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A vessel of volume V = 20 litre contains a mixture of $H_{2} a n d$ He at 20^oC and pressure P = 2 atm. The mass of the mixture is m = 5 g. The ratio of m₁ and m₂ (m₁ : m₂) where m₁ = mass of H₂ and m₂ = mass of He is ____________.

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Detailed Solution

By gas law we use $PV = nRT$

$\begin{array}{l} 2 \times 20 = (\frac{m_{1}}{2} + \frac{m_{2}}{4}) (0 .082) (293) \\ \Rightarrow 2 m_{1} + m_{2} = 6 .66 and m_{1} + m_{2} = 5 \end{array}$

Solving we get $m_{1} = 1 .66 g and m_{2} = 3 .34 g$

$\Rightarrow \frac{m_{1}}{m_{2}} = \frac{1}{2}$

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