Solution Koretsky ch05

Prévia do material em texto

Chapter 5 Solutions 
Engineering and Chemical Thermodynamics 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Wyatt Tenhaeff 
Milo Koretsky 
 
Department of Chemical Engineering 
Oregon State University 
 
koretsm@engr.orst.edu 
 
 
 
 2
5.1 
 
(a) 
Following the example given by Equation 5.5a in the text 
 
 dP
P
udT
T
udu
TP
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
(b) 
 ds
s
udT
T
udu
Ts
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
(c) 
 ds
s
udh
h
udu
hs
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
 
 3
5.2. 
The internal energy can be written as follows 
 
 dv
v
udT
T
udu
Tv
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Substituting Equations 5.38 and 5.40 
 
 v
v
c
T
u =⎟⎠
⎞⎜⎝
⎛
∂
∂ and ⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ P
T
PT
v
u
vT
 
 
into the above expression yields 
 
 dvP
T
PTdTcdu
v
v ⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂+= 
 
From the ideal gas law, we have 
 
 
v
R
T
P
v
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 dvP
v
RTdTcdu v ⎥⎦
⎤⎢⎣
⎡ −+= 
 
which upon noting that 
v
RTP = for an ideal gas, becomes 
 
 dTcdu v= 
 
 4
5.3 
The heat capacity at constant pressure can be defined mathematically as follows 
 
 
( )
PvPP
P T
vP
T
u
T
Pvu
T
hc ⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
For an ideal gas: 
 
 
P
R
T
v
P
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 R
T
uc
v
P +⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
One mathematical definition of du is 
 
 dP
P
udT
T
udu
TP
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
We can now rewrite 
vT
u ⎟⎠
⎞⎜⎝
⎛
∂
∂ : 
 
 v
vTvPv
c
T
P
P
u
T
T
T
u
T
u =⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
For an ideal gas: 
 
 0=⎟⎠
⎞⎜⎝
⎛
∂
∂
TP
u 
 
so 
 
 
P
v T
uc ⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Substituting this result into our expression for Pc gives 
 
 Rcc vP += 
 
 5
5.4 
In terms of P, v, and T, the cyclic equation is 
 
 
TPv P
v
v
T
T
P ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂=−1 
 
For the ideal gas law: 
 
 RTPv = 
 
so the derivatives become: 
 
 
v
R
T
P
v
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
R
P
v
T
P
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
P
v
P
RT
P
v
T
−=−=⎟⎠
⎞⎜⎝
⎛
∂
∂
2 
 
Therefore, 
 
 1−=⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂
P
v
R
P
v
R
P
v
v
T
T
P
TPv
 
 
The ideal gas law follows the cyclic rule. 
 
 6
5.5 
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., 
 
 
0
,
=⎟⎠
⎞⎜⎝
⎛
∂
∂
PTv
h 
 
 7
5.6 
Expansion of the enthalpy term in the numerator results in 
 
 
ss T
PvsT
T
h ⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
ss T
Pv
T
h ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂∴ 
 
Using a Maxwell relation 
 
 
Ps v
sv
T
h ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
PPs v
T
T
sv
T
h ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂∴ 
 
We can show that 
 
 
T
c
T
s P
P
=⎟⎠
⎞⎜⎝
⎛
∂
∂ (use thermodynamic web) 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−−=⎟⎠
⎞⎜⎝
⎛
∂
∂
32
221
v
ab
v
a
bv
RT
Rv
T
P
 (differentiate van der Waals EOS) 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛ −−−=⎟⎟⎠
⎞⎜⎜⎝
⎛ +−−=⎟⎠
⎞⎜⎝
⎛
∂
∂
v
b
vRT
a
bv
vc
v
ab
v
a
bv
RT
RT
vc
T
h
P
P
s
1222 32 
 
 
 8
5.7 
 
:
TP
h ⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
 v
P
sT
P
PvsT
P
h
TTT
+⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 ( )2''1 PCPB
P
R
T
v
P
s
PT
++−=⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 ( ) vvvPCPB
P
RT
P
h
T
+−=+++−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 2''1 
 0=⎟⎠
⎞⎜⎝
⎛
∂
∂
TP
h
 
 
 
:
sP
h ⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
 v
P
PvsT
P
h
ss
=⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 ( )2''1 PCPB
P
RT
P
h
s
++=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
 
:
PT
h ⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
 P
P
c
T
h =⎟⎠
⎞⎜⎝
⎛
∂
∂
 (Definition of cP) 
 
 
:
sT
h ⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
 
sss T
Pv
T
PvsT
T
h ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
( )2''1 PCPBR PTcvTTcsPTsTP PPPTPs ++=⎟⎠⎞⎜⎝⎛ ∂∂=⎟⎠⎞⎜⎝⎛ ∂∂⎟⎠⎞⎜⎝⎛ ∂∂−=⎟⎠⎞⎜⎝⎛ ∂∂ 
 9
 ( ) ( )( )2
2
2 ''1
''1
''1
1
PCPB
PCPBc
PCPBRT
Pvc
T
h
PP
s ++
++=++=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 P
s
c
T
h =⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
 10
5.8 
 
(a) 
A sketch of the process is provided below 
 
∂mwell�
insulated
∆s = 0T1�
P1 T2�
P2
 
 
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 
 
 WQU +=∆ 
 
Since the process is adiabatic, the energy balance reduces to 
 
 WU =∆ 
 
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 
 
 0>⎟⎠
⎞⎜⎝
⎛
∂
∂
sP
u 
 
 11
5.9 
(a) 
By definition: 
 
PT
v
v
⎟⎠
⎞⎜⎝
⎛= ∂
∂β 1 
and 
 
 
TP
v
v
⎟⎠
⎞⎜⎝
⎛ ∂−= ∂κ
1 
 
Dividing, we get: 
 
 
TP
T
P
v
P
T
v
P
v
T
v
⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−=
⎟⎠
⎞⎜⎝
⎛
⎟⎠
⎞⎜⎝
⎛
−= ∂
∂
∂
∂
∂
∂
∂
∂
κ
β 
 
where derivative inversion was used. Applying the cyclic rule: 
 
 
vTP P
T
v
P
T
v ⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=− ∂
∂
∂
∂
∂
∂1 
 
Hence, 
 
 
vT
P ⎟⎠
⎞⎜⎝
⎛= ∂
∂
κ
β 
 
(b) 
If we write T = T(v,P), we get: 
 
 dP
P
Tdv
v
TdT
vP
⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛= ∂
∂
∂
∂ (1) 
 
From Equations 5.33 and 5.36 
 
 dP
T
vdT
T
cdv
T
PdT
T
cds
P
P
v
v ⎟⎠
⎞⎜⎝
⎛−=⎟⎠
⎞⎜⎝
⎛+= ∂
∂
∂
∂ 
 
We can solve for dT to get: 
 
 12
 dP
T
v
cc
Tdv
T
P
cc
TdT
PvPvvP
⎟⎠
⎞⎜⎝
⎛
−+⎟⎠
⎞⎜⎝
⎛
−= ∂
∂
∂
∂ (2) 
 
For Equations 1 and 2 to be equal, each term on the left hand side must be equal. Hence, 
 
 
vvPP T
P
cc
T
v
T ⎟⎠
⎞⎜⎝
⎛
−=⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂ 
 
or 
 
 
PPv
vP T
vT
T
v
T
PTcc ⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=− ∂
∂
κ
β
∂
∂
∂
∂ 
 
where the result from part a was used. Applying the definition of the thermal expansion 
coefficient: 
 
 
 κ
β
∂
∂
∂
∂ 2Tv
T
v
T
PTcc
Pv
vP =⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=− 
 
 13
5.10 
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: 
 
Species ⎥⎥⎦
⎤
⎢⎢⎣
⎡×
mol
m 10
3
6v [ ]1-3K 10×β [ ]1-10 Pa 10×κ 
Acetone 73.33 1.49 12.7 
Benzene 86.89 1.24 9.4 
Copper 7.11 0.0486 0.091 
 
We can calculate the difference in heat capacity use the result from Problem 5.9b: 
 
 κ
β 2vTcc vP =− 
 
or 
 
 
( ) [ ]( )
[ ] ⎥⎦⎤⎢⎣⎡ ⋅=×
×⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡×
=− −
−−
Kmol
J 6.37
Pa 107.12
K 1049.1K 293
mol
m 1033.73
1-10
21-3
3
6
vp cc 
 
 
Species ⎥⎦
⎤⎢⎣
⎡
⋅− Kmol
J vp cc ⎥⎦
⎤⎢⎣
⎡
⋅ Kmol
J pc % difference 
Acetone 37.6 125.6 30% 
Benzene 41.6 135.6 31% 
Copper 0.5 22.6 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. 
 
 
 14
5.11 
We know from Equations 4.71 and 4.72 
 
 
PT
v
v
⎟⎠
⎞⎜⎝
⎛
∂
∂= 1β and 
TP
v
v
⎟⎠
⎞⎜⎝
⎛
∂
∂−= 1κ 
 
Maxwell relation: 
 
 
vT T
P
v
s ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Employing the cyclic rule gives 
 
 
PTv T
v
v
P
T
P ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
which can be rewritten as 
 
T
P
vT
v
P
v
T
v
v
T
P
v
s
⎟⎠
⎞⎜⎝
⎛
∂
∂−
⎟⎠
⎞⎜⎝
⎛
∂
∂
=⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
1
1
 
 
Therefore, 
 
 κ
β=⎟⎠
⎞⎜⎝
⎛
∂
∂
Tv
s
 
 
Maxwell Relation: 
 
 
PT T
v
P
s ⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
From Equation 4.71: 
 
 v
T
v
P
β=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
v
P
s
T
β−=⎟⎠
⎞⎜⎝
⎛
∂
∂
 15
5.12 
 
(a) 
An isochor on a Mollier diagram can be represented mathematically as 
 
 
vs
h ⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
This can be rewritten: 
 
 
vvv s
PT
s
PvsT
s
h ⎟⎠
⎞⎜⎝
⎛
∂
∂+=⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Employing the appropriate Maxwell relation and cyclic rule results in 
 
 
Tvv v
s
s
TvT
s
h ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂+=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
We know 
 
 
vv c
T
s
T =⎟⎠
⎞⎜⎝
⎛
∂
∂ and 
vT T
P
v
s ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
For an ideal gas: 
 
 
v
R
T
P
v
s
vT
=⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=+=⎟⎠
⎞⎜⎝
⎛
∂
∂
vvv c
RT
v
R
c
TvT
s
h 1 
 
