JEE Main 2023PhysicsKinetic Theory of GasesEasyMCQ

JEE Main 2023Kinetic Theory of Gases Question with Solution

JEE Main 2023 (12 Apr Shift 1)

Question

If the r.m.s speed of chlorine molecule is 490 m s-1 at 27oC, the r.m.s speed of argon molecules at the same temperature will be (Atomic mass of argon =39.9 u, molecular mass of chlorine =70.9 u)

Choose an option

Show full solutionCorrect option: B
Correct answer
B651.7 m s-1

Step-by-step explanation

The formula to calculate the rms speed of the argon molecules is given by

vAr=3RTMAr   ...1

The formula to calculate the rms speed of chlorine molecule is given by

vCl=3RTMCl   ...2

Divide equation (1) by equation (2) and simplify to obtain the required ratio.

vArvCl=MClMAr   ...3

Substitute the values of the known parameters into equation (3) and solve to calculate the required value.

vAr490 m s-1=7140vAr= 7140×490 m s-1651.7 m s-1

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Kinetic Theory of Gases chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2023, covering the Kinetic Theory of Gases 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.