Q.One mole of an ideal gas at a temperature of $300\ \text{K}$ expands isothermally and reversibly to twice its initial volume. Find the work done by the gas. (Take $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$, $\ln 2 = 0.693$.)

An isothermal process is one in which the temperature of a thermodynamic system stays constant while its pressure and volume are free to change. Since $PV=\mu RT$ and $T$ is fixed, the equation of state becomes $PV = \text{constant}$, so if the gas moves from $(P_1,V_1)$ to $(P_2,V_2)$, then $P_1V_1=P_2V_2$; on a P-V diagram this traces a hyperbola called an isotherm, with a curve closer to the origin corresponding to a lower temperature. Since $U$ for an ideal gas depends only on $T$, and $T$ is constant, $\Delta U = 0$ throughout an isothermal process, so the first law reduces to $Q=W$: all the heat supplied is used entirely to do external work (isothermal expansion, $Q,W>0$), or all the work done on the gas during compression flows straight out as heat ($Q,W<0$). The wor …