Question

A container is divided into two chambers by a partition. The volume of first chamber is $4.5$ litre and second chamber is $5.5$ litre. The first chamber contain $3.0$ moles of gas at pressure $2.0$ atm and second chamber contain $4.0$ moles of 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

Answer 1

By energy Conservation
$\frac{3}{2}n_{1}\left(RT\right)_{1}+\frac{3}{2}n_{2}\left(RT\right)_{2}=\frac{3}{2}\left(n_{1} + n_{2}\right)RT$
Using $PV=nRT$
$P_{1}V_{1}+P_{2}V_{2}=P\left(V_{1} + V_{2}\right)$
$P=\frac{P_{1} V_{1} + P_{2} V_{2}}{V_{1} + V_{2}}=\frac{2 \times 4 . 5 + 3 \times 5 . 5}{4 . 5 + 5 . 5}$
$P=\frac{9 + 16 . 5}{10}=\frac{25 . 5}{10}$
$\approx25\times 10^{- 1}atm$