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When an Ideal gas is left alone for a while, the atoms or molecules collide with each other, each time transferring a tiny amount of energy. So, essentially, the particles keep transferring energy and eventually all the atoms should have the same amount of energy. But this is not the case. The atoms resort to the maxwell- boltzmann distribution. But why is this so? Isn't the equilibrium state particles with equal energy?

What causes the maxwell-boltzmann distribution to be the distribution particles resort to?

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It's Boltzmann's H- Theorem. For an introduction, please read the Wikipedia page. The quantity $H$, a function of time $t$, is defined as :

$$ H(t) = \int_0^t f(E,t)log(\frac{f(E,t)}{\sqrt{E}}-1) $$

$f$ is the number of molecules having a kinetic energy between $E$ and $dE$. One sees, that this quantity is a measure of "information" in the gas.

Any given collision(s) between any two molecules of an ideal gas occurs with the molecules initially having uncorrelated velocity, angle of collision and starting points.

Now, for the gas molecules to run truly randomly, you want to make sure that the "information" - i.e. any regular pattern generated by them is minimal. To go back to your original question, if all molecules did eventually came to an uniform speed, then, starting from an initial fully random state, they will spontaneously start moving towards a particular direction. Order will rise spontaneously. This conflicts with second law.

At this point, please note, that the $H$-theorem is not perfectly rigorous in that sense - there are ongoing attempts to improve it.

Nevertheless for the purpose of this question, Boltzmann's argument is that the quantity $H$ must be minimized. And that can only happen when the velocity distribution converges to the Maxwell-Boltzmann distribution.

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  • $\begingroup$ So... is there any way to prove the H theorem? Where does this math all come from? What I mean is from which math is this derived from or how is the theory put into maths? Entropy? $\endgroup$ – Chandrahas Mar 19 '17 at 15:48
  • $\begingroup$ @Chandrahas I am on a train now, I will edit my answer with proof, when i come home. $\endgroup$ – Sean Mar 19 '17 at 16:47
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When a high energy particle has a low speed for example in x-direction, then a low energy particle can give it more speed in x-direction, increasing the high energy particle's energy.

That is how high energy particles appear to a gas whose high energy particles have been removed, when the gas is left alone for a while.

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  • $\begingroup$ Assume two molecules of same mass both running in the x direction. $Mol_1$ has velocity $v_1$, and $Mol_2$ has velocity $v_2$. If $Mol_1$ has a higher energy, then $\frac{1}{2} m v_1^2 > \frac{1}{2} m v_2^2 $ or $v_1 > v_2 $. In order for $Mol_2$ to transfer energy on $Mol_1$ former had to collide with the later from back. Will that ever happen? Experiment with billiard balls. $\endgroup$ – Sean Mar 19 '17 at 16:52
  • $\begingroup$ I learned that molecules or billiard balls with one degree of freedom will not resort to Maxwell-Boltzmann distribution by collisions. $\endgroup$ – stuffu Mar 19 '17 at 22:13

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