164 4 The Thermodynamics of Real Systems
c.For an isothermal process in a closed system,
∆S
∫
c
dS
∫V 2
V 1
(
∂S
∂V
)
rn
dV
∫V 2
V 1
[
R
Vm
+
R
Vm^2
(
B 2 +T
dB 2
dT
)]
dV
n
∫V 2
V 1
R
Vm
dVm+n
(
B 2 +T
dB 2
dT
)∫V 2
V 1
R
Vm^2
dVm
nRln
(
Vm2
Vm1
)
−n
(
B 2 +T
dB 2
dT
)(
1
Vm2
−
1
Vm1
)
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 .763JK−^1 + 0 .00729 J K−^1 5 .770JK−^1
The correction for nonideality, 0.007JK−^1 , is numerically almost insignificant in
this case.
We can also derive an expression for∆Sfor 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)
EXAMPLE 4.4
a.Find an expression for∆Sfor an isothermal pressure change on a pure liquid assuming
that the volume of the liquid is constant.
b.Find an expression for∆Sfor 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 )−κT(P−P 1 )]
whereαandκTare equal to constants and whereP 1 andT 1 are a reference pressure and
a reference temperature.
c.Find∆Sfor 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
∆S0 sinceα0ifVis constant.