268
4 Thermodynamics and Statistical Physics
∂
∂ y
∂U
∂ x
y
x
=
∂ T
∂ y
x
∂ S
∂ x
y
+ T
∂
∂ y
∂ S
∂ x
y
x
(15)
−
∂ P
∂ y
x
∂ V
∂ x
y
− P
∂
∂ y
∂ V
∂ x
y
x
∂
∂ x
∂U
∂ y
x
y
=
∂ T
∂ x
y
∂ S
∂ y
x
+ T
∂
∂ x
∂ S
∂ y
x
y
(16)
−
∂ P
∂ x
y
∂ V
∂ y
x
− P
∂
∂ x
∂ V
∂ y
x
y
Since the order of differentiation is immaterial, dU being a perfect differential, the left hand sides of (15) and (16) are equal. Further, since dS and dV
are perfect differentials.
∂
∂ y
∂ S
∂ x
y
x
=
∂
∂ x
∂ S
∂ y
x
y
(17)
and
∂
∂ y
∂ V
∂ x
y
x
=
∂
∂ x
∂ V
∂ y
x
y
(18)
Using (15), (16), (17), and (18),
∂ P
∂ x
y
∂ V
∂ y
x
−
∂ P
∂ y
x
∂ V
∂ x
y
=
∂ T
∂ x
y
∂ S
∂ y
x
−
∂ T
∂ y
x
∂ S
∂ x
y
(19)
Equation (19) can be written in the form of determinants
∂ P
∂ x
y
∂ P
∂ y
x
∂ V
∂ x
y
∂ V
∂ y
x
=
∂ T
∂ x
y
∂ T
∂ y
x
∂ S
∂ x
y
∂ S
∂ y
x
(20)
(a) Let the temperature and volume be independent variables. Put x = T and
y = V in (20). Then
∂ T
∂ x
y
=
∂ V
∂ y
x
= 1;
∂ T
∂ y
x
=
∂ V
∂ x
y
= 0
Since T and V are independent, we find
∂ S
∂ V
T
=
∂ P
∂ T
V
(21)