Solution for Question 28
Analysis:
The work done by the gas is used to push the piston up against atmospheric pressure, the weight of the piston, and the weight of the liquid. As the piston moves up, the liquid spills over the brim, reducing the height of the liquid column.
Let $x$ be the upward displacement of the piston ($0 \le x \le h$). The height of the remaining liquid is $(h-x)$.
The pressure opposing the gas at displacement $x$ is:
$$P(x) = p_0 + \frac{Mg}{A} + \rho g (h – x)$$
Work done for a small displacement $dx$ is $dW = P(x) A dx$. Integrating from $x=0$ to $x=h$ (when all liquid is expelled):
$$W = \int_{0}^{h} \left( p_0 + \frac{Mg}{A} + \rho g h – \rho g x \right) A \, dx$$
$$W = \int_{0}^{h} \left( (p_0 A + Mg + \rho g A h) – \rho g A x \right) \, dx$$
$$W = \left[ (p_0 A + Mg + \rho g A h)x – \frac{1}{2} \rho g A x^2 \right]_{0}^{h}$$
$$W = (p_0 A + Mg + \rho g A h)h – \frac{1}{2} \rho g A h^2$$
$$W = p_0 A h + M g h + \rho g A h^2 – 0.5 \rho g A h^2$$
$$W = p_0 h A + M g h + 0.5 \rho g h^2 A$$