Equal masses of \(\mathrm{H}_{2(\mathrm{~g})}\) and \(\mathrm{He}_{(\mathrm{g})}\) are enclosed in a container at constant temperature. The ratio of partial pressure of \(\mathrm{H}_2\) to \(\mathrm{He}\) is

Let the mass be ' \(x\) '.
\(
\mathrm{n}_{\mathrm{H}_2}=\frac{\mathrm{xg}}{2 \mathrm{~g} \mathrm{~mol}^{-1}}=\frac{\mathrm{x}}{2}
\)
\(
\begin{aligned}
& \mathrm{n}_{\mathrm{He}}=\frac{\mathrm{xg}}{4 \mathrm{~g} \mathrm{~mol}^{-1}}=\frac{\mathrm{x}}{4} \\
& \mathrm{n}_{\text {Total }}=\frac{3 \mathrm{x}}{4}
\end{aligned}
\)
Now,
\(
\begin{aligned}
& x_{\mathrm{H}_2}=\frac{\mathrm{n}}{\mathrm{n}_{\text {Total }}}=\frac{\mathrm{x} / 2}{3 \mathrm{x} / 4}=\frac{2}{3} \\
& x_{\mathrm{H}_e}=\frac{\mathrm{n}}{\mathrm{n}_{\text {Total }}}=\frac{\mathrm{x} / 4}{3 \mathrm{x} / 4}=\frac{1}{3}
\end{aligned}
\)
Now,
\(
\begin{gathered}
\mathrm{P}_{\mathrm{H}_2}=x_{\mathrm{H}_2} \times \mathrm{P}_{\text {Tótal }}=\frac{2}{3} \times \mathrm{P} \\
\mathrm{P}_{\mathrm{He}}=x_{\mathrm{He}} \times \mathrm{P}_{\text {Total }}=\frac{1}{3} \times \mathrm{P} \\
\mathrm{P}_{\mathrm{H}_2}: \mathrm{P}_{\mathrm{H}_{\mathrm{e}}}=\frac{2 / 3 \mathrm{P}}{1 / 3 \mathrm{P}}=2: 1
\end{gathered}
\)