Recall the formula of work done in terms of change in volume and pressure, in the case of isobaric process, where pressure is constant, W=PV2−PV1, now take the idea of ideal gas equation,
W=nRT2−nRT1=21×8.31(70)=291J.
JEE Main 2019 — Physics Thermodynamics
Half mole of an ideal monoatomic gas is heated at a constant pressure of 1atm from 20∘C to 90∘C. Work done by the gas is(Gas constant,R=8.21Jmol−1K−1)
Held on 10 Jan 2019 · Verified 6 Jul 2026.
73J
581J
291J
146J
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Work through every JEE Main Thermodynamics PYQ, year by year.