Consider a sufficiently rigid and sealed container completely filled with liquid (e.g. water), pressurized at 1 bar, at constant temperature. Inside the container there is a buoy partially filled with gas (e.g. air) and with lower part opened. Both buoy and container are cylindrical in shape.

It is assumed that absolute pressure at the top is $100000 \ \mathrm{Pa}$ and at the bottom it is $100000 + \rho g h \ \mathrm{Pa}$

What happens when the buoy is pushed downwards?

Let's first examine what would happen if the container's top was open to the atmosphere:

  • As the hydrostatic pressure increases with depth, the gas would be compressed and the liquid level inside the buoy would raise.
  • Gas pressure would equal the "head" pressure at the surface between gas and liquid
  • Liquid level at the top would be lowered a little

Now let's go back to completely sealed case:

  • Liquid level inside the buoy cannot raise because this would create a vacuum elsewhere in the container?
  • If the liquid level inside the buoy would raise it would mean it's density in the whole container has decreased?
  • If the liquid level stays same then the gas experiences increased pressure at the interface (due to hydrostatic pressure) so we end up with gas with higher pressure but same volume (which is against Boyle's Law or requires gas temperature increase)?

I'm really confused here and unable to fix that out. So, what really happens?

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  • 1
    $\begingroup$ I don't follow your argument that "If the liquid level stays same then the gas experiences increased pressure at the interface..." As the buoy drops, the liquid level and gas volume can remain constant, corresponding to a constant pressure of 1 bar. (Of course, the hydrostatic relation $P_0+\rho gh$ would still hold, resulting $P<1\,\mathrm{bar}$ at the top of the container, ultimately resulting in room-temperature boiling the buoy moves down enough.) Please clarify this part of your question. $\endgroup$ Commented May 19, 2023 at 15:33
  • $\begingroup$ @Chemomechanics I mean that the lower the buoy the higher hydrostatic pressure at the interface between those fluids and so the gas should somehow be at the same pressure as the liquid at that level, thus if the level of liquid inside the buoy was the same (relative to the buoy) then we got paradox having gas with higher pressure but same volume (as height of the gas column stays the same). I understand that may be wrong, but I was just hypothesizing. $\endgroup$
    – PSz
    Commented May 19, 2023 at 16:24

1 Answer 1


Assuming the container isn't kilometers high, let's ignore liquid water compression, especially relative to air compression, as the bulk moduli differ by five orders of magnitude. Let's also ignore barometric variations in the air because its density differs from that of water by three orders of magnitude. Assume slow operation, maintaining equilibrium.

Then the liquid water volume is constant as long as the water in the container is entirely liquid. Under this assumption, the air volume is also constant at its original volume $V_0$, as the container is rigid, with its own constant volume. Under the ideal gas law, if the temperature $T$ remains constant, then the air pressure remains constant too, at the original pressure $P_0$ of 1 bar.

The hydrostatic formula still applies; the pressure decreases linearly with increasing height above the buoy at a rate of $\rho_\text{water}g$. Note that the pressure at the top of the container is now unfixed because equilibrium with the atmosphere is absent.

We encounter a nuance when the pressure at the top of the container drops below the water vapor pressure $P_\text{vapor}$ (0.03 bar at room temperature, for instance). Boiling then occurs, and the pressure at the container top is now fixed at the vapor pressure. With further descent of the buoy at depth $h\ge \frac{P_0-P_\text{vapor}}{\rho_\text{water}g}$ below the water liquid–water vapor interface, the air now compresses under a pressure of $P_\text{vapor}+\rho_\text{water}gh$. The air volume is then $V=V_0\frac{P_0}{P_\text{vapor}+\rho_\text{water}gh}$. The water continues to boil with further descent; the water vapor volume is $V_0-V$, with $n=\frac{P_\text{vapor}(V_0-V)}{RT}$ moles of water having evaporated, applying the ideal gas law.


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