A container is divided into two chambers by a partition. The volume of the first chamber is $4.5 L$ and the second chamber is $5.5 L$ . The first chamber contains $3.0 mol$ of gas at pressure $2.0 atm$ and the second chamber contain $4.0 mol$ of identical gas at pressure $3.0 atm$ . After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is $x \times 10^{- 1} atm$ . Value of $x$ is _______. $($ up to two significant figures $)$

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Solution

By energy Conservation :

$\frac{n_{1} f R T_{1}}{2} + \frac{n_{2} f R T_{2}}{2} = \frac{(n_{1} + n_{2}) f R T}{2}$

Using, $P V = n R T$ , we get,

$P_{1} V_{1} + P_{2} V_{2} = P (V_{1} + V_{2})$

$\Rightarrow P = \frac{P_{1} V_{1} + P_{2} V_{2}}{V_{1} + V_{2}}$

$\Rightarrow P = \frac{2 \times 4.5 + 3 \times 5.5}{4.5 + 5.5}$

$\Rightarrow P = 2.55 atm = 25.5 \times 10^{- 1} atm \approx 26 \times 10^{- 1} atm$

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