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

Answer: 25

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 $PV=nRT$
$P_1{V}_1+P_2{V}_2=P\left(V_1+V_2\right)$
$P=\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}=\frac{2\times 4.5+3\times 5.5}{4.5+5.5}$
$P=\frac{9+16.5}{10}=\frac{25.5}{10}$
$\approx 25\times 10^{-1}atm$