The temperature of a gas is -50^{circ}C. To w | vedclass

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(C) The $rms$ speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Let the initial temperature be $T_1

Vedclass · Accepted Answer

$3295^{\circ}C$

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(C) The $rms$ speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Let the initial temperature be $T_1 = -50 + 273 = 223\,K$ and the initial speed be $v_1 = v$.The speed is increased by $3$ times,so the final speed $v_2 = v + 3v = 4v$.Using the relation $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$,we have $\frac{v}{4v} = \sqrt{\frac{223}{T_2}}$.Squaring both sides,$\frac{1}{16} = \frac{223}{T_2}$.Thus,$T_2 = 223 	imes 16 = 3568\,K$.Converting to Celsius,$T_2 = 3568 - 273 = 3295^{\circ}C$.

The temperature of a gas is $-50^{\circ}C$. To what temperature should the gas be heated so that the $rms$ speed is increased by $3$ times?

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