CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 80
Proof
According to the equation of state for an ideal gas (2.15), pressure is
p=
nRT
V
.
For the total differential of pressure it holds [see (3.25)]
dp=
(
∂p
∂T
)
V
dT+
(
∂p
∂V
)
T
dV =
nR
V
dT−
nRT
V^2
dV.
The condition (3.26) is fulfilled here:
nR
(
∂(1/V)
∂V
)
T
=−
nR
V^2
(
∂T
∂T
)
V
.
dpis thus a total differential and pressure is a function of state. A similar proof can be established
for pressure calculated using any equation of state.
Example
By substituting the expressionTdSfollowing from (3.6) for ̄dQinto equation (3.1) we obtain
̄dW= dU−TdS.
Prove that workW is not a state function.
Proof
Comparison with (3.23) shows thatM = 1,N=−T,x=U, andy=S. Hence
(
∂M
∂y
)
x
=
(
∂ 1
∂S
)
U
6 =
(
∂N
∂x
)
y
=
(
∂(−T)
∂U
)
S
because
(
∂T
∂U
)
S
is generally different from zero. This shows that ̄dWis not a total differential
and therefore work is not a state function.
