A vessel of volume 20L contains a mixture of hydrogen and helium at...

Let there are $\mathrm{n}_{1}$ moles of hydrogen and $\mathrm{n}_{2}$ moles of helium in the given mixture. As $\mathrm{Pv}=\mathrm{nRT}$ Then the pressure of the mixture $$\begin{array}{r} P=\frac{\mathrm{n}_{1} \mathrm{RT}}{\mathrm{V}}+\frac{\mathrm{n}_{2} \mathrm{RT}}{\mathrm{V}}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \frac{\mathrm{RT}}{\mathrm{V}}\\ 2 \times101.3 \times10^{3}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \times \frac{(8.3 \times 300)}{20 \times 10^{-3}} \end{array}$$ or, $$\begin{array}{l}\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)=\frac{2 \times 101.3 \times 10^{3} \times 20 \times 10^{-3}}{(8.3)(300)}\\ \text{ or, } \mathrm{n}_{1}+\mathrm{n}_{2}=1.62 \end{array}$$The mass of the mixture is (in grams) $$\begin{array}{l}\mathrm{n}_{1} \times₂⁺\mathrm{n}_{2} \times4=5 \\ \Rightarrow \quad\left(\mathrm{n}_{1}+2 \mathrm{n}_{2}\right)=2.5 \quad \ldots \end{array}$$ Solving the eqns. (1) and (2), we get \mathrm{n}_{1}=0.74$and$\mathrm{n}_{2}=0.88$Hence,$\frac{\mathrm{m}_{\mathrm{H}}}{\mathrm{m}_{\mathrm{He}}}=\frac{0.74 \times 2}{0.88 \times 4}=\frac{1.48}{3.52}=\frac{2}{5}$