For you chemistry and physics types: does the lifting power of helium change according to how small of a space you put it in?
Asked by
Nullo (
22033)
February 24th, 2010
Say you have two balloons. One will hold 1l of helium at STP. The other will hold ½l at STP. You fill the balloons with the same mass of gas.
Is there any difference in the amount of lift generated?
For the purposes of the exercise, the balloons weigh nothing on their own, and will not break.
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13 Answers
Yes. Basically the buoyancy of an object will depend on its volume and density.
Lift has to do with displacement of air and density of the gas displacing the air. In a rigid balloon you could get a lower density for the displaced air than a non-rigid balloon using the same mass of gas inside. This is due to atmospheric pressure on the non-rigid balloon preventing very low densities from occurring within the balloon.
The Vulcan Science Directorate has determined that the smaller balloon will have higher density and less buoyancy.
You don’t see bottles of compressed helium floating in the air.
Helium is only lighter than air because of its density. When you compress it, you eventually reach a point where it becomes heavier than air, thus, losing its buoyancy.
@ragingloli is correct (and more than just a little weird)
@Nullo Well, the containers are quite heavy themselves, seeing as the pressures exerted by highly compressed Helium can be quite large. However, the point still stands that it is the density of the gas which allows it to float.
Actually, nobody seems to have answered your question yet, so, here goes.
Helium, like all gasses, is compressible. That’s what it means when you have a gas. The atoms are scattered around somewhat randomly, unlike fluids, which have only limited compressibility, or solids which generally have even less. (I’m talking generalizations here that apply in the normal everyday universe, so let’s not get into things like neutronium.)
When you say 1 liter of helium versus ½ liter with the same mass of gas, that means that the 1 liter balloon has twice the volume and hence ½ the density of the other baloon. As other posters have pointed out, lifting power is a function of density. Helium is a lighter gas than air, which is made up of heavier elements, mostly nitrogen and oxygen. With less density, the 1 liter baloon will lift more than the ½ liter baloon.
Taking the example of the gas compressed into a container, although the container sits on the ground just fine, if you fill up a baloon big enough with the helium it will lift the container up, up, and away.
Scuba divers are familiar with a somewhat analogous phenomenon. A “cylinder” (i.e. breathing tank) full of compressed air is actually negatively buoyant, meaning that it will sink in water. Use up most of the air and it may actually be positively buoyant, meaning it will float.
Hope this teaches you how not to be an air head.
Two same masses of gas, both at STP will have the same volume. The question to me therefore is inerpreted as that both balloons are in ambient STP. For one mass to be in a volume that is twice as big as the other, means it will have half its density. For that reason the bigger balloon will have bigger lifting power, be it not twice as much, but a little more.
The bigger balloon will way the same, but displace twice the amount of air. The lifting power of the balloon equals the weight of the displaced air-/- the weight of the balloon. The displaced weight will be twice as high, the weight of the balloon will be the same.
I don’t think I understand the question the way it’s phrased. You’re saying you are putting the same molar amount of helium into a one liter balloon as a half liter balloon, and that they are both at the same pressure? For that to happen, one of the balloons has to be much colder than the other. That’s Boyle’s law. They CAN’T be at the same temperature. If they are, then one will be at a higher pressure because they have to be.
It’s a moot point, anyway. If you have the same mass occupying half the space, then it’s going to displace half as much air as the other. It will therefore not be as buoyant – that is Archimedes’ Principle.
I assumed Nullo meant the surrounding air was at STP.
@Shuttle128 Aye, that’s it.
Sigh To think that I spent weeks debating the wording of this question…
Thank you, and thank you Ragingloli, for providing the answer.
Nullo -
Don’t despair. I think your question was reasonably clear. The basic principle for you to get from this is that, because the atoms that form the gas in the larger balloon are spread out into a larger space, it is less dense than the smaller balloon, where the atoms are, of necessity, more crowded together. Less dense = more buyancy.
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