Nitrogen gas is at 300°C… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Thermodynamics

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Nitrogen gas is at $300^{\circ}C$ temperature. The temperature (in $K$ ) at which the rms speed of a ${H}_{2}$ molecule would be equal to the rms speed of a nitrogen molecule, is $\ldots \ldots \ldots .$ (Molar mass of ${N}_{2}$ gas $28g$ ).

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

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Solution

Given,

Temperature of Nitrogen gas is

${T}_{1}=300^{\circ}C=300+273=573K$

Molar mass of ${H}_{2}$ gas is ${M}_{{H}_{2}}=2g$

Molar mass of ${N}_{2}$ gas is ${M}_{{N}_{2}}=28g$

The rms speed of gas is given by

${V}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{M}}$

Now rms speed of hydrogen gas is equal to rms speed of nitrogen gas, i.e.; ${({V}_{rms})}_{{H}_{2}}={({V}_{rms})}_{{N}_{2}}$

$\Rightarrow \frac{{T}_{{H}_{2}}}{{T}_{{N}_{2}}}=\frac{{M}_{{H}_{2}}}{{M}_{{N}_{2}}}$

Substitute the given values i above equation, we get,

${T}_{{H}_{2}}=\frac{2}{28}\times 573$

${T}_{{H}_{2}}=41K$

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Solution

Given,

Temperature of Nitrogen gas is

${T}_{1}=300^{\circ}C=300+273=573K$

Molar mass of ${H}_{2}$ gas is ${M}_{{H}_{2}}=2g$

Molar mass of ${N}_{2}$ gas is ${M}_{{N}_{2}}=28g$

The rms speed of gas is given by

${V}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{M}}$

Now rms speed of hydrogen gas is equal to rms speed of nitrogen gas, i.e.; ${({V}_{rms})}_{{H}_{2}}={({V}_{rms})}_{{N}_{2}}$

$\Rightarrow \frac{{T}_{{H}_{2}}}{{T}_{{N}_{2}}}=\frac{{M}_{{H}_{2}}}{{M}_{{N}_{2}}}$

Substitute the given values i above equation, we get,

${T}_{{H}_{2}}=\frac{2}{28}\times 573$

${T}_{{H}_{2}}=41K$

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Solution

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Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics