An ideal gas expands isothermally from volume $V$ to $3V$. The ratio of the work done by the gas to the change in its internal energy is:

Solution

1. For an ideal gas, internal energy $U$ is a function of temperature $T$ only.
2. In an isothermal process, $\Delta T=0$, therefore $\Delta U=0$.
3. The work done $W=nRT\ln (3)$ is non-zero.
4. The ratio $$\begin{array}{rl}\frac{W}{\Delta U}& =\frac{W}{0}\\ & =\infty \end{array}$$.
5. Hence the answer is (A).