Solution 31
Initial State:
Both portions have height $h$ and temperature $T_0$. Top piston ($m$) supports pressure $P_1 = P_0 = \frac{mg}{A}$. Bottom piston ($m$) supports top gas + its own weight: $P_2 = P_0 + \frac{mg}{A} = 2P_0$. Mole ratio: Since $V$ and $T$ are same, $n \propto P$. $$n_{top} = n, \quad n_{bot} = 2n$$
Process:
Lower gas heated slowly. Pressure remains constant ($2P_0$) because the weights don’t change. When $T_{bot}$ reaches $2T_0$, the piston becomes a conductor. Heat flows to the top gas.
Conservation of Enthalpy:
Once conduction starts, the system is isolated from the heater (implied by “equilibrium re-established”). The pressures $P_0$ and $2P_0$ remain constant throughout the equilibration.
For an adiabatic, constant pressure process (for the combined system), the total enthalpy change is zero. Let the final equilibrium temperature be $T_f$. $$\Delta H_{total} = 0$$ $$n_{top} C_p (T_f – T_{initial,top}) + n_{bot} C_p (T_f – T_{initial,bot}) = 0$$
At the moment conduction starts: Top Gas: $T = T_0$ (Adiabatic until now) Bottom Gas: $T = 2T_0$ (Just finished heating)
$$n C_p (T_f – T_0) + 2n C_p (T_f – 2T_0) = 0$$ $$(T_f – T_0) + 2(T_f – 2T_0) = 0$$ $$3T_f – 5T_0 = 0 \implies T_f = \frac{5}{3}T_0$$
Calculation of Separation:
The separation between the pistons is the height of the top gas column ($h_{top}$). Since pressure is constant ($P_0$): $$\frac{h_{final}}{h_{initial}} = \frac{T_f}{T_{initial}}$$ $$h_{final} = h \cdot \frac{5/3 T_0}{T_0} = \frac{5h}{3}$$