An ideal gas undergoes a process where its molar heat capacity is given by $C=C_v+\frac{\alpha }{T}$, where $\alpha$ is a constant. If the gas expands from volume $V_0$ to $2V_0$ at a constant temperature $T_0$, what is the work done by the gas?

Solution

1. For an ideal gas undergoing an isothermal process at temperature $T_0$, the work done $W$ is given by the integral $W={\int }_{V_0}^{2V_0}PdV$.
2. Using the ideal gas law $PV=nR{T}_0$, we substitute $P=\frac{nR{T}_0}{V}$ into the integral.
3. Assuming one mole of gas ($n=1$), the integral becomes $W=R{T}_0{\int }_{V_0}^{2V_0}\frac{1}{V}dV$.
4. Evaluating the integral yields $$\begin{array}{rl}W& =R{T}_0[\ln (V){]}_{V_0}^{2V_0}\\ & =R{T}_0\ln (\frac{2V_0}{V_0})\end{array}$$.
5. Simplifying the expression results in $W=R{T}_0\ln (2)$.
6. Hence the answer is (A).