A gaseous reaction,
$3 A ⟶ 2 B$
is carried out in a 0.0821 litre closed container initially containing 1 mole of gas A. After sufficient time a curve of P (atm) vs T (K) is plotted and the angle with x-axis was found to be 42.95°. The degree of association of gas A is [Given : tan 42.95 = 0.8]

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0.6

0.8

0.5

0.4

answer is B.

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Detailed Solution

$\underset{1 - 3 x}{3 A} \to \underset{2 x}{2 B} ∴ Total moles = 1 - 3 x + 2 x = 1 - x PV = nRT P \times 0.0821 = 1 - x \times 0.0821 \times 7$

We know slope = tan(angle between P–T curve) = tan(42.95º) = 0.8

$\Rightarrow 1 - x = 0.8 x = 0.2 % conversion = \frac{n_{A (initial)} - n_{A (final)}}{n_{A (initial)}} = \frac{1 - (1 - 3 \times 0.2)}{1} = 0.6$

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