ∂ v T
 ∂ T v  ∂ v T 
However, from Equation (5.43), we have
dh =
( cP − β Pv ) dT + vdP
Since the second terms of the right hand sides must be equal:
∂h
v
−
 κv =
 ∂ v T
or
1
∂h

 = −
κ
 ∂ v T
Similarly, the first terms of the right hand sides must be equal:
 ∂h  ∂h
( cP − β Pv )

 +
 βv =
 ∂ T v  ∂ v T
So
βv
∂h
cP − β Pv +

 =
κ
 ∂ T v
Finally, we get
βv 
1

dh =cP − β Pv +  dT − dv
κ 
κ

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5.43
First, focus on the numerator of the second term of the expression given in the problem
statement. We can rewrite the numerator as follows:
(
)(
uTr , v r − uTideal
= uTr , v r − uTideal
− uTideal
− uTideal
,v
,v =∞
,v
,v =∞
r
r
r
r
r
r
r
r
)
For an ideal gas, we know
uTideal
− uTideal
=0
,v
,v =∞
r
r
r
r
Therefore,
uTr , v r − uTideal
= uTr , v r − uTideal
,v
,v =∞
r
r
r
r
Substitute this relationship into the expression given in the problem statement:
∆uTdep
,v
r
r
RTc
=
uTr , v r − uTideal
,v
r
r
RTc
=
uTr , v r − uTideal
,v =∞
r
r
RTc
. Note that the temperature is constant.
Now, we need to find an expression for uTr ,vr − uTideal
,v = ∞
r
r
Equation 5.35 reduces to the following at constant temperature:
  ∂P 

duT = T 
 − P  dv
  ∂T  v

The pressure can be written as
P=
zRT
v
and substituted into the expression for the differential internal energy
  RT  ∂z 
 RT 2  ∂z  
RT 
zRT 


=
dv
duT = T 
−
+

   dv
 
v
v 
v 
 ∂T  v 
  v  ∂T  v

v

Applying the Principle of Corresponding States
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 T 2  ∂z  
  dvr
=  r 
RTc  vr  ∂Tr  v 
r 

duTr
If we integrate the above expression, we obtain
v
duTr
∫ RTc
=
v =∞
uTr , v r − uTideal
v =∞
r,
v
RTc
 T 2  ∂z  
  dv r
= ∫  r 
 v r  ∂Tr  v 
v =∞ 
r 
vr
Therefore,
∆uTdep
,v
r
RTc
r
=
uTr , v r − uTideal
,v
r
RTc
r
=
uTr , v r − uTideal
,v =∞
r
r
RTc
vr  2
 
T 
 r  ∂z   dv r
=
 v  ∂T  
v = ∞  r  r vr 
∫
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5.44
We write enthalpy in terms of the independent variables T and v:
 ∂h 
 ∂h 
dh = 
 dT +   dv
 ∂v T
 ∂T v
using the fundamental property relation:
dh = Tds + vdP
At constant temperature, we get:
  ∂P 
 ∂P  
dhT = T 
 + v   dv
 ∂v T 
  ∂T  v
For the Redlich-Kwong EOS
R
a
1
 ∂P 
+
 =

3
/
2
 ∂T  v v − b 2 T
v(v + b )
a
a
− RT
 ∂P 
+
+
  =
 ∂v T (v − b )2 T 1 / 2 v 2 (v + b ) T 1 / 2 v(v + b )2
Therefore,
 RT

RTv
a
a
3
−
+
+
dhT = 
 dv
 v − b (v − b )2 2 T 1 / 2 v(v + b ) T 1 / 2 (v + b )2 
To find the enthalpy departure function, we can integrate as follows
 RT

RTv
a
a
3
=
−
+
+
dh

∫ T ∫  v − b (v − b )2 2 T 1 / 2 v(v + b ) T 1 / 2 (v + b )2  dv

v =∞
v =∞ 
v
∆h
dep
=
v
Since temperature is constant, we obtain
∆h dep =
3a
RTb
a
 v 
ln
+
+

v − b 2bT 1 / 2  v + b  T 1 / 2 (v + b )
To calculate the entropy departure we need to be careful. From Equation 5.53, we have:
(
)(
gas
ideal gas
gas
gas
sT , P − sTideal
= sT , P − sTideal
− sTideal
,P
, P = 0 − sT , P
, P =0
)
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However, since we have a P explicit equation of state, we want to put this equation in terms of v.
Let’s look at converting each state. The first two states are straight -forward
sT , P = sT ,v
and
gas
ideal gas
sTideal
, P = 0 = sT , v = ∞
For the third state, however, we must realize that the ideal gas volume v’ at the T and P of the
system is different from the volume of the system, v. In order to see this we can compare the
equation of state for an ideal gas at T and P
P=
RT
v'
to a real gas at T and P
P=
RT
a
−
v −b
T v(v + b )
The volume calculated by the ideal gas equation, v’, is clearly different from the volume, v,
calculated by the Redlich-Kwong equation. Hence:
gas
gas  ideal gas
gas 
sTideal
= s ideal' gas = sTideal
+s '
− sTideal

