Below it is proved that the internal energy of an ideal gas depends only on its temperature. Therefore,
$\begin{align*}\left.\frac{\partial E}{\partial T} \right|_V = \left.\frac{\partial E}{\partial T} \right|_P\end{align*}$
and therefore
$\begin{align*}C_P - C_V &= \left.\frac{d Q}{dT}\right|_P - \left.\frac{d Q}{dV}\right|_P \\&= P \left.\frac{\partial...$
The process is adiabatic, so from the first law of thermodynamics
$\begin{align*}0 &= d E - dW \\&= C_V dT + P dV \\&= C_V dT + \frac{T dV}{V} \left(C_P - C_V \right)\end{align*}$
which implies that
$\begin{align*}0 = \frac{dT}{T} + \frac{dV}{V} \left( \gamma -1 \right)\end{align*}$
Using integration by parts then gives that $T V^{\gamma -1}$ is constant. Therefore, answer (C) is correct.

Proof that the internal energy of an ideal gas depends only on its temperature:
$\begin{align*}dS &= \left.\frac{\partial S}{\partial T}\right|_V dT + \left.\frac{\partial S}{\partial V}\right|_T dV\end...$
implies that
$\begin{align*}dE &= T dS - P dV \\&= T \left.\frac{\partial S}{\partial T}\right|_V dT + T \left.\frac{\partial S}{\p...$
implies that
$\begin{align}\left.\frac{\partial E}{\partial V}\right|_T = T \left.\frac{\partial S}{\partial V}\right|_T - P \label{eqn:1}\...$
From
$\begin{align*}d(E-TS) = - S dT - P dV\end{align*}$
and the equality of mixed partial derivatives it follows that
$\begin{align*}\left.\frac{\partial S}{\partial V} \right|_T = \left.\frac{\partial P}{\partial T} \right|_V\end{align*}$
Therefore, equation 1 becomes
$\begin{align*}\left.\frac{\partial E}{\partial V}\right|_T = T \left.\frac{\partial P}{\partial T} \right|_V - P \end{align*}$
which equals $0$ for an ideal gas. This completes the proof.

Solution to 2008 Problem 21