In a vertical cylinder of length $2l+d$, a well-fitting piston $C$ can slide without friction; its thickness is $d$, and its weight is $P$. The top base plate $A$ and the bottom base plate $B$ are equipped with valves $a$ and $b$, respectively. We perform the following experiments:

$1^{0}$. When $C$ is equidistant from the base plates, we open the valves $a$ and $b$; then we close the valves and $C$ descends;

$2^{0}$. we then open the valve $b$.

Calculate how much $C$ descends in each case and what is the pressure in the $A C$ region in the first case (at the end of the experiment)?
$1^{0}$. Let $\pi$ denote the atmospheric pressure, $q$ the cross-sectional area of the piston, $x$ the displacement of the piston in the first experiment, $p_{1}$ the pressure in the $A C$ space, and $p_{2}$ the pressure in the $C B$ space. ($\pi, p_{1}$, and $p_{2}$ are expressed in the same units.)

Applying Boyle-Mariotte's law to the space above and below the piston. The former increased from $l q$ to $(l+x) q$ and the pressure changed from $\pi$ to $p_{1}$. The latter decreased from $l q$ to $(l-x) q$ and the pressure changed from $\pi$ to $p_{2}$. Therefore,

$$
\begin{aligned}
l \pi & =(l+x) p_{1} \ldots \\
l \pi & =(l-x) p_{2} \ldots
\end{aligned}
$$

The pressure $p_{2}$ in the lower space is greater than $p_{1}$, and in fact,

$$
q p_{2}-q p_{1}=P \ldots
$$

From 1) and 2), we express $p_{1}$ and $p_{2}$ and substitute them into 3); after rearranging,

$$
P x^{2}+q l \pi x-P l^{2}=0 \ldots
$$

This equation has real roots of opposite signs; among them, only the positive one is valid, which is smaller than $l$. Once we have calculated $x$, then

$$
p_{1}=\frac{l x}{l+x} \quad \text { and } \quad p_{2}=\frac{l \pi}{l-x}
$$

$2^{0}$. If we open the lower $b$ valve, the piston will descend by $y \mathrm{~cm}-$ rel. Let the pressure in the upper space be $\pi_{1}$; the atmospheric pressure, $\pi$, will prevail in the lower space. Therefore:

$$
l \pi=(l+y) \pi_{1} \quad \text { and } \quad q \pi_{1}+P=q \pi
$$

Thus,

$$
\pi_{1}=\frac{q \pi-P}{q}=\pi-\frac{P}{q} \quad \text { and } \quad y=\frac{l\left(\pi-\pi_{1}\right)}{\pi_{1}}=\frac{l P}{q \pi-P}
$$

Since $y<l$ must hold, it is necessary that $P<\frac{q \pi}{2}$. If $P \geq \frac{q \pi}{2}$, then the piston will fall to the bottom of the cylinder.