At what temperature is the rms speed of air m | vedclass

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(A) The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Let $v_1$ be the $rms$ spee

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$819^{\circ} C$

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(A) The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Let $v_1$ be the $rms$ speed at $NTP$ $(T_1 = 273 \ K)$ and $v_2$ be the $rms$ speed at temperature $T_2$.Given that $v_2 = 2v_1$,we have:$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$2 = \sqrt{\frac{T_2}{273}}$Squaring both sides:$4 = \frac{T_2}{273}$$T_2 = 4 	imes 273 = 1092 \ K$Converting to Celsius:$T_2(^{\circ}C) = 1092 - 273 = 819^{\circ} C$.

At what temperature is the $rms$ speed of air molecules doubled compared to its value at $NTP$?

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