0.5 mole of each of $H, S O_{2}$ and $C H_{4}$ , are kept in a container. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be:

P (C H_{4}) > P (S O_{2}) > P (H_{2})

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P (H_{2}) > P (C H_{4}) > P (S O_{2})

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P (S O_{2}) > P (C H_{4}) > P (H_{2})

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P (H_{2}) > P (S O_{2}) > P (C H_{4})

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Solution

The correct option is D $P (S O_{2}) > P (C H_{4}) > P (H_{2})$
$S O_{2}$ Molar mass= $32 + 32 = 64 g / m o l$
$C H_{4}$ Molar mass= $12 + 4 = 16 g / m o l$
$H_{2}$ Molar mass= $2 \times 1 = 2 g / m o l$
Rate of diffusion is inversely related to molecular weight. Lighter the gas, more is the diffusion. The rate of diffusion is given by,
$r_{H_{2}} > r_{C H_{4}} > r_{S O_{2}}$
Hence the order of partial pressure will be
$P_{S O_{2}} > P_{C H_{4}} > P_{H_{2}}$

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