,P
,v
,
v
T ,v
T
v
,


Thus,
(
)(
)
gas
gas
ideal gas
gas  ideal gas
gas 
sT , P − sTideal
= sT , v − sTideal
− sTideal
− sTideal

,P
, v = ∞ − sT , v
, v = ∞ −  sT , v '
,v

Using a Maxwell relation:
 ∂P 
 ds 

  =
 dv T  ∂T  v
Therefore,
 ∂P 
dsT = 
 dv
 ∂T  v
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For the Redlich-Kwong EOS
a
R
1
 ∂P 
+
 =

3
/
2
 ∂T  v v − b 2 T
v(v + b )
so
(s
) ∫  v R− b + 12 T
ideal gas
T , v − sT , v = ∞ =
v
v =∞ 

 dv
3/ 2
v(v + b )
a
For an ideal gas
R
 ∂P 
 =

 ∂T  v v
so
) ∫  Rv dv
(
gas
gas
− sTideal
sTideal
,v
,v =∞ =
v
v =∞
Finally:
v'
RT
dv
v'
ideal gas
ideal gas
− sT , v
=R
= R ln = R ln
s '
T ,v
Pv
v
v
v
∫
Integrating and adding together the three terms gives:
∆s dep = R ln
(v − b ) +
v
RT
 v 
ln
− R ln

Pv
2bT 3 / 2  v + b 
a
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5.45
Calculate the reduced temperature and pressure:
Tc = 647.3 [K ]
Pc = 220.48 [bar ]
(Table A.1.2)
w = 0.344
300 [bar ]
= 1.36
220.48 [bar ]
673.15 K
Tr =
= 1.04
647.3 K
Pr =
By double interpolation of data from Tables C.3 and C.4
 ∆h dep 
 Tr , Pr 
 RTc 


( 0)
= −2.921
 ∆h dep 
 Tr , Pr 
 RTc 


(1)
 ∆s dep 
 Tr , Pr 
 R 


(1)
= −1.459
From Tables C.5 and C.6:
 ∆s dep 
 Tr , Pr 
 R 


( 0)
= −2.292
= −1.405
Now we can calculate the departure functions
(1)
  dep  (0)
 ∆h dep  
∆hT , P

Tr , Pr  
r
r 
∆h dep = RTc  
+ w


 RTc  
  RTc 

 


 

 J 
 J 
(647.3 )(− 2.921 + 0.344(− 1.459 )) = −18421 
∆h dep =  8.314 

 mol ⋅ K  
 mol 

(1)
  dep  (0)
 ∆s dep  
s
∆

Tr , Pr 
Tr , Pr  
+ w
∆s dep = R  
 R  


 R 


 

 

 J 
 J 
(− 2.292 + 0.344(− 1.405)) = −23.07 
∆s dep =  8.314 

 mol ⋅ K 
 mol ⋅ K  

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To use the steam tables for calculating the departure functions, we can use the following
relationships.
∆h dep = hT , P − hTideal
,P
∆s dep = sT , P − sTideal
,P
From the steam tables
 kJ 
 kJ 
hT , P = 2151.0   and sT , P = 4.4728 

 kg 
 kg ⋅ K 
We need to calculate the ideal enthalpies and entropies using the steam tables’ reference state.
vap
(0 .01º C) +
hTideal
, P = ∆h
673.15 K
c ideal
dT
p
273.16 K
∫
 kJ 
from the steam tables and heat capacity data from Table A.2.2.
We can get ∆h vap = 45.1 
 mol 
Using this information, we obtain

 kJ  
hTideal
, P = 45.1 
 +  0.008314 
kJ  
 
 mol ⋅ K  
 mol  
673.15 K 
∫
3.47 + 1.45 × 10
273.16 K 
 kJ 
hTideal
, P = 59.14 
 mol 
Now, calculate the ideal entropy.
vap
(0 .01º C) +
sTideal
, P = ∆s
673.15 K c ideal
p
∫
273.16 K
T
P 
dT − R ln 2 
 P1 
From the steam tables:
 kJ 
∆s vap (0.01 º C ) = 0.165 
 mol ⋅ K 
Substitute values into the entropy expression:
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−3
0.121 × 10 5 
T+
 dT
T2

