Solution 29
Process Analysis:
- Initial Equilibrium: Gas at $T_0, P_0$. Since the system is vertical (implied by “piston rises”), $P_0 A = mg$.
- Heating (Isochoric): Temperature becomes $\eta T_0$. Internal energy increases to $U_{start} = nC_v(\eta T_0)$.
- Expansion: The gas expands adiabatically (insulated walls) against the weight of the piston until it stops at a new equilibrium.
- Final Equilibrium: The piston stops, so forces balance again. $P_{final} A = mg$. Thus, $P_{final} = P_0$.
Energy Balance:
We apply conservation of energy between the state just after heating (Temp $\eta T_0$) and the final state (Temp $T_f$). The decrease in internal energy equals the work done in lifting the piston.
$$-\Delta U = \Delta PE_{piston}$$ $$nC_v (\eta T_0 – T_f) = mg \Delta h$$
Since $P_{final} = P_{initial} = P_0 = mg/A$, we can relate height change to temperature change using the ideal gas law ($P \Delta V = nR \Delta T$): $$mg \Delta h = (P_0 A) \Delta h = P_0 \Delta V = nR(T_f – T_0)$$ Note: The volume change is relative to the volume at $T_0$, not $\eta T_0$, because the piston starts moving from the height corresponding to $T_0$.
For mono-atomic gas, $C_v = \frac{3}{2}R$. Substituting into the energy equation: $$n \frac{3}{2} R (\eta T_0 – T_f) = nR(T_f – T_0)$$ $$\frac{3}{2} (\eta T_0 – T_f) = T_f – T_0$$ $$3\eta T_0 – 3T_f = 2T_f – 2T_0$$ $$3\eta T_0 + 2T_0 = 5T_f$$ $$T_f = \frac{3\eta + 2}{5} T_0$$