Question

The container shown in figure has two chambers, separated by a partition, of volumes $V_1=2.0L$ and $V_2=2.0L$ The chambers contain $n_1=4.0$ and $n_2=5.0$ 𝑚𝑜𝑙𝑒 of an ideal gas at pressures $P_1=1.00\text{atm}$ and $P_2=1.00\text{atm}$. Calculate the pressure (in atm) after the partition is removed and the mixture attains equilibrium.

Answer 1

Consider the diagram


For chamber- $I{P}_1{V}_1=n_{1}R{T}_1$
For chamber-$II{P}_1{V}_2=n_{1}R{T}_2$
When the partition is removed the gases get mixed without any loss of energy. The mixture now attains a common equilibrium pressure, and the total volume of the system is sum of the volume of individual chambers $V_1$ and $V_2$.
$n=n_1+n_2\text{and}V=V_1+v_2$
From kinetic theory of gases.
The kinetic energy per mole,
$E=\frac{3}{2}RT$
For $n$ mole of gas
$E_{total}=\frac{3}{2}nRT=\frac{3}{2}PV$
Initially total energy
$E_{intial}=n_2{E}_1+n_2{E}_2$
$=\frac{3}{2}{P}_1{V}_1+P_2{V}_2=\frac{3}{2}(P_1{V}_1+P_2{V}_2)$
Final total energy
$E_{final}=\frac{3}{2}P(V_1+V_2)$
As energy remains constant
Hence $E_{intial}=E_{final}$
$\frac{3}{2}P(V_1+V_2)=\frac{3}{2}(P_1{V}_1+P_2{V}_2)$
$P=\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}$
$P=\left(\frac{1.00\times 2.0+2.00\times 3.0}{2.0+3.0}\right)\text{atm}$
$=1.60\text{atm}$
Final answer: (1.60)