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673.15 K 
5
kJ   
3.47
30


− 3 0.121 × 10


+
dT
1
.
45
10
ln
−
×
+





∫
 mol ⋅ K   273.16 K  T
 0.000613 
T3



sTideal
, P = 0.165 +  0.008314 

 J 
sTideal
, P = 107  mol ⋅ K 


Now, calculate the departure functions:
 kJ 
 kJ 
 kJ 
= −20.4 
∆h dep = 2151.0   (0.0180148 [kg/mol]) − 59.14 

 mol 
 mol 
 kg 
 kJ 
(0.0180148 [kg/mol]) − 0.107  kJ  = −0.0264  kJ 
∆s dep = 4.4728 

 mol ⋅ K 
 mol ⋅ K 
 kg ⋅ K 
Table of Results
Generalized
Percent Difference
Steam Tables
Tables
(Based on steam tables)
 kJ 
∆h dep 
 mol 
 kJ 
∆s dep 
 mol ⋅ K 
-18.62
-20.4
9.9
-0.0231
-0.0264
12.5
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5.46
State 1 is at 300 K and 30 bar. State 2 is at 400 K and 50 bar. The reduced temperature and
pressures are
30 [bar ]
= 0.616
48.74 [bar ]
300 K
T1, r =
= 0.982
305.4 K
P1, r =
50 [bar ]
= 1.026
48.74 [bar ]
400 K
= 1.31
T 2, r =
305.4 K
P2,r =
and
ω = 0.099
By double interpolation of data in Tables C.3 and C.4
 ∆h dep 
 T1, r , P1, r 
 RT 
c 


( 0)
 ∆h dep 
 T2 , r , P2 , r 

 RT
c 


(1)
= −0.825
 ∆h dep 
 T1, r , P1, r 
 RT 
c 


= −0.711
 ∆h dep 
 T2 , r , P2 , r 
 RT

c 


( 0)
= −0.799
(1)
= −0.196
Therefore,
 ∆h dep 
 T1, r , P1, r 
 RT  = −0.825 + 0.099(− 0.799 ) = −0.904
c 


 ∆h dep 
 T2 ,r , P2 ,r  = −0.711 + 0.099(− 0.196 ) = −0.730
 RTc 


The ideal enthalpy change from 300 K to 400 K can be calculated using ideal cP data from Table
A.2.1.
400 K
∆hTideal
=R
→T
1
2
−3
−6 2
∫1.131 + 19.225 ×10 T − 5.561×10 T dT = 717.39R
300 K
The total entropy change is
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dep
dep
∆h = −∆hT , P + ∆hTideal
+ ∆hT , P
→T
1, r
1, r
1
2
2,r
2,r
∆h = R[− (− )0.904TC + 717.39 − 0.730TC ]

 J 
∆h =  8.314 
 [0.904(305.4 K ) + 717.39 K − 0.730(305.4 K )]
 mol ⋅ K  

 J 
∆h = 6406.2 

 mol 
Using the data in Table C.5 and C.6
 ∆s dep 
 T1, r , P1, r  = −0.601 + 0.099(− 0.756 ) = −0.676


R


 ∆s dep 
 T2 , r , P2 , r 
 = −0.394 + 0.099(− 0.224 ) = −0.416

R


Substituting heat capacity data into Equation 3.22, we get
400 K 1.131 + 19.225 × 10 −3 T − 5.561 × 10 −6 T 2
 50 bar 
dT − ln
∆s ideal = R  ∫

T
 30 bar 
300 K
∆s ideal = 1.542 R
Therefore,
∆s = −∆sT , P + ∆s ideal + ∆sT , P = R(0.676 + 1.542 − 0.416 )
dep
1, r
dep
1, r
2,r
2,r
 J 
∆s = 14.98 

 mol ⋅ K 
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5.47
The turbine is isentropic. Therefore, we know the following
dep
dep
∆s = −∆sT , P + ∆s ideal + ∆sT , P = 0
1, r
1, r
2,r
2,r
Using the van der Waals EOS, we can find P1,r, which leaves one unknown in the above
equation: T2.
P1 =


