The r.m.s speed of oxygen molecules at 47 ° C is equal to that of the hydrogen molecules kept at \ \ \ \ ° C . (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

Physics · Kinetic Theory of Gases

JEE Main 2026 — 21 January, Evening Shift — Question 29

The r.m.s speed of oxygen molecules at $47^{\circ} \mathrm{C}$ is equal to that of the hydrogen molecules kept at $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$ . (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

Option A:
$-235$
Option B:
$-100$
Option C:
$-253$
Correct
Option D:
$-20$

Answer: C

Step-by-step solution

$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{V}_{\mathrm{rmsO}_{2}}=\mathrm{V}_{\mathrm{rmsH}_{2}}$ $\mathrm{T}_{\mathrm{O}_{2}}=273+47=320 \mathrm{~K}$ $\sqrt{\frac{3 \mathrm{RT}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{RT}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ $\frac{\mathrm{T}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}=\frac{\mathrm{T}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}$ $\frac{320}{32}=\frac{\mathrm{T}_{\mathrm{H}_{2}}}{2}$ $\mathrm{T}_{\mathrm{H}_{2}}=20 \mathrm{~K}$ $\mathrm{T}_{\mathrm{H}_{2}}=-253^{\circ} \mathrm{C}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Pressure of Gas and Different speeds of Gas Molecules