(b) 
In Part (a), we found 
 
 
vvv T
P
c
TvT
s
h ⎟⎠
⎞⎜⎝
⎛
∂
∂+=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
For a van der Waals gas: 
 
 16
 
bv
R
T
P
v −
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ⎟⎠
⎞⎜⎝
⎛
−+=⎟⎠
⎞⎜⎝
⎛
∂
∂
bv
v
c
RTT
s
h
vv
 
 
 
 17
5.13 
 
(a) 
The cyclic rule can be employed to give 
 
 
TPs P
s
s
T
P
T ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Substitution of Equations 5.19 and 5.31 yields 
 
 
PPs T
v
c
T
P
T ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
For an ideal gas: 
 
 
P
R
T
v
P
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 
PPs c
v
cP
RT
P
T ==⎟⎠
⎞⎜⎝
⎛
∂
∂ 1 
 
(b) 
Separation of variables provides 
 
 
P
P
c
R
T
T
P
∂=∂ 
 
Integration provides 
 
 Pc
R
P
P
T
T
⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛
1
2
1
2 lnln 
 
which can be rewritten as 
 
 Pc
R
P
P
T
T
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2
1
2 
 
The ideal gas law is now employed 
 
 18
 Pc
R
P
P
vP
vP
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2
11
22 
 1
1
12
1
2 vPvP
PP c
R
c
R
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
= 
 
where 
 
 
kc
c
c
Rc
c
R
P
v
P
P
P
11 ==−=− 
 
If we raise both sides of the equation by a power of k, we find 
 
 kk vPvP 1122 = 
 .constPvk =∴ 
 
(c) 
In Part (a), we found 
 
 
PPs T
v
c
T
P
T ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Using the derivative inversion rule, we find for the van der Waals equation 
 
 ( )( )23
3
2 bvaRTv
bvRv
T
v
P −−
−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ( )( )23
3
2
1
bvaRTv
bvRTv
cP
T
Ps −−
−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
 
 
 19
5.14 
The development of Equation 5.48 is analogous to the development of Equation E5.3D. We 
want to know how the heat capacity changes with pressure, so consider 
 
 
T
P
P
c ⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
which can be rewritten as 
 
 
PTTPT
P
P
h
TT
h
PP
c ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂=⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Consider the 
TP
h ⎟⎠
⎞⎜⎝
⎛
∂
∂ term: 
 
 v
T
vTv
P
sT
P
PvsT
P
h
PTTT
+⎟⎠
⎞⎜⎝
⎛
∂
∂−=+⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Substitution of this expression back into the equation for 
T
P
P
c ⎟⎠
⎞⎜⎝
⎛
∂
∂ results in 
 
PPT
P v
T
vT
TP
c
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛
∂
∂−∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
PPPT
P
T
v
T
vT
T
v
T
T
P
c ⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂−⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂
2
2
 
 
PT
P
T
vT
P
c
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂
2
2
 
 
Therefore, 
 
 ∫∫ ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂−=
real
ideal
real
P
ideal
P
P
P P
c
c
P dP
T
vTdc 2
2
 
 
and 
 
 ∫ ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂−=
real
ideal
P
P P
ideal
P
real
P dP
T
vTcc 2
2
 
 
 20
5.15 
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 
 
∫ ⎟⎠⎞⎜⎝⎛ ∂
∂=−=∆
bar 12
bar 10
12 dPP
ssss
T
 
 
Applying a Maxwell relation, we can relate the above equation to the change in molar volume: 
 
 ∫∫ ⎟⎠⎞⎜⎝⎛ ∂
∂−+=⎟⎠
⎞⎜⎝
⎛
∂
∂+=
bar 12
bar 10
1
bar 12
bar 10
12 dPT
vsdP
P
sss
PT
 
 
As 10 bar: 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅×=⎟⎠
⎞⎜⎝
⎛
∆
∆≅⎟⎠
⎞⎜⎝
⎛
∂
∂ −
Kkg
m 1060.5
3
4
PP T
v
T
v 
 
At 12 bar: 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅×=⎟⎠
⎞⎜⎝
⎛
∆
∆≅⎟⎠
⎞⎜⎝
⎛
∂
∂ −
Kkg
m 1080.4
3
4
PP T
v
T
v 
 
To integrate the above entropy equation, we need an expression that relates 
PT
v ⎟⎠
⎞⎜⎝
⎛
∂
∂ to pressure. 
Thus, we will fit a line to the data. We obtain 
 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅×+⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅⋅×−=⎟⎠
⎞⎜⎝
⎛
∂
∂ −−
Kkg
m106.9
PaKkg
m 100.4
3
4
3
10 P
T
v
P
 
 
Now integrate the equation to find the entropy: 
 
 ( )[ ] ⎥⎦⎤⎢⎣⎡ ⋅=⎥⎦⎤⎢⎣⎡ ⋅−=×−×+= ∫
×
×
−−
Kkg
kJ 392.5
Kkg
kJ 104.04960.5106.9100.4
Pa 101.2
Pa 100.1
410
12
6
6
dPPss 
 
 
 
 21
5.16 
A schematic of the process follows: 
Propane in
v1 = 600 cm3/mol 
T1= 350 oC
Turbine
P2 = 1 atm
ws
Propane
out 
 
We also know the ideal gas heat capacity from Table A.2.1: 
 
 263 10824.810785.28213.1 TT
R
cP −− ×−×+= 
 
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: 
 
Temperature
vo
lu
m
e
Ideal 
Gas
st
ep
 1
step 2
v1,T1
∆s=0
v2,T2
∆s1
∆s2
 
 
or in mathematical terms: 
 
 0=⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= dv
v
sdT
T
sds
Tv
 
 
However, From Equation 5.33: 
 
 
 22
 ds = cv
T
dT + ∂P∂T
⎛ 
⎝ 
⎞ 
⎠ v dv 
 
To get ∆s1P = RT
v − b −
a
v2
 so 
∂P
∂T
⎛ 
⎝ 
⎞ 
⎠ v =
R
v − b 
and 
 
 ∆s1 = d∫ s = ∂P∂T
⎛ 
⎝ 
⎞ 
⎠ v dvv1
v2∫ = R
v − b dv = R lnv1
v2∫ v2 − b
v1 − b
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
 
or, using the ideal gas law, we can put ∆s1 in terms of T2: 
 
 ∆s1 = R ln
RT2
P2
− b
v1 − b
⎡ 
⎣ 
⎢ 
⎢ 
⎤ 
⎦ 
⎥ 
⎥ 
 
 
For step 2 
 
 ∫ ∫ −− ×−×+==∆ 2
1
2
K 15.623
263
2
10824.810785.28213.0
T
T
T
v dT
T
TTRdT
T
c
s 
 
Now add both steps 
 
( ) ( )[ ]2226232
1
2
2
21
K 15.623
2
10824.8K 15.62310785.28
15.623
ln213.0ln
0
−×−−×+⎟⎠
⎞⎜⎝
⎛+
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−
=
=∆+∆=∆
−− TTT
bv
b
P
RT
sss
 
Substitute 
 
 K 15.6231 =T 
 [ ]/molcm 600 31 =v 
 atm12 =P 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅
⋅=
Kmol
atmcm 06.82
3
R 
 
 23
and solve for T2: 
 
 [ ]K 3.4482 =T 
 
(b) T, P as independent variables 
 
Choosing T and P as the independent variables, (and changing T under ideal gas conditions), we 
get: 
 
Temperature
Pr
es
su
re
Ideal 
Gas
st
ep
 1
step 2
P1,T1∆s=0
P2,T2
∆s1
∆s2
 
 
Mathematically, the entropy is defined as follows 
 
 0=⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= dP
P
sdT
T
sds
TP
 
 
Using the appropriate relationships, the expression can be rewritten as 
 
 0=⎟⎠
⎞⎜⎝
⎛
∂
∂−= dP
T
vdT
T
cds
P
P 
 
For the van der Waals equation 
 
 
( )
( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡ +−−
⎥⎦
⎤⎢⎣
⎡
−=⎟⎠
⎞⎜⎝
⎛
∂
∂
32
2
v
a
bv
RT
bv
R
T
v
P
 
Therefore, 
 
 24
 
( )
( )
0
2
2
1
2
1
32
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ +−−
⎥⎦
⎤⎢⎣
⎡
−−=∆ ∫∫ dP
v
a
bv
RT
bv
R
dT
T
cs
P
P
T
T
P 
 
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: 
 
 ( ) dvbv
RT
v
adP
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−= 23
2 
 
Substituting this into the entropy expression, we get 
 
 ( ) 0
10824.810785.28213.1 2
1
2
1
263
=−−
×−×+=∆ ∫∫ −− dvbv RdTT TTs
v
v
T
T
 
 
Upon substituting 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅
⋅=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
=
=
=
Kmol
atmcm 06.82
mol
cm 91b
atm) 1at ideally acts (gas 
cm 600
K 15.623
3
3
2
2
2
3
1
1
R
P
RT
v
v
T
 
 
we obtain one equation for one unknown. Solving, we get 
 
 K 3.4482 =T 
 
 25
5.17 
 
(a) 
Attractive forces dominate. If we examine the expression for z, we see that at any absolute 
temperature and pressure, .1<z The intermolecular attractions cause the molar volume to 
deviate negatively from ideality and are stronger than the repulsive interactions. 
 