 3
 82.06  cm ⋅ atm  (623.15 K )


 mol ⋅ K  


 3 
 3
 600  cm  − 91  cm  


 mol  
 mol 

−
 atm ⋅ cm 3 
91 × 10 5 

 mol 2 

 3 
 600  cm  


 mol  

P1 = 75.19 [atm] = 76.19 [bar ]
2
Calculate reduced temperature and pressures using data from Table A.1.1
76.19 [bar ]
= 1.8
42.44 [bar ]
623.15 K
T1, r =
= 1.68
370.0 K
P1, r =
P2,r =
1.013 [bar ]
= 0.024
42.44 [bar ]
Also,
ω = 0.152
From Tables C.5 and C.6:
 ∆s dep 
 T1, r , P1, r  = −0.327 + 0.152(− 0.102 ) = −0.343


R


Substituting heat capacity data into Equation 3.22, we get
∆s
ideal
 T2 1.213 + 28.785 × 10 − 3 T − 8.824 × 10 − 6 T 2

 1 atm 

dT − ln
=R

75.19 atm 
T
 ∫

623.15 K

Therefore,
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
 ∆s dep  
T2
1.213 + 28.785 × 10 −3 T − 8.824 × 10 −6 T 2

 T2 , r , P2 , r  
R 0.343 +
dT
4
.
32
+
+
∫
  = ∆s

T
R

623.15 K
 


We can solve this using a guess-and-check method
T2 = 600 K : T2,r = 1.62
 J 
∆s = 33.84 

 mol ⋅ K 
T2 = 450 K : T2,r = 1.22
 J 
∆s = 0.77 

 mol ⋅ K 
T2 = 446.6 K : T2,r = 1.21
 J 
∆s ≅ 0 
 mol ⋅ K 
Therefore,
T2 = 446.6 K
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5.48
A reversible process requires the minimum amount of work. Since the process is reversible and
adiabatic
∆s = 0
which can be rewritten as
dep
dep
∆s = −∆sT , P + ∆s ideal + ∆sT , P = 0
1, r
1, r
2,r
2,r
Calculate reduced temperature and pressures using data from Table A.1.1
1 [bar ]
= 0.0217
46.0 [bar ]
300 K
T1, r =
= 1.57
190.6 K
P1, r =
P2,r =
10 [bar ]
= 0.217
46.0 [bar ]
From Tables C.5 and C.6:
 ∆s dep 
 T1, r , P1, r 
= −0.00457 + 0.008(− 0.0028) = −0.0046

R 


Substituting heat capacity data into Equation 3.22, we get
∆s
ideal

 T2 1.702 + 9.081× 10−3T − 2.164 × 10−6 T 2
 10 bar 
dT − ln
= R ∫

T
300 K
 1 bar 


Therefore,

 ∆s dep  
T2
−3
−6 2
T
T
1
.
213
28
.
785
10
8
.
824
10
+
×
−
×

 T2 , r , P2 , r  
dT − 2.303 + 
∆s = R 0.0046 + ∫

T
R

300 K
 



We can solve using a guess-and-check method
T2 = 400 K : T2,r = 2.10
 J 
∆s = 4.98 

 mol ⋅ K 
T2 = 385 K :
T2,r = 2.02
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 J 
∆s = 1.42 

 mol ⋅ K 
T2 = 379 K :
T2,r = 1.99
 J 
∆s = −0.018 

 mol ⋅ K 
Therefore,
T2 ≅ 379 K
An energy balance reveals that
h2 − h1 = ∆h = ws
We can calculate the enthalpy using departure functions. From Tables C.3 and C.4:
 ∆h dep 
 T1, r , P1, r 
 RT  = −0.0965 + 0.008(− 0.011) = −0.0966
c 


dep

 ∆h
 T2 , r , P2 , r 
 = −0.0614 + 0.0089(0.015) = −0.0613
 RT
c


Ideal heat capacity data can be used to determine the ideal change in enthalpy

379 K
∆h ideal = R  ∫ 1.702 + 9.081 × 10 − 3 T − 2.164 × 10 − 6 T 2 dT 

300 K
Therefore,
379 K


 J  
−3
−6 2


(
)(
)

∆h =  8.314 
T
T
dT
190
.
6
K
0
.
0966
0
.
0613
1
.
702
9
.
081
10
2
.
164
10
−
+
+
×
−
×
∫

 mol ⋅ K   


300 K
 J 
w s = ∆h = 3034.2 

 mol 
and
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
 mol  
 J 
 3034.2 
W S = 1 / 30 

  = 101.1 W
 s  
 mol  

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5.49
Equation 4.32 states
1  ∂v 

v  ∂T  P
β= 
 ∂v 
∴ βv = 

 ∂T  P
This can be substituted into Equation 5.57 to give
µ JT =
v(βT − 1)
cP
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5.50
For an ideal gas
R
 ∂v 
 =

