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(D) For an ideal gas,the ideal gas equation is $PV = nRT$.Since $T$ and $V$ are constant,$n \propto P$.Let $n_X$ and $n_Y$ be the number of moles of g

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$2: 1$

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(D) For an ideal gas,the ideal gas equation is $PV = nRT$.Since $T$ and $V$ are constant,$n \propto P$.Let $n_X$ and $n_Y$ be the number of moles of gases $X$ and $Y$ respectively.For gas $X$: $n_X = \frac{10}{M_X} \propto 2 \ atm$  ---$(1)$When gas $Y$ is added,the total pressure becomes $6 \ atm$. The pressure due to gas $Y$ is $6 - 2 = 4 \ atm$.For gas $Y$: $n_Y = \frac{80}{M_Y} \propto 4 \ atm$  ---$(2)$Dividing equation $(1)$ by $(2)$:$\frac{10/M_X}{80/M_Y} = \frac{2}{4} \Rightarrow \frac{10}{M_X} 	imes \frac{M_Y}{80} = \frac{1}{2} \Rightarrow \frac{M_Y}{8M_X} = \frac{1}{2} \Rightarrow \frac{M_Y}{M_X} = 4$.The root mean square velocity is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$,so $V_{rms} \propto \frac{1}{\sqrt{M}}$.Therefore,$\frac{(V_{rms})_X}{(V_{rms})_Y} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{4} = 2$.The ratio is $2:1$.

$A$ closed vessel contains $10 \ g$ of an ideal gas $X$ at $300 \ K$,which exerts $2 \ atm$ pressure. At the same temperature,$80 \ g$ of another ideal gas $Y$ is added to it and the pressure becomes $6 \ atm$. The ratio of root mean square velocities of $X$ and $Y$ at $300 \ K$ is

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