(b) 
Energy balance: 
 
 qhh =− 12 
 
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: 
 
P [bar] 
T [K]
P,T1
step 1 st
ep
 3
step 2
ideal gas
300
q = ∆h 
50
0
500
P,T2
 ∆h1 
 ∆h2 
 ∆h3 
 
 
Choosing T and P as the independent properties: 
 
 dP
P
hdT
T
hdh
TP
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
or using Equation 5.46 
 
 dPv
T
vTdTcdh
P
P ⎥⎦
⎤⎢⎣
⎡ +⎟⎠
⎞⎜⎝
⎛
∂
∂−+= 
 
The given EOS can be rewritten as 
 
 ⎟⎠
⎞⎜⎝
⎛ += 2/11 aT
P
Rv 
 
 26
Taking the derivative gives: 
 
 5.05.0 −+=⎟⎠
⎞⎜⎝
⎛
∂
∂ aRT
P
R
T
v
P
 
 
so 
 
 ( )dPaRTdTcdh P 5.05.0+= 
 
For step 1 
 
 ( ) ⎥⎦⎤⎢⎣⎡=−==∆ ∫ molJ 2525.05.0
0
bar 50
5.0
1
5.0
11 PaRTdPaRTh 
For step 2 
 
 ( ) ⎥⎦⎤⎢⎣⎡=−×+=∆ ∫ −− molJ 7961875.01002.358.3
K 500
K 300
5.03
2 dTTTRh 
 
For step 3: 
 
 ( ) ⎥⎦⎤⎢⎣⎡−===∆ ∫ molJ 3235.05.0
bar 50
0
5.0
2
5.0
23 PaRTdPaRTh 
 
Finally summing up the three terms, we get, 
 
 ⎥⎦
⎤⎢⎣
⎡=∆+∆+∆=
mol
J 7888321 hhhq 
 
Alternative 2: real heat capacity 
For a real gas 
 
 realPch =∆ 
 
From Equation 5.48: 
 
 ∫ ⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂−=
real
ideak
P
P P
ideal
P
real
P dPT
vTcc
2
2
 
For the given EOS 
 
 27
 ⎟⎠
⎞⎜⎝
⎛ += 2/11 aT
P
Rv 
Therefore, 
 
 5.12
2
25.0 −−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂ aRT
T
v
P
 
and 
 [ ]( ) 5.01/2bar 50
bar 0
5.0
2
2
K 875.025.0 −
=
=
− =−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂ ∫∫ RTdPaRTdPT
vT
real
ideak
real
ideak
P
P
P
P P
 
 
We can combine this result with the expression for realPc and find the enthalpy change. 
 
 ( )dTTTRh ∫ −− −×+=∆ K 500
K 300
5.03 875.01002.358.3 
 ⎥⎦
⎤⎢⎣
⎡=∆=
mol
J 7888hq 
 
The answers is equivalent to that calculated in alternative 1
 28
5.18 
 
(a) 
Calculate the temperature of the gas using the van der Waals equation. The van der Waals 
equation is given by: 
 
 2v
a
bv
RTP −−= 
 
First, we need to find the molar volume and pressure of state 1. 
 
 [ ]( ) [ ]( )[ ] ⎥⎥⎦⎤⎢⎢⎣⎡==== molm 00016.0mol 250 m 4.0m 1.0
32
1
1 n
Al
n
Vv 
 
[ ]( )
[ ] [ ] [ ]Pa 1008.1Pa 1001325.1m 1.0 s
m 81.9kg 10000
65
2
2
1 ×=×+
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
=+= atmPA
mgP 
 
Substituting these equations into the van der Waals equation above gives 
 
 [ ] 23
3
3
5
3
1
6
mol
m 00016.0
mol
mJ .50
mol
m 104
mol
m 00016.0
Kmol
J 314.8
Pa 1008.1
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⋅
−
⎥⎥⎦
⎤
⎢⎢⎣
⎡×−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=×
−
T
 
 K 5.2971 =T 
 
Since the process is isothermal, the following path can be used to calculate internal energy: 
 
v 
T
v1,T1
v2,T2= T1
∆u 
∆s
 
 
Thus, we can write the change in internal energy as: 
 
 29
 dv
v
udv
v
udT
T
udu
TTv
⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Using Equation 5.40 
 
∫ ⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
2
1
v
v v
dvP
T
PTu 
 
For the van der Waals EOS: 
 
 2v
a
bv
RTP −−= 
so 
 
 
bv
R
T
P
v −
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ∫=∆ 2
1
2
v
v
dv
v
au 
 
We can assume the gas in state 2 is an ideal gas since the final pressure is atmospheric. 
Therefore, we calculate 2v , 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡==
mol
m 0244.0
3
2
2
2 P
RTv 
 
and 
 
 ⎥⎦
⎤⎢⎣
⎡=⎥
⎥
⎦
⎤
⎢⎢⎣
⎡ ⋅
=∆ ∫ molJ 5.3104
mol
mJ 5.00244.0
00016.0
2
3
dv
v
u 
or 
 
 [ ]kJ 1.776=∆U 
 
 
 
 30
(b) 
From the definition of entropy: 
 
 surrsysuniv sss ∆+∆=∆ 
 
First, let’s solve for syss∆ using the thermodynamic web. 
 
 dv
v
sdT
T
sds
Tv
sys ⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Since the process is isothermal, 
 
 dv
v
sds
T
sys ⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 ∫ ⎟⎠⎞⎜⎝⎛ ∂
∂=∆∴
2
1
v
v v
sys dvT
Ps 
 
Again, for the van der Waals equation, 
 
 
bv
R
T
P
v −
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Substitution of this expression into the equation for entropy yields 
 
 ∫ −=∆
2
1
v
v
sys dvbv
Rs 
 ⎥⎦
⎤⎢⎣
⎡
⋅=
⎥⎥⎦
⎤
⎢⎢⎣
⎡×−
⎥⎦
⎤⎢⎣
⎡
⋅=∆ ∫
− Kmol
J 17.44
mol
m 104
Kmol
J 314.80244.0
00016.0
3
5
dv
v
ssys 
 
K
J 5.11042 ⎥⎦
⎤⎢⎣
⎡=∆ sysS 
 
The change in entropy of the surroundings will be calculated as follows 
 
 
surr
surr
surr T
Q
s =∆ 
 
where 
 31
 
 QQsurr −= (Q is the heat transfer for the system) 
 
Application of thefirst law provides 
 
 WUQ −∆= 
 
We know the change in internal energy from part a, so let’s calculate W using 
 
 ∫−= 2
1
v
v
PdvnW 
 
Since the external pressure is constant, 
 
 [ ]( ) [ ]( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡−
⎥⎥⎦
⎤
⎢⎢⎣
⎡×−=
mol
m 00016.0
mol
m 0244.0Pa 1001325.1mol 250
33
5W 
 [ ]J 614030−=W 
 
Now calculate heat transfer. 
 
 [ ] ( ) [ ] [ ]J 1039.1J 614030J 776100 6×=−−=Q 
 
Therefore, 
 
 [ ][ ] ⎥⎦
⎤⎢⎣
⎡−=×−=∆
K
J 4672
K 5.297
J 1039.1 6
surrS 
 
and the entropy change of the universe is: 
 
 ⎥⎦
⎤⎢⎣
⎡=⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡=∆
K
J 5.6370
K
J 4672
K
J 5.11042univS 
 
 32
5.19 
First, calculate the initial and final pressure of the system. 
 
 [ ] [ ]( ) [ ]( )[ ] [ ]Pa 1092.4m05.0 m/s81.9kg 20000Pa1010 62 25 ×=+×=iP 
 [ ] [ ]( ) [ ]( )[ ] [ ]Pa 1089.6m05.0 m/s81.9kg 30000Pa1010 62 25 ×=+×=fP 
 
To find the final temperature, we can perform an energy balance. Since the system is well-
insulated, all of the work done by adding the third block is converted into internal energy. The 
energy balance is 
 
 wu =∆ 
 
To find the work, we need the initial and final molar volumes, which we can obtain from the 
given EOS: 
 
 [ ]/molm 1037.8 34−×=iv 
 ( ) [ ]/molm 103.22511089.6
314.8 35-
6
×+
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +×
=
f
f
f
T
T
v 
 
Now, calculate the work 
( ) ( ) ( ) ⎟⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜⎜
⎝
⎛
×−×+
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +×
×−=−−= −45-
6
6 1037.8103.2
2511089.6
314.8
Pa 1089.6
f
f
iff
T
T
vvPw 
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: 
 
v 
T
vi,Ti
vf,Tf
st
ep
 1
step 3
step 2
ideal gas
500 Tf
∆u = -Pf (vf -vi)
 
 33
 
For step 1, we calculate the change in internal energy as follows 
 
 dvP
T
PTdv
v
uu
low
i
low
i
PRTv
v v
PRTv
v T
∫∫
==
⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
//
1 
 ( ) ( ) ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
+=−⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
+=∆ ∫
=
bv
bPRT
aT
aRT
bv
dv
aT
aRTu
i
lowi
i
i
PRTv
v i
i
low
i
/ln2
2/
2
2
1 
 
Similarly, for step 3: 
 
 dvP
T
PTdv
v
uu
f
low
f
low
v
PRTv v
v
PRTv T
∫∫
==
⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
//
3 
 ( ) ( ) ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
+=−⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
+=∆ ∫= bPRT
bv
aT
aRT
bv
dv
aT
aRT
u
lowf
f
f
f
v
PRTv f
f
f
low
/
ln2
2
/
2
2
3 
 
Insert the expression for the final molar volume into the equation for 3u∆ : 
 
 ( ) ( )( )( )⎟⎟⎠
⎞
⎜⎜⎝
⎛
−+×+=∆ bPRTT
T
aT
aRT
u
lowff
f
f
f
//2511089.6
314.8
ln 62
2
3 
 
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. 
 
 ( )∫∫
==
+−==∆
f
i
f
i
T
T
T
T
v dTTRdTcu
K 500K 500
2 05.020 
 ( ) ( )222 500025.0500686.11 −+−=∆ ff TTu 
 
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: 
 
 34
 
( ) ( ) ( )
( ) ( )( )( )
( ) ( ) ⎟⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜⎜
⎝
⎛
×−×+
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +×
×−
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
−+×+
+−+−+⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
+
−45-
6
6
62
2
22
2
2
1037.8103.2
2511089.6
314.8
Pa 1089.6
//2511089.6
314.8
ln
500025.0500686.11/ln
f
f
lowff
f
f
f
ff
i
lowi
i
i
T
T
bPRTT
T
aT
aRT
TT
bv
bPRT
aT
aRT
 
 
Solving for Tf we get 
 
 K 2.536=fT 
 
The piston-cylinder assembly is well-insulated, so 
 
 sysuniv ss ∆=∆ 
 
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. 
 
