Solution to 1992 Problem 62

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.

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