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The following explanation is from my textbook:

Consider a small region within a gas where the temperature increases in the $x$ direction. If $\Delta t$ is the average time between collisions, some atoms will move from box 1 to 2 and some from box 2 to 1 during this period.

If the total energy of atoms in box 1 is $U_1$, then the energy crossing the dotted line from the left is roughly $U_1/2$ since only half of the molecules will have positive $x$ velocities at this moment. Also, $l$ is the mean free path of the particles which is the distance they travel during $\Delta t$.

Now here's what I don't understand:
1- Why is the energy crossing the dotted line $U_1/2$?
2- Why does it say that only half of the molecules will have positive $x$ velocities at this moment?

Thank you in advance

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This conclusion comes from the fact that, this being a collection of a large number of particles with no external forces, the gas's average particle velocity vector is zero. This means it is equally likely to find a particle moving in any direction. Crossing the boundary requires that the $x$-velocity is positive. The possible velocity orientations with positive $x$-velocity constitute half of the total number of orientations, so half of the particles are oriented to cross the barrier at any given point in time. Assuming that $\langle |\vec{v}|\rangle\Delta t>>\ell$, any particle that has a positive $x$-velocity should cross the barrier. Therefore half the energy leaves in the interval $\Delta t$.

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  • $\begingroup$ Thanx for the answer. I've added an edit. $l$ is the mean free free path of the particles. So I think we can't say $\langle |\vec{v}|\rangle\Delta t>>\ell$. Am I right? $\endgroup$ – Ali Jul 14 '17 at 6:58
  • $\begingroup$ @Ali I direct your attention to the word "roughly" in your question. It may have been that the textbook writer assumed that a particle with positive $x$-velocity will always cross the barrier. $\endgroup$ – probably_someone Jul 14 '17 at 7:06
  • $\begingroup$ I think I know what you mean. This would mean that the particles have ONLY $x$ components. Which doesn't seem to be a bad assumption. $\endgroup$ – Ali Jul 14 '17 at 7:37

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