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(D) Let the final equilibrium temperature be $T$. Since the container is adiabatic and the piston is conducting,heat flows from the hotter side to the

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$\frac{6}{5}P_0 A$

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(D) Let the final equilibrium temperature be $T$. Since the container is adiabatic and the piston is conducting,heat flows from the hotter side to the colder side until equilibrium is reached. The total internal energy of the system remains constant.Using the ideal gas law $PV = nRT$,the number of moles in each part is $n_L = \frac{P_0 V_0}{R(4T_0)}$ and $n_R = \frac{P_0 V_0}{RT_0}$.Since the total internal energy is conserved:$n_L C_V (4T_0 - T) = n_R C_V (T - T_0)$$\frac{P_0 V_0}{4RT_0} (4T_0 - T) = \frac{P_0 V_0}{RT_0} (T - T_0)$$\frac{4T_0 - T}{4} = T - T_0$$4T_0 - T = 4T - 4T_0$$8T_0 = 5T \implies T = \frac{8}{5}T_0$Since the piston is free to move but held at rest by an external force,the volumes remain $V_0$ in each part. The final pressures are:$P_{L,f} = \frac{n_L RT}{V_0} = \frac{P_0 V_0}{4RT_0} \cdot \frac{R(8T_0/5)}{V_0} = \frac{2}{5}P_0$$P_{R,f} = \frac{n_R RT}{V_0} = \frac{P_0 V_0}{RT_0} \cdot \frac{R(8T_0/5)}{V_0} = \frac{8}{5}P_0$The external force $F$ required to keep the piston at rest is:$F = (P_{R,f} - P_{L,f})A = (\frac{8}{5}P_0 - \frac{2}{5}P_0)A = \frac{6}{5}P_0 A$

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