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When we study kinetic theory of gases we find relationships between the processes of diffusion, heat conduction and viscosity, and collisions in the gas. In particular the diffusion coefficient is given approximately by $$ D \simeq \frac{1}{2} \lambda \bar{v} $$ where $\lambda$ is the mean free path and $\bar{v}$ is the mean speed of gas molecules. We can understand the diffusion process as a type of random walk. We can understand conduction of heat the same way. As each molecule moves around, it brings its energy with it, so this general idea suggests that the coefficient in the thermal diffusion equation, $\kappa / C$ (where $C$ is the heat capacity per unit volume) should be equal to the diffusion coefficient, and therefore $$ \kappa \simeq C D = n C_1 D $$ where $C_1$ is the heat capacity per molecule. As I understand it, this simple argument gets the answer wrong by a factor about 2, in that a more precise formula is $$ \kappa \simeq \frac{25}{12} n C_1 D. $$ My question is, why is the thermal diffusion process that much faster than one might have guessed? Is there an easy way to say why the first argument has to be multiplied by a factor of approximately 2? e.g. It could be something such as a mean free path for energy transport is longer than for particle transport, or perhaps a bigger influence of persistence of velocities for the thermal conduction as compared with particle diffusion. Any better informed suggestions out there?

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  • $\begingroup$ I would postulate the (hotter) higher energy molecules have sufficient velocity to penetrate more deeply into the fluid-bulk leaving cooler molecules to gain more energy at the interface of energy transfer. $\endgroup$
    – Jim Clark
    Commented Jul 12, 2021 at 19:45
  • $\begingroup$ Compared to a random walk. $\endgroup$
    – Jim Clark
    Commented Jul 12, 2021 at 19:53
  • $\begingroup$ @JimClark thanks; I suspect it is something like that; I know it has been calculated more fully but I have not (yet) read any more complete calculation $\endgroup$ Commented Jul 12, 2021 at 20:13
  • $\begingroup$ I think the diffusion analysis neglects energy exchange beween the two gas components. Not only high energy molecules diffise into the colder gas, but they also heat it. Note also that diffusion is Brownian motion, that can be analyzed on a level of a single molecule, whereas temperature is defined only for a macroscopically small volume. But I am just speculating. $\endgroup$
    – Roger V.
    Commented Jul 12, 2021 at 20:25
  • $\begingroup$ Does this have to do with how you construct the integrals to define the diffusion coefficient etc.? That is, when you compute collision frequencies between particles, there is often a bunch of extra numerical constants out front due to geometrical considerations during the integration... $\endgroup$ Commented Sep 8, 2022 at 14:29

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Since asking the question I have found out more about this. The answer is essentially that when two molecules collide, the energy passes on more readily than the molecules themselves. Consider, for example, a head-on collision of two molecules of the same mass, with one of them initially at rest. The incoming molecule stops completely, thus placing a limit on diffusion of molecules, but the kinetic energy is fully transferred, thus in such a collision there is no restriction on diffusion of kinetic energy. More generally, the effective mean free path for the diffusion of kinetic energy is considerably larger than the one for diffusion of particles. (The related effect in a solid is of course well known: particles don't diffuse but their energy does.)

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