A vertical cylinder closed at both ends is fitted with a smooth piston dividing the volume into two parts each containing one mole of air. At the equilibrium temperature of 320 K, the upper and lower parts are in the ratio 4 : 1. The ratio will become 3 : 1 at a temperature of

$\left(p_2-p_1\right)A=mg$

or                       $\frac{mg}{A}=\frac{R{T}_i}{V_1}-\frac{R{T}_i}{4V_1}=\frac{3R{T}_i}{4V_1}$                       …(i)

Similarly, in second case,

                          $\frac{mg}{A}=\frac{R{T}_f}{V_2}-\frac{R{T}_f}{3V_2}=\frac{2R{T}_f}{3V_2}$                       …(ii)

Further, 5V₁ = 4V₂
Equating Eqs. (i) and (ii), we get

                        $\frac{3T_i}{4V_1}=\frac{2T_f}{3V_2}$

$\Rightarrow$                        $T_f=\frac{9}{8}\times \frac{V_2}{V_1}\times T_i$

                                  $= \frac{9}{8} \times \frac{5}{4} \times 320$

                                   = 450 K