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Why does hot air rise?

I did an experiment:

I made a model of hot air and cool air with marshmallows and toothpicks.
The hot air had more volume than the cold air.
Then, I weighed it.
They both weighed the same, even though they had a different volume.

So I don't understand why does hot air rise?

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Consider each molecule.

As you say, masses $m$ are constant but volumes usually grow with higher temperature.

That hot air molecules will float up is not do to any changes in mass or so - but to changes in density $\rho$:

$$\rho=\frac{m}{V}$$

The fluid of highest density will seek the bottom, and this will be the colder air molecules:

$$\rho_{cold}=\frac{m}{V_{cold}} \text{ and } \rho_{hot}=\frac{m}{V_{hot}} \text{. Since } V_{hot}>V_{cold} \text{ then }\\ \rho_{hot}<\rho_{cold}$$

Think of putting light plastic balls in a bowl along with stones of equal sizes. Shake the bowl and the lighter balls will end up on the top and the heavy stones on the bottom.

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    $\begingroup$ I disagree, I don't find the lower "density" of hot air to be convincing because each hot molecule has the same density as a cold molecule. $\endgroup$ – CognisMantis Aug 7 '15 at 3:35
  • $\begingroup$ This answer is not correct - at least not correct in the case where the "air" is composed of hard-spheres. (Hard sphere model.) In this case, the only surfaces are the surfaces of the hard spheres. There is no "pressure" on any "layers of air" as such. Those things don't exist. $\endgroup$ – user3728501 Aug 20 '15 at 15:24
  • $\begingroup$ This is an incorrect answer: Only sufficiently hot air will rise. Particularly, only if the upwards entropy gradient is negative, then buoyancy will set in. This has to do a lot with the physics of instabilities, and nothing with such oversimplified density arguments. $\endgroup$ – AtmosphericPrisonEscape Mar 8 at 13:23
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Due to Charles law, the volume of the gas is directly proportional to the temperature of the gas,

$$ V \propto T $$

or

$$ \frac{V}{T} = k $$

Therefore, when the temperature of the gas is increasing, the volume of the gas is increasing as well. And due to the conservation of mass, the temperature will not affect the mass of the gas.

Since density is

$$ \rho = \frac{m}{V} $$

And the mass of the gas is constant

$$ m = const $$

That means the density of the gas always inversely proportional to the volume of the gas, and hence also inversely proportional to the temperature of the gas.

$$ \rho \propto \frac{1}{V} $$

and

$$ \rho \propto \frac{1}{T} $$

Conclusion: When the hot air expands, it becomes less dense than its surrounding air, then the air pressure will exert an upward buoyant force for the hot air to rise.

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