## Solution to 2001 Problem 36

 (A) This is the definition of an adiabatic process.(B) \begin{align*}dS = \frac{dQ}{T}\end{align*}Because the process is adiabatic, $dQ = 0$, so $dS = 0$, so the integral of $dS$ over the entire process also equals 0.(C) By the Second Law of Thermodynamics, \begin{align*}\Delta U = Q - W\end{align*}where $\Delta U$ is the change in the internal energy of the gas, $Q$ is the heat added to the gas, and $W$ is the work done BY the gas. Since the process is adiabatic, $Q = 0$ and $\Delta U = - W$. $W$ is given by\begin{align*}W = \int P dV\end{align*}Therefore, \begin{align*}\Delta U = - W = - \int P dV\end{align*} (D) As discussed in part (C),\begin{align*}W = \int P dV\end{align*}(E) The temperature of an ideal gas is related to the internal energy by\begin{align*}U = 3/2 N k T\end{align*}where $N$ is the number of gas molecules, $T$ is the temperature of the gas, and $k$ is Boltzmann's constant. The internal energy of the gas is not constant (see part (C)), so the temperature cannot stay constant.Therefore, answer (E) is correct.