Calculating Average Distance Diffused

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The discussion centers on calculating the average distance a molecule of insulin diffuses in water using its diffusion coefficient of 8.2 × 10−7 cm²/s over one hour. Participants question whether to incorporate mean free path and rms speed equations or simply use the diffusion length formula. The proposed calculation suggests substituting D and time (3600 seconds) into the diffusion length equation, yielding approximately 0.0768 cm. Additionally, a more complex approach involving concentration functions is mentioned for determining average distance traveled. The conversation highlights the need for clarity in interpreting the problem's requirements.
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Homework Statement


In a simple model of diffusion in one dimension the quantity D can be shown to be given by half the product of the mean free path and the rms speed of the diffusing particle. Calculate the average distance a molecule of insulin will diffuse in water in one hour if its diffusion coefficient is 8.2 × 10−7 cm^2/s.

Homework Equations



J = −D(dc/dx)
Diffusion length = (2 x D x t)^0.5 -Einsteins

Do I need the mean free path and rms speed equations here?

The Attempt at a Solution


[/B]
The thing confusing me is the wording of the problem, do I need to use the mean free path and rms speed equations or is the question just making a statement? Can I get away with subbing in D and t = 3600s? Is that finding the same thing - in that case the answer would be Diffusion Length = (2 x 8.2 × 10−7 cm2/s. x 3600s)2 = 0.07683749085cm/hour?
Thanks in advance.
 
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Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
It would interpret this problem differently. If the initial concentration is zero, and, if C0 is the concentration at the boundary, then the concentration as a function of time and distance is given by:
C=C_0erfc\left(\frac{x}{2\sqrt{Dt}}\right)
To get the average distance traveled, use:
x_{ave}=\frac{\int_0^\infty{xCdx}}{\int_0^\infty{Cdx}}

Chet
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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