At what temperature rms speed of air molecules is doubled of that at NTP?

819^{o} C

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719^{o} C

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909^{o} C

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None of these

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Solution

The correct option is A $819^{o} C$
$v_{r m s} \propto \sqrt{T}$
$\Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{T_{1}}{T_{2}}$
$\Rightarrow \frac{v^{2}}{4 v^{2}} = \frac{273}{T_{2}}$
$\Rightarrow T_{2} = 1092 K$
$= 819^{o} C$

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