What would the water pressure be at this point in a static incompressible fluid?

I have a bowl full of water. I invert a glass and place it upside down in the water, leaving a small pocket of air.

My question: what would the water pressure at positions P1 and P2 be? I know P1 = P2 because there's no hydrostatic effect as the water is level, but what is this value?

*This may seem rather trivial but I can't decide between two lines of reasoning. From where my confusion arises:

The atmospheric pressure acts to push the water level in the bowl down and up into the glass. The small pocket of air in the glass has the opposite effect, acting to push the water level in the glass down and increase the level in the bowl. Since the air pressures are equal, the water is at the same level in the bowl as in the glass.

1) Now, since the fluid is in equilibrium, P0 must equal the pressure at P1 for no flow to occur. Likewise, then P2 must equal P0 in the glass. Then P1 = P2 = P0

2) Or the water pressure at P1 and P2 completely zero? And this would be because the pressure from the air in the glass counters the pressure from the air outside acting on the bowl?

Thank you in advance, sorry for the length of this.

• I dont see how the water level will be the same. The gas is rarified inside the glass, so the pressure will be less than $p_o$ and the water level in the glass will be above that in the bowl – Wolphram jonny Dec 22 '19 at 20:19
• Actually, the gas in the glass will be compressed, so the gas pressure in the glass will be higher than outside, and the level of water in the glass will below that of the water outside the glass. – Chet Miller Dec 22 '19 at 23:34

Let L be the total length of the glass (assuming constant cross sectional area), z be the depth that the lower lip of the glass is inserted below the water surface, and h be the height that water rises inside the glass above the lower lip. Then the pressure of the air trapped in the glass is given by:$$p_a=p_{atm}+\rho g z-\rho g h$$This trapped air pressure is also given by the ideal gas law: $$p_a=p_{atm}\frac{L}{L-h}$$So, from these two equations,$$p_{atm}+\rho g z-\rho g h=p_{atm}\frac{L}{L-h}$$This equation is non-linear in h. But, if we assume that h << L, we can linearize the equation to obtain:$$p_{atm}+\rho g z-\rho g h=p_{atm}+p_{atm}\frac{h}{L}$$The solution to this equation for h is $$h=\frac{z}{\left[1+\frac{p_{atm}}{\rho g L}\right]}$$which is significantly less than z but greater than zero.

• So, just to confirm, if the pressure of the air contained in the glass is Pa and P0 is the pressure of the air outside, then this means P2 = Pa and P1 = P0? – S H Dec 23 '19 at 1:50
• Yes, in my analysis $P_2$ is what I call $P_a$ – Chet Miller Dec 23 '19 at 2:23

What you have is a water barometer. The bowl starts out full of water. When it is inverted and immersed, the water level in the bowl falls if the bowl has enough height which would be about 10 meters. If the bowl has less height, it will stay full of water. It does not fall to the level of the water outside, but will be at a height above the outside water level. This is because the gas pocket in the bowl formed by the falling water level will be solely water vapor. There is no air pocket in this problem. Its pressure will be the vapor pressure of water at whatever temperature the system is, which is not given.

Assuming the temperature is not at or above the boiling point, the vapor pressure will be below atmospheric and the liquid-vapor interface in the bowl will be a degree of vacuum. We know that the head provided by the column of water in the bowl will be such that the pressure of the water in the bowl is 1 atm at the same level as the water outside the bowl.

In the figure, the pressure will be the same all the way across.

The water pressure at both points $$P_1 = P_2 = 1 \mathrm{atm}$$.

If you placed the entire glass under the water surface, inverted it, and then raised it, there would be no air in the pocket (the one above $$P_2$$). The water level would rise to $$1 \mathrm{atm}$$, or about 10 meters.

Conversely, if you inverted the glass and then put it into the water, then there would be air in the pocket, but the pressure of that air + the water column would also be $$1 \mathrm{atm}$$. In this scenario, it's quite possible the water level in the glass and outside of it are the same.