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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