Kinetic Theory

JEE Main level, numerical answer, medium

55 of 168

Question 55 of 168

Nitrogen gas is at 300∘C300^{\circ}C temperature. The temperature (in KK ) at which the rms speed of a H2{H}_{2} molecule would be equal to the rms speed of a nitrogen molecule, is ……….\ldots \ldots \ldots . (Molar mass of N2{N}_{2} gas 28g28g ).

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Answer: 41

Given,Temperature of Nitrogen gas is
T1=300∘C=300+273=573K{T}_{1}=300^{\circ}C=300+273=573K
Molar mass of H2{H}_{2} gas is MH2=2g{M}_{{H}_{2}}=2g
Molar mass of N2{N}_{2} gas is MN2=28g{M}_{{N}_{2}}=28g
The rms speed of gas is given by
Vrms=3RTM{V}_{rms}=\sqrt{\dfrac{3RT}{M}}
Now rms speed of hydrogen gas is equal to rms speed of nitrogen gas, i.e.; (Vrms)H2=(Vrms)N2{\left({V}_{rms}\right)}_{{H}_{2}}={\left({V}_{rms}\right)}_{{N}_{2}}
⇒TH2TN2=MH2MN2\Rightarrow \dfrac{{T}_{{H}_{2}}}{{T}_{{N}_{2}}}=\dfrac{{M}_{{H}_{2}}}{{M}_{{N}_{2}}}
Substitute the given values i above equation, we get,
TH2=228×573{T}_{{H}_{2}}=\dfrac{2}{28}\times 573
TH2=41K{T}_{{H}_{2}}=41K

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