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.