Initially a gas of diatomic molecules is contained in a cylinder of volume $V_{1}$ at a pressure $P_{1}$ and temperature $250 K$ . Assuming that $25 %$ of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature $2000 K$ , when contained in a volume $2 V_{1}$ is given by $P_{2}$ . The ratio $P_{2} / P_{1}$ is

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Solution

Using ideal gas equation, $P V = n R T$
Let number of moles initially was $n$
$\Rightarrow P_{1} V_{1} = n R \times 250 ∵ [T_{1} = 250 K] . . . (i)$
and, after dissociation number of moles become $0.75 n + 2 \times 0.25 n = 1.25 n$
$\Rightarrow P_{2} (2 V_{1}) = (1.25 n) R \times 2000 [∵ T_{2} = 2000 K] . . . . (i i)$
Dividing eq.(i) by (ii),
$\frac{P_{1}}{2 P_{2}} = \frac{n R \times 250}{1.25 n R \times 2000} \Rightarrow \frac{P_{1}}{P_{2}} = \frac{1}{5}$

$∴ \frac{P_{2}}{P_{1}} = 5$

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