JEE Main 2022ChemistryStates of MatterMediumNumerical

JEE Main 2022 — States of Matter Question with Solution

JEE Main 2022 (27 Jul Shift 2)

Question

for a real gas at 25 °C temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Van 1der Waal's constant 'b' should be_____×10-2 L mol-1 (Given R=0.083 L bar K-1 mol-1)

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Show full solutionCorrect answer: 25
Correct answer
25

Step-by-step explanation

For real gas under high pressure we can neglect pressure corrections due to intermolecular forces. But since volume will be low, we can not neglect volume corrections due to molecular size.

Z=1+PbRT = 2 ⇒b=RTP

=0.083×29899

=0.25×10-2 L mol-1

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About this question

This is a previous-year question from JEE Main 2022, covering the States of Matter chapter of Chemistry. 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.