Initially, the root mean square (rms) velocity of \(\mathrm{N_2}\) molecules at certain temperature is \(u\). If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new rms velocity will be :
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- \(u/2\)
- \(2u\)
- \(4u\)
- \(14u\)
The Correct Option is B
Solution and Explanation
After dissociation \(\mathrm{N_2} \rightarrow 2\mathrm{N}\) the molar mass becomes 14 and \(T' = 2T\).
\(u' = \sqrt{\frac{3R(2T)}{14}} = \sqrt{\frac{3RT}{28}\times4} = 2u\).
Answer: (2) \(2u\)
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