JEE Main 2022PhysicsThermodynamicsEasyMCQ

JEE Main 2022 — Thermodynamics Question with Solution

JEE Main 2022 (25 Jul Shift 1)

Question

A certain amount of gas of volume V at 27°C temperature and pressure 2×107N m-2 expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use γ=1.5)

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Show full solutionCorrect option: B
Correct answer
B3.536×106 Pa

Step-by-step explanation

Given here, P1=2×107 N m-2=2×107 Pa

Using Boyle's law,

P1V1=P2V2

Since V2=2V1.

Hence, P2=P12 (isothermal expansion)

P2=1×107 Pa

Now, during adiabatic expansion, 

P2V2γ=P32V2γ

P3=1×10721.5=3.536×106 Pa

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About this question

This is a previous-year question from JEE Main 2022, 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.