# How to figure out the pressure when you pick up a cup upside down in water

## Main Question or Discussion Point

Pressure of water in upside down cup

We all played with a cup in a body of water, you know when you put it in water, turn it upside down and lift above the water level and the cup holds the water in, untill you get above the water surface... So say you did that in a fish tank, could the fish swim up the cup, if you held it still, or is there too much pressure, and how would you measure that pressure? Please help I have no idea how to figure this out.

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The pressure in the top of the cup is actually lower than atmospheric pressure. At the bottom of the cup, the pressure would be equal to atmospheric pressure (assuming the top of the cup is on the surface of the water). A fish should be able to swim into the cup with no problems.

The pressure at the top of the cup can be calculated by
$\displaystyle P=101 kPA - \rho g h$
h is the height of the cup, 101 kPA is atmospheric pressure, and ρ is the density of water (1000 kg/m3). g, of course, is the gravitational acceleration on earth's surface.

CWatters
Homework Helper
Gold Member
I can't beat that reply but perhaps worth adding that the equation implies a maximium height before a vacuum forms at the top. The fish might have a problem approaching that limit that as there might be no disolved oxygen in the water?

It might be fun building an aquarium with an arch shaped tube that would allow the fish to travel up one side and back down the other.

That aquarium idea does sound quite interesting.

I wonder though, with the water column (tube-arch) constantly drawing a negative (gauge) pressure, would it do as you said & cause the dissolved oxygen in the water to come out here?

That would cause an air bubble to form, slowly displacing the water in the arch, while also de-oxygenating the water.

If so, you'd need a good bubbler to keep up (keep the fish happy), as well as a way of periodically removing the trapped air (syringe & tubing).

Might want to do a simple test setup first, as a proof of concept.

That aquarium idea does sound quite interesting.
I wonder though, with the water column (tube-arch) constantly drawing a negative (gauge) pressure, would it do as you said & cause the dissolved oxygen in the water to come out here?
That would cause an air bubble to form, slowly displacing the water in the arch, while also de-oxygenating the water.
If so, you'd need a good bubbler to keep up (keep the fish happy), as well as a way of periodically removing the trapped air (syringe & tubing).
Might want to do a simple test setup first, as a proof of concept.
At room temperature, the vapor pressure of water is about 2 kPA. Doing a quick calculation, this means that in order to form vapor in the tube, the tube would need to be about 10.1 meters high; so unless you plan on having an extraordinarily high tube for the fish to swim through, you won't need to worry about forming vapor at the top.

This is the same reason I neglected to mention the maximum height for the equation I posted above. I figured the cup isn't going to be anywhere near 10 meters tall, which is when the pressure at the top starts to approach 0.

I believe dissolved oxygen works slightly differently than the vapour partial pressure.

While I agree (& thank you for looking up the value) that you wouldn't be forming a large vapour bubble (essentially room temperature steam) for a short tube, I still think an air bubble may form (albiet slowly).

With a bubbler constantly trying to add air to the water (keeping it at the saturation point for room temperature/pressure), a low-pressure area would seem to invite rejecting disolved gas (like coming up from a dive).

If I'm completely off base with the above, I'd still say fish breath could collect there over time, so design for maintenance :)

LURCH