A closed vessel contains $10 \mathrm{~g}$ of an ideal gas $\mathbf{X}$ at $300 \mathrm{~K}$, which exerts 2…
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
$2 \sqrt{2}: \sqrt{3}$
$2 \sqrt{2}: 1$
$1: 2$
$2: 1$
Solution
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)
$\because \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}}}$
$\therefore \quad \frac{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{X}}}{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{Y}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{x}}}}=\sqrt{\frac{4}{1}}=\frac{2}{1}$
!