4.3 Solutions 271
u=T
3
∂u
∂T−
u
3ordu
u+ 4
dT
T= 0
Integrating,
lnu=4lnT+lna=lnaT^4
where lnais the constant of integration. Thus,
u=aT^44.27 S=f(T,V)
whereTandVare independent variables.
dS=(
∂S
∂T
)
VdT+(
∂S
∂V
)
TdV
(
∂S
∂T)
p=
(
∂S
∂T
)
V+
(
∂S
∂V
)
T(
∂V
∂T
)
P
Multiplying out byTand re-arrangingT(
∂S
∂T
)
p−T
(
∂S
∂T
)
V=T
(
∂S
∂V
)
T(
∂V
∂T
)
P
Now,T(
∂S
∂T
)
p=Cp; T(
∂S
∂T
)
ν=Cνand from Maxwell’s relation,
(
∂S
∂V)
T=
(
∂P
∂T
)
ν
Therefore,Cp−Cν=T(
∂P
∂T
)
V(
∂V
∂T
)
P(1)
For one mole of a perfect gas,PV=RT. Therefore
(
∂P
∂T)
V=
R
V
and(
∂V
∂T
)
P=
R
P
It follows that
Cp−Cν=RT4.28
(
P+
a
V^2)
(V−b)=RT (1)Neglectingbin comparison withV,P=RT
V
−
a
V^2