Examine the following diagram:

enter image description here

If the pressure vessel, initially out of the water containing only air at a pressure of 1atm, is then submerged in the water as shown in the diagram, will the air pressure within the pressure vessel be equal to the pressure at the depth of the bottom of the pressure vessel?

We know that the pressure at a depth of fluid is given by $$ P = \rho g h $$ where $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the depth of the fluid. Therefore, the pressure at the bottom of the container is some pressure $P$. Since water is incompressible, and air is compressible, will the water compress the air to the point at which its pressure matches the pressure $P$?

I feel as if my logic is sound, but I would like confirmation.

  • $\begingroup$ According to the details of your question, the pressure of the water at the air-water interface inside the vessel is lower than P. Why, then, would the air be compressed to the higher value P? Please clarify. $\endgroup$ May 9, 2023 at 3:20
  • $\begingroup$ @Chemomechanics would submerging the container with the open bottom not pressurize the air? Pressure increases as depth increases, so there is a pressure higher than atmosphere at the depth h which is at the bottom of the pressure vessel. Would that water pressure not compress the air which was initially at 1atm? $\endgroup$
    – Kalcifer
    May 9, 2023 at 3:52
  • $\begingroup$ I didn't say the air wouldn't be pressurized; I'm asking why it would be pressurized at pressure P if (1) P is the pressure at the bottom of the vessel, (2) the water pressure varies with vertical distance, and (3) the interface isn't at the bottom of the vessel. These four statements can't all be simultaneously true. It might be useful to express the pressure of both the water and air at various important locations, using the barometric formula you give. $\endgroup$ May 9, 2023 at 4:48
  • $\begingroup$ If pressure in a submarine were the same pressure as outside hydrostatic, the occupants would not last long. $\endgroup$
    – Bill Watts
    May 9, 2023 at 5:38

1 Answer 1


Yes, the inverted box with air will be pressurized to pressure but the pressure is not $\rho g h$ but $$ p = \rho g (h-\Delta h) $$ This is the pressure with which the liquid at the bottom of the box pushes the air.

Its the weight of the water above that produces hydrostatic pressure. The depth of the ocean where you run the experiment ($h$) does not matter, it it the depth of submersion $h-\Delta h$ that determines the pressure.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.