 ∂T  P P
Therefore,

 RT
− v

P
 = (v − v ) = 0
µ JT = 
cP
cP
This result could also be reasoned from a physical argument.
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The van der Waals equation is given by:
P=
RT
a
−
v − b v2
(1)
The thermal expansion coefficient is given by:
1  ∂v 
 1   ∂T 
 =   
v  ∂T  P  v   ∂v  P
β= 
(2)
Solving Equation 1 for T:
a  v − b 

T =  P + 2 

v  R 

Differentiating by applying the chain rule,

a  1  v − b  2a Pv 3 − av + 2ab
 ∂T 
=

  =  P + 2  − 
 ∂v  P 
v  R  R  v3
Rv 3
Substitution into Equation 2 gives
β=
Rv 2
Pv 3 − av + 2ab
Substituting Equation 1 for P gives b in terms of R, T, v, a , and b:
β=
Rv 2 (v − b )
2
RTv 3 − 2a(v − b )
The isothermal compressibility is given by:
1  ∂v 
 1   ∂P 
 = −   
v  ∂P  T
 v   ∂v  T
κ =− 
From the van der Waals equation:
RT
2a − RTv 3 + 2a(v − b )
 ∂P 
+
=
  =−
2
(v − b )2 v 3
v 3 (v − b )
 ∂v T
2
so
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(3)
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κ=
v 2 (v − b )
2
RTv 3 − 2a(v − b )
2
For the Joule-Thomson coefficient, we can use Equation 5.57:
  ∂v 

T  ∂T  − v 
  P

µ JT =
Preal
2
 ∂ v 
c Pideal − ∫ T  2   dP
Pideal 
  ∂T  P 
Substituting the van der Waals equation into Equation 3 gives
RTv 3 − 2a (v − b )
T
2a (v − b )
 ∂T 
=
−
  =
3
(v − b ) Rv 3
(v − b )Rv
 ∂v  P
2
(4)
Thus, the second derivative becomes:
 ∂ 2T 
1  ∂T 
2a 6a(v − b )
T
 2  = −
+
  − 3+
2
Rv 4
(v − b ) v − b  ∂v  P Rv
 ∂v  P
or simplifying using Equation 4,
 ∂ 2T 
2a(v − 3b )
 2  =
Rv 4
 ∂v  P
(5)
Substituting Equations 5 and 4 into Equation 5.57 gives:
− bRTv 3 + 2av(v − b )
2
RTv 3 − 2a(v − b )
µ JT =
Preal
 RTv 4 
c Pideal − ∫ 
 dP
(
)
a
v
b
−
2
3


Pideal
2
At a given temperature the integral in pressure can be rewritten in terms of volume using the van
der Waals equation to give:
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− bRTv 3 + 2av(v − b )
2
RTv 3 − 2a (v − b )
µ JT =
vreal
 RTv  RTv 3 − 2a (v − b )2
ideal
dv
cP + ∫ 
2a (v − 3b ) 
(v − b )2
videal 
2
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5.52
We can solve this problem by using the form of the Joule-Thomson coefficient given in Equation
5.57. The following approximation can be made
 ∆vˆ 
 ∂vˆ 


 ≅
 ∂T  P  ∆T  P
At 300 ºC,
vˆ(350 º C,1MPa ) − vˆ(250 º C,1MPa )
 ∆vˆ 

 =
350 − 250 º C
 ∆T  P
 m3 
 m3 
0.28247 

 − 0.23268 
 ∆vˆ 
 kg 
 kg 
 =

350 − 250 º C
 ∆T  P
 m3 
 m3 
 ∂vˆ 
∴
 = 0.0005 

 = 0.0005 
 ∂T  P
 kg ⋅ º C 
 kg ⋅ K 
A similar process was followed to find cP.
 ∂hˆ 
 ˆ
 ≅  ∆h 
cˆ P = 



 dT  P  ∆T  P
At 300 ºC,
 ∆hˆ 
hˆ(350 º C,1MPa ) − hˆ(250 º C,1MPa )

 =


350 − 250 º C
 ∆T  P
 kJ 
 kJ 
3157.7   − 2942.6  
 kg 
 kg 
=
350 − 250 º C
 ∆hˆ 


 ∆T 
P

 ∂hˆ 
 ˆ




 ≅  ∆h  = 2.15  kJ  = 2.15  kJ 
cˆ P = 



 kg ⋅ K 
 kg ⋅ º C 
 dT  P  ∆T  P
Now, µ JT can be found.
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3 
3 


 (573.15 K ) 0.0005  m   − 0.25794  m  
  ∂vˆ 





T 


  ∂T  − vˆ  
kg
⋅
kg
K



 




