The process is isothermal, so the ideal gas law $PV = n R T$ must hold (where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas, $T$ is the temperature in Kelvin, and $R$ is the universal gas constant. Therefore, the work done by the gas during the expansion is
$\begin{align*}\int P dV = R T_0 \int_{V_0}^{V_1} \frac{dV}{V} = \boxed{ R T_0 \ln \left(V_1/V_0 \right)}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1992 Problem 62