Thermodynamics

JEE Main level, single correct, medium

148 of 190

Question 148 of 190

Half mole of an ideal monoatomic gas is heated at a constant pressure of 1atm1 atm from 20∘C20^{\circ}C to 90∘C90^{\circ}C. Work done by the gas is((Gas constant,R=8.21Jmol−1K−1R=8.21J{mol}^{-1}{K}^{-1})

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Answer: (C) 291J291 J

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−PV1W=P{V}_{2}-P{V}_{1}, now take the idea of ideal gas equation,
W=nRT2−nRT1W=nR{T}_{2}-nR{T}_{1}=12×8.31(70)=\dfrac{1}{2}\times 8.31\left(70\right)=291J=291J.

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