Solution 30
Initial State (State 1):
Piston mass $m + \Delta m$. Height $h_0$. $$P_1 A = (m + \Delta m)g$$ $$T_1 = \frac{P_1 V_1}{nR} = \frac{(m+\Delta m)g h_0}{nR}$$
Final State (State 2):
Mass $\Delta m$ removed. Piston rises and settles at height $h$. $$P_2 A = mg$$ $$T_2 = \frac{P_2 V_2}{nR} = \frac{mg h}{nR}$$
Conservation of Energy:
The system is adiabatic. The work done by the gas in expanding equals the change in potential energy of the piston plus the change in internal energy.
$$\Delta U + \Delta PE_{grav} = 0$$ $$nC_v(T_2 – T_1) + mg(h – h_0) = 0$$
Using $C_v = \frac{R}{\gamma – 1}$:
$$\frac{nR}{\gamma – 1} \left( \frac{mgh}{nR} – \frac{(m+\Delta m)gh_0}{nR} \right) + mg(h – h_0) = 0$$
Cancel $nR$ inside the bracket and divide the whole equation by $mg$:
$$\frac{1}{\gamma – 1} \left( h – \left(1 + \frac{\Delta m}{m}\right)h_0 \right) + (h – h_0) = 0$$ $$h – h_0 – \frac{\Delta m}{m}h_0 + (\gamma – 1)(h – h_0) = 0$$ $$h – h_0 – \frac{\Delta m}{m}h_0 + \gamma h – \gamma h_0 – h + h_0 = 0$$ $$\gamma h = \gamma h_0 + \frac{\Delta m}{m}h_0$$ $$h = h_0 \left( 1 + \frac{\Delta m}{\gamma m} \right)$$