0
$\begingroup$

Suppose an ideal airtight room, filled with air, no changes in volume, amount of air or energy.

On the one hand, "hot air rises" (less dense, flotation, convection...) ((there's gravity, we are on Earth))

On the other hand, temperature tends to equalise (thermodynamics, heat flow, conduction, convection, diffusion, ...)

Do these two effects end up in a dynamic equilibrium in the end, resulting in some heat gradient? Or does the first effect just disappear in this ideal scenario?

From real life, and what I've absorbed from my vague unscrutinised encounters with thermodynamics over the years, I feel the answer should be the latter, but my understanding votes for the former. Maybe it's the former but to a negligible degree?

My confusion comes from the undersanding that, even in equilibrium, some molecules have higher energies than others (that's how evaporation works IIUC). Wouldn't they tend to rise?

$\endgroup$
  • $\begingroup$ What is the mean free path of a molecule of air at room temperature? Do collisions between molecules allow for exchange of kinetic energy between molecules? $\endgroup$ – Chet Miller Nov 4 '18 at 12:30
  • $\begingroup$ @ChesterMiller From Wiki, it's ~70nm. And IIUC, yes, collisions do allow for exchange of kinetic energy. $\endgroup$ – Woe Nov 4 '18 at 14:16
  • $\begingroup$ Then do you really think that all the high energy molecules are going to make it to the top of the container before then are slowed down by collision with lower energy molecules? $\endgroup$ – Chet Miller Nov 4 '18 at 14:42
  • $\begingroup$ @ChesterMiller I don't see it yet, I don't see clearly why that's necessary. I would think all we need is for the high energy molecule to be sliightly more likely to collide with a molecule on top of it to end up in a gradient situation (due to gravity or whatever the microscopic explanation of flotation is). But that doesn't seem to be the case, and that's why I'm asking. $\endgroup$ – Woe Nov 10 '18 at 11:11
0
$\begingroup$

I think a small change in wording may answer your question: it’s not hot air that rises but hotter air. As an example: consider a hot air balloon. The air in the balloon is hotter than the surrounding air. Thus it’s less dense. This produces the buoyant force required to compensate for the weight of the balloon and its passengers.

Concerning convection the situation is a little bit different. Convection sets in when there exists a temperature stratification, the physical term is temperature gradient. For this gradient to occur it is required to have a heat source from below. But even then, only when the temperature gradient is big enough, convection will occur. This minimum gradient is called adiabatic gradient. As long as convection is ongoing, the temperature gradient will adjust to the adiabatic gradient but it will not completely disappear.

Bottom line: it takes two things. Firstly a temperature difference or gradient and secondly gravitational force that produces the buoyancy. Gravitation alone will not do the job. Thus, in the situation you describe in your question no convection or similar effects occur.

$\endgroup$
0
$\begingroup$

Do these two effects end up in a dynamic equilibrium in the end, resulting in some heat gradient?

In equilibrium, the temperature will be the same everywhere in the room, which corresponds to the maximum entropy. So, there won't be any temperature gradient.

Or does the first effect just disappear in this ideal scenario?

Yes, the first effect, i.e., convection or any other forms of heat transfer on macro level will disappear, since the temperature everywhere will be the same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.