Calculating Osmotic Pressure Of Red Blood Cells

[devon]

Homework Statement

i) A human erythrocyte (red blood cell) may be approximated as a disc around 2 microns in height and with a diameter of 10 microns. If the pressure inside were to become large enough for the cell membrane to rupture, where would you expect it to fail, assuming the membrane is of uniform thickness and composition?

ii) The concentration of impermeant ions inside an erythrocye is approximately 2 × 10^26 m^−3. Calculate the osmotic pressure, assuming the cell is at 37◦C.

Homework Equations

Van't Hoff's law: Osmotic Pressure = nRT/V
n= moles of solute particles in solution of Volume V
v= volume of RBC = (pi x 5um^2 x 2um) = 157.0796327um^3 = 1.570796327x10^-16 m^3
R= Gas Constant 8.3145 m^3 Pa/Kelvin x Mol
T= Temperature in Kelvin

The Attempt at a Solution

[/B]
i) I don't really know what theoretical knowledge I need to answer this but I'm saying that the sides of the blood vessel would rupture (fail) as there is less surface area along them compared to the 10um diameter circle at the top of the cylinder which there is a higher chance for more pressure to act on a specific spot on the sides of the blood vessel which means more chance for that specific spot to fail? Is this correct?

ii) So I assume I just whack everything into the equation?
Osmotic Pressure = ((8.3145 m^3 Pa/Kelvin x Mol) x 310Kelvin x(2x10^26m^3 )/(1.570796327x10^-16 m^3)
= 3.28x10^45 (m^3 Pa/mol)

Those units look really wrong so I doubt I worked out the answer correctly. Did I do something wrong? Osmotic pressure normally has just pressure units right so I guess I need to convert some stuff (figure out a way to make the impermeant ion concentration into mols?)

Thanks for the help.

 

[Borek]

Van't Hoff's law: Osmotic Pressure = nRT/V

Doesn't look to me like the formula for the osmotic pressure. More like ideal gas equation solved for p.

 

[devon]

Is this the equation I'm after? The units for the impermeat ion's concentration are throwing me of (m^3).
M is the molar concentration of dissolved species (units of mol/L).
R is the ideal gas constant (0.08206 L atm mol-1 K-1, or other values depending on the pressure units).
T is the temperature on the Kelvin scale.

The only reason I used the other equation is because it used the volume of the RBC and I assumed that them giving the diameter and height was to be used for this reason

Also was I right about the sides of the blood vessel rupturing?

 

[Borek]

Is this the equation I'm after?

Yes. There is nothing strange with length given sometimes in m, sometimes in L and sometimes in mL - you have to convert them, but it is not difficult.

Sorry, no idea about the rupture part.

 

[devon]

Oh so am I assuming that the ion concentration given is actually mol/m^3? That makes way more sense. So I convert the m^3 to L to get mol/L (2x10^26/1000L) = 2x10^23mol/L.

So the answer would be: Osmotic Pressure = (2x10^23mol/L) x (8.3145LPa/Kmol) x 310K = 5.15499x10^26 Pa?
I assume that the height and diameter given are probably used for the first part of the question (rupture)?
Thanks so much for all of the help by the way, I'm actually learning how to do stuff as opposed to just copying it down from my teacher.

 

[Borek]

Beware! The concentration is not given as moles/m³, but as ions/m³. You have to convert it to moles.

Think about it this way - molar concentrations above 10 M are always suspicious, there are not many substances than are that soluble. Pure water is just 55.5 M.

 

[fishes]

Wow I need more practice, I got something completely different to that Devon and your working seems to be right.

 

[devon]

Ah that makes complete sense! So I just need to divide that number by avogadro's number and then do all the conversion? (2x10^23ion/L)/6.022x10^23 = 0.3321155762mol/L?
Osmotic Pressure = 0.3321155762mol/L x (8.3145LPa/Kmol) x 310K = 856.0262371 Pa?

 

[Borek]

8.3145LPa/Kmol

Check these units.

 

[devon]

Ah its LkPa/kmol. So that would make my final answer 856.0262371 kPa?

 

[Borek]

I am not saying that's right - all I can say is I don't see any more problems ;)

Actually it is $\frac {L \times kPa}{K \times mol}$ (K×mol not kmol).

 

[devon]

I am not saying that's right - all I can say is I don't see any more problems ;)

Actually it is $\frac {L \times kPa}{K \times mol}$ (K×mol not kmol).

Thank you so much for all of your help and putting up with my 1 million questions!

 

[rude man]

Doesn't look to me like the formula for the osmotic pressure. More like ideal gas equation solved for p.

They are the same.

"The osmotic pressure of a dilute solution is equal to the pressure exerted by an ideal gas at the same temperature and occupying the same volume as the solution and containing a number of moles equal to the number of moles of the solute(s)^* dissolved in the solution".

- Enrico Fermi, Thermodynamics, Dover 1936

* I added the "(s)" since Dr. Fermi analyzed a solution wth possibly more than one solute.

 

[devon]

They are the same.

But doesn't the volume being divided on the bottom give me a vastly different answer?
This is how the equation is given to me in my textbook:

Do I need to divide my final answer with the volume of the blood cell I found in my first attempt?

 

[rude man]

Homework Statement

i) A human erythrocyte (red blood cell) may be approximated as a disc around 2 microns in height and with a diameter of 10 microns. If the pressure inside were to become large enough for the cell membrane to rupture, where would you expect it to fail, assuming the membrane is of uniform thickness and composition?

ii) The concentration of impermeant ions inside an erythrocye is approximately 2 × 10^26 m^−3. Calculate the osmotic pressure, assuming the cell is at 37◦C.

