JEE Main 2020PhysicsThermodynamicsEasyMCQ

JEE Main 2020Thermodynamics Question with Solution

JEE Main 2020 (05 Sep Shift 1)

Question

Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are labelled as AB,AC and AD. The change in internal energies during these process are taken as EAB,EAC and EAD and the work done as WAB,WAC and WAD. The correct relation between these parameters are:

Choose an option

Show full solutionCorrect option: B
Correct answer
BEAB=EAC=EAD,WAB>0,WAC=0,WAD<0

Step-by-step explanation

EAB=EAC=EAD

dU=nfR2 Tf-Ti

WAB>0(+) as V

WAC=0 as V=constant

WAD<0(-) as V

ΔT is same for EAB=EAC=EAD
 

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About this question

This is a previous-year question from JEE Main 2020, 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.