\[\int_C \frac{1}{T} dU + \frac{p}{T} dV\]

is independent of the path.Since the integral is indeoendent of the path we can define a scalar function

\[S = \int_C \frac{1}{T} dU + \frac{p}{T} dV \]

This scalar is called the entropy and is a measure of how disordered the gas is and the integral gives the change in

\[S\]

along \[C\]

.\[U=U(p,V) \rightarrow dU = \frac{\partial U}{\partial p} dp + \frac{\partial U}{\partial V} dV \]

Hence

\[\begin{equation} \begin{aligned} S &= \int_C \frac{1}{T} dU + \frac{p}{T} dV = \\ &= \int_C (\frac{1}{T} (\frac{\partial U}{\partial p} dp + \frac{\partial U}{\partial V} dV ))+ \frac{p}{T} dV \\ &= \int_C \frac{1}{T} \frac{\partial U}{\partial p} dp + \frac{1}{T} (\frac{\partial U}{\partial V} + p ) dV \end{aligned} \end{equation}\]