JEE Main 2021ChemistryThermodynamics (C)HardNumerical

JEE Main 2021 — Thermodynamics (C) Question with Solution

JEE Main 2021 (25 Feb Shift 2)

Question

Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is ___ kJmol-1. (Rounded-off to the nearest integer)

[ Use R=8.314 J mol-1 K-1]

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Show full solutionCorrect answer: 3
Correct answer
3

Step-by-step explanation

n=5, T=293 K=const,ΔU=0

P1=2.1 MPa,P2=1.3 MPa

Pext=4.3 MPa=const.

W=-PextV2-V1=-PextnRTP2-nRTP1

or, W=-PextnRT1P2-1P1

=-4.3×5×8.314×29311.3-12.1

=-4.3×5×8.314×2932.1-1.31.3×2.1

=-15347.7 J

or, W=-15.35 kJ

ΔU0=q+W

∴  q=-W

or, q=15.35 kJ (for 5 moles)

∴  q/mole=15.355=3 kJ mol-1

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About this question

This is a previous-year question from JEE Main 2021, covering the Thermodynamics (C) chapter of Chemistry. 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.