The temperature of a gas is -80^{circ} C. To | vedclass

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(C) Initial temperature $T_1 = -80^{\circ} C = -80 + 273 = 193 \ K$.The root mean square (r.m.s.) speed of a gas is given by $v_{rms} = \sqrt{\frac{3R

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$1464$

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(C) Initial temperature $T_1 = -80^{\circ} C = -80 + 273 = 193 \ K$.The root mean square (r.m.s.) speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Let the initial speed be $v_1$ and the final speed be $v_2$. We are given that the speed is increased by $2$ times,meaning $v_2 = v_1 + 2v_1 = 3v_1$.Using the relation $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$,we have $\frac{3v_1}{v_1} = \sqrt{\frac{T_2}{T_1}}$.$3 = \sqrt{\frac{T_2}{193}} \implies 9 = \frac{T_2}{193}$.$T_2 = 9 	imes 193 = 1737 \ K$.Converting to Celsius: $T_2 = 1737 - 273 = 1464^{\circ} C$.

The temperature of a gas is $-80^{\circ} C$. To what temperature should the gas be heated so that the r.m.s. speed is increased by $2$ times (in $^{\circ} C$)?

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