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(D) The formula for the r.m.s. speed of gas molecules is $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$.Given the initial temperature $T_1 = 100^{\circ} C =

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(D) The formula for the r.m.s. speed of gas molecules is $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$.Given the initial temperature $T_1 = 100^{\circ} C = 373 	ext{ K}$.Thus, $v = \sqrt{\frac{3R(373)}{M}}$.We want to find the temperature $T_2$ such that the new speed $v' = \sqrt{3}v$.Since $v_{	ext{rms}} \propto \sqrt{T}$, we have $\frac{v'}{v} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values, $\sqrt{3} = \sqrt{\frac{T_2}{373}}$.Squaring both sides, $3 = \frac{T_2}{373}$.$T_2 = 3 	imes 373 = 1119 	ext{ K}$.Converting back to Celsius, $T_2 = 1119 - 273 = 846^{\circ} C$.

The r.m.s. speed of the molecules of a gas at $100^{\circ} C$ is $v$. The temperature at which the r.m.s. speed will be $\sqrt{3} v$ is: (in $^{\circ} C$)

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