Homework Equations

Van't Hoff's law: Osmotic Pressure = nRT/V
n= moles of solute particles in solution of Volume V
v= volume of RBC = (pi x 5um^2 x 2um) = 157.0796327um^3 = 1.570796327x10^-16 m^3
R= Gas Constant 8.3145 m^3 Pa/Kelvin x Mol
T= Temperature in Kelvin

The Attempt at a Solution

[/B]
i) I don't really know what theoretical knowledge I need to answer this but I'm saying that the sides of the blood vessel would rupture (fail) as there is less surface area along them compared to the 10um diameter circle at the top of the cylinder which there is a higher chance for more pressure to act on a specific spot on the sides of the blood vessel which means more chance for that specific spot to fail? Is this correct?

ii) So I assume I just whack everything into the equation?
Osmotic Pressure = ((8.3145 m^3 Pa/Kelvin x Mol) x 310Kelvin x(2x10^26m^3 )/(1.570796327x10^-16 m^3)
= 3.28x10^45 (m^3 Pa/mol)

Those units look really wrong so I doubt I worked out the answer correctly. Did I do something wrong? Osmotic pressure normally has just pressure units right so I guess I need to convert some stuff (figure out a way to make the impermeant ion concentration into mols?)

Thanks for the help.

part (ii)
Looks like you have the wrong n. n is number of moles and you are using number of ions.

part (i): sorry, not a clue.

 

[devon]

I figured out the correct n down the bottom of this thread, I just divided the ion concentration by avagadro's number to get 0.3321155762mol/L. So do I have to divide my final answer with the volume of 1 blood cell? I'm getting a bit confused.

 

[rude man]

I figured out the correct n down the bottom of this thread, I just divided the ion concentration by avagadro's number to get 0.3321155762mol/L. So do I have to divide my final answer with the volume of 1 blood cell? I'm getting a bit confused.

If you'll read Dr. Fermi's statement in post 13 ...

 

[Borek]

No need to divide anything. These equations will be equivalent when you use volume that contains given number of ions - that is, 1 cubic meter.

Red cell contains much less of them, so you can get the correct answer finding how many ions are in the red cell and dividing by the red cell volume. That means multiplying by the red cell volume and dividing by the red cell volume. They cancel out.

 

[rude man]

I figured out the correct n down the bottom of this thread, I just divided the ion concentration by avagadro's number to get 0.3321155762mol/L. So do I have to divide my final answer with the volume of 1 blood cell? I'm getting a bit confused.

n = no. of gms. of ions in 1 cell/gm-mol. wt. of ions.
Simplifies to n = 2e26V/6.022e23. Which I think is what borek implied.
I got p ~ 8.7 atmospheres.

(gm-mol wt. = mass of avogadro's number of ions.
This assumes everything other than the ions is permeable to the membrane.)

 

[fishes]

n = no. of gms. of ions in 1 cell/gm-mol. wt. of ions.
Simplifies to n = 2e26V/6.022e23. Which I think is what borek implied.
I got p ~ 8.7 atmospheres.

(gm-mol wt. = mass of avogadro's number of ions.
This assumes everything other than the ions is permeable to the membrane.)

I got 856kpa, which converts to roughly 8.4 atm so I guess I got the right thing in the end?

 

[devon]

I get either 856kPa or 8.4 atm as well so I guess I'm done as long as ~8.7atm = 8.4atm! Thanks so much for all of the help!

 

[rude man]

I got 856kpa, which converts to roughly 8.4 atm so I guess I got the right thing in the end?

I believe so! Osmosis is very interesting. I use a "reverse osmosis" system to remove impurities from the water supply. The osmotic pressure is overcome by reverse pressure so that the solvent/solute mixture is forced backwards thru the membrane, leaving only solvent at the output end.

BTW in part (i) I would guess that the most likely rupture point would be in the middle of either the top or bottom of the right circular cylindrically shaped cell. Just a guess though.

 

[fishes]

I believe so! Osmosis is very interesting. I use a "reverse osmosis" system to remove impurities from the water supply. The osmotic pressure is overcome by reverse pressure so that the solvent/solute mixture is forced backwards thru the membrane, leaving only solvent at the output end.

BTW in part (i) I would guess that the most likely rupture point would be in the middle of either the top or bottom of the right circular cylindrically shaped cell. Just a guess though.

What's your reasoning for the rupture point being the centre? I worked out the surface area for both the rectangle (2xpix5umx2um = 63um) and the top and the bottom (157um together). I was thinking that as the side had less surface area than either the top or bottom that it would be easier for the pressure to act on one particular spot causing it to fail (rupture)? Does that make sense?

 

[rude man]

What's your reasoning for the rupture point being the centre? I worked out the surface area for both the rectangle (2xpix5umx2um = 63um) and the top and the bottom (157um together). I was thinking that as the side had less surface area than either the top or bottom that it would be easier for the pressure to act on one particular spot causing it to fail (rupture)? Does that make sense?

I would argue that the sides, being supported by the top and bottom, are less prone to rupture. The top & bottom would absorb side forces by expansion. Whereas the top & bottom have no nearby support structure, esp. near their centers. The centers of the top & bottom are most removed from any stress support so they would go first.

In other words, yes, the area of the side, and thus the total force on it, is larger than that of the top or bottom, but it is more closely supported all around by tension stress applied radially to the top & bottom.