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Using a kinetic model for gases, is it possible to derive a relationship between pressure and thermal conductivity?

From what I have read, I thought pressure was independent of thermal conductivity?

$$\kappa = \frac{n\langle v\rangle\lambda c_V }{3N_A} \tag{1}$$

And $$\lambda =\frac{kT}{\sqrt 2\pi d^2 p} \tag{2}$$ and then just plug this into eq. $(1)$? .

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    $\begingroup$ Just go to the extreme of a "zero" pressure vacuum. There will be no thermal conductance through convection since there is no medium though which heat is exchanged. Hence there must be a dependency. $\endgroup$ – Jan Bos Feb 22 '16 at 1:22
  • $\begingroup$ It is a way of measuring pressure: en.wikipedia.org/wiki/Pirani_gauge $\endgroup$ – Pieter Jul 22 at 8:46
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It is always a good first step to think about the limiting cases. The simple limiting case for the thermal conduction of a gas in dependence of the pressure is the thermal conduction of a vacuum (which is the limit $p \to 0$). This tells you that thermal conductivity depends strongly on pressure (at least for some pressures).

You can derive a heat conductivity formula for a gas akin to the Drude formula for the electric conductivity of a classical electron gas (source: Hyperphysics on heat conductivity):

$$ \kappa = \frac{n\langle v \rangle \lambda c_V}{3N_A} $$

Where $\langle v \rangle$ is mean particle speed (that is $\langle \sqrt{ \vec v \cdot \vec v}\rangle$, $n$ is the particle density, $c_V$ is the molar heat capacity at constant volume and $\lambda$ is the mean-free path.

At very low pressures (high vacuum) the dependence on pressure is linear, as the mean free path is limited by the system size and not by collisions in the gas, but the density $n$ is proportional to the pressure (due to $pV = NRT$, which means $n = p/RT$).

For higher pressures the collision probability rises and dominates the mean free path, then $\lambda$ has a dependence approximately like $\lambda \propto 1/p$, which means that the heat conductivity saturates and only slightly rises with a further increase of pressure.

Note, that for gases in an external field (gravity) the dominating heat transport process is usually convection, not conduction. Convection depends even stronger on pressure as can be shown by a crude approximation. We assume a warm object in a large pool of fluid. A first approximation is the the viscosity of a dense gas is independent of the pressure. Convection is driven by density differences, assuming laminar flow, the flow rate is proportional to the driving pressure difference: $$ I \propto \Delta p. $$ The pressure difference that drives the convection is due to the difference in hydrostatic pressure exerted by the heated column above the heat source and the non-heated column. As the hydrostatic pressure is given by $p = \rho h g$ and the density is given by $\rho(p) = \frac{p \rho(p_0)}{p_0}$, we get that $\Delta p \propto \Delta \rho \propto p$. The transferred heat is proportional to the temperature difference of the object and the fluid, the particle number flux and the heat capacity per particle. Combining the above results we get $$\dot Q \propto p^2.$$

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  • $\begingroup$ I appreciate your response, but I am still not seeing a derivation that allows for a direct relationship between thermal conductivity and pressure. You talk about electrical conductivity. Remember, I would like to use the kinetic theory of gases to derive this relationship. Are you ultimately saying that Q dot depends on the pressure squared, and we know thermal conductivity can be related to Q? $\endgroup$ – Jackson Hart Feb 22 '16 at 2:17
  • $\begingroup$ (a) I am not talking about electrical conductivity. The quantity $\dot Q$ denotes the rate of heat exchange and earlier I say "akin to the Drude formula" (because it is also a classical transport equation not because it has anything to do with electrical conductivity). $n$ has dimension $1/\text{Volume}$. (b) There is not simple relationship, except in some limiting cases. (c) The $p^2$ behavior is for convection not conductivity. $\endgroup$ – Sebastian Riese Feb 22 '16 at 3:11
  • $\begingroup$ Can't I use the equation for thermal conductivity here: hyperphysics.phy-astr.gsu.edu/hbase/thermo/thercond.html and then use the equation for mfp as a function of pressure? $\endgroup$ – Jackson Hart Feb 22 '16 at 3:16
  • $\begingroup$ This is what I did, which gives the result, that as a first approximation the heat conductivity is independent of pressure for sufficiently large systems/sufficiently dense gas. But there are pressure dependent corrections to the heat capacity (due to interaction effects) and the mean free path formulas are not exact. One can for example relate the mean free path to the pressure and the viscosity with $\lambda \propto \mu / p$ – then you have to calculate the viscosity in dependence of the pressure. This leads to non-universal corrections. $\endgroup$ – Sebastian Riese Feb 22 '16 at 3:27
  • $\begingroup$ Why can't I use this equation for mfp: hyperphysics.phy-astr.gsu.edu/hbase/kinetic/menfre.html You can see it has pressure in it. I can just plug that into thermal conductivity eq right? $\endgroup$ – Jackson Hart Feb 22 '16 at 3:28
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Early work in the upper atmosphere showed that electrical conductivity of air depends upon the pressure; the quantitative relationship is known as Paschen's Law, https://en.wikipedia.org/wiki/Paschen%27s_law.

But what about thermal conductivity? As pressure is reduced, so is the density of the gas particles, which changes the mean free path statistics. Thus continuing to reduce the pressure will eventually result in changes in the thermal conductivity: http://www.electronics-cooling.com/2002/11/the-thermal-conductivity-of-air-at-reduced-pressures-and-length-scales/.

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  • $\begingroup$ this is not exactly a derivation it does not seem. $\endgroup$ – Jackson Hart Feb 22 '16 at 2:11
  • $\begingroup$ It's up to you to do the derivation - I've provided the element you were missing: the increase in the mean free path, and the physics behind it. $\endgroup$ – Peter Diehr Feb 22 '16 at 2:23
  • $\begingroup$ The first link you provided does not mention the word "conductivity" a single time. The second link has a single equation. Forgive me for not directly seeing the way to proceed $\endgroup$ – Jackson Hart Feb 22 '16 at 2:26
  • $\begingroup$ Could I use the equation for thermal conductivity here, hyperphysics.phy-astr.gsu.edu/hbase/thermo/thercond.html and the equation for mfp as a function of temp and pressure, and just plug that in for mfp in thermal conductivity equation? $\endgroup$ – Jackson Hart Feb 22 '16 at 2:45

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