272 4 Thermodynamics and Statistical Physics
(
∂P
∂T)
V=
R
V
(3)
Re-writing (1)PV+a
V=RT
DifferentiatingVwith respect toT, keepingPfixedP(
∂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 Iff(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)
ButCP−CV=T(
∂P
∂T
)
V(
∂V
∂T
)
P