A container has two chambers of volumes V₁ = 2 litres and V₂ = 3 litres separated by a partition made of a thermal insulator. The chambers contains n₁ = 5 and n₂ = 4 moles of ideal gas at pressures p₁ = 1 atm and p₂ = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

The correct answer is 1.6 atm.

To find the equilibrium pressure of the mixture after the partition is removed, we can apply the principles of conservation of energy and the ideal gas law.

Step 1: Understand the initial state of the two chambers
For Chamber 1:
Volume, $V_1=2\text{L}$
Number of moles, $n_1=5$
Pressure, $p_1=1\text{atm}$
Using the ideal gas equation $pV=nRT$, the temperature of the first chamber $T_1$ satisfies:
$p_1{V}_1=n_{1}R{T}_1$
Similarly, for Chamber 2:
Volume, $V_2=3\text{L}$
Number of moles, $n_2=4$
Pressure, $p_2=2\text{atm}$
Its temperature $T_2$ satisfies:
$p_2{V}_2=n_{2}R{T}_2$

Step 2: Apply Conservation of Energy
Since the container is thermally insulated, no heat is exchanged with the surroundings ($Q=0$). Additionally, no work is performed by or on the gas mixture ($W=0$). Therefore, the total internal energy of the system remains constant.
Let $C_v$ be the molar heat capacity at constant volume for the ideal gas. The initial total internal energy is the sum of the internal energies of the two chambers:
$U_{\text{initial}}=n_1{C}_v{T}_1+n_2{C}_v{T}_2$
Upon removing the partition, let the system reach a final equilibrium temperature $T_f$ and pressure $p_f$. The final total internal energy is:
$U_{\text{final}}=(n_1+n_2)C_v{T}_f$
Equating the initial and final internal energy:
$(n_1+n_2)T_f=n_1{T}_1+n_2{T}_2$

Step 3: Relate Temperature to Pressure and Volume
From the ideal gas relation, we substitute $n_i{T}_i=\frac{p_i{V}_i}{R}$ for each state:
$\frac{p_f{V}_f}{R}=\frac{p_1{V}_1}{R}+\frac{p_2{V}_2}{R}$
Since the total volume after removing the partition is $V_f=V_1+V_2$, we get:
$p_f(V_1+V_2)=p_1{V}_1+p_2{V}_2$
Rearranging the equation to solve for the final pressure $p_f$:
$p_f=\frac{p_1{V}_1+p_2{V}_2}{V_1+V_2}$

Step 4: Calculate the final numerical value
Substitute the given values into the derived formula:
$p_f=\frac{(1\text{atm}\times 2\text{L})+(2\text{atm}\times 3\text{L})}{2\text{L}+3\text{L}}$
$p_f=\frac{2+6}{5}=\frac{8}{5}=1.6\text{atm}$
Thus, the final equilibrium pressure is 1.6 atm.