Imagine a stainless steal container with an unsealed lid that contains normal air, inside a room containing normal air. Now, pump a steady flow of CO2 in that container, bringing it to positive pressure versus the environment outside. This system (inside the container) should eventually reach a steady state. Let's look at the Oxygen content of the system.

By Fick's law of diffusion, since the partial pressure of O2 outside is greater than inside, O2 should start to seep in until the partial pressure of O2 inside and outside has reached equilibrium.

However, by the second law of thermodynamics, energy should always flow toward the lower entropy system.

In a less theoretical mental model, we can visualize the CO2 exiting the container in a equal manner from the junction of the lid and the container. The flow being faster than the rate of diffusion, even if the partial pressure of O2 in the container has reached zero, I have a hard time imagining how there could be O2 ingress in the container.

Asked in a different way: How would you model O2 ingress of the container in the example ?

EDIT: I am referring to the passage about Fick's law in this book

  • $\begingroup$ You are describing a situation in which oxygen is diffusing in the direction opposite to that of the bulk flow. Is this correct? $\endgroup$ Aug 9, 2018 at 0:50
  • $\begingroup$ @ChesterMiller Correct $\endgroup$
    – m4mush
    Aug 9, 2018 at 0:57
  • $\begingroup$ I would think that the bulk flow would prevent O2 from flowing in. This is a concern in beer brewing, where O2 is avoided in certain situation. $\endgroup$
    – m4mush
    Aug 9, 2018 at 1:13
  • $\begingroup$ Let's simplify and see whether we can get a handle on this. Suppose you have 1D bulk gas flow in the positive x direction with a velocity V, and, initially, the oxygen concentration is 0 in the region x <0 and concentration $C_0$ in the region x>0. The diffusion coefficient of oxygen in the gas is D. You would like to know the concentration of oxygen as a function of time and position, particularly in the region x < 0. Would the answer to this simplified model help give you a handle on the answer to your question? $\endgroup$ Aug 9, 2018 at 1:15
  • $\begingroup$ There is no reason to expect an O2 ingress into a container that is at a higher pressure than ambient. I am sure that Fick worked on systems that did NOT contain pressure differences. $\endgroup$ Aug 9, 2018 at 1:18

1 Answer 1


For the problem I posed in my comment, the solution for the concentration as a function of time and position in the case where the fluid velocity V is zero is as follows: $$C=\frac{C_0}{2}\left[1+erf \left(\frac{x}{2\sqrt{Dt}}\right)\right]$$where erf is the error function. For finite fluid velocity, the solution is then:$$C=\frac{C_0}{2}\left[1+erf \left(\frac{x-Vt}{2\sqrt{Dt}}\right)\right]$$For negative values of x, this becomes: $$C=\frac{C_0}{2}\left[1-erf \left(\frac{|x|+Vt}{2\sqrt{Dt}}\right)\right]=\frac{C_0}{2}\ erfc \left(\frac{|x|+Vt}{2\sqrt{Dt}}\right)$$ Do you want to do calculations of your own to evaluate this as a function of x and t, or would you like me to continue with the analysis?

  • $\begingroup$ It bothers me that there is no \erf and \erfc operators for MathJax. I was going to edit that in, but to my surprise I doesn't exist! $\endgroup$
    – Kyle Kanos
    Aug 9, 2018 at 14:46
  • $\begingroup$ There's always Abramowitz and Stegan $\endgroup$ Aug 9, 2018 at 15:39
  • $\begingroup$ @ChesterMiller Yes, please do. I appreciate you taking the time to help me out. My goal is to see if C > 0 for any t > 0 when x < 0 $\endgroup$
    – m4mush
    Aug 10, 2018 at 1:33
  • $\begingroup$ It would be of interest to people bottling beer to also see if C > 0 when V = 0. My attempts at modeling this in Mathematica have been unsuccessful so far. $\endgroup$
    – m4mush
    Aug 10, 2018 at 1:39
  • $\begingroup$ OK. If you actually did calculations using this equation, the results would be as follows: For the case where V=0, C would be equal to $C_0/2$ at x = 0 and all times t > 0. At all locations x <0, C would start out at a value of 0, and then gradually rise over time until, at long times, it approached $C_0/2$. In other words, it would be > 0 at all times >0, although far from x = 0, its value would be very small until long enough times for trace species to diffuse there. I'll continue with V>0 in the next comment. $\endgroup$ Aug 10, 2018 at 4:55

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