Q 11-11-164JEE MainJEE Main 2017 (8 Apr)Medium
An engine operates by taking $n$ moles of an ideal gas through the cycle $ABCDA$ shown in figure. The thermal efficiency of the engine is (take $C_v = 1.5R$, where $R$ is gas constant):
Answer: (B) $0.15$
Net work = area of the rectangle $= (2P_0 - P_0)(2V_0 - V_0) = P_0V_0$.
Heat is absorbed in $AB$ (isochoric, pressure rising) and $BC$ (isobaric, volume rising). Using $nR\,\Delta T = \Delta(PV)$ and $C_p = 2.5R$:
$$Q_{AB} = nC_v\Delta T = 1.5(2P_0V_0 - P_0V_0) = 1.5P_0V_0$$
$$Q_{BC} = nC_p\Delta T = 2.5(4P_0V_0 - 2P_0V_0) = 5P_0V_0$$
$$\eta = \frac{W}{Q_{\text{in}}} = \frac{P_0V_0}{6.5P_0V_0} \approx 0.15$$
Solution by Sreeraj P, M.Sc Physics