A container has two chambers of volumes $V_1=2$ litres and $V_2=3$ litres separated by a partition made of a thermal insulator. The chambers contain $n_1=5$ and $n_2=4$ moles of ideal gas at pressures $p_1=1\mathrm{atm}$ and $p_2=2\mathrm{atm}$, respectively. When the partition is removed, the mixture attains an equilibrium pressure of$$

**Step 1 – List the given data**

Volumes: V₁ = 2 L, V₂ = 3 L

Moles: n₁ = 5 mol, n₂ = 4 mol

Initial pressures: p₁ = 1 atm, p₂ = 2 atm

The total volume after the partition is removed is

$V=V_1+V_2$ = 2 L + 3 L = 5 L

Total moles of gas

$n=n_1+n_2$ = 5 mol + 4 mol = 9 mol

**Step 2 – Use the ideal‑gas law for each chamber before mixing**

For an ideal gas, 

$\mathrm{pV}=\mathrm{nRT}$

$T_1=\frac{p_1{V}_1}{n_{1}R}$

$T_2=\frac{p_2{V}_2}{n_{2}R}$

**Step 3 – Apply energy conservation (adiabatic, insulated container)**

Because the container is thermally insulated, the total internal energy stays constant. For an ideal gas the internal energy depends only on temperature, so the final temperature (T_f) satisfies

$n_1{T}_1+n_2{T}_2=n{T}_{\mathrm{f}}$

Substituting the expressions for T₁ and T₂ and cancelling the universal gas constant R gives

$T_{\mathrm{f}}=\frac{p_1{V}_1+p_2{V}_2}{n\times R}$

where n = n₁ + n₂ = 9 mol.

**Step 4 – Find the final equilibrium pressure**

Using the ideal‑gas law again for the whole container:

$\mathrm{p\_f}=\frac{nR{T}_{\mathrm{f}}}{V}$

Insert the expression for T_f and simplify (the factors n R cancel):

$\mathrm{p\_f}=\frac{p_1{V}_1+p_2{V}_2}{V_1+V_2}$

Plug in the numbers:

$\mathrm{p\_f}=\frac{1\times 2+2\times 3}{2+3}$

This evaluates to

$\mathrm{p\_f}=\frac{8}{5}=1.6atm$

**Conclusion** – The equilibrium pressure after the partition is removed is **1.6 atm**, which matches the given correct option.