4
$\begingroup$

I have some Oxygen at Temp A in one container and some Nitrogen at Temp B in another container. If I mix these two containers eventually both the Oxygen and Nitrogen will be at the same temperature. Why is that?

Specifically why do the two species converge to have the same distribution of kinetic energy rather than the same distributions of momentum or any other parameter?

$\endgroup$
  • $\begingroup$ What happens when two bodies of different mass but similar velocity collide? Will they equalize their momenta, i.e. will the small body speed up (potentially a lot, if the mass ratio is large) and the large body slow down, or will they be more likely to equalize their relative velocities? $\endgroup$ – CuriousOne Apr 29 '16 at 21:35
  • 1
    $\begingroup$ "If I mix these two containers eventually both the Oxygen and Nitrogen will be at the same temperature. Why is that?" Because we observe it in experiment. $\endgroup$ – lucas Apr 29 '16 at 21:38
1
$\begingroup$

In a system of many particles, we essentially observe the most probable configuration, and relative fluctuations around it are negligible. Here I will prove that the most probable state of a 2-particle system is this with equal energies.

The probability of a state is proportional to the volume of the corresponding part of phase space. If a particle has kinetic energy between $E$ and $E+\mathrm dE$, its velocity is between $v$ and $v+\mathrm dv$, with $E=\frac 12mv^2$ and $\mathrm dE=mv\mathrm dv$, that is $\mathrm dv=\mathrm dE/\sqrt{2mE}$. The corresponding volume in phase space is $4πv^2\mathrm dv∝\sqrt E\mathrm dE$.

Now for two particles of masses $m$ and $m'$, and of kinetic energies $E$ and $E'$. At ordinary temperatures, collisions are elastic, thus total kinetic energy is conserved: $E+E'=\text{const.}=K$. If the first particle has kinetic energy between $E$ and $E+\mathrm dE$, the second one is between $K-E$ and $K-E-\mathrm dE$. The corresponding volume in phase space is thus $∝\sqrt E\sqrt{K-E}\,(\mathrm dE)^2$. The function $\sqrt{E(K-E)}$ is maximum at $E=K/2$.

$\endgroup$
0
$\begingroup$

I think, the comment of lucas is the best answer; I'll elaborate a bit

If I mix these two containers eventually both the Oxygen and Nitrogen will be at the same temperature. Why is that?

By temperature you mean the mean energy per particle. We observe experimentally, that if you mix two heaps of particles with different mean energies, after some time the energies will have the same mean. Even if the particles are completely different (see disclaimer below), i.e. have very different masses as oxygen and hydrogen.
I mean: the mean taken over the particles of one species will be equal to the mean taken over the other species. I'd say, that this is not at all trivial (correct me if you see something), if you consider them in the mechanical model as balls that collide all the time.

I'd like to stress: only this fact allows us to define temperature as the mean energy. This would not be a meaningful quantity, if the energies of a mixture would not equate!


Of course it is more correct to say, that the energy per degree of freedom will be at equillibrium. In the case of $O_2$ and $N_2$ this coinsides with an equal energy per particle.

$\endgroup$
  • $\begingroup$ The mean energy is well defined for any mixtures, even for those far from equilibrium. Out of equilibrium, there is a non stationary energy exchange governed with Boltzmann equations. $\endgroup$ – Vladimir Kalitvianski May 1 '16 at 20:20
  • $\begingroup$ Yes, it is well defined; just as the mean mass ... or the mean wordlength for the names of the particles is well defined. Nonetheless, the mean energy would not be a meanigful quantity without this property. Maybe the mean energy per mass would be, if this were the quantity that was equal in equillibrium. Or anything other. $\endgroup$ – Ilja May 1 '16 at 20:27
  • $\begingroup$ I agree with you. I just want to precise that we deal with things that appear in our equations. Their appearance makes them meaningful. Any other (imaginary) combinations are possible to construct, but they are useless since they are not involved in our routine calculations. In other words, our equations determine what we deal with. $\endgroup$ – Vladimir Kalitvianski May 2 '16 at 7:22
0
$\begingroup$

If you mix two gases, their atoms/molecules exchange with energy, momentum, etc. It is not surprising that a hot gaz heats a cold one. It (heating) can only stop when the mutual energy exchange becomes equal to both gases. So the energy distributions of "subsystems" are equal in the end, but the momentum distributions are still not.

P.S. The subsystem energies are not obliged to be equal. In dynamical equilibrium it is the heat fluxes who should be equal.

$\endgroup$
  • $\begingroup$ Well, very few thins are surprising a posteriori :) but if one doesn't know, there is no simple reason (that I see) to think that the mean energy should not depend on the mass for example. $\endgroup$ – Ilja May 1 '16 at 20:08
  • $\begingroup$ @Ilja: the energy $E$ is self sufficient and does not need any mass in it. The momentum on average is zero so it is not clear why momentum exchange should lead to heating/cooling of subsystems. $\endgroup$ – Vladimir Kalitvianski May 1 '16 at 20:12

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.