Diffusion and why we look at only one species

[hongiddong]

Hello physicsforums,

I was wondering in the case of CO2 exchange in the capillary and alveolar system, if there is less CO2 in the alveolar, I understand that CO2 will move in the direction to the alveolar from the capillary through diffusion.

My question is, if we are looking at brownian motion of co2, why don't we also look at the other particles that are around co2. To further clarify with examples, in the alveolar, there is also a lot of oxygen as well as co2, so why would CO2 go into the direction of lower pressure despite the fact that there are other particles it might have to rub against? Why do we only look at CO2 by itself in this case of diffusion?

I hope I made sense. Thank you.

 

[sophiecentaur]

All diffusion is affected by the surrounding medium. CO₂ will diffuse much faster in air than in water but it still goes 'downhill' overall.
CO₂ is pretty soluble in water and our bodies don't need to give it a helping hand to move about via the blood. Otoh, Oxygen can only be carried around in sufficient quantities by the use of the Haemoglobin molecule, which attaches to and releases an Oxygen molecule easily enough to achieve gaseous exchange through the capillaries (with the existing concentration gradient) at both ends of its journey through the blood.
I learned recently that foetal haemoglobin is different because the gradients across the placental interface are lower than in the alveoli. This accounts for the Jaundice that's common in newborns; they have to renew their red cells PDQ.
The solubility of CO₂ could be looked upon as (yet another) bit of luck, for nature.

 

[hongiddong]

Hi Sophiecentaur,

When looking at this down hill effect, why do we usually ignore the other particles that are within the system? For example, say that there is 10CO2 inthe capillaries and 5 o2 in the capillaries, but in the alveolar, there is 5CO2 and 40o2. Why would the 10co2 in the capillaries move to the region where there is 5co2 when there is 40o2 that might hinder the brownian motion of CO2 diffusing to a place of least concentration?

 

[sophiecentaur]

why do we usually ignore the other particles that are within the system

I'm not sure that we do. Don't we just consider the gradients in the medium as a whole?

 

[jim mcnamara]

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2856815/
Provides a model used for CO₂ in discussing pulmonary(lung) inflammation and 0₂ and CO₂ diffusion.
Hope this will help clarify.

 

[Andy Resnick]

I hope I made sense. Thank you.

I'm not exactly sure what you are asking about: partial pressures? concentration gradients? osmolarily

 

[hongiddong]

Hi Andy Resnick,

In this scenario, I am looking at the particulate concentrations.

 

[Drakkith]

The basic idea is that with or without the other molecules present the CO2 would spread from an area of higher concentration to an area of lower concentration. This is obvious is you just consider a grouping of CO2 molecules in an otherwise empty container. The group will quickly spread out to fill the container with a roughly equal concentration of molecules.

However, the basic idea is the same when CO2 is present with other particles. All that random jiggling about tends to spread all of the molecules about equally. This is also true of larger things, like M&M's in a jar. I can fill a jar with layers of different colored M&M's, and when I shake it all of the colors will tend to get mixed together such that the concentration of anyone color is approximately equal everywhere in the jar.

If you're still concerned with how this happens, then you'd probably need to look at the motion of a couple of individual molecules. You'll find that if you clump them together, they will always tend to spread apart over time. Call it randomness or chaos or something.

We don't have to include the other molecules in our basic models because the end result is the same. The CO2 will move from an area of higher concentration to an area of lower concentration. Any difference in the rate of diffusion (perhaps because of the presence of other molecules that might impede CO2 diffusion) is likely negligible or not high enough to bother modeling unless you get very in-depth.

 

[Andy Resnick]

Hi Andy Resnick,

In this scenario, I am looking at the particulate concentrations.

As others have mentioned, each molecular species establishes it's own independent density equilibrium by diffusion. Imagine two scenarios using a divided box: one is full of water, each side has a differently colored dye while the other is full of air and has differently colored smoke. Open the partition and in either case the tracer particles will diffuse independently of the other. The *rate* of diffusion will depend on the viscosity of the underlying medium. Note also, the air/water molecules are also diffusing, but since they are not 'tagged' you ignore that (the Gibbs paradox is a different topic).

 

[hongiddong]

I am starting to see a better picture: 1. In the experiment with water: let says in space A we have a 10 o2 molecules and 10 CO2 molecules and are separated by a semipermeable membrane to only CO2 molecules in which space B we have 100 o2 molecules but 5 CO2 molecules. Why would the effect of the 100 o2 not effect the diffusion capacity of the 10 CO2 from space A to move into space B. Why do they have independent equilibrium when in respect to looking at this problem as motions vibrating and bouncing off of each other?

Thank you for all your help. This has been keeping me up at night loll

 

[Andy Resnick]

I am starting to see a better picture: 1. In the experiment with water: let says in space A we have a 10 o2 molecules and 10 CO2 molecules and are separated by a semipermeable membrane to only CO2 molecules in which space B we have 100 o2 molecules but 5 CO2 molecules. Why would the effect of the 100 o2 not effect the diffusion capacity of the 10 CO2 from space A to move into space B. Why do they have independent equilibrium when in respect to looking at this problem as motions vibrating and bouncing off of each other?

Thank you for all your help. This has been keeping me up at night loll

This is a confusing setup, but at equilibrium, there would be 8 CO2 molecules on each side (I added an extra CO2 when you weren't looking. Now, since the total amount of solute (the osmolarity) is imbalanced across the membrane at equilibrium, the (perhaps surprising) result is that the osmotic imbalance results in a hydrostatic imbalance- one one side, the water is higher than the other, by an amount that balances the osmotic pressure difference:

https://en.wikipedia.org/wiki/Osmotic_pressure

I caution you not to think too deeply about dissolved CO2 as this generates carbonic acid:

https://en.wikipedia.org/wiki/Carbonic_acid

The charge separation makes these simple arguments more complicated- you should be discussing the Nernst-Planck equation and Donnan equilibrium.

 

[hongiddong]

Thanks everyone for helping me with this. This makes more sense now. The diffusion depends on the volumes and pressures of the particles in question. In the case of the human body, there is enough volume for other particulate pressures to not be able to interfere with the individual diffusion equilibrium(the tendency for a molecule to spread out into a place of lesser concentration of itself.)