# Calculating Osmotic Pressure Of Red Blood Cells

• devon
In summary, 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 of solute particles equal to the concentration of the solute.
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.

devon said:
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.

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?

Last edited:
devon said:
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.

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.

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.

Last edited:
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?

fishes said:
8.3145LPa/Kmol

Check these units.

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

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
Borek said:
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!

Borek said:
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.

Last edited:
rude man said:
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?

devon said:

## 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.

Last edited:
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.

Last edited:
devon said:
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 ...

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.

devon said:
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.)

devon and fishes
rude man said:
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?

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!

fishes said:
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.

rude man said:
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?

fishes said:
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.

## 1. What is osmotic pressure and why is it important to calculate it for red blood cells?

Osmotic pressure is the pressure that is exerted by a solution in order to prevent the flow of water into or out of a cell. It is important to calculate osmotic pressure for red blood cells because it helps to maintain the proper balance of fluids and electrolytes within the cells, which is crucial for their normal functioning.

## 2. How is osmotic pressure of red blood cells calculated?

Osmotic pressure of red blood cells can be calculated using the Van't Hoff equation, which is π = iMRT, where π is the osmotic pressure, i is the van't Hoff factor, M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin.

## 3. What factors can affect the osmotic pressure of red blood cells?

The osmotic pressure of red blood cells can be affected by the concentration of solutes inside and outside the cells, as well as the temperature and the presence of any membrane impermeable molecules that may alter the osmotic gradient.

## 4. How does the osmotic pressure of red blood cells change in different environments?

The osmotic pressure of red blood cells will increase in a hypertonic environment, where the concentration of solutes outside the cell is higher, causing water to flow out of the cell. It will decrease in a hypotonic environment, where the concentration of solutes outside the cell is lower, causing water to flow into the cell. In an isotonic environment, where the concentration of solutes is equal inside and outside the cell, there will be no net movement of water and the osmotic pressure will remain constant.

## 5. What is the significance of measuring the osmotic pressure of red blood cells in medical research?

The osmotic pressure of red blood cells can provide valuable information about the health and functionality of these cells. Abnormal osmotic pressure can indicate various conditions, such as dehydration, kidney disease, and certain genetic disorders. Measuring osmotic pressure can also help in understanding the effects of different drugs and treatments on red blood cells.

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