What is the barometric height formula for a mixture of gases?

The main problem I can see, the curves describing the minimal energy (kinetic + gravitational/centrifugal) won't be any more a simple derivative, but a functional equation.

I think, this or similar equations could be useful in very separate areas (isotope separation, estimating high atmosphere components of planets, etc.) Despite I couldn't find anything with google about this.

So, the question is: we have a high gas mixture in constant g gravity and T temperature. We know the $p_1$, $p_2$, ... partial pressures of the gases on the $h=0$. Of course we know every relevant constants of the gases, too. What is their minimal-energy, isothermic, height-dependent partial pressure function $p_n(h)$?

  • $\begingroup$ Shouldn't you just be able to insert the molecular Mass of each gas $p(x) = p(0)e^{-\frac{xMg}{RT}}$ and calculate ratios? $\endgroup$ – John H. K. Jan 21 '15 at 21:27
  • $\begingroup$ @JohnH.K. Why? The pressure (the whole) is created by the weight of the whole gas mix over the the actually examinatex hight level. Not only the individual components are working independently. $\endgroup$ – peterh - Reinstate Monica Jan 23 '15 at 0:14
  • $\begingroup$ With p(0) and p(x), I meant the partial pressure of one specific component. $\endgroup$ – John H. K. Jan 23 '15 at 6:00
  • $\begingroup$ @JohnH.K. I understood what you think, but I think it is not okay. p(x) [the whole, not only the partial] is coming from the weight of the whole mix over h>x . Not only from a single component. $\endgroup$ – peterh - Reinstate Monica Jan 23 '15 at 7:21

The barometric height formula is defined as:

$$ p(z) = p(0)\exp(-Mgz/(RT))$$

For a centrifuge the upper expression is transformed to: $$ p(r) = p(0)\exp(M\omega^2r^2/(2RT))$$

As you see the $g$ is transformed to the radial acceleration $a = \omega^2 r$. The factor $2$ stems from the integration which you have to do during the derivation of the barometric height formula (see barometric height formula, derivation). And of course, the minus sign disappeared because the force points outward.

Both formulas hold also for a mixture of gases, I cite "Kemp, R. Scott. "Gas Centrifuge Theory and Development: A Review of US Programs." Science and Global Security 17 (2009): 1-19.":

When the rotor contains a mixture of gases, the distribution holds independently for each species.

So, as a first approximation, it is as I suggested in the comment section. The cited source gives the following equations for a two-isotope gas:

$$ p_A(r) = p_A(0)\exp(M_A\omega^2r^2/(2RT))$$ $$ p_B(r) = p_B(0)\exp(M_B\omega^2r^2/(2RT))$$

A separation factor can be calculated:

$$ \alpha_0 = \frac{p_A(0)}{p_B(0)}/\frac{p_A(r)}{p_B(r)} =\exp((M_B- M_A)\omega^2r^2/(2RT))$$

Keep in mind that all formulas here don't include convection of any kind. In earth's atmosphere, we have too much convection and temperature differences for those formulas to work. Furthermore, in centrifuges, to improve their performance, there is a current-flow introduced (see the cited source).

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