Question

A cylindrical adiabatic container of total volume $2V_{0}$ 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 $4T_{0}$ and $T_{0}$ respectively. An external force is applied on the piston of area $A$ to keep the piston at rest. The value of external force required when thermal equilibrium is reached is

• A. $\frac{8}{5}P_{0}A$
• B. $\frac{2}{5}P_{0}A$
• C. $\frac{5}{6}P_{0}A$
• D. $\frac{6}{5}P_{0}A$

Answer 1

Answer: (D)

If final temp is $T$
$\frac{P_{0} V_{0}}{4 \left(RT\right)_{0}}C_{v}\left(4 T_{0} - T\right)=\frac{P_{0} V_{0}}{\left(RT\right)_{0}}C_{v}\left(T - T_{0}\right)$
$\frac{4 T_{0} - T}{4}=T-T_{0}$
$4T_{0}-T=4T-4T_{0}$
$8T_{0}=5T$
$T=\frac{8 T_{0}}{5}$
Final pressure
In left $P_{f}=\frac{T_{f}}{T_{0}}P_{i}=\frac{\frac{8}{5} T_{0}}{4 T_{0}}P_{0}=\frac{2}{5}P_{0}$
In right $P_{f}=\frac{\frac{8}{5} T_{0}}{T_{0}}P_{0}=\frac{8}{5}P_{0}$
$Force=\left(\frac{8 P_{0}}{5} - \frac{2 P_{0}}{5}\right)A=\frac{6 P_{0} A}{5}$