Two identical containers each of volume V₀ are joined by a small pipe. The containers contain identical gases at temperature T₀ and pressure P₀. One container is heated to temperature 2T₀ while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature2T₀. Then

Initially for container A,

 ${\mathrm{P}}_0{\mathrm{V}}_0={\mathrm{n}}_0{\mathrm{RT}}_0$

For container B,

$\begin{array}{l}{\mathrm{P}}_0{\mathrm{V}}_0={\mathrm{n}}_0{\mathrm{RT}}_0\\ \therefore {\mathrm{n}}_0=\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{{\mathrm{RT}}_0}\end{array}$

Total number of moles $={\mathrm{n}}_0+{\mathrm{n}}_0=2{\mathrm{n}}_0$

Since even on heating, the total number of moles is conserved,

We have ${\mathrm{n}}_1+{\mathrm{n}}_2=2{\mathrm{n}}_0$   …….(i)

Let P be the common pressure. Then for container A,

 ${\mathrm{PV}}_0={\mathrm{n}}_1\mathrm{R}2{\mathrm{T}}_0$

$\therefore {\mathrm{n}}_1=\frac{{\mathrm{PV}}_0}{2{\mathrm{RT}}_0}$

And for container

$\mathrm{B},{\mathrm{PV}}_0={\mathrm{n}}_2{\mathrm{RT}}_0$

${\mathrm{n}}_2=\frac{{\mathrm{PV}}_0}{{\mathrm{RT}}_0}$

Substituting the values of n₀, n₁ and n₂ in equation (i), we get

$\frac{{\mathrm{PV}}_0}{2{\mathrm{RT}}_0}+\frac{{\mathrm{PV}}_0}{{\mathrm{RT}}_0}=\frac{2\cdot {\mathrm{P}}_0{\mathrm{V}}_0}{{\mathrm{RT}}_0}\Rightarrow \mathrm{P}=\frac{4}{3}{\mathrm{P}}_0$

Number of moles in container A (at temperature 2T₀)

 $={\mathrm{n}}_1=\frac{{\mathrm{PV}}_0}{2{\mathrm{RT}}_0}=\left(\frac{4}{3}{\mathrm{P}}_0\right)\frac{{\mathrm{V}}_0}{2{\mathrm{RT}}_0}=\frac{2}{3}\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{{\mathrm{RT}}_0}\left[\text{ As }\mathrm{P}=\frac{4}{3}{\mathrm{P}}_0\right]$