Given p₁ = 5 bar, V₁ = 1 m³, T₁ = 400 K

So, $${n_1} = {5 \over {400R}}$$ (from pV = nRT)

Similarly, p₂ = 1 bar, V₂ = 3 m³, T₂ = 300 K,

$${n_2} = {3 \over {300R}}$$

Let at equilibrium the new volume of A will be (1 + x)

So, the new volume of B will be (3 $$-$$ x)

Now, from the ideal gas equation,

$${{{p_1}{V_1}} \over {{n_1}R{T_1}}} = {{{p_2}{V_2}} \over {{n_2}R{T_2}}}$$

and at equilibrium (due to conduction of heat)

$${{{p_1}} \over {{T_1}}} = {{{p_2}} \over {{T_2}}}$$

So, $${{{V_1}} \over {{n_1}}} = {{{V_2}} \over {{n_2}}}$$ or $${V_1}{n_2} = {V_2}{n_1}$$

After putting the values

$$(1 + x) \times {3 \over {300R}} = (3 - x) \times {5 \over {400R}}$$

or $$(1 + x) = {{(3 - x)5} \over 4}$$ or $$4(1 + x) = 15 - 5x$$

or $$4 + 4x = 15 - 5x$$ or $$x = {{11} \over 9}$$

Hence, new volume of A i.e., (1 + x) will comes as $$1 + {{11} \over 9} = {{20} \over 9}$$ or 2.22.