DIY liquid piston .
Hi everyone,
I'm working on a DIY project and experimenting with the arrangements shown in the image below. Basically, we have a container full of water that is exposed to the atmosphere. There is also an empty cylinder that is sealed on one end, where a pressure gauge is attached.

I'm not a physicist, but I do know that for a gas in a closed container, reducing the volume increases the pressure (regardless of whether this is being done isothermally or adiabatically).
Accordingly, case #3 is puzzling me. In case 3, the volume of air inside the cylinder is obviously shrinking, but surprisingly the pressure stays the same!  Regardless of how deep I move the tube, the pressure gauge doesn't change.  The water height doesn't change either, both outside and inside the cylinder, so the pressure in the cylinder is indeed equal to 1 atm
I also thought the air might be dissolving in the water due to the pressure increase, but the solubility of air in water (at room temperature, and up to 40 psig) is less than 5% (volume-wise), so this can't be the reason.
The gauge is working (tested with a compressor), and the seals are good (tested with soap+water).
 A: You must have a leak. The water levels inside and outside will not remain even if there is no leak. Take a clear drinking glass or a clear jar that you are sure has no holes in it. invert it and push it down in the water. You will see that the levels do not remain even. With no leaks the pressure inside the cylinder will increase by about 1.47 PSI per meter of depth. This will give you half the volume of air and 2 atm pressure  in the cylinder as is shown or expected in figure 3 but at a depth of about 10 meters, not level with the surface.
A: There must be a leak, and the description below attempts to prove it.
One thing we can be almost sure of not being wrong is the gauge.
Because if the pressure inside the cylinder was indeed $1 \ \text{atm}$ more than outside, you would feel a significant upward force even for a cylinder of very small cross section area. E.g. for a cylinder of radius 5 cm, the upward force would be:
$$\begin{align}
F &= \Delta P \times area\\
&=1atm \times \pi \times.05 \times.05\\
&=101325 \times 3.14 \times .0025\\
&\approx795 \ \text{Newtons} \\& \approx 75 \ \text{kg !}
\end{align}
$$
When we push something like a cylinder in your question in water, say by depth $d$, the air does get compressed, but this compression is just enough to increase the pressure so that the new pressure matches the pressure in water at the depth $d$. That's the necessary condition for the equilibrium of water level.
Let's say your cylinder has length $L$, cross-sectional area $A$, the new length of air column in the cylinder is $l$, and density of water $\rho$, we have:
$$\begin{align}
P_{atm} \times V_{initial} &= P_{new} \times V_{new} \\
\text{Or, } P_{atm} \times L \times A &  = (P_{atm} + \rho dg) \times l \times A \\
\text{Hence, } l &= \frac{P_{atm} \times L}{P_{atm} + \rho dg} \\
 &= \frac{L}{1 + \frac{\rho dg}{P_{atm}}}
\end{align}
$$
So, for $l$ to be half of $L$, $\rho dg$ should be equal to $P_{atm}$. On plugging in the values, we get $d$ to be about $10m$.
And, in that case, the pressure inside the cylinder should be $2atm$, in which case you should have observed three things:

*

*A considerable upward force on the cylinder;

*The cylinder would be submerged by about $10m$; and

*Water rising up to half the length of the cylinder.

If these are not happening, and in fact, the water level is not changing at all, there must be a leak.
Note: There is an approximation used in the calculation above.
The statement "...compression is just enough to increase the pressure so that the new pressure matches the pressure in water at the depth $d$" assumes that the rise of water column ($x$) inside the cylinder will be small enough so that we can ignore it.
In exact terms, "the new pressure of the air should match the water pressure at a depth of $(d-x)$". Shall be grateful if someone can present the exact calculations.
