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, dUbeing a perfect differ-
ential, the left hand sides of (15) and (16) are equal. Further, since dSand 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. Putx=Tand
y=Vin (20). Then
(
∂T
∂x)
y=
(
∂V
∂y)
x=1;
(
∂T
∂y)
x=
(
∂V
∂x)
y= 0
SinceTandVare independent, we find
(
∂S
∂V)
T=
(
∂P
∂T
)
V