Solution to 2008 Problem 34

\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*}
\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.

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