272 4 Thermodynamics and Statistical Physics
(
∂P
∂T
)
V
=
R
V
(3)
Re-writing (1)
PV+
a
V
=RT
DifferentiatingVwith respect toT, keepingPfixed
P
(
∂V
∂T
)
P
−
a
V^2
(
∂V
∂T
)
P
=R
or
(
∂V
∂T
)
P
=
R
P−a/V^2
(4)
Now,
Cp−Cν=T
(
∂P
∂T
)
V
(
∂V
∂T
)
P
(5)
(By Problem 4.27)
Using (3) and (4) in (5)
Cp−Cν=
R^2 T
V(P−a/V^2 )
=R
(P+a/V^2 )
(P−a/V^2 )
≈R(1+ 2 a/PV^2 )
=R
(
1 +
2 a
RT V
)
4.29 If f(x,y,z)=0, then it can be shown that
(
∂x
∂y
)
z
(
∂y
∂z
)
x
(
∂z
∂x
)
y
=−1(1)
Thus, iff(P,V,T)= 0
(
∂P
∂V
)
T
(
∂V
∂T
)
P
(
∂T
∂P
)
V
=−1(1)
or
(
∂P
∂T
)
V
=−
(
∂P
∂V
)
T
(
∂V
∂T
)
P
(2)
and
(
∂V
∂T
)
P
=−
(
∂P
∂T
)
V
(
∂V
∂P
)
T
(3)
But
CP−CV=T
(
∂P
∂T
)
V
(
∂V
∂T
)
P