## Integrals Independent of the Path in the Temperature Volume Plane for Ideal Gases

Theorem
In the
$TV$
plane the integral
$\int_C S dT + p dV$
is independent of

, the curve between given initial and final states of an ideal gas.
Proof
The entropy of the system is a function of the volume and pressure, and
$S=S(V,p), dS= \frac{1}{T} dU + \frac{p}{T} dV$

Similarly the internal energy of the system is a function of the volume and pressure,
$U=U(V,T), dU= \frac{\partial U}{\partial T} dT + \frac{\partial U}{\partial p} dp$

Then
$dS =\frac{1}{T} \frac{\partial U}{\partial T} dT + (\frac{1}{Y} \frac{\partial U}{\partial T} + \frac{p}{T}) dV$

Since
$S$
is a function of
$V,T$
,
$S=S(V,T) \rightarrow dS = \frac{\partial S}{\partial V} dV + \frac{\partial S}{\partial T} dT$

Equating coefficients of partial derivatives gives us
$\frac{\partial S}{\partial T}= \frac{1}{T} \frac{\partial U}{\partial T}, \: \frac{\partial S}{\partial V}= \frac{1}{T} \frac{\partial U}{\partial V} + \frac{p}{T}$

Differentiating the first of these with respect to
$V$
and the second with respect to
$T$
gives
$\frac{\partial^2 S}{\partial V \partial T}= \frac{1}{T} \frac{\partial^2 U}{\partial V \partial T}, \: \frac{\partial^2 S}{\partial T\partial V}= - \frac{1}{T^2} \frac{\partial U}{\partial V} + \frac{1}{T} \frac{\partial^2 U}{\partial T \partial V}- \frac{p}{T^2} + \frac{1}{T} \frac{\partial p}{\partial T}$

If
$S, U$
have continuous first and second partial derivatives, then
$\frac{\partial^2 S}{\partial V \partial T} = \frac{\partial^2 S}{\partial T \partial V} , \: \frac{\partial^2 U}{\partial V \partial T} = \frac{\partial^2 U}{\partial T \partial V}$
.
Hence
$\frac{1}{T} \frac{\partial^2 U}{\partial V \partial T}= - \frac{1}{T^2} \frac{\partial U}{\partial V} + \frac{1}{T} \frac{\partial^2 U}{\partial T \partial V}- \frac{p}{T^2} + \frac{1}{T} \frac{\partial p}{\partial T} \rightarrow - \frac{1}{T^2} \frac{\partial U}{\partial V} + \frac{p}{T^2} + \frac{1}{T} \frac{\partial p}{\partial T} =0$

Now multiply by
$T$
to give
$- \frac{1}{T} \frac{\partial U}{\partial V} + \frac{p}{T} + \frac{\partial p}{\partial T} =0$

Now use
$\frac{\partial S}{\partial V}= \frac{1}{T} \frac{\partial U}{\partial V} + \frac{p}{T}$
from above to give
$\frac{\partial S}{\partial V} = \frac{\partial p}{\partial T}$
. This last result implies independence of path.