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

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.