Solution
System Analysis:
We have a piston of mass $m$ confining a mono-atomic ideal gas ($\gamma = 5/3$). The piston is released from rest. Since heat loss is negligible, the expansion is adiabatic ($Q = 0$).
The problem asks for the maximum speed of the piston. As the piston rises, the pressure inside decreases. The piston accelerates as long as the upward force from the gas is greater than the downward force of gravity.
1. Condition for Maximum Speed:
The speed is maximum when the acceleration is zero. Before this point, the upward gas force exceeds gravity (acceleration upwards). After this point, gravity exceeds the gas force (deceleration).
At max speed, Net Force $F_{net} = 0$.
$$P_{final} S – mg = 0 \implies P_{final} = \frac{mg}{S}$$2. Conservation of Energy:
Since the process is adiabatic ($Q=0$), we apply the Work-Energy Theorem on the piston-gas system.
$\text{Work done by gas} + \text{Work done by gravity} = \Delta K_{piston}$
From the First Law of Thermodynamics for the gas: $W_{gas} = -\Delta U_{internal}$ (since $Q = \Delta U + W_{gas} = 0$).
Therefore:
$$-\Delta U_{internal} + W_{gravity} = K_f – K_i$$Where:
- $K_i = 0$ (released from rest)
- $K_f = \frac{1}{2}mv^2$ (at height $h$)
- $W_{gravity} = -mgh$ (work done against gravity moving up distance $h$)
- $\Delta U_{internal} = \frac{f}{2} (P_f V_f – P_i V_i)$
3. Calculating Internal Energy Change:
For mono-atomic gas, $f=3$.
- Initial State: $P_i = p_0$, $V_i = S h_0$
- Final State: $P_f = \frac{mg}{S}$, $V_f = S(h_0 + h)$
4. Solving for Velocity ($v$):
Substitute the values back into the energy equation:
$$-\Delta U – mgh = \frac{1}{2}mv^2$$ $$-\frac{3}{2} [ mg(h_0 + h) – p_0 S h_0 ] – mgh = \frac{1}{2}mv^2$$Multiply entire equation by 2 to clear fractions:
$$-3 [ mg(h_0 + h) – p_0 S h_0 ] – 2mgh = mv^2$$ $$-3mg(h_0 + h) + 3p_0 S h_0 – 2mgh = mv^2$$ $$3p_0 S h_0 – 3mgh_0 – 3mgh – 2mgh = mv^2$$ $$3p_0 S h_0 – 3mgh_0 – 5mgh = mv^2$$Rearranging the terms:
$$mv^2 = 3p_0 S h_0 – mg(3h_0 + 5h)$$ $$v^2 = \frac{3p_0 S h_0}{m} – g(3h_0 + 5h)$$