JEE Main 2019PhysicsThermodynamicsMediumMCQ

JEE Main 2019 — Thermodynamics Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is -180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc . The work done by the gas along the path abc is:

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Show full solutionCorrect option: A
Correct answer
A130 J

Step-by-step explanation

In a cyclic process, ΔU=0
From 1st law of thermodynamics
Q=ΔU+W
Q=W
Qab+Qbc+Qca=Wabc+Wca
Wabc=Qab+Qbc+Qca-Wca
Wabc=Qab+Qbc+Uca
Wabc=250+60-180=130 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.