Five moles of an ideal gas has pressure $p_0$ volume $V_0$ and temperature $T_0$. The gas is expanded to volume $3 V_0$ along a path, so that the pressure $p$ is changed as function of volume $V$ as $p=p_0\left(V / V_0\right)$. The pressure is then reduced to $p_0$ maintaining the volume constant. The gas undergoes an isobaric compression till the volume and temperature become $V_0$ and $T_0$, respectively. The total work done by the gas during the entire process is

• A

  $p_0 V_0 / 3$
• B

  $3 p_0 V_0$
• C

  $5 p_0 V_0 / 3$
• Correct answer
  D

  $2 p_0 V_0$

Solution

The given situation is shown in the following diagram
Work done in expansion (process I) $=\int_{V_i}^{V_f} p d V=\int_{V_0}^{3 V_0} \frac{p_0 V}{V_0} \cdot d V$ $=\frac{p_0}{V_0}\left[\frac{V^2}{2}\right]_{V_0}^{3 V_0}=\frac{p_0}{2 V_0}\left(9 V_0^2-V_0^2\right)=4 p_0 V_0$ In process $\mathrm{II}$, there is no work done as $V=$ constant. In process III, work done is given as $W=p_0 \int_{V_i}^{V_f} d V=p_0 \int_{3 V_0}^{V_0} d V$ $=p_0\left(V_0-3 V_0\right)=-2 p_0 V_0$ So, total work done in entire process is $W_{\text {Total }}=4 p_0 V_0+\left(-2 p_0 V_0\right)=2 p_0 V_0$

Hence, the correct answer is option: 'D'.