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 contain 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

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

Let the initial states of the two chambers be:
Chamber 1: Volume $V_1=2\text{L}$, pressure $P_1=1\text{atm}$, and moles $n_1=5$.
Chamber 2: Volume $V_2=3\text{L}$, pressure $P_2=2\text{atm}$, and moles $n_2=4$.

Since the container is insulated, there is no heat exchange with the surroundings ($Q=0$). Also, no external work is performed during the mixing process ($W=0$). According to the first law of thermodynamics, the total internal energy of the system remains conserved:
$U_{\text{total}}=U_1+U_2$

For an ideal gas, the internal energy $U$ is given by:
$U=\frac{f}{2}nRT=\frac{f}{2}PV$
where $f$ is the degrees of freedom of the gas. Assuming the gases are of the same type (or have the same degrees of freedom), the constant factor $\frac{f}{2}$ cancels out. Thus, conservation of internal energy simplifies to:
$P_f{V}_f=P_1{V}_1+P_2{V}_2$

Here, $P_f$ is the final equilibrium pressure and $V_f$ is the final total volume. The final volume is the sum of the volumes of both chambers:
$V_f=V_1+V_2=2\text{L}+3\text{L}=5\text{L}$

Now, we can substitute the known values into the conservation equation to solve for $P_f$:
$P_f\left(5\right)=\left(1\right)\left(2\right)+\left(2\right)\left(3\right)$
$P_f\left(5\right)=2+6$
$P_f\left(5\right)=8$
$P_f=\frac{8}{5}=1.6\text{atm}$

Therefore, the mixture attains an equilibrium pressure of 1.6 atm.