P
=
µ JT = 
cˆ P
 kJ 
2.15 

 kg ⋅ K 
 m3 ⋅ K 

 kJ 
µ JT = 0.0133 
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5.53
At the inversion line, the Joule-Thomson coefficient is zero. From Equation 5.57:
  ∂v 

T  ∂T  − v 
  P

=0
µ JT =
Preal
2

∂ v 
ideal
c P − ∫ T  2   dP
  ∂T  P 
Pideal 
This is true when the numerator is zero, i.e.,
  ∂v 

T   − v  = 0
  ∂T  P

For the van der Waals equation, we have
P=
RT
a
− 2
v −b v
Solving for T:
a  v − b 

T =  P + 2 

v  R 

so
a  1  v − b  2a Pv 3 − av + 2ab

 ∂T 
=

  = P + 2  −
Rv 3
v  R  R  v3
 ∂v  P 
Substituting for P:
RTv − 2a(v − b )
 ∂T 
  =
(v − b )Rv 3
 ∂v  P
3
2
Hence,
− bRTv + 2av(v − b )
 ∂v 
T  − v = 0 =
2
RTv 3 − 2a(v − b )
 ∂T  P
3
2
Solving for T:
2av(v − b )
bRv 3
2
T=
(1)
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Substituting this value of T back into the van der Waals equation gives
P=
2av(v − b ) a a(2v − 3b )
− 2 =
v
bv 2
bv 3
(2)
We can solve Equations 1 and 2 by picking a value of v and solving for T and P. For N2, the
critical temperature and pressure are given by Tc = 126.2 [K] and Pc = 33.84 [bar], respectively.
Thus, we can find the van der Waals constants a and b:
2

RT 
 Jm 3 
a = 27  c  = 0.137 
2
64 Pc
 mol 
b=
 m3 
RTc
= 3.88 ×10 -5 

8 Pc
 mol 
Using these values in Equations (1) and (2), we get the following plot:
P [bar]
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5.54
We can solve this problem using departure functions, so first find the reduced temperatures and
pressures.
50 [bar ]
= 0.99
50.36 [bar ]
273.15 K
T1, r =
= 0.967
282.4 K
P1, r =
P2,r =
10 [bar ]
= 0.2
50.36 [bar ]
Since the ethylene is in two-phase equilibrium when it leaves the throttling device, the
temperature is constrained. From the vapor-liquid dome in Figure 5.5:
T2,r ≅ 0.76
∴T2 = 214.6
The process is isenthalpic, so the following expression holds
dep
dep
∆h = −∆hT , P + ∆hTideal
+ ∆hT , P = 0
→T
1, r
1, r
1
2
2,r
2,r
Therefore,
dep
dep
∆hT , P = ∆hT , P − ∆hTideal
→T
2,r
2,r
1, r
1, r
1
2
From Table A.2.1:
∫ [1.424 + 14.394 ×10 T − 4.392 ×10
214.6 K
∆hTideal
=R
1 → T2
−3
−6 2
]
T dT
273.15 K
From Tables C.3 and C.4 (ω = 0.085) :

 ∆h dep
 T1,r , P1,r  = −3.678 + 0.085(− 3.51) = −3.976

 RT
c


Now we can solve for the enthalpy departure at state 2.
∆hTdep, P
2,r
RTc
2,r
214.6 K


1
−3
−6 2

= − 3.976 −
∫1.424 + 14.394 ×10 T − 4.392 ×10 T dT 
282.4 K

273.15 K


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∆hTdep, P
2,r
RTc
2,r
= −3.01
We can calculate the quality of the water using the following relation
∆hTdep, P
2,r
2,r
RTc
= (1 − x )
∆hTdep,, liq
P
2,r
2,r
RTc
+x
∆hTdep,, vap
P
2,r
2,r
RTc
where x represents the quality. From Figures 5.5 and 5.6:
∆hTdep,, liq
P
2,r
2,r
RTc
∆hTdep,, vap
P
2,r
2,r
RTc
= −4.6 + 0.085(− 5.5) = −5.068
= −0.4 + 0.0859(− 0.75) = −0.464
Thus,
x = 0.447
55.3% of the inlet stream is liquefied.
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5.55
Density is calculated from molar volume as follows:
ρ=
MW
v
2
Substitute the above into the expression for Vsound
:




∂P 
1  ∂P 
2



=
Vsound =
  MW  
MW  ∂ (1 / v )  s

 ∂
  v  s
The following can be shown using differentials:
∂v
1
∂  = −
v
v2
Therefore,
 ∂P 
v 2  ∂P 
2
Vsound
=   = −
 
