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(C) The root mean square speed of a gas molecule is given by the formula $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas const

Vedclass · Accepted Answer

$\frac{\sqrt{7}}{2}$

Vedclass · Answer

(C) The root mean square speed of a gas molecule is given by the formula $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.Since the temperature $T$ is the same for both gases in the mixture, the ratio of their root mean square speeds depends only on their molar masses.The ratio is given by $\frac{v_{	ext{rms}}^{	ext{Ar}}}{v_{	ext{rms}}^{	ext{Cl}}} = \frac{\sqrt{3RT/M_{	ext{Ar}}}}{\sqrt{3RT/M_{	ext{Cl}}}} = \sqrt{\frac{M_{	ext{Cl}}}{M_{	ext{Ar}}}}$.Given the atomic mass of argon $M_{	ext{Ar}} = 40.0 	ext{u}$ and the molecular mass of chlorine $M_{	ext{Cl}} = 70.0 	ext{u}$.Substituting these values, we get $\frac{v_{	ext{rms}}^{	ext{Ar}}}{v_{	ext{rms}}^{	ext{Cl}}} = \sqrt{\frac{70}{40}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$.Therefore, the correct option is $C$.

Column-$I$	Column-$II$
$(a)$ ${v_{rms}} = \sqrt {\frac{3P}{\rho}}$	$(i)$ For $1 \text{ mole ideal gas}$
$(b)$ ${v_{rms}} = \sqrt {\frac{3RT}{M_0}}$	$(ii)$ For one molecule of gas
$(c)$ ${v_{rms}} = \sqrt {\frac{3{k_B}T}{m}}$	$(iii)$ On the basis of kinetic theory

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