JEE Main 2019PhysicsThermodynamicsEasyMCQ

JEE Main 2019Thermodynamics Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

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=8.21 J mol-1 K-1)

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Show full solutionCorrect option: C
Correct answer
C291J

Step-by-step explanation

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=12×8.3170=291 J

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About this question

This is a previous-year question from JEE Main 2019, 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.