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

For Ideal Gas
\(\mathrm{PV}=\mathrm{nRT}\)
\(\therefore \mathrm{n} \propto \mathrm{P}\) at constant \(\mathrm{T} \& \mathrm{~V}\).
\(\because \quad\) mole \(=\frac{\text { Mass }}{\text { Molar mass }}\)
For gas \(\mathrm{X}: \frac{10}{\mathrm{M}_{\mathrm{X}}} \propto 2 \mathrm{~atm}\) \(\quad\) \(\ldots\)(1)
For gas \(\mathrm{X} \& \mathrm{Y}: \frac{10}{\mathrm{M}_{\mathrm{X}}}+\frac{80}{\mathrm{M}_{\mathrm{Y}}} \propto 6 \mathrm{~atm}\) \(\ldots\)(2)
From \((2)-(1)\)
\(\frac{80}{\mathrm{M}_{\mathrm{y}}} \propto 4\) \(\ldots\)(3)
On dividing (1) by (3)
\(\frac{\mathrm{M}_{\mathrm{Y}}}{8 \mathrm{M}_{\mathrm{X}}}=\frac{1}{2}\)
\(\therefore \quad \frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{X}}}=4\) \(\ldots\)(4)
\(\because \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}}}\)
\(\therefore \quad \frac{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{X}}}{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{Y}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{x}}}}=\sqrt{\frac{4}{1}}=\frac{2}{1}\)