Why does the liquid stay in the straw?

itsthemac

This might sound trivial, but I've always been curious as to why exactly when you put your finger on the top of a straw in a drink and lift it up, the liquid stays in it. We're studying pressure and fluids right now in my physics class, and I have a general idea that it has to do with the pressure of the outside air pushing up on the liquid in the straw. What I don't understand is exactly how to describe the whole thing. Shouldn't the pressure of the air in between the liquid and your finger be the same as it was before you put your finger on the straw? Does that pressure actually change at all just from enclosing it? And if it doesn't change, then wouldn't the same forces be acting on the liquid that were acting on it before you put your finger on it? thanks in advance

Pengwuino

The air pressure comes from the fact that there's the equivalent of 15 pounds per square inch air pushing down due to the air pressure. When you put your finger on the straw, you've cut off the liquid in the straw from all but a few inches of air (whatever sits between your finger and the liquid). The air pressure from the top drops to almost nothing whereas the air pressure from the bottom is still the ol 1atm pushing up.

itsthemac

But if the air pressure at the top drops to almost nothing, shouldn't the water in the straw move upwards towards your finger a little bit the second you put your finger on the straw? I would also think that this would mean that if you tipped the straw upside-down (while keeping your finger on it), that the water would fall towards your finger. But this doesn't happen.

Dr Lots-o'watts

There's air between your finger and the water, it can't be crushed anymore than if it were in a balloon!

Pengwuino

Wait wait no no, I didn't think that one through. The pressure doesn't drop to 0, it stays the same. Except now it's in equilibrium being pushed from below at 1atm and from above at 1atm.

itsthemac

but if they're both at 1 atm, then shouldn't the force caused by the pressure pushing up on the liquid from the bottom and the force caused by the pressure pushing down on the liquid from the air in between the liquid and your finger cancel out? which would mean that the net force acting on the liquid in the vertical direction would just be the force of gravity? which should just pull it down and out of the straw?

Andy Resnick

This is a good puzzle.

First, let's say there is *no* air space- the water completely fills the straw. Then, you can pull the straw up about 400 inches (atmospheric pressure is 408 inH2O = 760 mmHg = etc.) before a small vapor bubble will appear in the top: that's how Torricelli's experiment worked.

The key concept is that the pressure in the water anywhere along the length of the straw is given simply by the height. Once the pressure gets to zero (or the vapor pressure of water), the water will vaporize. Drawing the column up further will only make a larger region of vacuum.

Now we do this again, but there is a small region of air already present. There's no substantive difference between this case and the above- the pressure in the air bubble will go down as the straw is raised, but the water column will rise for the same reason it did before. Conversely, the pressure in the bubble will increase if the straw is submerged. The size of the bubble will change as the straw is raised and lowered because the pressure changes and air is compressible.

DaveC426913

The moment the water starts to drop, it creates a low pressure area at the top. The force on the water is no longer in equilibrium. There is 1atm pushing up, but none pushing down. This is enough to compensate for gravity.

rcgldr

Note, if there's air in the straw, then the water level doesn't quite rise as much as the straw, because the air in the straw expands as it's pressure decreases. It's a very small amount in the case of a straw, but if this were a huge tube, 20 meters tall, there would be a large decrease in pressure and a bit increase in expansion of the air.

The pressure of the water inside the straw at the same height as the top of the water outside the straw is 1 atm, and the pressure above that point decreases according to the formula, pressure(height) = 1 atm - ( density (of the water) x gravity (eg, 9.8m/s^2) x height (above 1 atm point) ).