Verified Solution • 2025–26 Session moderate

QJEEM13Ph191d: JEE Main 2013 Physics - Class 11 JEEM13Ph191d Solution: The above - diagram represents the thermodynamic cycle of an…

JEE Main 2013 (Offline) Physics - Heat and Thermodynamics Question 370 English

The above pp - vv diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is

Multiple Choice Options

  • A p0v0{p_0}{v_0}
  • B (132)p0v0\left( {{{13} \over 2}} \right){p_0}{v_0}
  • C (112)p0v0\left( {{{11} \over 2}} \right){p_0}{v_0}
  • D 4p0v04{p_0}{v_0}

Step-by-Step Verified Solution

Along path DA, volume is constant.

Hence, Δ\Delta QDA = nCv Δ\Delta T = nCv(TA – TD)

∴\therefore Δ\Delta QDA = n(32R)[2p0v0nR−p0v0nR]=32p0v0n\left( {{3 \over 2}R} \right)\left[ {{{2{p_0}{v_0}} \over {nR}} - {{{p_0}{v_0}} \over {nR}}} \right] = {3 \over 2}{p_0}{v_0}

Along the path AB, pressure is constant.

Hence Δ\Delta QAB = nCp Δ\Delta T = nCp(TB – TA)

∴\therefore Δ\Delta QAB = n(52R)[2p02v0nR−2p0v0nR]=102p0v0n\left( {{5 \over 2}R} \right)\left[ {{{2{p_0}2{v_0}} \over {nR}} - {{2{p_0}{v_0}} \over {nR}}} \right] = {{10} \over 2}{p_0}{v_0}

∴\therefore The amount of heat extracted from the source in a single cycle is

Δ\Delta Q = Δ\Delta QDA + Δ\Delta QAB

=32p0v0+102p0v0= {3 \over 2}{p_0}{v_0} + {{10} \over 2}{p_0}{v_0} = (132)p0v0\left( {{{13} \over 2}} \right){p_0}{v_0}

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