(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.

Solution to 2001 Problem 36