Fig. 4.2.1

The water moves across the semi-permeable membrane from the solution containing a low concentration of solute to the solution with a high concentration of solute until the weight of the column of water is sufficient to stop any further transport. Thus, the pressure exerted on the membrane by the raised column of water at equilibrium is said to be equal and opposite to the ‘osmotic pressure’ of the solution.

where µ_{0} is the chemical potential of water at its
standard state, *V*_bar is the molal volume, *P* is the hydrostatic pressure
and *a _{w}* is the activity of the water.

If we consider the U tube apparatus shown above, then the osmotic
pressure (*P*) is the hydrostatic pressure required to stop the osmotic
flow. Although we’ve shown it as an increase in hydrostatic pressure
(due to the gravitational potential of the raised column of solution) it
could just as easily be a negative hydrostatic pressure (tension) on the
opposite side of the membrane. At this pressure, the change in chemical
potential of the solution (in this case the osmotic potential) is equal
to the hydrostatic pressure difference across the membrane.

If we consider that the activity of the water is reduced by adding solutes, then we can relate the decrease in chemical potential to the increase in hydrostatic pressure needed to obtain equilibrium. For a situation in which a semi-permeable membrane separates pure water from a solution, then

For a solution in which there are no interactions between the solvent
and the solute (an ideal solution), the activity can be described by the
mole fraction of solvent molecules (X_{w})

where n_{w} is the number of water molecules and n_{s}
is the number of solute molecules. If n_{w} >> n_{s}
then we can use the approximation

we can get an expression for the osmotic pressure

where X_{s} is the mole fraction of solute, m is the molal concentration of solute,
and c is the molar concentration
of solute. This is the van’t Hoff relation for osmotic pressure.

For water at 20°C, this pressure would be over 1300 atmospheres if there
were no intermolecular forces. Obviously the forces between water
molecules are appreciable. If we consider the semi-permeable membrane to
be constructed of water filled pores, then the mole fraction of water on
the solute side will drop from 1 to X_{w} at the pore openings. Since the
solute molecules will not be able to transfer their momentum to the
water molecules inside the pores, then those water molecules at the pore
openings will be bombarded by fewer water molecules from the solution
side than from the pure water side. This will give a pressure on the
solution side of

Thus the osmotic pressure is given by the difference in kinetic pressure in the pore openings

The gas equation of van der Waals takes two factors into account. First,
the volume occupied by the molecules of the gas reduces the total volume
by a certain amount (the constant b represents the volume of the gas molecules,
so we have *P*(*V-b*)=n*RT* ). Second, the intermolecular forces
reduce the
momentum that the molecules impart to the walls of the container through
collisions. This second fact is due to the attraction between a molecule
about to collide with the container wall and the other gas molecules
within the container.

Fig. 4.2.2

Since this attractive force only counteracts momentum transfer to the walls of the container (within the bulk of the gas, the attractive forces are symmetrical and therefore cancel), the pressure will be reduced by an amount inversely proportional to the square of the volume . Van der Waals equation is

where *a* and *b* are constants (*a* is not related
to the thermodynamic activity).

We can extend this reasoning to the case for a liquid by considering the
intermolecular forces between solvent molecules and between solvent and
solute molecules. The equipartition theorem says that solvent and solute
molecules will have the same mean kinetic energy; the important
difference being that the membrane is impermeable for the solute. At the
surfaces, the intermolecular forces are such that there is no transfer
of momentum from the molecules to the container. Within the liquid,
symmetry guarantees that each molecule is attracted equally on all
sides. Thus the kinetic pressure does not have a component corresponding
to *P* in the van der Waals equation and is only given by
*a/V ^{2}*. If we consider the membrane pores to be so
small that water molecules must pass single-file, then we can see that
symmetry once again eliminates the intermolecular forces.

Fig. 4.2.3

In an ideal solution, the volume is given by the sum of the volume of solvent and the volume of solute, thus the volume correction for the space taken up by solute molecules is already included in the molal volume of water. So we can see why the equation for osmotic pressure corresponds better with the equation for an ideal gas than does the equation for the pressure of a real gas. It is because the momentum deficiency which causes flow is only important at the pore openings where the intermolecular forces are canceled by symmetry.

