Chemistry › States of Matter › Mixture of gases
Equal masses of $\mathrm{H}_{2(\mathrm{~g})}$ and $\mathrm{He}_{(\mathrm{g})}$ are enclosed in a container…
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
$1: 1$ $1: 2$ $2: 1$ $1: 4$ Solution
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,
$\begin{gathered}
\mathrm{P}_{\mathrm{H}_2}=x_{\mathrm{H}_2} \times \mathrm{P}_{\text {Tótal }}=\frac{2}{3} \times \mathrm{P} \\
\mathrm{P}_{\mathrm{He}}=x_{\mathrm{He}} \times \mathrm{P}_{\text {Total }}=\frac{1}{3} \times \mathrm{P} \\
\mathrm{P}_{\mathrm{H}_2}: \mathrm{P}_{\mathrm{H}_{\mathrm{e}}}=\frac{2 / 3 \mathrm{P}}{1 / 3 \mathrm{P}}=2: 1
\end{gathered}$
Asked in: MHT CET 2023 (12 May Shift 2)
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