 
P
T
Pi,Ti
Pf,Tf
step 1
st
ep
 3
step 2
ideal gas
500 536
Plow
4.9
6.9
 
 
For steps 1 and 3 
 
 ∫∫ ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
low
i
low
i
P
P P
P
P T
dP
T
vdP
P
ss1 
 35
 ∫∫ ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
f
low
f
low
P
P P
P
P T
dP
T
vdP
P
ss3 
 
We can differentiate the given EOS as required: 
 
 
( )
( )
( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
+−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+
+−=∆ ∫
i
low
i
ii
P
P i
ii
P
P
Ta
TaRTdP
PTa
TaRTs
low
i
ln
22
221 
 
( )
( )
( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
+−=⎥⎥⎦
⎤
⎢⎢⎣
⎡
+
+−=∆ ∫
low
f
f
ff
P
P f
ff
P
P
Ta
TaRT
dP
PTa
TaRT
s
f
low
ln
22
223 
 
For step 2 
 
 ∫∫∫ ⎟⎠⎞⎜⎝⎛ +==⎟⎠⎞⎜⎝⎛ ∂
∂=∆
f
i
f
i
f
i
T
T
T
T
P
T
T P
dT
T
dT
T
cdT
T
ss 05.0202 
 ( )if
i
f TT
T
T
s −+⎟⎟⎠
⎞
⎜⎜⎝
⎛=∆ 05.0ln202 
 
Sum all of the steps to obtain the change in entropy for the entire process 
 
 321 sssss sysuniv ∆+∆+∆=∆=∆ 
( )
( ) ( )
( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
+−−+⎟⎟⎠
⎞
⎜⎜⎝
⎛+⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
+−=∆
low
f
f
ff
if
i
f
i
low
i
ii
univ P
P
Ta
TaRT
TT
T
T
P
P
Ta
TaRTs ln
2
05.0ln20ln2 22 
Arbitrarily choose Plow (try 100 Pa), substitute numerical values, and evaluate: 
 
 
( ) ⎥⎦
⎤⎢⎣
⎡=⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=∆
⎥⎦
⎤⎢⎣
⎡
⋅=∆
K
J 766.0
Kmol
J 388.0mol 2
Kmol
J 388.0
univ
univ
S
s
 
 
 
 
 36
5.20 
A schematic of the process is given by: 
 
V = 1 L
T = 500 K
well 
insulated
Vacuum
V = 1 L
nCO=1 mole
State i
V = 2 L
T = ?
nCO=1 mole
State f 
 
(a) 
The following equation was developed in Chapter 5: 
 
 dv
T
PTcc
v
v v
ideal
v
real
v
ideal
∫ ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂+= 2
2
 
 
For the van der Waals EOS 
 
 02
2
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
T
P 
 
Therefore, 
 
 idealv
real
v cc = 
 
From Appendix A.2: 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡
⋅=−⎟⎟⎠
⎞
⎜⎜⎝
⎛ −×+= −
Kmol
J 0.22
500
31005001057.5376.3 2
4 RRcrealv 
 
(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. 
 
 
 37
(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: 
 
v 
T
vi,Ti
vf,Tf
st
ep
 2
step 1
500Tf
∆u2 
∆s2
∆u1, ∆s1
 
 
For the first section of the path, we have 
 
 ∫∫ ==∆
f
i
f
i
T
T
ideal
v
T
T
real
v dTcdTcu1 
 
 
( )
( ) ( ) 4.105074.2577375.191032.2
31001057.5376.2
23
1
K 500
2
4
1
−++×=∆
⎥⎦
⎤⎢⎣
⎡ −×+=∆
−
−∫
f
ff
T
T
TTu
dT
T
TRu
f
i 
 
For the second step, we can use the following equation 
 
 ∫ ⎟⎠⎞⎜⎝⎛ ∂
∂=∆
f
i
v
v T
dv
v
uu2 
 
If we apply Equation 5.40, we can rewrite the above equation as∫ ⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
f
i
v
v v
dvP
T
PTu2 
 
For the van der Waals EOS, 2v
a
bv
RTP −−= , 
 
 38
 
bv
R
T
P
v −
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡==⎥⎦
⎤⎢⎣
⎡ −−=∆ ∫∫
=
=
=
= mol
J7.73
002.0
001.0
2
002.0
001.0
1
f
i
f
i
v
v
v
v
dv
v
advP
bv
RTu 
 
Now set the sum of the two internal energies equal to zero and solve for Tf: 
 ( ) 07.734.105074.2577375.191032.2 2321 =+−++×=∆+∆ −
f
ff T
TTuu 
 K 497=fT 
 
(d) 
Since the system is well-insulated 
 
 sysuniv ss ∆=∆ 
 
To solve for the change in entropy use the following development: 
 
 dv
v
sdT
T
sds
Tv
sys ⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Using the thermodynamic web, the following relationships can be proven 
 
T
c
T
s v
v
=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
vT T
P
v
s ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
For the van der Waals EOS 
 
 ( )bv
R
T
P
v −
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Now we can combine everything and calculate the change in entropy 
 
 39
 ( )∫∫ −+=∆
002.0
001.0
K 497
K 500
dv
bv
RdT
T
c
s vsys 
 
( ) ( )
⎥⎦
⎤⎢⎣
⎡
⋅=∆=∆
⎥⎥⎦
⎤
⎢⎢⎣
⎡
×−+⎥⎦
⎤⎢⎣
⎡ −×+=∆ ∫∫ −−
Kmol
J 80.5
1095.3
31001057.5376.2
002.0
001.0
5
K 497
K 500
3
4
sysuniv
sys
ss
v
dvdT
TT
Rs
 
 
 
 40
5.21 
A schematic of the process is given by: 
 
V = 0.1 m3
T = 300 K
well 
insulated
Vacuum
nA=400 moles
State i
T = ?
State f
V = 0.1 m3
nA=400 moles
V = 0.2 m3
 
 
Energy balance: 
 
 0=∆u 
 
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: 
 
v [m3/mol] 
T [K]
vi,Ti
vf,Tf
st
ep
 1
step 3
step 2
ideal gas
300Tf
∆u 
0.2/400
0.1/400
 
 
For the first section of the path, we have 
 
 ∫
∞=
⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
v
v Ti
dv
v
uu1 
 
If we apply Equation 5.40, we can rewrite the above equation as 
 
 ∫
∞=
⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
v
v vi
dvP
T
PTu1 
 41
 
For the van der Waals EOS 
 
 22vT
a
bv
R
T
P
v
+−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡==⎥⎦
⎤⎢⎣
⎡ −+−=∆ ∫∫
∞=
×=
∞=
×= −− mol
J11202
44 105.2
2
105.2
21
v
v i
v
v ii
dv
vT
advP
Tv
a
bv
RTu 
 
Similarly for step 3: 
 
 
 ⎥⎦
⎤⎢⎣
⎡
⋅
−==⎥⎦
⎤⎢⎣
⎡ −+−=∆ ∫∫
−− ×=
∞=
×=
∞= Kmol
J1680002
44 105
2
105
23 f
v
v f
v
v
T
dv
vT
advP
Tv
a
bv
RTu
ff
 
 
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: 
 
 ( )K 300
2
3
2 −=∆ fTRu 
 
If we set sum of the changes in internal energy for each step, we obtain one equation for one 
unknown: 
 
 ( ) 0168000K 300
Kmol
J314.8
2
3
mol
J 1120321 =−+−⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅+⎥⎦
⎤⎢⎣
⎡=∆+∆+∆
f
f T
Tuuu 
 
Solve for Tf: 
 
 K 6.261=fT 
 
 42
5.22 
A schematic of the process is shown below: 
 
Ethane 3 MPa; 500K
Initially: 
vacuum
Tsurr = 293 K
 
 
(a) 
Consider the tank as the system. Since kinetic and potential energy effects are negligible, the 
open system, unsteady-state energy balance (Equation 2.47) is 
 
 ∑∑ ++−=⎟⎠⎞⎜⎝⎛ out soutoutin ininsys WQhnhndt
dU &&&& 
 
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 
 
 inin
sys
hn
dt
dU &=⎟⎠
⎞⎜⎝
⎛ 
 
Integration must now be performed 
 
 ∫∫ = t inin
U
U
dthndU
0
2
1
& 
ininin
t
inin hnnhndtnhunun )( 12
0
1122 −===− ∫ & 
 
Since the tank is initially a vacuum, n1=0, and the relation reduces to: 
 
 inhu =2 
 
 43
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: 
 
 ininin vPuu +=2 
 
or 
 
 ( ) ( ) ininin vPuTu =− K 500 MPa, 3at MPa, 3at 2 (1) 
 
From the equation of state: 
 
( ) ( ) ( )[ ] ⎥⎦⎤⎢⎣⎡=××−⎟⎟⎠⎞⎜⎜⎝⎛ ⎥⎦⎤⎢⎣⎡ ⋅=+= − molJ 800,3Pa 103108.21K 552KmolJ 314.8'1 68PBRTvP inin (2) 
 
The change in internal energy can be found from the following path: 
 
Plow
T
step 1 st
ep
 3
step 2 ideal gas
500 K T2
∆u1
3 MPa
∆u3
∆u2
P
 
 
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: 
 
 
TPTTT P
vP
T
vT
P
vP
P
sT
P
u ⎟⎠
⎞⎜⎝
⎛
∂
∂−⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂−⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
From the equation of state 
 
 ( ) RTB
P
RTPB'P
P
RT
P
u '
T
−=⎟⎠
⎞⎜⎝
⎛−−+−=⎟⎠
⎞⎜⎝
⎛
∂
∂
21 
 
 44
So for step 1: 
 
 [J/mol] 349
0
'
0
'
1 −==−=⎟⎠
⎞⎜⎝
⎛
∂
∂=∆ ∫ ∫ in
Pin PinT
PRTBRTdPBdP
P
uu (3) 
 
and for step 3: 
 
 TPRTBRTdPBdP
P
uu
P P
T
7.02
2
0
'
2
0
'
3 =−=−=⎟⎠
⎞⎜⎝
⎛
∂
∂=∆ ∫ ∫ (4) 
 
For step 2 
 
 [ ]dTRcdT
T
Pv
T
hdT
T
uu
T
P
T
PP
T
P
∫∫∫ −=⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂−⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
500500500
2 
 
or 
 
 [ ]∫
=
−− ×−×+=∆
2
1 K 500
263
2 10561.510225.19131.0
T
T
dTTTRu (5) 
 
Substituting Equations 2, 3, 4, and 5 into 1 and solving for T gives: 
 
 K 5522 =T 
 
Alternative 2: real heat capacity 
Starting with: 
 
 inhu =2 
 
 
The above equation is equivalent to 
 
 inhvPh =− 222 
 222 vPhh in =−∴ 
 
To calculate the enthalpy difference, we can use the real heat capacity 
 
 dP
T
vTcc
P
P P
ideal
P
real
P
ideal
∫ ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂−= 2
2
 
 
For the truncated viral equation, 
 45
 
 02
2
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
PT
v
 
 
Therefore, 
 
 idealP
real
P cc = 
 
Now, we can calculate the change in enthalpy and equate it to the flow work term. 
 
