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Is there an equation for how much water can rise in a vacuum?


Science Student
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As a thought experiment, I imagine having a full cup of water sealed with plastic saran wrap. I think that if I turned it upside down in a shallow puddle of water in a petri dish and slipped off the saran wrap, then the water would probably come out of the glass causing the petri dish to overflow.

 

So how much water do I need for the atmospheric pressure to be enough to keep the water in the glass.

Edited by Science Student
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Actually, you seem to have that the wrong way round. Atmospheric pressure would keep the water in the cup ... until the cup got really big! It is easier to test this by putting a filled bottle under water and slowly pulling it out (with the opening under the water). You can see that the water stays in the bottle.

 

The pressure caused by a column of water is density x height x g. If this pressure is greater than air pressure then it will no longer be supported by the atmosphere and you will get a gap at the top (which will actually be a vacuum). This requires a tube about 10 metres high so tricky to do in practice. Easier with mercury, which only requires a height of about 1 m.

 

barometer.png

http://onlinephys.com/pressure2.html

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Actually, you seem to have that the wrong way round. Atmospheric pressure would keep the water in the cup ... until the cup got really big! It is easier to test this by putting a filled bottle under water and slowly pulling it out (with the opening under the water). You can see that the water stays in the bottle.

 

The pressure caused by a column of water is density x height x g. If this pressure is greater than air pressure then it will no longer be supported by the atmosphere and you will get a gap at the top (which will actually be a vacuum). This requires a tube about 10 metres high so tricky to do in practice. Easier with mercury, which only requires a height of about 1 m.

 

I have read what you have in your second paragraph before, but I don't understand it.

 

I understand that the atmospheric pressure acts on the surface of the water in the petri dish. So then why would water come out of an open test tube if the test tube was not submerged in the water of the petri dish? Does it have something to do with the surface of the water that the atmospheric pressure acts on? If so, then wouldn't this "10m rule" require a certain area of water that the atmospheric pressure must act on?

 

The equation that I was hoping for to make sense of all of this would have a water surface variable included.

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I understand that the atmospheric pressure acts on the surface of the water in the petri dish. So then why would water come out of an open test tube if the test tube was not submerged in the water of the petri dish?

 

It is because water can pour out of the bottom and let air in. If the tube is narrow enough, then there will be sufficient surface tension to stop this happening. I have no idea know how one would work that out, but it may involve surface area...

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It is because water can pour out of the bottom and let air in. If the tube is narrow enough, then there will be sufficient surface tension to stop this happening. I have no idea know how one would work that out, but it may involve surface area...

Oh I see.

 

What about the size of the petri dish; do you think that the more surface for the atmosphere to push on, the more water weight that can be held up in the down glass upside down?

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What about the size of the petri dish; do you think that the more surface for the atmosphere to push on, the more water weight that can be held up in the down glass upside down?

 

This is a bit counter-intuitive, I suppose. It puzzled me when I was young.

 

One thing to note is that we are dealing with pressure. This is force (weight) per unit area. So the area is already (implicitly) taken into account. If you expressed the equation terms of force (not pressure) then you would need to include area in the equation. But then you could simplify it by replacing force/are with pressure. Back to square 1! (And, actually, I'm not sure how you would get force, other than multiplying pressure by area. And then the area would just cancel out.)

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That's basically the unit of pressure anyway: kg/m² (Pa), or pounds/in² (psi). So the surface has already been taken into account. The mass pressing down on the surface area has already been normalized to 1 m² or in².

Unit of pressure is [math]\frac{N}{m^2}=\frac{kg*m}{s^2*m^2}=\frac{kg}{s^2*m}[/math]

Edited by Sensei
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This is a bit counter-intuitive, I suppose. It puzzled me when I was young.

 

One thing to note is that we are dealing with pressure. This is force (weight) per unit area. So the area is already (implicitly) taken into account. If you expressed the equation terms of force (not pressure) then you would need to include area in the equation. But then you could simplify it by replacing force/are with pressure. Back to square 1! (And, actually, I'm not sure how you would get force, other than multiplying pressure by area. And then the area would just cancel out.)

There is something deeply wrong with how I am trying to understand this. I know that the surface of the water exposed to the atmosphere doesn't matter because I have read it in multiple sources. But I don't understand why.

 

When I look at the picture that you posted at the top, I see that there is a certain amount of surface area of water in the petri dish that is exposed to atmospheric pressure. So, the equation pressure = force/area leads me to believe that force = area*pressure which seems to mean that the more surface there is the more force will be applied to the water.

 

Or as an example, if I apply x pressure to a 2m^2 side of a rock, then the force applied = 2m^2*x. But if I apply x pressure to a different side of the rock that is 3m^2, the force is greater f = 3m^2*x; this might move the rock where the other force might not.

 

This is driving me crazy because I know I am wrong, but I feel like I am right.

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There is something deeply wrong with how I am trying to understand this. I know that the surface of the water exposed to the atmosphere doesn't matter because I have read it in multiple sources. But I don't understand why.

How many water molecules is in [math]1 m^3[/math] volume?

[math]1 m^3[/math] volume of water has mass [math]1000 kg = 10^6 g[/math]

H2O has molar mass ~18 g/mol.

 

[math]\frac{1,000,000g}{18\frac{g}{mol}}=55555.556 mol[/math]

 

[math]1 mol = 6.022141*10^{23} molecules[/math]

 

 

[math]55555.556 mol * 6.022141*10^{23}=3.34563389*10^{28}[/math]

 

1 Liter of gas has approximately 0.0446 mol = 2.686*10^22 molecules

1 m^3 volume has 1000 L, so 44.6 mol = 2.686*10^25 molecules (at normal pressure, room temperature)

 

That's approximately 1245 times less than in water with the same volume.

 

Only tiny tiny fraction of molecules of water has direct contact with molecules of air, on surface of water.

 

Try to slice these volumes/areas to smaller and smaller pieces. 1m^3 volume with 1 m^2 area on top. Then 1mm^3 with 1mm^2 on top and so on so on, until you will receive single molecules at nano and pico scale.

 

Air molecule don't let water molecule to fly away from container.

If we will place glass of water in hermetic container, and pomp out whole air from it, nothing will be pressing at water from above, and water molecules will fly away from glass, becoming gas.

 

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So, the equation pressure = force/area leads me to believe that force = area*pressure which seems to mean that the more surface there is the more force will be applied to the water.

 

Indeed. That total force is distributed evenly over the entire surface of the water. And there is an equal and opposite force in the opposite direction (at every point).

 

So if we want to work out how much force is acting to hold up the water in the tube, we can take that total force and divide it by the fraction represented by surface area of the tube (as a proportion of total area). This gives us the total force pushing up on the column of water.

 

Now we need to know the total force pushing down due to the column of water (these two obviously have to be equal). This force downwards is given by: density x height x g x area. And there is your equation containing area!

 

Note that if you increase the area of the tube, you will increase the force pushing up and the force pushing down by the same amount. So the height is still the same.

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