$\begin{align*}C_V &= \left(\frac{dQ}{dT} \right)_V = \left(\frac{dE + P dV}{dT} \right)_V = \left(\frac{\partial E}{\part...$
For an ideal gas,
$\begin{align*}E &= \frac{3}{2} N k T \\V &= \frac{N k T}{P}\end{align*}$
$E = \frac{3}{2} N k T$ implies that $\left(\frac{\partial E}{\partial T} \right)_V$ and $\left(\frac{dE}{dT} \right)_P$ are both equal to $(3/2) N k$ . $V = \frac{N k T}{P}$ implies that
$\begin{align*}\left(\frac{dV}{dT}\right)_P = \frac{Nk}{P}\end{align*}$
So,
$\begin{align*}C_V &= (3/2) N k \\ C_P &= (5/2) N k\end{align*}$
Now, let $Q'$ be the heat required to produce the temperature change $\Delta T$ under conditions of constant pressure. Then,
$\begin{align*}Q'/C_P = Q/C_V \Longrightarrow Q' = \frac{C_P}{C_V} Q = \boxed{\frac{5}{3}Q}\end{align*}$
Therefore, answer (C) is correct.

Solution to 2008 Problem 34