Use the formula $dS = dQ/T$ . Entropy is a state function, so if we compute the integral of $dS$ for a different process with the same initial and final states, the result should be the same. So, we choose a reversible isothermal expansion. The original process was adiabatic and involved no work done by the gas. So, the temperature of the gas cannot change. So, the initial and final temperatures are the same, so a reversible isothermal process will indeed have the same initial and final states as the actual process. For a reversible isothermal process, the internal energy of the gas does not change (because $U = 2/3 N k T$ ). So, $dU = 0$ . Therefore, by the second law of thermodynamics $dQ = dW$ . Therefore, if we let $T$ be the temperature of the gas (which does not change) and $V$ denote the initial volume of the gas, then
$\begin{align*}\Delta S = \int dQ/T = \int dW/T = \frac{1}{T} \int dW = \frac{1}{T} \int P dV = \frac{N k T}{T} \int \frac{dV}...$
Therefore, answer (B) is correct.

\section*{alternate solution:}
Use the statistical definition of entropy, $S = k \ln W$ where $W$ is the number of microscopic states of the system. We assume that the number of locations that each particle could be in (i.e. the number of single particle states) is proportional to the volume $V$ , i.e. $W = C V$ where $C$ is a constant. Then the total number of states is given by $W = (C V)^N$ where $N$ is the number of particles. Thus,
$\begin{align*}S_{\text{final}} - S_{\text{initial}} = k \ln (2 C V)^N - k \ln (C V)^N = N k \ln 2 = \boxed{n R \ln 2}\end{ali...$
Therefore, answer (B) is correct.

Solution to 2001 Problem 47