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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2020
  • Medium
  • Numerical

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........ (Molar mass of \(N _{2}\) gas \(28\, g\) )

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Answer & Solution

Correct Answer

(B) \(41\)

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Detailed explanation

\(V _{ rms }=\sqrt{\frac{3 RT }{ M }}\)

\(V _{ N _{2}}= V _{ H _{2}}\)

\(\sqrt{\frac{3 RT _{ N _{2}}}{ M _{ N _{2}}}}=\sqrt{\frac{3 RT _{ H _{2}}}{ M _{ H _{2}}}}\)

\(\frac{573}{28}=\frac{ T _{ H _{2}}}{2} \Rightarrow T _{ H _{2}}=40.928\)

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