JEE Main 2021PhysicsThermal Properties of MatterMediumNumerical

JEE Main 2021Thermal Properties of Matter Question with Solution

JEE Main 2021 (26 Feb Shift 1)

Question

A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x×10-1 atm. Value of x (nearest integer) is              

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Show full solutionCorrect answer: 26
Correct answer
26

Step-by-step explanation

Let common equilibrium pressure of mixture is P atmp. then

f2P1 V1+f2P2 V2=f2PV1+V2

f224.5+f235.5=f2P4.5+5.5

P=2.55=x×10-1 atmp

So x=25.526 (Nearest integer)

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About this question

This is a previous-year question from JEE Main 2021, covering the Thermal Properties of Matter 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.