Q23 / 90 · JEE Main 2021 (25 Feb Shift 2) · Chemistry

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]

Correct answer: 3

Solution

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

Hence, the correct answer is 3.

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