This last point shows why osmosis is unique to the liquid state. Osmotic flow only occurs if there is some mechanism for generating tension within the solvent. The cohesive intermolecular forces that exist in a liquid pull the solvent molecules through the pores like a train. Thus although the intermolecular forces do not create the osmotic potential, they are essential for flow to occur in response to that potential.

Fig. 4.2.4

A semipermeable membrane separates a solution, on the left side of the
apparatus, from pure solvent on the right side. A pressure difference is
then set up to balance the osmotic pressure across the membrane. Since
the vapour pressure above the pure solvent is higher than the vapour
pressure above the solution (eq. 4.12), there is a pressure gradient in
the top of the apparatus between the left and right sides. This gradient
will drive a flow of vapour from the right side to the left which is
used to drive a turbine blade. The vapour condenses on the surface of the
solution on the left side, thereby diluting that solution and reducing
its osmotic pressure. This causes solvent to flow back through the semi-
permeable membrane into the right side to restore the original
concentration. Thus we extract energy from the machine without having to
put anything in.

[Stop here and think about this before proceeding]

The hydrostatic pressure of the column of solution on the left hand side is given by

The column of vapor that exists over the solvent on the right hand side is given by the difference in vapor pressure between the top and bottom of the column

where *M/V* is the density of the water vapor. If water vapor obeys the
ideal gas law then we can combine these relations to get an equation for
the pressure

If we recall that Raoult's law states

leaving us with

However, we can refine this by noting that the gas above the water column is not incompressible, and will hence vary in density as the height is increased. This means that the weight of this column should be given by

Resulting in

Now we invoke the laws of thermodynamics to tell us that perpetual motion machines don’t work so we realize that the pressure at equilibrium must be equal to the osmotic pressure

If we assume that each solute molecule binds a certain number of solvent
molecules (a) which are then removed from the mole fraction of
solvent molecules, we can derive an expression relating this factor to
the mole fraction of solvent that is available for the osmotic
potential. The number of solvent molecules is thus reduced from *n _{w}* to
(

Resulting in an expression for the osmotic pressure.

Unlike the thermodynamic parameter activity, which is dependent on concentration, the binding constant has no such dependencies (i.e. it’s specific for a given solute) and describes osmotic pressure for any solution.

Where f is the molal osmotic coefficient. For non-electrolytes (e.g. glucose), f > 1 in 'ordinary' concentrations. For electrolytes, f < 1 due to the electrical interactions between the ions (e.g. NaCl at physiological concentrations has an osmotic coefficient of 0.93). For macromolecules, the deviation becomes much more dramatic; hemoglobin, for example, has an osmotic coefficient of 2.57.

If we consider the cell to be a spherical semipermeable membrane that holds water and solutes, then we can write:

Where *b* is the non-solvent component of the cytoplasm or
"osmotically inactive volume". We can define the osmotic pressure of the
cytoplasm using the Van't Hoff equation:

Noting that the change from molality to molarity introduces a temperature dependence. Substituting:

At constant temperature,

Where *k* is a constant. Under isotonic (normal) conditions, we can
write:

If we change the external osmotic
conditions, we will get water movement until p
_{i} = p_{e}, so:

So

This equation was first developed by Lucke and McCutcheon in 1932.

Leaving us with the Boyle-Van't Hoff Relation:

If we plot the cell volume vs. (p_{0}/p_{e}), we get a straight line in which the
y-intercept is the osmotically inactive volume and the slope is the
isotonic water volume. Boyle-Van't Hoff plots are used extensively in
cryobiology to determine the osmotic parameters of given cell types. The
assumption of constant temperature that is part of the derivation
implies that the relation is temperature dependent.

Typical Boyle-Van't Hoff plot showing the relative volume changes of
fertilized mouse ova (Leibo S., Water Permeability of Mouse Ova. *J
Membrane Biol* **53**: 179-88. 1980.)

The water volume that results from this relation has been found to be different from the water volume that results from other methods of measurement, so it is assumed that there is a compartment of water that is "bound" (unavailable for osmotic exchange). It should be noted, however, that any given water molecule within the cell is available for osmosis, but there is always a residual compartment (that is freely exchangeable with bulk water) that does not participate in osmosis.

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Document last updated Oct. 13, 1998.

Copyright © 1998, Ken Muldrew.