 ∫
=
=
2
1 K 500
22
T
T
ideal
P vPdTc 
 [ ] ( )∫
=
−− +==×−×+
2
1 K 500
2222
263 '110561.510225.19131.1
T
T
PBRTvPdTTTR 
 
Integrate and solve for T2: 
 
 K 5522 =T 
 
(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 
 
 ( )
( ) ( )[ ]Pa 103108.21
Pa103
K 552
Kmol
J 314.8
'1 686 ××−×
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=+= −PB
P
RTv 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
mol
m 0014.0
3
v 
 
This quantity will not change as the tank cools, so now we can calculate the final pressure. 
 
 
( )
( ) 28
3
2
108.21
K 293
Kmol
J 314.8
mol
m 0014.0
P
P
−×−=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
 
 
Solve for 2P : 
 46
 
 Pa 1066.1 62 ×=P 
 
The entropy change of the universe can be expressed as follows: 
 
 surrsysuniv SSS ∆+∆=∆ 
 
To solve for the change in entropy of the system start with the following relationship: 
 
dP
P
sdT
T
sds
TP
sys ⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Alternative 1: path through ideal gas state 
Using the proper relationships, the above equation canbe rewritten as 
 
 dP
T
vdT
T
cds
P
P
sys ⎟⎠
⎞⎜⎝
⎛
∂
∂+= 
 
We can then use the following solution path: 
 
Plow
T
step 1
st
ep
 3
step 2 ideal gas
500 K T2
∆s1
3 MPa
∆s3
∆s2
P
1.66 MPa
 
 
Choosing a value of 1 Pa for Plow, for step 1: 
 
( )∫∫ +−=⎟⎠⎞⎜⎝⎛ ∂∂=∆
Pa 1
MPa 66.1
'
Pa 1
MPa 66.1
1 1 dPPBP
RdP
T
vs
P
 
 
For step 3, 
 
 ( )∫∫ +−=⎟⎠⎞⎜⎝⎛ ∂∂=∆
MPa 3
Pa 1
'
MPa 3
Pa 1
1 1 dPPBP
RdP
T
vs
P
 
 47
 
For step 2: 
 
 ∫∫ ⎥⎦⎤⎢⎣⎡ ×−×+==∆ −−
K 293
K 552
63
K 293
K 552
2 10561.510225.19
131.1 dTT
T
RdT
T
cs P 
 
Adding together steps 1, 2 and 3: 
 
 ⎥⎦
⎤⎢⎣
⎡
⋅−=∆ Kmol
J9.46syss 
______________________________________________________________________________ 
 
Alternative 2: real heat capacity 
Using the proper relationships, the above equation can be rewritten as 
 
 dP
T
vdT
T
cds
P
real
P
sys ⎟⎠
⎞⎜⎝
⎛
∂
∂+= 
 
For the truncated virial equation 
 
 ⎟⎠
⎞⎜⎝
⎛ +=⎟⎠
⎞⎜⎝
⎛
∂
∂ '1 B
P
R
T
v
P
 
 
 
Now, substitute the proper values into the expression for entropy and integrate: 
 
 
∫∫
×
×
−− ⎟⎠
⎞⎜⎝
⎛ ++⎥⎦
⎤⎢⎣
⎡ ×−×+=∆
Pa 1066.1
Pa 103
K 293
K 552
63
6
^
'110561.510225.19131.1 dPB
P
RdTT
T
Rssys
⎥⎦
⎤⎢⎣
⎡
⋅−=∆ Kmol
J9.46syss 
______________________________________________________________________________ 
 
In order to calculate the change in entropy of the surroundings, first perform an energy balance. 
 
 qu =∆ 
 
Rewrite the above equation as follows 
 
 ( ) qPvh =∆−∆ 
 
 48
Since the real heat capacity is equal to ideal heat capacity and the molar volume does not change, 
we obtain the following equation 
 
 ( )∫ =−−f
i
T
T
if
ideal
P qPPvdTc 
 
[ ] ( )∫=
=
−− =××⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡−×−×+
K293
K 552
66
3
263 Pa 103-Pa 1066.1
mol
m 0014.010561.510225.19131.1
f
i
T
T
qdTTTR
 
 ⎥⎦
⎤⎢⎣
⎡−=
mol
J 15845q 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡=
mol
J 15845surrq 
 
and 
 
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 08.54surrs 
 
Before combining the two entropies to obtain the entropy change of the universe, find the 
number of moles in the tank. 
 
 [ ] mol 7.75
mol
m 0014.0
m 05.0
3
3
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=n 
 
Now, calculate the entropy change of the universe. 
 
 ( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅−+⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 9.46
Kmol
J08.54mol 7.75univS 
 ⎥⎦
⎤⎢⎣
⎡=∆
K
J 544univS 
 
 49
5.23 
First, focus on the numerator of the second term of the expression given in the problem 
statement. We can rewrite the numerator as follows: 
 
 ( ) ( )idealvTidealvTidealvTvTidealvTvT rrrrrrrrrrrr uuuuuu ∞=∞= −−−=− ,,,,,, 
 
For an ideal gas, we know 
 
 0,, =− ∞=idealvTidealvT rrrr uu 
 
Therefore, 
 
 idealvTvT
ideal
vTvT rrrrrrrr
uuuu ∞=−=− ,,,, 
 
Substitute this relationship into the expression given in the problem statement: 
 
 
c
ideal
vTvT
c
ideal
vTvT
c
dep
vT
RT
uu
RT
uu
RT
u
rrrrrrrrrr ∞=−=
−
=
∆ ,,,,, 
 
Now, we need to find an expression for idealvTvT rrrr uu ∞=− ,, . Note that the temperature is constant. 
Equation 5.41 reduces to the following at constant temperature: 
 
 dvP
T
PTdu
v
T ⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
The pressure can be written as 
 
 
v
zRTP = 
 
and substituted into the expression for the differential internal energy 
 
 dv
T
z
v
RTdv
v
zRT
v
RT
T
z
v
RTTdu
vvv
T ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎥⎥⎦
⎤
⎢⎢⎣
⎡ −⎟⎟⎠
⎞
⎜⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛
∂
∂=
2
 
 
Applying the Principle of Corresponding States 
 
 50
 r
vrr
r
c
T dv
T
z
v
T
RT
du
r
r
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂=
2
 
 
If we integrate the above expression, we obtain 
 
 ∫∫
∞=∞=
∞=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂=
−
=
r
r
vrrrr
v
v
r
vrr
r
v
v c
ideal
vTvT
c
T dv
T
z
v
T
RT
uu
RT
du 2,
, 
 
Therefore, 
 
 ∫
∞=
∞=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂=
−
=
−
=
∆ r
r
rrrrrrrrrr
v
v
r
vrr
r
c
ideal
vTvT
c
ideal
vTvT
c
dep
vT dv
T
z
v
T
RT
uu
RT
uu
RT
u 2,,,,, 
 
 51
5.24 
We write enthalpy in terms of the independent variables T and v: 
 
 dv
v
hdT
T
hdh
Tv
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
using the fundamental property relation: 
 
 vdPTdsdh += 
 
At constant temperature, we get: 
 
 dv
v
Pv
T
PTdh
Tv
T ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
For the Redlich-Kwong EOS 
 
 ( )bvvT
a
bv
R
T
P
v +
+−=⎟⎠
⎞⎜⎝
⎛
∂
∂
2/32
1
 
 ( ) ( ) ( )22/122/12 bvvT
a
bvvT
a
bv
RT
v
P
T +
+++−
−=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
Therefore, 
 
 ( ) ( ) ( ) dvbvT
a
bvvT
a
bv
RTv
bv
RTdhT ⎥⎥⎦
⎤
⎢⎢⎣
⎡
++++−−−= 22/12/12 2
3 
 
To find the enthalpy departure function, we can integrate as follows 
 
 ( ) ( ) ( )∫∫ ∞=∞= ⎥⎥⎦
⎤
⎢⎢⎣
⎡
++++−−−==∆
v
v
v
v
T
dep dv
bvT
a
bvvT
a
bv
RTv
bv
RTdhh 22/12/12 2
3 
 
Since temperature is constant, we obtain 
 
 ( )bvT
a
bv
v
bT
a
bv
RTbhdep ++⎟⎠
⎞⎜⎝
⎛
++−=∆ 2/12/1 ln2
3
 
 
To calculate the entropy departure we need to be careful. From Equation 5.64, we have: 
 
 ( ) ( )gas ideal 0,gas ideal,gas ideal 0,,gas ideal,, == −−−=− PTPTPTPTPTPT ssssss 
 52
 
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 
 
 vTPT ss ,, = 
 
and 
 
 gas ideal,
gas ideal
0, ∞== = vTPT ss 
 
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 
 
 'v
RTP = 
 
to a real gas at T and P 
 
 ( )bvvT
a
bv
RTP +−−= 
 
The volume calculated by the ideal gas equation, v’, is clearly different from the volume, v, 
calculated by the Redlich-Kwong equation. Hence: 
 
 ⎟⎠
⎞⎜⎝
⎛ −+== gas ideal,gas ideal,
gas ideal
,
gas ideal
,
gas ideal
, '' vTvTvTvTPT
sssss 
 
Thus, 
 
 ( ) ( ) ⎟⎠⎞⎜⎝⎛ −−−−−=− ∞=∞= gas ideal,gas ideal,gas ideal,gas ideal,gas ideal,,gas ideal,, ' vTvTvTvTvTvTPTPT ssssssss 
 
Using a Maxwell relation: 
 
 
vT T
P
dv
ds ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛ 
 
Therefore, 
 
 dv
T
Pds
v
T ⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
 53
For the Redlich-Kwong EOS 
 
 ( )bvvT
a
bv
R
T
P
v +
+−=⎟⎠
⎞⎜⎝
⎛
∂
∂
2/32
1 
 
so 
 
 ( ) ( )∫∞=∞= ⎥⎥⎦
⎤
⎢⎢⎣
⎡
++−=−
v
v
vTvT dvbvvT
a
bv
Rss 2/3
gas ideal
,, 2
1 
 
 
For an ideal gas 
 
 
v
R
T
P
v
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
so 
 
 
 ( ) ∫
∞=
∞= ⎥⎦
⎤⎢⎣
⎡=−
v
v
vTvT dvv
Rss gas ideal,
gas ideal
, 
 
Finally: 
 
