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A cylindrical adiabatic container of total volume $2 V_{0}$ is divided into two equal parts by a conducting piston which is free to move as shown in the figure. Each part contains identical gas at pressure $P_{0} .$ Initially the temperatures of the left and right parts are $4 T_{0}$ and $T_{0}$ respectively. An external force is applied on the piston of area A to keep the piston at rest at initial position. The value of external force required when thermal equilibrium is reached is

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$\frac{8}{5} P_{0} A$

$\frac{2}{5} P_{0} A$

$\frac{5}{6} P_{0} A$

$\frac{6}{5} P_{0} A$

answer is D.

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Detailed Solution

If final temp is T $\frac{P_{0} V_{0}}{4 R T_{0}} C_{v} (4 T_{0} - T) = \frac{P_{0} V_{0}}{R T_{0}} C_{v} (T - T_{0})$

$\frac{4 T_{0} - T}{4} = T - T_{0}$

$4 T_{0} - T = 4 T - 4 T_{0}$

$8 T_{0} = 5 T$

$T = \frac{8 T_{0}}{5}$

Final pressure In left $P_{f} = \frac{T_{f}}{T_{i}} P_{i} = \frac{\frac{8}{5} T_{0}}{4 T_{o}} P_{0} = \frac{2}{5} P_{0}$

In right $P_{f} = \frac{\frac{8}{5} T_{0}}{T_{o}} P_{0} = \frac{8}{5} P_{0}$ ,

Force $= (\frac{8 P_{0}}{5} - \frac{2 P_{0}}{5}) A = \frac{6 P_{0} A}{5}$

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