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4 The Thermodynamics of Real Systems
c. For an isothermal process in a closed system,
∆S
c
dS
V 2
V 1
∂S
∂V
r n
dV
V 2
V 1
R
V m
+
R
V 2
m
B 2 + T
dB 2
dT
dV
n
V 2
V 1
R
V m
dV m + n
B 2 + T
dB 2
dT
V 2
V 1
R
V 2
m
dV m
nR ln
V m2
V m1
− n
B 2 + T
dB 2
dT
1
V m2
−
1
V m1
d.
∆S (1.000 mol)(8.3145 J K −1 mol −1 ) ln
0.05000
0.02500
− (1.000 mol)(8.3145 J −1 mol −1 )
×
− 15.8 × 10 −6 m 3 mol −1
+ (298.15 K)(0.20 × 10 −6 m 3 mol −1 K −1 )
×
1
0.05000 m 3 mol −1 −
1
0.02500 m 3 mol −1
5.763 J K −1 + 0.00729 J K −1 5.770 J K −1
The correction for nonideality, 0.007 J K −1 , is numerically almost insignificant in
this case.
We can also derive an expression for ∆S for an isothermal pressure change, using
another Maxwell relation:
∆S
P 2
P 1
∂S
∂P
T ,n
dP −
P 2
P 1
∂V
∂T
P,n
dP
(4.2-24)
E X A M P L E 4.4
a. Find an expression for ∆S for an isothermal pressure change on a pure liquid assuming
that the volume of the liquid is constant.
b. Find an expression for ∆S for an isothermal pressure change on a pure liquid, assuming
that the volume of the liquid is given by
V (T , P) V (T 1 , P 1 )[1 + α(T − T 1 ) − κ T (P − P 1 )]
where α and κ T are equal to constants and where P 1 and T 1 are a reference pressure and
a reference temperature.
c. Find ∆S for pressurizing 1.000 mol of liquid water isothermally at 298.15 K from a
pressure of 1.00 atm to a pressure of 100.00 atm.
Solution
a.
∆S
P 2
P 1
∂S
∂P
T ,n
dP −
P 2
P 1
∂V
∂T
P,n
dP −
P 2
P 1
V αdP 0
∆S 0 since α 0 if V is constant.