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The dissociation equilibrium of a gas ${AB}_{2}$ can be represented as
$2 {AB}_{2} (g) ⇌ 2 AB (g) + B_{2} (g)$
The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant $K_{P}$ and total pressure P is

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${(2 K_{P} / P)}^{1 / 2}$

${(2 K_{P} / P)}^{1 / 3}$

$(K_{P} / P)$

$(2 K_{P} / P)$

answer is A.

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Detailed Solution

$2 {AB}_{2} (g) ⇌ 2 AB (g) + B_{2} (g)$

1 0 0

1 - x x x/2

$∴ K_{P} = \frac{x^{2} . x}{2 {(1 - x)}^{2}} . [\frac{P}{1 + \frac{x}{2}}] = \frac{x^{3} . P}{2}$

$(1 - x \approx 1 and 1 + \frac{x}{2} \approx 1, since x < < 1)$

Or $P = \sqrt[3]{\frac{2 K_{P}}{P}}$

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