Table of Contents

• Developing the Equations
• 1: Review of diffusion equations
• 2: The two-compartment model
• 3: What does continuous mean?
• 4: Examples of permeability
• 5: A formula for permeability
• 6: The area of the membrane
• 7: The final equations
• Examples
• 8: Calculating molecular weight
• 9: Molarity of sugar-goo
• 10: Practicing conversions
• 11: Rate of flux
• 12: Review
• Discrete vs. Continuous
• 13: A metaphor
• 14: Continuous curves
• 15: Discrete curves
• 16: Iterating the discrete curve

We compared the shape of the curves given by the two models. Solving the continuous model here is beyond the scope of this module (that would get into a lot of calculus), but we can do the discrete model with a modest amount of algebra.

Let's go back to sugar water, with the following assumptions:

• Initial concentration on left = 1 M
• Initial concentration on right = 0 M
• P = 0.1 cm/sec
• A = 1 cm²
• equal volumes on both sides of membrane

We want to know what the final concentrations will be over time , i.e., at time 1 sec, 2 sec, 3 sec, 4 sec....... So lets set up our equation from above we know that :

Δn/Δt = 0.1 cm/sec x 1 cm² x (1M - 0M) = 0.1 moles/sec

So in the first second, approximately 0.1 moles change sides (from the more concentrated to the less concentrated side). That leaves

on the left side: 1 moles - 0.1 moles= 0.9 moles/liter

on the right side: 0 moles + 0.1 moles= 0.1 moles/liter

How about the concentrations at time 2 seconds? We need to readjust our estimate of diffusion rate by plugging in the new concentrations. Everything else stays the same:

Δn/Δt = .1 cm/sec x 1 cm² x (.9M - .1M) = .08 moles/sec

and the new concentrations are 0.82 and 0.18M.

on the left side: 0.9 moles - 0.08 moles= 0.82 moles/liter

on the right side: 0.1 moles + 0.08 moles= 0.18 moles/liter

Continuing on like this,

Δn/Δt at 3 seconds = 0.074 moles/sec,
C on left = 0.746 M, C on right = 0.254 M

Δn/Δt t at 4 seconds = 0.049 moles/sec,
C on left = 0.697 M, C on right = 0.303 M

Δn/Δt t at 5 seconds = 0.039 moles/sec,
C on left = 0.658 M, C on right = 0.342 M

Δn/Δt at 6 seconds = 0.032 moles/sec,
C on left = 0.626 M, C on right = 0.374M

This process is called "iteration" -- you iterate, or repeat, the equation over and over, each time substituting in the values calculated in the last iteration. You can see that, if we continued iterating long enough, we would eventually reach a point where the two concentrations were equal -- in other words, an equilibrium. (It would take a long time though, better use a spreadsheet rather than a calculator!) If you graph the two sets of concentrations, you'll see something like this:

Congratulations, you've reached the end of the diffusion through a membrane section! You can now proceed to take the quiz...

 

 

 

If you want a printer-friendly version of this module, you can find it here in a Microsoft Word document. This printer-friendly version should be used only to review, as it does not contain any of the interactive material, and only a skeletal version of problems solved in the module.