The relation between root mean square speed ( v _ rms ) and most probable speed ( v _ p ) for the molar mass M of oxygen gas molecule at the temperature of 300K will be

Correct answer: C. v _ rms = (3)/(2) v _ p RMS speed of O _ 2 gas is v _ rms = (3RT)/(M) = (3RT)/(32) and most probable speed of O _ 2 gas is v _ mostprobable = (2RT)/(M) = (2RT)/(32) . Here temperature is same. Thus, v _ rms v _ mostprobable = (3RT)/(32) (2RT)/(32) = (3)/(2) Hence, v _ rms = (3)/(2) v _ p

JEE Main 202225 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 25 June
Shift: Morning Shift
Subject: Physics
Chapter: Kinetic Theory of Gases

JEE Main 2022 · 25 June · Morning Shift · Physics · Kinetic Theory of Gases

The relation between root mean square speed ( v _ rms ) and most...

The relation between root mean square speed

\left({v}_{rms}\right)

and most probable speed

\left({v}_{p}\right)

for the molar mass

M

of oxygen gas molecule at the temperature of

300K

will be

Options

A ${v}_{rms} = \sqrt{\dfrac{2}{3}}{v}_{p}$
B ${v}_{rms} = \sqrt{\dfrac{1}{3}}{v}_{p}$
C ${v}_{rms} = \sqrt{\dfrac{3}{2}}{v}_{p}$
D ${v}_{rms} = {v}_{p}$

Correct answer

C. ${v}_{rms} = \sqrt{\dfrac{3}{2}}{v}_{p}$

Step-by-step solution

RMS speed of ${O}_{2}$ gas is ${v}_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3RT}{32}}$ and most probable speed of ${O}_{2}$ gas is ${v}_{mostprobable} = \sqrt{\dfrac{2RT}{M}} = \sqrt{\dfrac{2RT}{32}}$ .

Here temperature is same.

Thus, $\dfrac{{v}_{rms}}{{v}_{mostprobable}} = \dfrac{\sqrt{\dfrac{3RT}{32}}}{\sqrt{\dfrac{2RT}{32}}} = \sqrt{\dfrac{3}{2}}$

Hence, ${v}_{rms} = \sqrt{\dfrac{3}{2}}{v}_{p}$

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