At T(K), in A 10 L vessel equal masses of $O_{2}, C H_{4}$ and $S O_{2}$ are present. If P is the total pressure of the mixture in bar, the partial pressure of $S O_{2}$ (inn bar) is

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$\frac{2 P}{7}$

$\frac{P}{7}$

$\frac{4 P}{7}$

$\frac{9 P}{7}$

answer is A.

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Detailed Solution

$p_{{SO}_{2}} = \frac{n_{{SO}_{2}}}{n_{T}} x P_{T} p_{{SO}_{2}} = \frac{x / 64}{x / 32 + x / 16 + x / 64} x P_{T} p_{{SO}_{2}} = \frac{x / 64}{\frac{2 x + 4 x + x}{64}} x P_{T} p_{{SO}_{2}} = \frac{x}{7 x} x P_{T} p_{{SO}_{2}} = \frac{P_{T}}{7}$

Where x is the mass of each gas.

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