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A solid A dissociates to give B and C gases as shown,
$A (g) ⇌ 2 B (g) + 3 C (g) .$
At equilibrium some B(g) is introduced keeping volume constant so that pressure of B at n equilibrium become equal to $\frac{8}{5 \sqrt{2}}$ times original total pressure calculate ratio of initial equilibrium pressure of C to that of its final pressure.

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1 : 2

2 : 1

8 : $\sqrt{2}$

$\sqrt{2} : 8$

answer is B.

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Detailed Solution

The reaction of A, B and C is represented as;

$A (s) ⇌ 2 B (g) + 3 C (g)$

Old eq., $2 P_{0} 3 P_{0}; K_{P} = 2^{2} \cdot 3^{3} \cdot P_{0}^{5}$

New eq., $\frac{8}{5 \sqrt{2}} \times 5 P_{0} P_{C} = ?$

$\begin{array}{l} \Rightarrow {(\frac{8}{\sqrt{2}})}^{2} \times P_{0}^{2} \times P_{C}^{3} \\ = 2^{2} \cdot 3^{3} \cdot P_{0}^{5} \\ P_{C} = \frac{3 P_{0}}{2} \\ \Rightarrow \frac{3 P_{0}}{3 \frac{P_{0}}{2}} \\ = \frac{2}{1} \end{array}$

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