Solution 34
Since the piston C is movable and the external pressure is $p_0$, the pressure of the gas between B and C remains constant at $p_0$. Because the process is slow (quasi-static) and partitions equalize pressure, the pressure everywhere is $p_0$. Thus, the entire process occurs at constant pressure.
- Gas 1 (between A and B): Volume $V_1$ is fixed (Isochoric). Moles $n_1 = \frac{p_0 V_1}{R T_0}$.
- Gas 2 (between B and C): Pressure is constant (Isobaric). Moles $n_2 = \frac{p_0 V_2}{R T_0}$.
Since partitions A and B are conducting, both gases reach the same final temperature $T = T_0 + \Delta T$. The total heat supplied $\Delta Q$ is distributed between the internal energy change of Gas 1 and the enthalpy change of Gas 2 (since Gas 2 expands at constant P).
$$ \Delta Q = (n_1 C_v \Delta T) + (n_2 C_p \Delta T) $$Substituting values for a monoatomic gas ($C_v = \frac{3}{2}R, C_p = \frac{5}{2}R$):
$$ \Delta Q = \left[ \frac{p_0 V_1}{R T_0} \cdot \frac{3R}{2} + \frac{p_0 V_2}{R T_0} \cdot \frac{5R}{2} \right] \Delta T $$ $$ \Delta Q = \frac{p_0 \Delta T}{2 T_0} (3V_1 + 5V_2) \implies \Delta T = \frac{2 T_0 \Delta Q}{p_0 (3V_1 + 5V_2)} $$Final Temperature:
Heat Transferred through B:
The heat passing through partition B is the heat absorbed by Gas 2.
$$ \Delta Q_2 = n_2 C_p \Delta T = \frac{p_0 V_2}{R T_0} \frac{5R}{2} \left( \frac{2 T_0 \Delta Q}{p_0 (3V_1 + 5V_2)} \right) $$