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Graham's Law of Effusion

Equal mass of...

Question

Equal mass of $S O_{2}$ and $C H_{4}$ are introduced in two different vessels - $S O_{2}$ in a 4 litre vessel and $C H_{4}$ in a 2 litre vessel at the same temperature. Both vessels have identical pin holes. What is the ratio of the rates of diffusion of $S O_{2}$ and $C H_{4}$ , if the rate is calculated in $m o l / s$ ?

A

1:2

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B

1:8

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C

1:16

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D

1:4

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Solution

The correct option is C 1:16
$\frac{r a t e_{S O_{2}}}{r a t e_{C H_{4}}} = \frac{P_{S O_{2}}}{P_{C H_{4}}} \times \sqrt{\frac{M_{C H_{4}}}{M_{S O_{2}}}}$
$P = \frac{w R T}{M_{w} V}$ ,
Since the masses are the same,
$\frac{r a t e_{S O_{2}}}{r a t e_{C H_{4}}} = \frac{(M V)_{C H_{4}}}{(M V)_{S O_{2}}} \times \sqrt{\frac{M_{C H_{4}}}{M_{S O_{2}}}} = (\frac{16}{64})^{\frac{3}{2}} \times \frac{2}{4} = \frac{1}{16}$

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Similar questions

Q. Equal mass of

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