A vessel of volume 20 L contains a mixture of hydrogen and helium at temperature of 27^oC and pressure2.0 atm. The mass of the mixture is 5 g. Assuming the gases to be ideal, the ratio of the mass of hydrogen to that of helium in the given mixture will be : 

2 : 5

Suppose there are n1 moles of hydrogen and n2 moles of helium in the given mixture.

Then the pressure of the mixture will be : P

\(P = \frac{n_1RT}{V} + \frac{n_2RT}{V} \)

\(P = (n_1 + n_2) \frac{RT}{V}\) ... (i)

2 x 101.3 x 10³ = \((n_1 + n_2) \frac{8.3*300}{20*10^{-3}}\)

\((n_1 + n_2)\) = \(\frac{2*101.3*10^3*20^10^{-3}}{8.3*300}\) = 1.62 ... (ii)

The mass of the mixture is (in g) : \(n_1*2 + n_2*4 = 5\)

\(\Rightarrow n_1 + 2n_2 = 2.5\)

Solving eq (i) and (ii) : n₁ = 0.74 ; n₂ = 0.88

\(\Rightarrow \frac{m_{H}}{m_{He}} = \frac{0.74*2}{0.88*4} \)

\(\Rightarrow \frac{m_{H}}{m_{He}} = \frac{1.48}{3.52}\) = 2 : 5