Solution 28
System Analysis:
Consider the nitrogen gas (diatomic, $C_v = \frac{5}{2}R$) and the piston as the system. The container is insulated (adiabatic) and placed in a vacuum ($P_{ext} = 0$).
Initial State: The piston is in equilibrium. $$P_0 A = mg \implies P_0 = \frac{mg}{A}$$ The piston is given a downward velocity $u$.
Final State: The piston oscillates and eventually stops due to damping. The final pressure must again balance the weight of the piston, so $P_{final} = P_0$. Since pressure is constant and $PV=nRT$, the volume change is proportional to the temperature change.
Energy Conservation:
Since the walls are adiabatic, the loss in kinetic energy of the piston is converted into the change in internal energy of the gas and the change in gravitational potential energy of the piston.
$$\Delta K + \Delta U_{gas} + \Delta PE_{piston} = 0$$
Let $h$ be the upward displacement of the piston. $$0 – \frac{1}{2}mu^2 + nC_v(T_f – T_i) + mgh = 0$$ $$\frac{1}{2}mu^2 = mgh + n\left(\frac{5}{2}R\right)\Delta T$$
Since the process ends at the same pressure $P_0 = \frac{mg}{A}$, the work done by the gas is $P_0 \Delta V = P_0 (Ah) = mgh$. Using the Ideal Gas Law for constant pressure process: $$nR \Delta T = P_0 \Delta V = mgh$$
Substitute $nR \Delta T$ into the energy equation: $$\frac{1}{2}mu^2 = mgh + \frac{5}{2}(mgh)$$ $$\frac{1}{2}mu^2 = \frac{7}{2}mgh$$ $$u^2 = 7gh \implies h = \frac{u^2}{7g}$$