Solution for Question 13
1. Initial State:
The piston (mass $m$) is in equilibrium in the middle of the cylinder. The gas supports the atmospheric pressure plus the weight of the piston.
$$P_{initial} A = P_0 A + mg$$
$$P_{initial} = P_0 + \frac{mg}{A}$$
Initial Volume $V_i = \frac{V_{total}}{2}$
2. Condition to Lift the Cylinder:
To lift the cylinder (mass $M$), the upward force from the atmosphere on the bottom rim/area must overcome gravity and the downward push of the internal gas pressure.
$$P_0 A \ge P_{gas} A + Mg$$
$$P_{gas} \le P_0 – \frac{Mg}{A}$$
3. Process:
The piston is pulled up slowly (isothermal process). To maximize the mass $M$ that can be lifted, we need to reduce the internal gas pressure $P_{gas}$ as much as possible. We can pull the piston to the very top of the cylinder, doubling the volume ($V_{final} = 2V_i$).
Using Boyle’s Law ($P_i V_i = P_f V_f$):
$$P_{initial} \cdot V_i = P_{final} \cdot (2V_i)$$
$$P_{final} = \frac{P_{initial}}{2} = \frac{1}{2} \left( P_0 + \frac{mg}{A} \right)$$
4. Solving for M:
Substitute this minimum pressure into the lift condition:
$$\frac{1}{2} \left( P_0 + \frac{mg}{A} \right) = P_0 – \frac{Mg}{A}$$
Multiplying by $2A$:
$$P_0 A + mg = 2 P_0 A – 2Mg$$
$$2Mg = P_0 A – mg$$
$$M = \frac{P_0 A – mg}{2g}$$