2 moles of He and 6 moles of $O_{2}$ exert a pressure of 16 atm in a vessel. If a small hole is made, then the ratio of He to $O_{2}$ . Effusing out initially is $2 \sqrt{2} : n$ . Find the value of n.

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answer is 3.

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Detailed Solution

$\begin{array}{r} P = 16 atm \\ P_{He} = \frac{2}{2 + 6} \times 16 = 4 atm \end{array}$

He to $O_{2}$ are at different pressures

$P_{o_{2}} = 16 - 4 = 12 atm$

$\begin{array}{r} \frac{r_{He}}{r_{O_{2}}} = \frac{P_{He}}{P_{o 2}} \sqrt{\frac{M_{O_{2}}}{M_{He}}} = \frac{4}{12} \sqrt{\frac{32}{4}} = 2 \sqrt{2} / 3 \\ n = 3 . \end{array}$

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