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Q & A: Weight of air

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How can I figure out how many pounds one cubic foot of air weighs?
- Kathy (age 16)
There are several ways you can do this:

1) (easiest) -- look it up in a book. My CRC Handbook of Chemistry and Physics gives the density of dry air at 20C at 760 mm of mercury (one atmosphere of pressure) to be 1.204 milligrams per cubic centimeter. I'll leave it to you to convert to pounds and cubic feet. The density of air depends on pressure and humidity.

2) You can weight it on a scale. But be very careful about this! If you fill a container (ball, balloon) with air and put it on a scale, you are measuring the weight of the ball plus the weight of the air, *minus* the weight of the volume of air that's displaced by the inflated ball (this is the buoyant force). The buoyant force is quite noticeable if you put helium in a balloon, because the total weight of the helium plus the balloon minus the buoyant force can be negative.

You could take two equally heavy metal containers (check with a balance, and you can add small weights on one side if they aren't equally heavy). You can measure their weight difference on a balance (again, by adding weights to one side until they balance out), when one container has air in it and the other has a vacuum. But then the tricky part is getting a vacuum in one of the containers.

(Maybe you could weigh a vacuum-packed can of nuts, then open the lid a bit and weigh it with air inside. You could figure out the air volume by seeing how much water you have to pour in to fill the can to the brim. Then dry off the nuts quickly, since they're too good to waste. I'm guessing that the initial vacuum is pretty good. / mbw)

An alternative is to put two atmospheres worth of pressurized air into one container and one atmosphere of air in the other and then the weight difference will be what you want.

You can go back to the balloons -- inflate one balloon with air and one with helium and find their difference in weight (you'll have to tie the helium balloon to the scale however), and you'll still need the density of helium to get your final answer.

If your container is special with a piston which can be moved in and out and held there (maybe you can do this with screw threads), you can put one atmosphere of air in a two cubic-foot container, weight it, and then squeeze the air inside down to one cubic foot and weigh that. The container and the air inside will weigh the same, but the difference in the buoyant force will be the weight of one cubic foot of air.

Can you think of any other ways to do this?

Tom J.

(republished on 08/01/06)

Follow-Up #1: weighing air

My son and I are working on this for his science fair project. Our attempts to fill a 1liter Gatorade bottle with air from a bicycle pump (we put an inner tube valve on the lid) and weigh it against an 'empty' bottle on a homemade balance scale failed catastrophically when the bottle exploded from too much pressure. It just occurred to me that my next door neighbor is a scuba diver. Maybe we could go to her workplace and ask to weigh an empty canister, then weigh it after it has been filled. What do you think? I think failure and the questions it brings is the most fun part of an experiment, but my son is losing his enthusiasm for the project.
- Marilyn Corbin (age 50)
Yeah, exploding bottles can be alarming. I think that your final weight should be taken with atmospheric pressure inside. That way you can start with either vacuum or high pressure inside and just open a little leak to atmosphere to see the change. Starting with atmospheric pressure and then pumping air in or out means too much messing with the container during the experiment- not to mention nasty surprises such as you encountered when you get too far from equilibrium.

You could take that scuba tank, filled, and put it on a good scale. Open the valve a crack, let the air out slowly, and watch the weight drop. If it's say a 3 L tank filled to about 200 atm, you could see roughly a 1 kg drop, which should be easy to measure even if the tank weighs 10 kg or more.

Mike W.

(published on 03/09/11)

Follow-up on this answer.