JEE Main 2023PhysicsThermodynamicsMediumMCQ

JEE Main 2023 — Thermodynamics Question with Solution

JEE Main 2023 (31 Jan Shift 2)

Question

Heat energy of 735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:

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Show full solutionCorrect option: A
Correct answer
A525 J

Step-by-step explanation

For isobaric process and diatomic gas we can write, ΔQ=nCPΔT=735 J

⇒ n7R2ΔT=735 J⇒nRΔT=210

Now internal energy will be,

ΔU=nCVΔT=n5R2ΔT=52×210

=525 J

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About this question

This is a previous-year question from JEE Main 2023, 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.