B What is the pressure acting on the air?

[Lotto]

TL;DR Summary

If I take a glass and dip it (perpendicularly to a water surface) with its opened part into water, there should be a thin layer of compressed air. How to determine its pressure?

Let us say that the glass isn't whole under the water. If the height of the air layer under the water is $\Delta x$, then the pressure the water acts on the gas is $p=p_{\mathrm {atm}}+\Delta x \rho g$. But my confusion is why the "hydrostaic pressure" is $\Delta x \rho g$, not $L\rho g$, where $L$ is the total lenght of a dipped glass. It would be more intuitive for me. I imagine it the way that the hydrostatic pressure $L\rho g$ pushes water up into the glass. It seems strange to me that it is different.

How to explain it?

 

[erobz]

TL;DR Summary: If I take a glass and dip it (perpendicularly to a water surface) with its opened part into water, there should be a thin layer of compressed air. How to determine its pressure?

Let us say that the glass isn't whole under the water. If the height of the air layer under the water is $\Delta x$, then the pressure the water acts on the gas is $p=p_{\mathrm {atm}}+\Delta x \rho g$. But my confusion is why the "hydrostaic pressure" is $\Delta x \rho g$, not $L\rho g$, where $L$ is the total lenght of a dipped glass. It would be more intuitive for me. I imagine it the way that the hydrostatic pressure $L\rho g$ pushes water up into the glass. It seems strange to me that it is different.

How to explain it?

Something like this?

 

[Lotto]

View attachment 337132

Something like this?

Yes, exactly. My question is why the part of the total pressure is $(h-\delta)\rho g$ instead of $h\rho g$.

 

[erobz]

Yes, exactly. My question is why the part of the total pressure is $(h-\delta)\rho g$ instead of $h\rho g$.

The absolute pressures of the gas above section 1-1 and the absolute hydrostatic pressure of the fluid below 1-1 must balance at their interface (if this is happening as a quasistatic equilibrium process).

 

[Lotto]

The absolute pressures of the gas above section 1-1 and the absolute hydrostatic pressure of the fluid below 1-1 must balance at their interface (if this is happening as a quasistatic equilibrium process)

Yes, I know, but why is the part of the pressure $(h-\delta)\rho g$? It is just unintuitive to me.

 

[erobz]

Yes, I know, but why is the part of the pressure $(h-\delta)\rho g$? It is just unintuitive to me.

I'm not getting the issue. It is that, because that is where the interface is - depth of section 1-1 relative to surface. The fluid is incompressible (virtually).

 

[Lotto]

I'm not getting the issue. It is that, because that is where the interface is - depth of section 1-1 relative to surface. The fluid is incompressible (virtually).

Ah, OK, I can see my weird thoughts now.

 

[erobz]

Ah, OK, I can see my weird thoughts now.

Maybe if the container of water is relatively small surrounding the cup (such that the height of water inside the container would significantly change as you dipped the cup) you can argue there is some less obvious relationship there. It still would be measured from the surface, but the surface height would have some dependency on $\delta$.

Edit: Disregard, I got myself twisted up.

 

[Lotto]

If the total pressure is $p$, I would derive it from

$p_{atm}+h\rho g=p+\delta\rho g$.

 

[kuruman]

You don't need to mess with all that. Just use the ideal gas law assuming that the temperature is constant. If $A$ is the area of the vessel and no air escapes, the gauge pressure inside the vessel will be given by $$p_1V_1=p_2V_2\implies p_{atm}~\cancel{A}~l=p~\cancel{A}~(l-\delta)\implies p=\left(\frac{l}{l-\delta}\right)p_{atm}.$$I assume that by "total" pressure you mean absolute pressure.

 

[erobz]

You don't need to mess with all that. Just use the ideal gas law assuming that the temperature is constant. If $A$ is the area of the vessel and no air escapes, the gauge pressure inside the vessel will be given by $$p_1V_1=p_2V_2\implies p_{atm}~\cancel{A}~l=p~\cancel{A}~(l-\delta)\implies p=\left(\frac{l}{l-\delta}\right)p_{atm}.$$I assume that by "total" pressure you mean absolute pressure.

Let us say that the glass isn't whole under the water.

In order to find $P(h)$ we need to first find $\delta(h)$

$$ P_{atm} \cancel{A} \ell = P \cancel{A} ( \ell - \delta ) \tag{1} $$

$$ P_{atm} + \rho g h = P + \rho g \delta \tag{2} $$

$(1)$ and $(2)$ combine to give $\delta (h)$

Then sub back into $(1)$ for $P(h)$