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(D) The $r.m.s.$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}.$Since $R$ and $M$ are constant,$v_{rms} \propto \sqrt

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(D) The $r.m.s.$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}.$Since $R$ and $M$ are constant,$v_{rms} \propto \sqrt{T}.$Given initial temperature $T_1 = 0^\circ C = 273 \ K$ and initial velocity $(v_{rms})_1 = 0.5 \ km \ sec^{-1}.$Final temperature $T_2 = 819^\circ C = 819 + 273 = 1092 \ K.$Using the ratio formula: $\frac{(v_{rms})_1}{(v_{rms})_2} = \sqrt{\frac{T_1}{T_2}}.$Substituting the values: $\frac{0.5}{(v_{rms})_2} = \sqrt{\frac{273}{1092}}.$Since $\frac{1092}{273} = 4,$ we have $\frac{0.5}{(v_{rms})_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}.$Therefore,$(v_{rms})_2 = 0.5 	imes 2 = 1 \ km \ sec^{-1}.$

The $r.m.s.$ speed of the molecules of a gas at a pressure $10^5 \ Pa$ and temperature $0^\circ C$ is $0.5 \ km \ sec^{-1}.$ If the pressure is kept constant but temperature is raised to $819^\circ C,$ the velocity will become ........ $km \ sec^{-1}.$

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