Q 11-11-136JEE MainJEE Main 2020 (6 Sep, Shift 2)Medium
An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to ______.
Numerical value type. Enter your answer.
Answer: 19
Work done per cycle = area enclosed $= (3P_0 - P_0)(2V_0 - V_0) = 2P_0V_0$.
Heat is absorbed in A→B (isochoric) and B→C (isobaric). For a monatomic gas $C_v = \tfrac32R$, $C_p = \tfrac52R$, and $nR\Delta T = \Delta(PV)$:
A→B: $Q_1 = \tfrac32(3P_0V_0 - P_0V_0) = 3P_0V_0$
B→C: $Q_2 = \tfrac52(6P_0V_0 - 3P_0V_0) = 7.5P_0V_0$
$$\eta = \frac{2P_0V_0}{10.5P_0V_0} \approx 0.19 = 19\%$$
Solution by Sreeraj P, M.Sc Physics