The temperature of a gas is -68^{circ} C. To | vedclass

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(C) The initial temperature $T_1 = -68^{\circ} C = -68 + 273 = 205 \ K$.The root mean square (r.m.s.) velocity of gas molecules is given by $v_{rms} =

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$547$

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(C) The initial temperature $T_1 = -68^{\circ} C = -68 + 273 = 205 \ K$.The root mean square (r.m.s.) velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Let the initial velocity be $(v_{rms})_1$ and the final velocity be $(v_{rms})_2 = 2(v_{rms})_1$.Using the ratio: $\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $2 = \sqrt{\frac{T_2}{205}}$.Squaring both sides: $4 = \frac{T_2}{205}$.Therefore,$T_2 = 4 	imes 205 = 820 \ K$.Converting back to Celsius: $T_2 = 820 - 273 = 547^{\circ} C$.

The temperature of a gas is $-68^{\circ} C$. To what temperature should it be heated,so that the r.m.s. velocity of the molecules be doubled (in $^{\circ} C$)?

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