 # A container is divided into two chambers by a partition. `
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.

Solution:

Let common equilibrium pressure of mixture is $\mathrm{P}$ atmp. then

$\mathrm{U}_{1}+\mathrm{U}_{2}=\mathrm{U}_{\text {mixutre }}$

$\frac{\mathrm{f}}{2} \mathrm{P}_{1} \mathrm{~V}_{1}+\frac{\mathrm{f}}{2} \mathrm{P}_{2} \mathrm{~V}_{2}=\frac{\mathrm{f}}{2} \mathrm{P}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)$

$\frac{\mathrm{f}}{2}(2)(4.5)+\frac{\mathrm{f}}{2}(3)(5.5)=\frac{\mathrm{f}}{2} \mathrm{P}(4.5+5.5)$

$\Rightarrow P=2.55=x \times 10^{-1}$ atmp

So $x=25.5 \approx 26$ (Nearest integer)