JEE Main 2021PhysicsThermodynamicsHardMCQ

JEE Main 2021 — Thermodynamics Question with Solution

JEE Main 2021 (24 Feb Shift 2)

Question

If one mole of an ideal gas at P1,V1 is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value B→C. Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to:

Choose an option

Show full solutionCorrect option: D
Correct answer
DRTln2-12γ-1

Step-by-step explanation

A-B= isothermal process

WAB=P1V1 ln2V1 V1=P1V1ln2

B-C→ Isochoric process

WBC=0

C-A→Adiabatic process

WCA=P1V1-P14×2V11-γ=P1V11-121-γ=P1V121-γ

Wnet=WAB+WBC+WCA  P1V1=RT

=P1V1 ln2+0+P1V121-γ

Wnet=RT ln2-12γ-1

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About this question

This is a previous-year question from JEE Main 2021, covering the Thermodynamics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.