 
Pv
RTR
v
vR
v
dvRss
v
v
vTvT
lnln
'
gas ideal
,
gas ideal
,
'
' ===− ∫ 
 
Integrating and adding together the three terms gives: 
 
 
( )
Pv
RTR
bv
v
bT
a
v
bvRsdep lnln
2
ln 2/3 −⎟⎠
⎞⎜⎝
⎛
++
−=∆ 
 
 54
5.25 
Calculate the reduced temperature and pressure: 
 
 
[ ]
[ ]
344.0
bar 48.220
K 3.647
=
=
=
w
P
T
c
c
 (Table A.1.2) 
 
 
[ ]
[ ]
04.1
K 647.3
K 15.673
36.1
bar 20.482
bar300
==
==
rr
T
P
 
 
By double interpolation of data from Tables C.3 and C.4 
 
 921.2
)0(
, −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 459.1
)1(
, −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 
From Tables C.5 and C.6: 
 
 292.2
)0(
, −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
R
sdepPT rr 405.1
)1(
, −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
R
sdepPT rr 
 
Now we can calculate the departure functions 
 
 
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
+
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
=∆
)1(
,
)0(
,
c
dep
PT
c
dep
PT
c
dep
RT
h
w
RT
h
RTh rrrr 
 ( ) ( )( ) ⎥⎦
⎤⎢⎣
⎡−=−+−⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=∆ mol
J 18421459.1344.0921.23.647
Kmol
J 314.8deph 
 
 
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
+
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
=∆
)1(
,
)0(
,
R
s
w
R
s
Rs
dep
PT
dep
PTdep rrrr 
 ( )( ) ⎥⎦
⎤⎢⎣
⎡
⋅−=−+−⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 07.23405.1344.0292.2
Kmol
J 314.8deps 
 
 55
 To use the steam tables for calculating the departure functions, we can use the following 
relationships. 
 
 idealPTPT
dep hhh ,, −=∆ 
 idealPTPT
dep sss ,, −=∆ 
 
From the steam tables 
 
 ⎥⎦
⎤⎢⎣
⎡=
kg
kJ 0.2151,PTh and ⎥⎦
⎤⎢⎣
⎡
⋅= Kkg
kJ 4728.4,PTs 
 
We need to calculate the ideal enthalpies and entropies using the steam tables’ reference state. 
 
 ( ) ∫+∆=
K 673.15
K 273.16
, C.01º 0 dTchh
ideal
p
vapideal
PT 
 
We can get ⎥⎦
⎤⎢⎣
⎡=∆
mol
kJ 1.45vaph from the steam tables and heat capacity data from Table A.2.2. 
Using this information, we obtain 
 
∫ ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×+×+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅+⎥⎦
⎤⎢⎣
⎡= −
K 673.15
K 273.16
2
5
3
,
10121.01045.147.3
Kmol
kJ 008314.0
mol
kJ 1.45 dT
T
ThidealPT
 ⎥⎦
⎤⎢⎣
⎡=
mol
kJ 14.59,
ideal
PTh 
 
Now, calculate the ideal entropy. 
 
 ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛−+∆= ∫
1
2
K 673.15
K 273.16
, lnC.01º 0 P
PRdT
T
c
ss
ideal
pvapideal
PT 
 
From the steam tables: 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
kJ 165.0Cº 01.0vaps 
 
Substitute values into the entropy expression: 
 
 56
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎠
⎞⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×+×+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅+= ∫ − 000613.0 30ln10121.01045.147.3KmolkJ 008314.0 165.0
K 673.15
K 273.16
3
5
3
, dT
TT
sidealPT
⎥⎦
⎤⎢⎣
⎡
⋅= Kmol
J 107,
ideal
PTs 
 
Now, calculate the departure functions: 
 
 [ ]( ) ⎥⎦
⎤⎢⎣
⎡−=⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡=∆
mol
kJ 4.20
mol
kJ 14.59kg/mol 0180148.0
kg
kJ 0.2151deph 
[ ]( ) ⎥⎦
⎤⎢⎣
⎡
⋅−=⎥⎦
⎤⎢⎣
⎡
⋅−⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
kJ 0264.0
Kmol
kJ 107.0kg/mol 0180148.0
Kkg
kJ 4728.4deps 
 
Table of Results 
 Generalized Tables Steam Tables
Percent Difference 
(Based on steam tables) 
⎥⎦
⎤⎢⎣
⎡∆
mol
kJ deph -18.62 -20.4 9.9 
⎥⎦
⎤⎢⎣
⎡
⋅∆ Kmol
kJ deps -0.0231 -0.0264 12.5 
 
 
 
 57
5.26 
State 1 is at 300 K and 30 bar. State 2 is at 400 K and 50 bar. The reduced temperature and 
pressures are 
 
 
[ ]
[ ]
982.0
K 305.4
K 300
616.0
bar 74.48
bar30
,1
,1
==
==
r
r
T
P
 
[ ]
[ ]
31.1
K305.4
K 400
026.1
bar 74.48
bar50
,2
,2
==
==
r
r
T
P
 
 
and 
 
 099.0=ω 
 
By double interpolation of data in Tables C.3 and C.4 
 
 825.0
)0(
, ,1,1 −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 799.0
)1(
, ,1,1 −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 
 711.0
)0(
, ,2,2 −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 196.0
)1(
, ,2,2 −=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 
Therefore, 
 
 ( ) 904.0799.0099.0825.0,1,1 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 ( ) 730.0196.0099.0711.0,2,2 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 
The ideal enthalpy change from 300 K to 400 K can be calculated using ideal cP data from Table 
A.2.1. 
 
 RdTTTRhidealTT 39.71710561.510225.19131.1
2
K 400
K 300
63
21
=×−×+=∆ ∫ −−→ 
 
The total entropy change is 
 
 58
 depPT
ideal
TT
dep
PT rrrr
hhhh
,2,221,1,1 ,,
∆+∆+∆−=∆ → 
 ( )[ ]CC TTRh 730.039.717904.0 −+−−=∆ 
 ( ) ( )[ ]K 4.305730.0K 39.717K 4.305904.0
Kmol
J314.8 −+⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=∆h 
 ⎥⎦
⎤⎢⎣
⎡=∆
mol
J 2.6406h 
 
Using the data in Table C.5 and C.6 
 
 ( ) 676.0756.0099.0601.0,1,1 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
R
sdepPT rr 
 ( ) 416.0224.0099.0394.0,2,2 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
R
sdepPT rr 
 
Substituting heat capacity data into Equation 3.62, we get 
 
 ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎠
⎞⎜⎝
⎛−×−×+=∆ ∫ −− bar 30 bar 50ln10561.510225.19131.1
K 400
K 300
263
dT
T
TTRsideal 
 Rsideal 542.1=∆ 
 
Therefore, 
 
 ( )416.0542.1676.0
,2,2,1,1 ,,
−+=∆+∆+∆−=∆ Rssss depPTidealdepPT rrrr 
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 98.14s 
 
 
 59
5.27 
The turbine is isentropic. Therefore, we know the following 
 
 0
,2,2,1,1 ,,
=∆+∆+∆−=∆ depPTidealdepPT rrrr ssss 
 
Using the van der Waals EOS, we can find P1,r, which leaves one unknown in the above 
equation: T2. 
 
 
( )
23
2
3
5
33
3
1
mol
cm 600
mol
cmatm 1091
mol
cm 19
mol
cm 600
K 15.623
Kmol
atmcm 06.82
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⋅×
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅
⋅
=P 
 [ ] [ ]bar 19.76atm 19.751 ==P 
 
Calculate reduced temperature and pressures using data from Table A.1.1 
 
 
[ ]
[ ] 8.1bar 44.42
bar19.76
,1 ==rP [ ][ ] 024.0bar 2.444
bar013.1
,2 ==rP 
 68.1
K 370.0
K 23.156
,1 ==rT 
 
Also, 
 
 152.0=ω 
 
From Tables C.5 and C.6: 
 
 ( ) 343.0102.0152.0327.0,1,1 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
R
sdepPT rr 
 
Substituting heat capacity data into Equation 3.62, we get 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎠
⎞⎜⎝
⎛−×−×+=∆ ∫ −− atm 5.197 atm 1ln10824.810785.28213.1
2
K 23.156
263
dT
T
TTRs
T
ideal 
Therefore, 
 
 60
s
R
s
dT
T
TTR
dep
PT
T
rr ∆=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
++×−×++ ∫ −− ,2,22 ,
K 23.156
263
32.410824.810785.28213.1343.0 
We can solve this using a guess-and-check method 
 
 K 6002 =T : 62.1,2 =rT 
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 84.33s 
 
 K 4502 =T : 22.1,2 =rT 
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 77.0s 
 
 K 6.4462 =T : 21.1,2 =rT 
 ⎥⎦
⎤⎢⎣
⎡
⋅≅∆ Kmol
J 0s 
 
Therefore, 
 
 K 6.4462 =T 
 
 
 61
5.28 
A reversible process requires the minimum amount of work. Since the process is reversible and 
adiabatic 
 
 0=∆s 
 
which can be rewritten as 
 
 0
,2,2,1,1 ,,
=∆+∆+∆−=∆ depPTidealdepPT rrrr ssss 
 
Calculate reduced temperature and pressures using data from Table A.1.1 
 
 
[ ]
[ ] 0217.0bar 0.46
bar1
,1 ==rP [ ][ ] 217.0bar 6.04
bar10
,2 ==rP 
 57.1
K 190.6
K 003
,1 ==rT 
 
From Tables C.5 and C.6: 
 
 ( ) 0046.00028.0008.000457.0,1,1 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
R
sdepPT rr 
 
Substituting heat capacity data into Equation 3.62, we get 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎠
⎞⎜⎝
⎛−×−×+=∆ ∫ −− bar 1 bar 01ln10164.210081.9702.1
2
K 003
263
dT
T
TTRs
T
ideal 
 
Therefore, 
 
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
+−×−×++=∆ ∫ −− R
s
dT
T
TTRs
dep
PT
T
rr ,2,2
2 ,K 003
263
303.210824.810785.28213.10046.0 
We can solve using a guess-and-check method 
 
 K 4002 =T : 10.2,2 =rT 
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 98.4s 
 
