[blenhart (OP)]

im doing a project for my chemistry class, and i am doing it on how atmospheric pressure can be used to crush a can with boiling water inside of it. im not sure what law explains this phenomenon though. is it the ideal gas law? combined gas law? help would be greatly appreciated pleaseeeeeee 

[damocles]

Probably the best approach to an explanation of this demonstration is in terms of equilibrium vapour pressure. If you have liquid water, there is always a little bit of gas phase water (water vapour) in equilibrium with it. At the freezing point the amount of water vapour corresponds to a pressure of about 0.5% of atmospheric. It mixes in with the rest of the atmosphere. But the equilibrium vapour pressure increases rapidly with increasing temperature, so that at laboratory temperatures it is around 3% of atmospheric.

The boiling point of water is the temperature at which the equilibrium vapour pressure becomes equal to atmospheric pressure. So if you vigourously boil water in a vessel with a small opening, the vapour will come off at atmospheric pressure, and drive the other air out of the vessel before it mixes much. When the vessel is sealed and allowed to cool, there is almost no air inside, and the water vapour inside condenses back to liquid water, and drops its pressure as it does so.

When the vessel has returned to room temperature, you have a pressure of 100 kilopascal from the air on the outside of the vessel, but only a pressure of about 3 kilopascal trying to balance it from the inside. Many vessels will collapse under this sort of pressure difference.

Atmospheric pressure of about 100 kilopascal is equal in more familiar units to about 1 kg. wt. per sq. cm., or about 14.5 lb. wt. per sq. in. for those who still use imperial units.

The gas laws only enter into this sort of explanation in the most indirect way. But the demonstration is often used to illustrate just what a large pressure the atmosphere really exerts.