Understanding Pressure Changes: Air and Water in Containers

[NCC-1701]

In the pic I have 2 containers a green one and a red one within, the blue is water and air is white of course, the inside chamber is opened at bottom ,now if I added air to larger container for the sake of argument of 5 psi would the pressure in the interior chamber match the pressure or would it be slightly less or no change at all.
I would believe that it would be the same but then I think of a cave under water which with air trapped would be at a constant pressure because the cave walls take the pressure, I'm confused.
Someone please help me see the logic of the correct answer.

 

[Mark44]

I would believe that it would be the same but then I think of a cave under water which with air trapped would be at a constant pressure because the cave walls take the pressure

The situation with the cave is somewhat different from the scenario in your drawing. In the cave scenario, the atmosphere and sea are essentially open systems, but for the scenario in your drawing, you are increasing the pressure in a closed system. The increased air pressure will expand the air portion, and cause the water in the red tube to exert pressure on the air trapped in it. If you increase the air pressure in the green container by 5 psi, that will increase air pressure in the red tube by the same amount.

 

[Dale]

The air in the red container will be roughly at the pressure of the water at the bottom of the green container (assuming the density of air is negligible relative to the density of water). If you add 5 psi to the top of the green container then that will add 5 psi to the bottom of the green container which will add 5 psi to the red container.

 

[FactChecker]

Clearly, the air pressure at the bottom is the same as the water pressure. At the top, the pressure is reduced by the weight of the column below it, down to the bottom. The weight of the air column is much smaller than the weight of the water column but is not zero.

 

[Lnewqban]

Welcome!

If you consider the pressure created by the weight of the column of water above the open bottom of the small chamber, the pressure of air inside it should be equal to that weight induced pressure plus the pressure of the air in the top of the large container.
As you increase the air pressure, some water will enter the open bottom, acting as a piston that reduces the volume and compresses the air that is trapped in the small chamber.

 

[NCC-1701]

Ok I'm happy with the answers and I'll base my design of other components that connect to it accordingly, Thanks

 

[jbriggs444]

Someone please help me see the logic of the correct answer.

As I understand this, the right hand and left hand sides of the outer volume are connected both above and below the water line. Although depicted as separate volumes, the head space on the right and on the left are both part of a single connected volume.

As I understand it, the head space in the inner volume is sealed. Air cannot escape (unless you reduce pressure, in which case, some would bubble out the bottom).

Having forced 5 psi of over pressure in the outer chamber you will, all things being equal, have forced water into the inner chamber. The difference in water height between inside and outside will have been reduced. That means that the pressure differential between the inside chamber head space and outside chamber head space will also have been reduced.

Therefore, the increase in pressure in the inner chamber will be less than 5 psi.

 

[etotheipi]

Therefore, the increase in pressure in the inner chamber will be less than 5 psi.

It will be $\Delta P = 5\text{psi} - \rho g(x_1 + x_2)$ if $x_1$ and $x_2$ are the distances that the top surface moved down and the surface inside the inner chamber moved up. If we knew some cross sectional areas and made some assumptions about the gas we could estimate a numerical value.

 

[gmax137]

There used to be a little toy like this, except that the inner cylinder was freely floating, neutrally buoyant in the liquid. When you pressed down on the (rubber?) lid, the increased pressure pushed a little water up into the floating tube (as mentioned above); this reduced its buoyancy, so the little tube would sink. Release the pressure on the lid, and the little tube would rise. I don't remember what it was called, but the idea was you could make the "diver" go up and down.

 

[etotheipi]

There used to be a little toy like this, except that the inner cylinder was freely floating, neutrally buoyant in the liquid. When you pressed down on the (rubber?) lid, the increased pressure pushed a little water up into the floating tube (as mentioned above); this reduced its buoyancy, so the little tube would sink. Release the pressure on the lid, and the little tube would rise. I don't remember what it was called, but the idea was you could make the "diver" go up and down.

Like this?

 

[JT Smith]

I kind of remember a toy like that as a kid and being fascinated by it. Simple physics but somehow not what normal everyday experience prepares you to expect.