 K 3852 =T : 02.2,2 =rT 
 62
 ⎥⎦
⎤⎢⎣
⎡
⋅=∆ Kmol
J 42.1s 
 
 K 3792 =T : 99.1,2 =rT 
 ⎥⎦
⎤⎢⎣
⎡
⋅−=∆ Kmol
J 018.0s 
 
Therefore, 
 
 K 3792 ≅T 
 
An energy balance reveals that 
 
 swhhh =∆=− 12 
 
We can calculate the enthalpy using departure functions. From Tables C.3 and C.4: 
 
( ) 0966.0011.0008.00965.0,1,1 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 ( ) 0613.0015.00089.00614.0,2,2 , −=+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 
Ideal heat capacity data can be used to determine the ideal change in enthalpy 
 
 ⎥⎥⎦
⎤
⎢⎢⎣
⎡
×−×+=∆ ∫ −− dTTTRhideal
K 379
K 003
263 10164.210081.9702.1 
 
Therefore, 
 
( )( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡
×−×++−⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
⋅=∆ ∫ −− dTTTh
K 379
K 003
263 10164.210081.9702.10613.00966.0K 6.190
Kmol
J 314.8
 ⎥⎦
⎤⎢⎣
⎡=∆=
mol
J 2.3034hws& 
and 
 W1.101
mol
J 2.3034
s
mol 30/1 =⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡=SW& 
 
 63
5.29 
Equation 4.71 states 
 
 
PT
v
v
⎟⎠
⎞⎜⎝
⎛
∂
∂= 1β 
 
PT
vv ⎟⎠
⎞⎜⎝
⎛
∂
∂=∴ β 
 
This can be substituted into Equation 5.75 to give 
 
 ( )
P
JT c
Tv 1−= βµ 
 
 64
5.30 
For an ideal gas 
 
 
P
R
T
v
P
=⎟⎠
⎞⎜⎝
⎛
∂
∂
 
 
Therefore, 
 
 ( ) 0=−=⎟⎠
⎞⎜⎝
⎛ −
=
PP
JT c
vv
c
v
P
RT
µ 
 
This result could also be reasoned from a physical argument. 
 
 
 65
5.31 
The van der Waals equation is given by: 
 
 2v
a
bv
RTP −−= (1) 
 
The thermal expansion coefficient is given by: 
 
 
PP v
T
vT
v
v
⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛= ∂
∂
∂
∂β 11 (2) 
 
Solving Equation 1 for T: 
 
 ⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ +=
R
bv
v
aPT 2 
 
Differentiating by applying the chain rule, 
 
 3
3
32
221
Rv
abavPv
v
a
R
bv
Rv
aP
v
T
P
+−=⎟⎠
⎞⎜⎝
⎛ −−⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=⎟⎠
⎞⎜⎝
⎛
∂
∂ (3) 
 
Substitution into Equation 2 gives 
 
 
abavPv
Rv
23
2
+−=β 
 
Substituting Equation 1 for P gives b in terms of R, T, v, a , and b: 
 
 ( )( )23
2
2 bvaRTv
bvRv
−−
−=β 
 
The isothermal compressibility is given by: 
 
 
TT v
P
vP
v
v
⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−=⎟⎠
⎞⎜⎝
⎛−= ∂
∂
∂
∂κ 11 
 
From the van der Waals equation: 
 
 ( )
( )
( )23
23
32
22
bvv
bvaRTv
v
a
bv
RT
v
P
T −
−+−=+−−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
so 
 66
 
 ( )( )23
22
2 bvaRTv
bvv
−−
−=κ 
 
For the Joule-Thomson coefficient, we can use Equation 5.75: 
 
 ∫ ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
=
real
ideal
P
P P
ideal
P
P
JT
dP
T
vTc
v
T
vT
2
2
∂
∂
∂
∂
µ
 
 
Substituting the van der Waals equation into Equation 3 gives 
 
 ( )( ) ( )
( )
33
23 22
Rv
bva
bv
T
Rvbv
bvaRTv
v
T
P
−−−=−
−−=⎟⎠
⎞⎜⎝
⎛
∂
∂ (4) 
 
Thus, the second derivative becomes: 
 
 ( )
( )
4322
2 621
Rv
bva
Rv
a
v
T
bvbv
T
v
T
PP
−+−⎟⎠
⎞⎜⎝
⎛
−+−−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂ 
 
or simplifying using Equation 4, 
 
 ( )42
2 32
Rv
bva
v
T
P
−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂ (5) 
 
Substituting Equations 5 and 4 into Equation 5.75 gives: 
 
 
 
( )
( )
( )∫ ⎥⎦
⎤⎢⎣
⎡
−−
−−
−+−
=
real
ideal
P
P
ideal
P
JT
dP
bva
RTvc
bvaRTv
bvavbRTv
32
2
2
4
23
23
µ 
 
At a given temperature the integral in pressure can be rewritten in terms of volume using the van 
der Waals equation to give: 
 67
 
( )
( )
( )
( )
( )∫ − −−⎥⎦
⎤⎢⎣
⎡
−+
−−
−+−
=
real
ideal
v
v
ideal
P
JT
dv
bv
bvaRTv
bva
RTvc
bvaRTv
bvavbRTv
2
23
23
23
2
32
2
2
µ 
 
 
 68
5.32 
We can solve this problem by using the form of the Joule-Thomson coefficient given in Equation 
5.75. The following approximation can be made 
 
 
PP T
v
T
v ⎟⎠
⎞⎜⎝
⎛
∆
∆≅⎟⎠
⎞⎜⎝
⎛
∂
∂ ˆˆ 
 
At 300 ºC, 
 
 
( ) ( )
Cº 250350
C,1MPaº250ˆC,1MPaº350ˆˆ
−
−=⎟⎠
⎞⎜⎝
⎛
∆
∆ vv
T
v
P
 
 
Cº 250350
kg
m 23268.0
kg
m 28247.0
ˆ
33
−
⎥⎥⎦
⎤
⎢⎢⎣
⎡−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
=⎟⎠
⎞⎜⎝
⎛
∆
∆
PT
v 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅=⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅=⎟⎠
⎞⎜⎝
⎛
∂
∂∴
K kg
m 0005.0
Cº kg
m 0005.0
ˆ 33
PT
v 
 
A similar process was followed to find cP. 
 
 
PP
P T
h
dT
hc ⎟⎟⎠
⎞
⎜⎜⎝
⎛
∆
∆≅⎟⎟⎠
⎞
⎜⎜⎝
⎛ ∂= ˆˆˆ 
 
At 300 ºC, 
 
 
( ) ( )
Cº 250350
C,1MPaº250ˆC,1MPaº350ˆˆ
−
−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∆
∆ hh
T
h
P
 
 
Cº 250350
kg
kJ 6.2942
kg
kJ 7.3157ˆ
−
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∆
∆
PT
h 
 ⎥⎦
⎤⎢⎣
⎡
⋅=⎥⎦
⎤⎢⎣
⎡
⋅=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∆
∆≅⎟⎟⎠
⎞
⎜⎜⎝
⎛ ∂=
K kg
kJ 15.2
Cº kg
kJ 15.2
ˆˆ
ˆ
PP
P T
h
dT
hc 
 
Now, JTµ can be found. 
 
 69
 
( )
⎥⎦
⎤⎢⎣
⎡
⋅
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡−⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⋅
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛
∂
∂
=
K kg
kJ 15.2
kg
m 25794.0
K kg
m 0005.0K 15.573
ˆ
ˆˆ
33
P
P
JT c
v
T
vT
µ 
 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⋅=
kJ
K m 0133.0
3
JTµ 
 
 70
5.33 
At the inversion line, the Joule-Thomson coefficient is zero. From Equation 5.75: 
 
 
0
2
2
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛
=
∫real
ideal
P
P P
ideal
P
P
JT
dP
T
vTc
v
T
vT
∂
∂
∂
∂
µ
 
 
This is true when the numerator is zero, i.e., 
 
 0=⎥⎦
⎤⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛ v
T
vT
P∂
∂
 
 
For the van der Waals equation, we have 
 
 2v
a
bv
RTP −−= 
 
Solving for T: 
 
 ⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ +=
R
bv
v
aPT 2 
so 
 
 3
3
32
221
Rv
abavPv
v
a
R
bv
Rv
aP
v
T
P
+−=⎟⎠
⎞⎜⎝
⎛ −−⎟⎠
⎞⎜⎝
⎛ +=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Substituting for P: 
 
 ( )( ) 3
23 2
Rvbv
bvaRTv
v
T
P −
−−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Hence, 
 
 ( )( )23
23
2
20
bvaRTv
bvavbRTvv
T
vT
P −−
−+−==−⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Solving for T: 
 
 ( )3
22
bRv
bvavT −= (1) 
 71
 
Substituting this value of T back into the van der Waals equation gives 
 
 
 ( ) ( )223 322 bv
bva
v
a
bv
bvavP −=−−= (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
32
mol
Jm 0.137=
64
27
c
c
P
RT
a 
 
 ⎥⎦
⎤⎢⎣
⎡×=
mol
m 103.88=8
3
5-
c
c
P
RTb 
 
Using these values in Equations (1) and (2), we get the following plot: 
 
Joule-Thomson inversion line
0
200
400
600
800
1000
0 100 200 300 400
T [K]
 
 
 72
5.34 
We can solve this problem using departure functions, so first find the reduced temperatures and 
pressures. 
 
 [ ][ ] 99.0bar 0.365
bar50
,1 ==rP [ ][ ] 2.0bar 0.365
bar10
,2 ==rP 
 967.0
K 282.4
K 15.273
,1 ==rT 
 
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: 
 
 
6.214
76.0
2,2
=∴
≅
T
T r 
 
The process is isenthalpic, so the following expression holds 
 
 0
,2,221,1,1 ,,
=∆+∆+∆−=∆ → depPTidealTTdepPT rrrr hhhh 
 
Therefore, 
 
 idealTT
dep
PT
dep
PT hhh rrrr 21,1,1,2,2 ,, →∆−∆=∆ 
 
From Table A.2.1: 
 
 [ ]∫ −−→ ×−×+=∆ K 6.214
K 15.273
263 10392.410394.14424.1
21
dTTTRhidealTT 
 
From Tables C.3 and C.4 ( )085.0=ω : 
 
 ( ) 976.351.3085.0678.3,1,1 , −=−+−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡∆
c
dep
PT
RT
h
rr 
 
Now we can solve for the enthalpy departure at state 2. 
 
