4.3 Solutions
271
u =
T
3
∂u
∂ T
−
u
3
or
du
u
+ 4
dT
T
= 0
Integrating,
ln u = 4 ln T + ln a = ln aT
4
where ln a is the constant of integration. Thus,
u = aT
4
4.27 S = f (T, V )
where T and V are independent variables.
dS =
∂ S
∂ T
V
dT +
∂ S
∂ V
T
dV
∂ S
∂ T
p
=
∂ S
∂ T
V
+
∂ S
∂ V
T
∂ V
∂ T
P
Multiplying out by T and re-arranging
T
∂ S
∂ T
p
− T
∂ S
∂ T
V
= T
∂ S
∂ V
T
∂ V
∂ T
P
Now,
T
∂ S
∂ T
p
= C p ; T
∂ S
∂ T
ν
= C ν
and from Maxwell’s relation,
∂ S
∂ V
T
=
∂ P
∂ T
ν
Therefore,
C p − C ν = 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
C p − C ν = RT
4.28
P +
a
V 2
(V − b) = RT
(1)
Neglecting b in comparison with V ,
P =
RT
V
−
a
V 2
(2)