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 $T$
$\frac{P_0{V}_0}{4{\left(RT\right)}_0}{C}_v\left(4T_0-T\right)=\frac{P_0{V}_0}{{\left(RT\right)}_0}{C}_v\left(T-T_0\right)$
$\frac{4T_0-T}{4}=T-T_0$
$4T_0-T=4T-4T_0$
$8T_0=5T$
$T=\frac{8T_0}{5}$
Final pressure
In left $P_f=\frac{T_f}{T_0}{P}_i=\frac{\frac{8}{5}{T}_0}{4T_0}{P}_0=\frac{2}{5}{P}_0$
In right $P_f=\frac{\frac{8}{5}{T}_0}{T_0}{P}_0=\frac{8}{5}{P}_0$
$Force=\left(\frac{8P_0}{5}-\frac{2P_0}{5}\right)A=\frac{6P_{0}A}{5}$