Let there are \(n_1\) moles of hydrogen and \(n_2\) moles of helium in the given mixture. As \(Pv = nRT\)

Then the pressure of the mixture

\(P =\frac{ n _1 RT }{ V }+\frac{ n _2 RT }{ V }=\left( n _1+ n _2\right) \frac{ RT }{ V }\)

\(\Rightarrow 2 \times 101.3 \times 10^3=\left( n _1+ n _2\right) \times \frac{(8.3 \times 300)}{20 \times 10^{-3}}\)

\(\text { or, } \left( n _1+ n _2\right)=\frac{2 \times 101.3 \times 10^3 \times 20 \times 10^{-3}}{(8.3)(300)}\)

\(\text { or, } \quad n _1+ n _2=1.62\)

The mass of the mixture is (in grams) \(n _1 \times 2+ n _2 \times 4=5\)

\(\Rightarrow \quad\left(n_1+2 n_2\right)=2.5 \quad \ldots(2)\)

Solving the eqns. (1) and (2), we get \(n _1=0.74\) and \(n _2=0.88\)

Hence, \(\frac{ m _{ H }}{ m _{ He }}=\frac{0.74 \times 2}{0.88 \times 4}=\frac{1.48}{3.52}=\frac{2}{5}\)