4.3 Solutions
267
Therefore,
dU = T ds − PdV
(6)
where U is the internal energy, Q the heat absorbed, W the work done by the
system, S the entropy, P the pressure and T the Kelvin temperature.
Let the independent variables be called x and y. Then
U = U (x, y); V = V (x, y); S = S(x, y)
( 7 )
Now,
d f =
∂ f
∂ x
y
dx +
∂ f
∂ y
x
dy
(8)
Therefore
dU =
∂U
∂ x
y
dx +
∂U
∂ y
x
dy
(9)
dV =
∂ V
∂ x
y
dx +
∂ V
∂ y
x
dy
(10)
dS =
∂ S
∂ x
y
dx +
∂ S
∂ y
x
dy
(11)
Eliminating internal energy U and substituting (9), (10) and (11) in (6)
∂U
∂ x
y
dx +
∂U
∂ y
x
dy = T
∂ S
∂ x
y
dx +
∂ S
∂ y
x
dy
−P
∂ V
∂ x
y
dx +
∂ V
∂ y
x
dy
(12)
Equating the coefficients of dx and dy
∂U
∂ x
y
= T
∂ S
∂ x
y
− P
∂ V
∂ x
y
(13)
∂U
∂ y
x
= T
∂ S
∂ y
x
− P
∂ V
∂ y
x
(14)
Differentiating (13) with respect to y with x fixed, and differentiating (14)
with respect to x with y fixed