PrepRung
Sign inSign up free

JEE Main 2017 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

One mole of a monoatomic ideal gas is taken around a rectangular cycle on a P-V diagram: from (P0, V0) at constant volume to (2P0, V0), then at…

Single correct option

One mole of a monoatomic ideal gas is taken around a rectangular cycle on a PP-VV diagram: from (P0,V0)(P_0, V_0) at constant volume to (2P0,V0)(2P_0, V_0), then at constant pressure to (2P0,2V0)(2P_0, 2V_0), then at constant volume to (P0,2V0)(P_0, 2V_0) and finally at constant pressure back to (P0,V0)(P_0, V_0). The efficiency of this cycle as a heat engine is
  1. A. 213\dfrac{2}{13}
  2. B. 16\dfrac{1}{6}
  3. C. 211\dfrac{2}{11}
  4. D. 14\dfrac{1}{4}
Show answer and solution

Answer: (A) 213\dfrac{2}{13}

Net work == area of the rectangle =(2P0−P0)(2V0−V0)=P0V0= (2P_0 - P_0)(2V_0 - V_0) = P_0V_0. Heat is absorbed only on the two heating steps. Constant volume, P0→2P0P_0\to2P_0: Q1=ΔU=32Δ(PV)=32(2P0V0−P0V0)=1.5P0V0Q_1 = \Delta U = \tfrac32\Delta(PV) = \tfrac32(2P_0V_0 - P_0V_0) = 1.5P_0V_0. Constant pressure, V0→2V0V_0\to2V_0 at 2P02P_0: Q2=52Δ(PV)=52(4P0V0−2P0V0)=5P0V0Q_2 = \tfrac52\Delta(PV) = \tfrac52(4P_0V_0 - 2P_0V_0) = 5P_0V_0. Hence η=P0V06.5P0V0=213≈15%\eta = \dfrac{P_0V_0}{6.5P_0V_0} = \dfrac{2}{13} \approx 15\%. Counting only the isobaric heat gives 1/51/5, and ignoring Q1Q_1's partner step gives 1/61/6.
Report a mistake

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake