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If I have a 1.0m long cylindrical cup, and fill it up to $h$ = 0.5m with water, then the 0.5m of air above it will be at atmospheric pressure. But if I now place a perfect seal on the cylinder, and then invert the cylinder, what is the pressure of the air itself in the cylinder? There is now no air molecules above the 0.5m of air in the cylinder, and thus no net force downwards due to all the air molecules from the atmosphere. Thus there should no longer be this large net pressure on the air (i.e. the air in the cylinder is no longer at atmospheric pressure), and I would assume the only pressure within the cup at the seal is given by $p = p_{water} + p_{air} = \rho_{water}gh + nRT/V$ where $n$ is the density of air in the cylinder, $T$ is the temperature, and $V$ is $(1.0 - h)A_{cylinder}$ (the latter is the area of the cylinder).

This is based on a similar post.

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  • $\begingroup$ So, the air in the cylinder is initially at atmospheric pressure, when the top is open. Just to be clear, your saying that if you seal the top and turn the cylinder upside-down the air pressure will be greatly reduced? $\endgroup$ – M. Enns May 4 '18 at 1:44
  • $\begingroup$ That net pressure in the air inside the cylinder could have been transmitted to the external environment via collisions with the cylindrical cup surface and whatever is supporting the cylinder (e.g. person holding the inverted cup from above). Pressure is force per area, so what will permit the particles to continue having a net acceleration downwards equal to atmospheric pressure if the cylinder shields it from the atmospheric air? If it was simply conservation of momentum, I would agree the air in the cylinder should remain constant. But I have not heard about conservation of pressure. $\endgroup$ – Mathews24 May 4 '18 at 1:55
  • $\begingroup$ "There is now no air molecules above the 0.5m of air in the cylinder . . ." I don't understand this part. The air molecules that were above the air in the cylinder are still there, and they are still above the now inverted cup. They exert the same pressure on the top of the cup that they did before it was inverted, and that pressure is communicated throughout the contents of the cup, yes? $\endgroup$ – Rodney Dunning May 4 '18 at 1:56
  • $\begingroup$ Yes, but besides the volume of air in the cylinder, it is not in contact with external atmosphere. I assume the cylinder is supported (e.g. a person holding the following cup but with air in it above the water, and it should be sealed). $\endgroup$ – Mathews24 May 4 '18 at 2:00
  • $\begingroup$ The above is similar to case 2 from the following post but the answer appears to insufficiently explain the air pressure in the cylinder contributing to the total pressure on the cardboard since this experiment should not work if the cylindrical cup is now opened from above while it is inverted (i.e. the cardboard remains below on the water interface, but a hole is now inserted above). $\endgroup$ – Mathews24 May 4 '18 at 2:11
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then the 0.5m of air above it will be at atmospheric pressure.

Correct. You don't give the width of the cylinder, so the exact height isn't that relevant. But we can say there is a volume of air $V$ remaining in the cylinder.

But if I now place a perfect seal on the cylinder, and then invert the cylinder, what is the pressure of the air itself in the cylinder?

As the volume has not changed (and we presume the temperature is also constant), then $P = \frac{nRT}{V}$, so the pressure is the same as it was before.

There is now no air molecules above the 0.5m of air in the cylinder, and thus no net force downwards due to all the air molecules from the atmosphere.

That doesn't matter. The vessel is rigid. The air molecules inside bump against the walls. The walls respond with a normal force. That force over their area is a pressure. The pressure is the same as the atmosphere because it started at that pressure. Assuming it was strong enough, you could take it into outer space and the same pressure would remain on the inside.

The water is just a complication. It has nearly constant volume, so the air is constrained to a nearly constant volume in the vessel. The same would happen if you sealed it without water inside.

why can cardboard (or any sufficiently low-mass sealant) remain stuck to the cup?

Because the cardboard doesn't have to drop very far to equalize the pressure. This is normally done with a cup. Let's assume a cup 8cm tall, with 4cm of water in it.

When we turn it upside down, what pressure differential will hold the water in place? That would be equivalent to the static pressure at the bottom of the water.

$$P = \sigma g h$$ $$P = (\text{g}/\text{cm}^3)(9.81\text{m}/\text{s}^2)(4\text{cm}) = 392\text{Pa}^2$$

So what volume change lowers the atmospheric pressure that much? $$PV = P'V'$$ $$V' = \frac{P}{P'}V$$ $$V' = \frac{101325\text{Pa}}{100933\text{Pa}}V$$ $$V' = 1.0039 V$$

Since the sides are rigid, the entire volume change is accommodated by a change in height.

$$h' = 1.0039h = 4.016\text{cm} = h + 0.16\text{mm}$$

So the pressure change is met by lowering a bit more than a tenth of a millimeter. As we can see water remain in a straw that is several millimeters wide, that's no problem for the water to remain in place.

Really, the limiting case will be where one side of the card remains tight against the cup, while the other side drops to take the entire change. On average this will double the height that might appear, so assume you really cause it to drop by a third of a millimeter. Still not a problem.

Your calculation neglected the mass of the sealant (e.g. if it's 10kg).

Yes, I'm assuming a light card that can be neglected. The calculation just needs to add in the cross-sectional density of that material if it's significant. You can certainly pick materials that are light enough to ignore.

But the cup's cross-section does matter since it is the pressure difference acting across this cross-section on the sealant that support it from falling against gravity.

Nope. The wider the cup, the heavier the water, but also the greater the force from the atmosphere. These two effects both scale with the cross section so there's no change with area. That's why I didn't have to specify a particular area for the cup, just know the volumetric density of water and the atmospheric pressure.

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  • $\begingroup$ Then for case 2, why can cardboard (or any sufficiently low-mass sealant) remain stuck to the cup? Your answer would indicate the pressure in the cup is Patm + Pwater (and thus there's a pressure gradient into the cup) which would make this scenario unstable and thus the sealant fall if any air is present in the cup, but that is not observed. $\endgroup$ – Mathews24 May 4 '18 at 4:07
  • $\begingroup$ The cardboard drops slightly (increasing the available volume in cup). With the increased volume, the pressure drops and the differential keeps it from lowering further. $\endgroup$ – BowlOfRed May 4 '18 at 4:10
  • $\begingroup$ How far do you suggest the cardboard drops? That does not appear to be the case since the pressure gradient across the cardboard would have to support it against gravity. But the cardboard is initially already on the outer surface of the cup (i.e. not inside the cup). If the cardboard drops any farther, it is not in contact with the cup any longer. If this occurs, the water can fall out via the Rayleigh-Taylor instability. $\endgroup$ – Mathews24 May 4 '18 at 4:16
  • $\begingroup$ You need a fair amount of space before that is a problem. The tiny gap around the cup is not big enough. As an example, you can hold water in a fairly thick drinking straw with zero support. Only above a certain size does the instability overcome the surface tension. $\endgroup$ – BowlOfRed May 4 '18 at 4:18
  • $\begingroup$ This would indicate that wider cups have greater difficulty supporting the sealant since the sealant's mass would now be greater with the larger cross-sectional area (A). Thus h' - h would have to similarly increase since (P - P')A > mg for no instability to occur, where m is the mass of the sealant. But for a cup (or straw) of radius 10cm, water surface tension would likely be insufficient to support a drop of even a fraction of a millimetre. Thus is there generally a maximum height change that can be accommodated by the sealant? It seems unlikely the above is sufficient. $\endgroup$ – Mathews24 May 4 '18 at 5:32

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