MW  ∂v  s
 ∂ρ  s
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5.56
From Problem 5.55:
 ∂P 
2
Vsound
=  
v 2  ∂P 
=−
 
MW  ∂v  s
 ∂ρ  s
The thermodynamic web gives:
 ∂T   ∂s   ∂P   ∂s 
 ∂T   ∂P 
 ∂P   ∂s 
 ∂P 

     
 = −
 
  = −    = 
 ∂s  v  ∂v T  ∂s T  ∂T  P
 ∂s  v  ∂v  P  ∂v  s  ∂T  s
 ∂v  s
 T  ∂s   ∂P   c 
 c  ∂P   ∂T 
 ∂P 

 
  = −      P  = − P 
 ∂v  s
 cv  ∂v T  ∂s T  T 
 cv  ∂T  v  ∂v  P
If we treat air as an ideal gas consisting of diatomic molecules only
R
 ∂P 
 =

 ∂T  v v
cP 7
=
cv 5
P
 ∂T 
 =

 ∂v  P R
Therefore,
 7  P 
 ∂P 
  = −  
 5  v 
 ∂v  s
and
Vsound =
v 2  7  P 
   =
MW  5  v 
7  RT 


5  MW 
Vsound = 343 [m/s]
The lightning bolt is 1360 m away.
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5.57
From Problem 5.55:
 ∂P 
2
Vsound
=  
v 2  ∂P 
=−
 
MW  ∂v  s
 ∂ρ  s
The thermodynamic web gives:
 ∂P 
 ∂P   ∂s 
 ∂T   ∂P 
 ∂T   ∂s   ∂P   ∂s 
 = −
 
     

  = −    = 
 ∂v  s
 ∂s  v  ∂v  P  ∂v  s  ∂T  s
 ∂s  v  ∂v T  ∂s T  ∂T  P
so
 T  ∂P   ∂T   c P 
 ∂P 
  = − 
  
 
 ∂v  s
 cv  ∂T  v  ∂v  P  T 
For liquids c P ≈ cv water at 20 ºC, so
 ∂P   ∂T 
 ∂P 

 
  = −
 ∂T  v  ∂v  P
 ∂v  s
However, the cyclic rule gives:
 ∂P   ∂T   ∂v 
−1 = 
  
 
 ∂T v  ∂v  P  ∂P T
So
 ∂P 
 ∂P 
  = 
 ∂v  s  ∂v T
From the steam tables, for saturated water at 20 oC:
 m3 
ˆ
P = 2.34 kPa and v = .001002 

 kg 
For subcooled water at 20 oC:
 m3 

P = 5 MPa and vˆ = .0009995 
kg


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So
 kg kPa 
5000 − 2.34
 ∂P 
 ∆P 
 ∂P 
=
  =  ≈


 ∂vˆ  s  ∂vˆ T  ∆vˆ  .0009995 − .001002  m 3 
and
 ∂P 
Vsound = vˆ 2   = 1414 [m/s]
 ∂vˆ  s
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5.58
(a)
The fundamental property relation for internal energy is
dU = δQrev + δWrev
Substituting the proper relationships for work and heat, we obtain
dU = TdS + Fdz
The fundamental property relation for the Helmholtz energy is
dA = dU − d (TS ) = dU − TdS − SdT
Substitute the expression for the internal energy differential:
dA = Fdz − SdT
(b)
First, relate the entropy differential to temperature and length.
 ∂S 
 ∂S 
dS = 
 dT +   dz
 ∂Z T
 ∂T  z
Now we need to find expressions for the partial derivatives.
 ∂S 
 T∂S + F∂z 
 ∂u 
nc z = n

 = T
 =
∂T
z
 ∂T  z
 ∂T  z 
Therefore,
nc

a
 ∂S 
 = z = n + b 

T

T
 ∂T  z
The following statement is true mathematically (order of differentiation does not matter):
∂  ∂A  
∂  ∂A  
  =
  

∂Z  ∂T  z  T ∂T  ∂Z T  z
Furthermore,
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∂  ∂A  
 ∂S 
  = −


∂Z  ∂T  z  T
 ∂Z T
∂  ∂A  
 ∂F 
  =


∂T  ∂Z T  z  ∂T  z
 ∂S 
 ∂F 
  = −
 = −k (z − z0 )
 ∂Z T
 ∂T  z
Substituting the expressions for the partial derivatives into the expression for the entropy
differential, we obtain

a
dS = n + b dT − k (z − z 0 )dz

T
(c)
First, start with an expression for the internal energy differential:
 ∂U 
 ∂U 
dU = 
 dz
 dT + 
 ∂Z T
 ∂T  z
From information given in the problem statement:
 ∂U 
 = n(a + bT )

