At room temperature following reaction goes to completion
$2 AB (g) + B_{2} (g) \to 2 {AB}_{2} (s)$
$A B_{2}$ is solid with negligible vapour pressure below $0^{\circ} C.$ at $300 K,$ the $A B$ in the smaller flask exerts a pressure of $3 a t m$ and in the larger flask $B_{2}$ exerts a pressure of 1" " atm at $400 K$ when they are separated out by a close valve. The gases are mixed by opening the stop cock and after the end of the reaction the flask are cooled to $250 K$
The final pressure is:

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$3.2 atm$

$0.156 atm$

$0.3125 atm$

0.625 atm

answer is C.

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Detailed Solution

$n_{AB} = \frac{3 \times 8.21}{0.0821 \times 300} = 1, n_{B_{2}} = \frac{1 \times 8.21}{0.0821 \times 400} = 0.25$

$Initial moles 2 AB + B_{2} \overset{}{\to} 2 {AB}_{2}$

$After reaction 0.5 0.25$

$P_{AB} (8.21 \times 2) = 0.5 \times 0.0821 \times 250 \Rightarrow P_{AB} = 0.625 atm$

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