A gas is at a temperature of 0^{circ}C. To wh | vedclass

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(C) The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\f

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$819$

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(C) The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Given $v_2 = 2v_1$,so $\frac{2v_1}{v_1} = \sqrt{\frac{T_2}{T_1}}$,which implies $2 = \sqrt{\frac{T_2}{T_1}}$.Squaring both sides,$4 = \frac{T_2}{T_1}$,so $T_2 = 4T_1$.The initial temperature $T_1 = 0^{\circ}C = 273 \ K$.Therefore,$T_2 = 4 	imes 273 \ K = 1092 \ K$.To convert back to Celsius: $t_2 = 1092 - 273 = 819^{\circ}C$.

$A$ gas is at a temperature of $0^{\circ}C$. To what temperature in $^{\circ}C$ must the gas be heated so that the $rms$ speed of its molecules becomes double?

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