1. Dec 20, 2015

### msanx2

Let's imagine that a flask is initially opened and in contact with the atmosphere. I am thinking that when the flask is closed with a lid, the air density inside will be kept the same as outside. As so, the pressure inside should remain Patm:

P = (n/V).RT (n/V constant)

However, shouldn't this be the same rationale of the straw experiment? When one, after submerging the straw outlet, closes the top with the finger and the liquid remains inside the straw? The air pressure on top is no more Patm, but less.

2. Dec 20, 2015

### Staff: Mentor

The difference is that the liquid in the straw feels a downward force from gravity and has an open path to leave the straw. In a flask, the liquid is supported by the bottom of the container, whereas in the straw it is the difference in air pressure at the top and bottom of the column of liquid that supports the liquid. Note that the pressure at the top of the column of liquid is initially the same as atmospheric pressure. It's only once you remove the straw from the water that a difference in pressure develops.

3. Dec 20, 2015

### CWatters

That's not correct. The sequence is...

Bottom of straw submerged.
Top closed.
Pressure inside is still Patm.
Straw lifted out of liquid with some still in straw.
Pressure inside now less than Patm.

4. Dec 20, 2015

### msanx2

Thank you both. I think I got it. It's the fact that some small amount of liquid is lost that makes the pressure decrease. Like for example when a container is filled with water and has an outlet valve. If the lid is closed, liquid will flow out due to Patm, but at some point vacuum is created and the flow stops.

5. Dec 21, 2015

### msanx2

Just as another point. Imagine that I have an open tank, at 20ÂșC, in contact with the atmosphere. I fill it with water until half-volume an then I seal the tank. I am guessing that, at that moment, P = Pair + Pvap (water) = 101.325 kPa (because it was in mechanical equilibrium with the atmosphere before).

If then I start pumping water out (excluding any possible cavitation problems), until gas has 50x more available volume than before, then pressure should drop to around 2 kPa and the water vapor pressure at that temperature is 2.3 kPa. Then I assume water should start boiling.

Is this entire thinking correct?

6. Dec 21, 2015

### Staff: Mentor

I'm not sure what you mean. If your tank is half full of water, and you start pumping water out, the maximum available volume for the gas is only 2x the initial volume.

However, if the tank is 99% full, then pumping water out until the gas has 50x more volume available than before, then I believe the remaining water would indeed begin to boil.

7. Dec 21, 2015

### msanx2

Yes you're right.

8. Dec 21, 2015

### CWatters

Boiling is really only rapid vaporization. Should be possible to reduce the pressure slowly enough so that it doesn't visibly "boil", but the end result will be the same.