 ∂T  z
Using the expression for internal energy developed in Part (a) and information from Part (b)
 ∂U 
 ∂S 
 = T   + F = T (− k (z − z 0 )) + kT (z − z 0 ) = 0

 ∂Z T
 ∂z T
Therefore,
dU = [n(a + bT )]dT + 0dz = [n(a + bT )]dT
(d)
We showed in Part (c) that
 ∂U 
FU = 
 =0
 ∂z T
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 ∂S 
Using the expression for   developed in Part B, we obtain
 ∂z T
 ∂S 
FS = −T   = kT ( z − z 0 )
 ∂z T
(e)
First, perform an energy balance for the adiabatic process.
dU = δW
Substitute expressions for internal energy and work.
[n(a + bT )]dT = Fdz = kT (z − z 0 )dz
Rearrangement gives
dT kT (z − z 0 )
=
dz [n(a + bT )]
The right-hand side of the above equation is always positive, so the temperature increases as the
rubber is stretched.
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5.59
The second law states that for a process to be possible,
∆suniv ≥ 0
To see if this condition is satisfied, we must add the entropy change of the system to the entropy
change of the surroundings. For this isothermal process, the entropy change can be written
ds =
cv
 ∂P 
 ∂P 
dT +   dv =   dv
T
 ∂T  v
 ∂T  v
Applying the van der Waals equation:
ds =
R
dv
v −b
Integrating
v
 J 
∆s sys = R ln 2 = 11.5 
v1
 mol K 
For the entropy change of the surroundings, we use the value of heat given in Example 5.2:
 J 
q = −q surr = 600 
 mol 
Hence the entropy change of the surroundings is:
∆s surr =
q surr − 600
 J 
=
= −1.6 
Tsurr
373
 mol K 
and
 J 
∆suniv = ∆s sys + ∆s surr = 9.9 
 mol K 
Since the entropy change of the universe is positive we say this process is possible and that it is
irreversible.
Under these conditions propane exhibits attractive intermolecular forces (dispersion). The closer
they are together, on average, the lower the energy. That we need to put work into this system
says that the work needed to separate the propane molecules is greater than the work we get out
during the irreversible expansion.
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5.60
A schematic of the process is given by:
The energy balance for this process is provided below:
∆h = wS
Because the gas is not ideal under these conditions, we have to create a hypothetical path that
connects the initial and final states through three steps. One hypothetical path is shown below:
For the first section of the path, we have
P =0
∆h1 =
 ∂h 
∫  ∂P T dP
P1
If we apply Equation 5.45 *new we can rewrite the above equation as
P =0
∆h1 =

 ∂v 

∫ − Ti  ∂T  P + v dP
Pi
For the given EOS:
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R aP
 ∂v 
 = − 2

 ∂T  P P Ti
Therefore,
P =0
v =0

 2aP


aP
 J 
∆h1 =
∫ 5 − P + Ti + v dP = ∫ 5  Ti + b dP = −2467  mol 
Pi =100×10
Pi =100×10
Similarly for step 3
Pf = 20×10 5
∫
∆h3 =
P =0
 2aP

 J 

+ b dP = 250 
 Tf

 mol 


For step 2, the pressure is zero, so we can use the ideal heat capacity given in the problem
statement to calculate the enthalpy change.
Tf
445 K
Ti
600 K
∆h2 = ∫ c P dT =
 J 
∫ (30 + 0.02T )dT = −6270  mol 
Now sum each part to find the total change in enthalpy:
 J 
∆h = ∆h1 + ∆h2 + ∆h3 = −8487 
 mol 
 J 
ws = −8487 
 mol 
In other words, for every mole of gas that flows through the turbine, 8487 joules of work are
produced.
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5.61
(a) Ethane has attractive London interactions; when the membrane ruptures it takes energy to
“pull apart” the ethane so ∆U>0
 ∂u 
 ∂u 
 ∂u 
(b) du = 
 dT +   dv =   dv
 ∂T v
 ∂v T
 ∂v T
  ∂P 

a
= T 
= 2 dv
du
 − P  dv
v
  ∂T v

27 ( RTc )  1 1 
a 2
=n
Q =∆U =n∆u =− n
 −  =106 J
64 Pc  v2 v1 
v v1
2
v
(c)
Q
Q
J
S surr =surr =
− =
−0.284
K
T
T
R
J
=
∆S sys n ∫ =
dv 25.0
v −b
K
∆Suniv = ∆S sys + ∆S surr = 24.7
(d)
J
K
For n-hexane the London interactions are larger so ∆U will increase
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