n mole of a perfect gas… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Thermodynamics

2021mcqhard

$n$ mole of a perfect gas undergoes a cyclic process $ABCA$ (see figure) consisting of the following processes.

$A\rightarrow B:$ Isothermal expansion at temperature $T$ so that the volume is doubled from ${V}_{1}$ to ${V}_{2}=2{V}_{1}$ and pressure changes from ${P}_{1}$ to ${P}_{2}$

$B\rightarrow C:$ Isobaric compression at pressure ${P}_{2}$ to initial volume ${V}_{1}.$

$C\rightarrow A:$ Isochoric change leading to change of pressure from ${P}_{2}$ to ${P}_{1}$

Total work done in the complete cycle $ABCA$ is:

JEE Main 2021 Physics, Thermodynamics — question figure

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

Options

A
$nRT(\mathrm{ln}2-\frac{1}{2})$
B
$nRT\mathrm{ln}2$
C
$nRT(\mathrm{ln}2+\frac{1}{2})$
D
$0$

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Solution

${W}_{\mathrm{isothermal}}=nRT\mathrm{ln}(\frac{{V}_{2}}{{V}_{1}})$ \

${W}_{\mathrm{isobaric}}=P\Delta V=nR\Delta T$

${W}_{\mathrm{isochoric}}=0$

JEE Main 2021 Physics, Thermodynamics — solution figure

${W}_{1}=nRT\mathrm{ln}(\frac{2V}{V})=nRT\mathrm{ln}2$

${W}_{2}=nR(\frac{T}{2}-T)=-nR\frac{T}{2}$

${W}_{3}=0$

$\Rightarrow {W}_{\mathrm{net}}={W}_{1}+{W}_{2}+{W}_{3}$

${W}_{\mathrm{net}}=nRT(\mathrm{ln}2-\frac{1}{2})$

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Solution

${W}_{\mathrm{isothermal}}=nRT\mathrm{ln}(\frac{{V}_{2}}{{V}_{1}})$ \

${W}_{\mathrm{isobaric}}=P\Delta V=nR\Delta T$

${W}_{\mathrm{isochoric}}=0$

JEE Main 2021 Physics, Thermodynamics — solution figure

${W}_{1}=nRT\mathrm{ln}(\frac{2V}{V})=nRT\mathrm{ln}2$

${W}_{2}=nR(\frac{T}{2}-T)=-nR\frac{T}{2}$

${W}_{3}=0$

$\Rightarrow {W}_{\mathrm{net}}={W}_{1}+{W}_{2}+{W}_{3}$

${W}_{\mathrm{net}}=nRT(\mathrm{ln}2-\frac{1}{2})$

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics