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JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Kinetic Theory of Gases

JEE Main · Physics · Kinetic Theory of Gases

The temperature at which a molecule of nitrogen will have the same...

The temperature at which a molecule of nitrogen will have the same rms velocity as a molecule of oxygen at

127 ° C

is

Options

A $457℃$
B $273℃$
C $350℃$
D $77℃$

Correct answer

D. $77℃$

Step-by-step solution

Root mean square velocity

{v}_{rms} = \sqrt{\dfrac{3RT}{M}}

where

R

is gas constant,

T

the temperature and

M

molecular weight.

Given,

{MN}_{2} = 28, {MO}_{2} = 32, {TO}_{2} = 127℃ = 127 + 273 = 400

K

∴ \dfrac{{{v}_{O}}_{2}}{{{v}_{N}}_{2}} = \sqrt{\dfrac{{{T}_{O}}_{2}}{{{M}_{O}}_{2}} \times \dfrac{{{M}_{N}}_{2}}{{{T}_{N}}_{2}}} = \sqrt{\dfrac{400}{32} \times \dfrac{28}{{{T}_{N}}_{2}}} = 1

⇒ {TN}_{2} = 350K = 77℃

.

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