JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
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 \(28\, g\) )
- A \(45\)
- B \(41\)
- C \(38\)
- D \(49\)
Answer & Solution
Correct Answer
(B) \(41\)
Step-by-step Solution
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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