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Diffusion problem

  • Thread starter superwolf
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1. Homework Statement

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2. The attempt at a solution

a) The steady state concentration is the concentration when t --> infitine, right? How can I find that when I don't know c(x, 0)?
 

CarlB

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You have two fluxes, one due to diffusion, the other due to the potential. In steady state, these cancel each other. So try writing those equations down and solving them.
 
Flux due to diffusion = [tex]D\frac{dc}{dx}[/tex], right?

How about the flux due to the potential? Is that simply [tex]ax^2[/tex]?

It that is true, I get [tex]c=\frac{a}{3D}x^3[/tex]

I still don't know what to do with b) and c) though...
 
Last edited:

CarlB

Science Advisor
Homework Helper
1,212
8
No, you need to convert the potential into a force. You do that by taking the gradient.

Now is it the case that [tex]D\frac{dc}{dx}[/tex] is a force? If it is, then maybe the next step would be to set these forces equal, diffusion and that from the potential. As far as getting a force from diffusion, would that have something to do with "pressure"? I hope there is something in your text or notes that will further you on this.

By the way, I think Einstein was the one who originally solved this question for Brownian motion. One of the applications is to calculate the density of the earth's atmosphere as a function of altitude. In this case, the potential is due to gravity, which is balanced against diffusion. (I.e. that's why we can breathe at even high altitude places a couple miles above the lowest points on the planet.) Here's an article:
http://psas.pdx.edu/RocketScience/PressureAltitude_Derived.pdf [Broken]
 
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