Why do those compressed air cans freeze when you hold them upside down?

[dpk777]

just wondering...

 

[conjur]

Ask Dr. Science

 

[Brymo]

its compressed air... and when you have it down below it can not blow out, which makes it freeze over..

in fact i have no f'ing clue...

but its CO2 in a can, so it will get cold even if it is right up

 

[Rastus]

The can freezes as the pressure inside is released. Holding it upside-down makes the pressure release faster, because it is liquid that is coming out.

 

[silverpig]

Uh, they won't freeze if you just hold them upside down (or do they? I've never tried).

If you release all the air from them, then there's a simple explanation. I'll simplify it as much as possible. A pressurized gas will decrease in temperature as the pressure is released, and increase in temperature as the pressure is increased. This is how most cooling systems work. A gas is compressed causing an increase in temperature. It is then passed through a radiator to bring it to room temperature. Then it gets pumped where it is needed and the pressure is reduced. The gas cools, and draws in heat from it's surroundings, thereby cooling them. The now near room temperature gas is compressed, heating it up, and is passed it back to the radiator again to continue the cycle.


edit: clarity

 

[Soybomb]

Not all do, it depends on the gas inside

 

[amnesiac]

I don't know, but I do know that turning those cans upside down and spraying bugs is a good way to get them to stop....er.....bugging you.

 

[Hayabusa Rider]

Yep, not all do, but many cans of "compressed air" aren't. They are cans of propellent that are liquid under pressure, but gas at STP (you can google it boys )

If the liquid in the can is in two layers with the propellent at the top, inverting it would cause the propellent to be expelled instead of the usually desired substance. As the propellent escapes, it boils, removing heat and making the can or nozzle cold.

 

[glen]

Physics:
The entropy of the air increases and the temperature decrease by a corresponding amount.