Absolute Entropy in an isolated gas

[Austin0]

Given an ideally isolated volume of a single species of gas that has reached internal equilibrium , would the individual molecules :

[A] Retain the range of individual velocities and thermal energies [if present] and keep a merely statistically constant average?
OR
Would they equalize those energies over time becoming virtually identical within a range of quantum indeterminancy?

[C] Somewhere in between??

I hope this makes sense and is addressable in its generallity .
Ie: It is not a ,"Well , you see,it all depends" , kind of situation.
Thanks Cheers

 

[Llewlyn]

Sorry i didnt understand your question.
The distribution of velocity of a perfect gas in thermodinamic equilibrium is the Maxwellian. What's the point?

Ll.

 

[Austin0]

Sorry i didnt understand your question.
The distribution of velocity of a perfect gas in thermodinamic equilibrium is the Maxwellian. What's the point?

Ll.

Hi Thanks for your responce. My question is trying to understand the limits of entropy.
Whether given sufficiently long time the distribution of velocities would be completely equalized Ie: every molecule having the same velocity or whether the Maxwellian distribution
would pertain over ANY timespan.

From your responce I take it that the statistical range would endure indefinitely.
Logically that is what I assumed but I may have been thinking too much in terms of Hawking magic boxs. Thanks

 

[adriank]

This intuitive argument might help a bit.

The second law of thermodynamics implies that the number of microstates increases over time. If all your speeds are the same, then they are all in states with the same speed; there is only way to pick the speeds of the particles to do this. However, if your speeds follow a Maxwell-Boltzmann distribution, then the speeds are spread out and this gives many possible ways to pick the speed of each particle.

In short, with all speeds equal the system would have a much lower entropy than it would if the speeds were spread out. It's the Maxwell-Boltzmann distribution that gives the greatest entropy; since entropy cannot decrease, the speeds will stay in this distribution forever.

 

[Austin0]

This intuitive argument might help a bit.

The second law of thermodynamics implies that the number of microstates increases over time. If all your speeds are the same, then they are all in states with the same speed; there is only way to pick the speeds of the particles to do this. However, if your speeds follow a Maxwell-Boltzmann distribution, then the speeds are spread out and this gives many possible ways to pick the speed of each particle.

In short, with all speeds equal the system would have a much lower entropy than it would if the speeds were spread out. It's the Maxwell-Boltzmann distribution that gives the greatest entropy; since entropy cannot decrease, the speeds will stay in this distribution forever.

Thanks --that makes complete sense and was a real help

 

[Llewlyn]

the Maxwellian distribution would pertain over ANY timespan.
From your responce I take it that the statistical range would endure indefinitely.

Well, not indefinitely : a gas returns to its initial condition due to Poincarè recurrence theorem.
But this happens in a HUGE time scale, so huge that the second law of thermodinamic holds in physics.

Ll.

 

[Austin0]

Well, not indefinitely : a gas returns to its initial condition due to Poincarè recurrence theorem.
But this happens in a HUGE time scale, so huge that the second law of thermodinamic holds in physics.

Ll.

_________________________________________________________________________
I will look up that Poincare theorem.
SO for all practicle considerations , the Maxwell distribution will hold.
From this am I correct in assuming that even if we created a situation where all the velocities were unform, that over time, the uncertainties of interactions would effect a non-uniform distribution? Maximum entropy.
Thanks for your help, it has been spot on for resolving my question.