 73
⎥⎥⎦
⎤
⎢⎢⎣
⎡
×−×+−−=
∆ ∫ −−
K 6.214
K 15.273
263, 10392.410394.14424.1
K 4.282
1976.3,2,2 dTTT
RT
h
c
dep
PT rr
 01.3,2,2
, −=
∆
c
dep
PT
RT
h
rr 
 
We can calculate the quality of the water using the following relation 
 
 ( )
c
vapdep
PT
c
liqdep
PT
c
dep
PT
RT
h
x
RT
h
x
RT
h
rrrrrr
,
,
,
,, ,2,2,2,2,2,2 1
∆
+
∆
−=
∆
 
 
where x represents the quality. From Figures 5.5 and 5.6: 
 
 ( ) 068.55.5085.06.4
,
, ,2,2 −=−+−=
∆
c
liqdep
PT
RT
h
rr 
 ( ) 464.075.00859.04.0
,
, ,2,2 −=−+−=
∆
c
vapdep
PT
RT
h
rr 
 
Thus, 
 
 447.0=x 
 
55.3% of the inlet stream is liquefied. 
 
 74
5.35 
Density is calculated from molar volume as follows: 
 
 
v
MW=ρ 
 
Substitute the above into the expression for 2soundV : 
 
 ( ) s
s
sound v
P
MW
v
MW
PV ⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
⎟⎠
⎞⎜⎝
⎛∂
∂=
/1
12 
 
The following can be shown using differentials: 
 
 2
1
v
v
v
∂−=⎟⎠
⎞⎜⎝
⎛∂ 
 
Therefore, 
 
 
ss
sound v
P
MW
vPV ⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
2
2
ρ 
 
 75
5.36 
From Problem 5.35: 
 
 
ss
sound v
P
MW
vPV ⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
2
2
ρ 
 
The thermodynamic web gives: 
 
 
PTTvssPvs T
s
s
P
v
s
s
T
T
P
v
T
v
s
s
P
v
P ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Pvv
PP
TTvs v
T
T
P
c
c
T
c
s
P
v
s
c
T
v
P ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
If we treat air as an ideal gas consisting of diatomic molecules only 
 
 
5
7=
v
P
c
c 
v
R
T
P
v
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
R
P
v
T
P
=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−=⎟⎠
⎞⎜⎝
⎛
∂
∂
v
P
v
P
s 5
7 
and 
 
 ⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
MW
RT
v
P
MW
vVsound 5
7
5
72 
 
 [ ]m/s 343=soundV 
 
The lightening bolt is 1360 m away. 
 
 
 76
5.37 
From Problem 5.35: 
 
 
ss
sound v
P
MW
vPV ⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
2
2
ρ 
 
The thermodynamic web gives: 
 
 
 
PTTvssPvs T
s
s
P
v
s
s
T
T
P
v
T
v
s
s
P
v
P ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
so 
 ⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=⎟⎠
⎞⎜⎝
⎛
∂
∂
T
c
v
T
T
P
c
T
v
P P
Pvvs
 
 
For liquids vP cc ≈ water at 20 ºC, so 
 
 
Pvs v
T
T
P
v
P ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
However, the cyclic rule gives: 
 
 
TPv P
v
v
T
T
P ⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂⎟⎠
⎞⎜⎝
⎛
∂
∂=−1 
 
So 
 
 
Ts v
P
v
P ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
From the steam tables, for saturated water at 20 oC: 
 
 P = 2.34 kPa and 
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 001002.ˆ
3
v 
 
For subcooled water at 20 oC: 
 
 P = 5 MPa and ⎥⎥⎦
⎤
⎢⎢⎣
⎡=
kg
m 0009995.ˆ
3
v 
 77
So 
 
 ⎥⎦
⎤⎢⎣
⎡
−
−=⎟⎠
⎞⎜⎝
⎛
∆
∆≈⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
3m
kPa kg 
001002.0009995.
34.25000
ˆˆˆ v
P
v
P
v
P
Ts
 
 
and 
 [ ]m/s 1414
ˆ
ˆ2 =⎟⎠
⎞⎜⎝
⎛
∂
∂=
s
sound v
PvV 
 
 78
5.38 
 
(a) 
The fundamental property relation for internal energy is 
 
 revrev WQdU δδ += 
 
Substituting the proper relationships for work and heat, we obtain 
 
 FdzTdSdU += 
 
The fundamental property relation for the Helmholtz energy is 
 
 ( ) SdTTdSdUTSddUdA −−=−= 
 
Substitute the expression for the internal energy differential: 
 
 SdTFdzdA −= 
 
(b) 
First, relate the entropy differential to temperature and length. 
 
 dz
Z
SdT
T
SdS
Tz
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Now we need to find expressions for the partial derivatives. 
 
 
zzz
z T
ST
T
zFST
T
unnc ⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂+∂=⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
Therefore, 
 
 ⎟⎠
⎞⎜⎝
⎛ +==⎟⎠
⎞⎜⎝
⎛
∂
∂ b
T
an
T
nc
T
S z
z
 
 
The following statement is true mathematically (order of differentiation does not matter): 
 
 
zTTz Z
A
TT
A
Z ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂=⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂ 
 
Furthermore, 
 
 79
 
TTz Z
S
T
A
Z
⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂ 
 
zzT T
F
Z
A
T
⎟⎠
⎞⎜⎝
⎛
∂
∂=⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂ 
 
 ( )0zzkT
F
Z
S
zT
−−=⎟⎠
⎞⎜⎝
⎛
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Substituting the expressions for the partial derivatives into the expression for the entropy 
differential, we obtain 
 
 ( )dzzzkdTb
T
andS 0−−⎟⎠
⎞⎜⎝
⎛ += 
 
(c) 
First, start with an expression for the internal energy differential: 
 
 dz
Z
UdT
T
UdU
Tz
⎟⎠
⎞⎜⎝
⎛
∂
∂+⎟⎠
⎞⎜⎝
⎛
∂
∂= 
 
From information given in the problem statement: 
 
 ( )bTan
T
U
z
+=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Using the expression for internal energy developed in Part (a) and information from Part (b) 
 
 ( )( ) ( ) 000 =−+−−=+⎟⎠
⎞⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂ zzkTzzkTF
z
ST
Z
U
TT
 
 
Therefore, 
 
 ( )[ ] ( )[ ]dTbTandzdTbTandU +=++= 0 
 
(d) 
We showed in Part (c) that 
 
 0=⎟⎠
⎞⎜⎝
⎛
∂
∂=
T
U z
UF 
 
 80
Using the expression for 
Tz
S ⎟⎠
⎞⎜⎝
⎛
∂
∂
 developed in Part B, we obtain 
 
 ( )0zzkTz
STF
T
S −=⎟⎠
⎞⎜⎝
⎛
∂
∂−= 
 
(e) 
First, perform an energy balance for the adiabatic process. 
 
 WdU δ= 
 
Substitute expressions for internal energy and work. 
 ( )[ ] ( )dzzzkTFdzdTbTan 0−==+ 
 
Rearrangement gives 
 
 
( )
( )[ ]bTan
zzkT
dz
dT
+
−= 0 
 
The right-hand side of the above equation is always positive, so the temperature increases as the 
rubber is stretched. 
 
 81
5.39 
The second law states that for a process to be possible, 
 
 0≥∆ univs 
 
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 
 
 dv
T
Pdv
T
PdT
T
cds
vv
v ⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛+= ∂
∂
∂
∂ 
 
Applying the van der Waals equation: 
 
 dv
bv
Rds −= 
 
Integrating 
 
 ⎥⎦
⎤⎢⎣
⎡==∆
K mol
J 5.11ln
1
2
v
vRssys 
 
For the entropy change of the surroundings, we use the value of heat given in Example 5.2: 
 
 
 ⎥⎦
⎤⎢⎣
⎡=−=
 mol
J 600surrqq 
 
Hence the entropy change of the surroundings is: 
 
 ⎥⎦
⎤⎢⎣
⎡−=−==∆
K mol
J 6.1
373
600
surrsurr
surr T
qs 
 
and 
 ⎥⎦
⎤⎢⎣
⎡=∆+∆=∆
K mol
J 9.9surrsysuniv sss 
 
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.
 82
5.40 
A schematic of the process is given by: 
 
Gas A in
Pi = 100 bar 
Ti = 600 K
Turbine
Pf = 20 bar 
Tf = 445 K
ws
Gas A
out
 
 
The energy balance for this process is provided below: 
 
 Swh =∆ 
 
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: 
 
P [bar] 
T [K]
Pi,Ti
Pf,Tf
step 1
st
ep
 3
step 2
ideal gas
600445
∆h
100
20
Plow
 
 
For the first section of the path, we have 
 
 ∫
=
⎟⎠
⎞⎜⎝
⎛
∂
∂=∆
0
1
1
P
P T
dP
P
hh 
 
If we apply Equation 5.45 we can rewrite the above equation as 
 
 ∫
=
⎥⎦
⎤⎢⎣
⎡ +⎟⎠
⎞⎜⎝
⎛
∂
∂−=∆
0
1
P
P P
i
i
dPv
T
vTh 
 
For the given EOS: 
 
 83
 2
iP T
aP
P
R
T
v −=⎟⎠
⎞⎜⎝
⎛
∂
∂ 
 
Therefore, 
 
 ⎥⎦
⎤⎢⎣
⎡−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=⎥⎦
⎤⎢⎣
⎡ ++−=∆ ∫∫
=
×=
=
×= mol
J 24672
0
10100
0
10100
1
55
v
P i
P
P i ii
dPb
T
aPdPv
T
aPPh 
 
Similarly for step 3 
 
 ⎥⎦
⎤⎢⎣
⎡=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=∆ ∫
×=
= mol
J 2502
51020
0
3
fP
P f
dPb
T
aPh 
 
 
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. 
 
 ( ) ⎥⎦
⎤⎢⎣
⎡−=+==∆ ∫∫ molJ 627002.030
K 445
K 600
2 dTTdTch
f
i
T
T
P 
 
Now sum each part to find the total change in enthalpy: 
 
 ⎥⎦
⎤⎢⎣
⎡−=∆+∆+∆=∆
mol
J 8487321 hhhh 
 ⎥⎦
⎤⎢⎣
⎡−=
mol
J 8487sw 
 
In other words, for every mole of gas that flows through the turbine, 8487 joules of work are 
produced.