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Figure shows a cylindrical adiabatic container of total volume $2V_0$ divided into two equal parts by a conducting piston (which is free to move). Each part containing identical gas at pressure $P_0$ . Initially temperature of left and right part is $4T_0$ and $T_0$ respectively. An external force is applied on the piston to keep the piston at rest. Find the value of external force required when thermal equilibrium is reached. ( $A =$ Area of piston)

$\frac{8}{5}{P_0}A$
$\frac{2}{5}{P_0}A$
$\frac{5}{6}{P_0}A$
$\frac{6}{5}{P_0}A$
Solution
If final temp is $T.$
$\frac{\mathrm{P}_{0} \mathrm{V}_{0}}{\mathrm{R} 4 \mathrm{T}_{0}} \mathrm{C}_{\mathrm{V}}\left(4 \mathrm{T}_{0}-\mathrm{T}\right)=\frac{\mathrm{P}_{0} \mathrm{V}_{0}}{\mathrm{RT}_{0}} \mathrm{C}_{\mathrm{V}}\left(\mathrm{T}-\mathrm{T}_{0}\right)$
$\frac{4 \mathrm{T}_{0}-\mathrm{T}}{4}=\mathrm{T}-\mathrm{T}_{0}$
$4 \mathrm{T}_{0}-\mathrm{T}=4 \mathrm{T}-4 \mathrm{T}_{0}$
$8 \mathrm{T}_{0}=5 \mathrm{T}$
$\mathrm{T}=\frac{8 \mathrm{T}_{0}}{5}$
Final pressure
In left $\quad \mathrm{P}_{\mathrm{f}}=\frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{0}} \mathrm{P}_{1}=\frac{\frac{8}{5} \mathrm{T}_{0}}{4 \mathrm{T}_{0}}=\frac{2}{5} \mathrm{P}_{0}$
In right $\quad \mathrm{P}_{\mathrm{f}}=\frac{\frac{8}{5} \mathrm{T}_{0}}{\mathrm{T}_{0}}=\mathrm{P}_{0}=\frac{8}{5} \mathrm{P}_{0}$
Force $=\left(\frac{8 \mathrm{P}_{0}}{5}-\frac{2 \mathrm{P}_{0}}{5}\right) \mathrm{A}=\frac{6 \mathrm{P}_{0} \mathrm{A}}{5}$