If a hot-air balloon was approaching an active volcano where the air temperature outside was presumably much hotter than the air inside the balloon itself, would it start to fall into the volcano or still be able to fly?


2 Answers 2


What really matters is the total weight of the gas inside of the balloon (could be air, helium, etc) versus the mass of the air that would otherwise occupy the space if the balloon wasn't there. A hotter gas with a certain number of molecules tends to occupy a greater volume, thus being less dense and weighing less than a colder gas occupying the same expanded volume. Since it weighs less, a balloon experiences lift-off!

Once the balloon approaches an area where the density of the air below it is less than the density of the air inside of the balloon, it will begin to drop. I would indeed expect it to start to drop into the volcano, as you suspect.

The one caveat may be that the air on top of a volcano is not always stagnant. Instead, as air escapes from the earth, it rushes upward (there may be a wind of things like carbon dioxide). This will provide some lifting force that may help the balloon stay aloft, if the force is strong enough and the relative densities (or relative temperatures) of the air inside versus outside of the balloon large enough.


I'm ignoring the effect of any solid particles that the volcano might be introducing.

Note that the situation you are describing, low-density air parcels adjacent to high-density air parcels, is unstable. The low-density parcels will experience buoyancy lifting forces in the same way the balloon does.

This air parcel should therefore be driven upward. A balloon inside will experience both buoyancy and drag forces. It is quite possible that the balloon could be less dense than the air surrounding, yet still rise relative to the surface because of the drag from the rising air.


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