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At what temperature will the rms speed of air molecules be double that NTP?
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The correct answer is 1092K or 819°C 

Explanation :: 

We have \(V_{rms}={\sqrt{3RT\over M}}\)  --at T=\(T_0 (NTP)\) 

\(V_{rms}={\sqrt{3RT_0\over M}}\) 

But at temperature T 

\(V_{rms}={2\sqrt{3RT\over M}}\) 

 

\(\therefore \sqrt{3RT\over M}=2{\sqrt{3RT_0\over M}}\) 

\(\sqrt T = \sqrt{4T_0}\) 

T= 4× 273 k 

T=1092 K  

Or 

T=819°C

 

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Thanks the answer is correct varified by team Brainiak

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