Explanation:
D Given, $\mathrm{P}_{1}=\mathrm{P}, \mathrm{V}_{1}=\mathrm{V}, \mathrm{V}_{2}=4 \mathrm{~V}, \gamma=1.5$
For isothermal expansion,
$\mathrm{P}_{1} \mathrm{~V}_{1}=\mathrm{P}_{2} \mathrm{~V}_{2}$
$P_{2}=\frac{P_{1} V_{1}}{V_{2}}=\frac{P V}{4 V}$
$\mathrm{P}_{2}=\frac{\mathrm{P}}{4}$
For adiabatic process, $\mathrm{V}_{3}=\mathrm{V}$
$\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}=\mathrm{P}_{3} \mathrm{~V}_{3}^{\gamma}$
$P_{3} =P_{2}\left(\frac{V_{2}}{V_{3}}\right)^{\gamma} \quad \text { [from equation(i)] }$
$=\frac{P}{4}\left(\frac{4 V}{V}\right)^{\gamma}$
$=\frac{P}{4}(4)^{1.5}$
$=\frac{P}{4}(4)^{3 / 2}$
$=\frac{P}{4} \times 8=2 P$
$\mathrm{P}_{3}=2 \mathrm{P}$