This is a sketch of the common osmosis with low solute concentration at thermal equilibrium,
which can be found in several science books; e. g. in
Understanding Engineering Thermo from Octave Levenspiel, ISBN 0135312035,
in Chapter 22.C.
We have a system of a pure solvent A on the left side, e. g. water, a solution B,
e. g. with an organic salt or glass microspheres, on the right side and
a room above them, which is filled with the solvent vapor C, and maybe some gas.
On the bottom the solvent A and the solution B are separted by a semiperable
membrane M which is permeable to A alone.
The osmotic pressure, which causes the height difference z2-z1, is given by the
van't Hoff
law :
Posmosis = n' *k *T/V
where n' is the number of solved particles, V the volume of the solution
(right column) and k the Boltzmann constant.
A density difference between A and B modifies z2 -z1
but the classical theory neglects this, because it's a good approximation for
small z2 and z1 because in this case the osmotic pressure
is much larger than the modification by the different densities.
Above the pure liquid A the vapor pressure is given by the
barometric formula:
P(z) = P0 * e-m*g*(z-z1)/(k*T)
Where P0 is the vapor pressure of A at thermal equilibrium.
The classical theory neglects that the different vapor pressures on the left
and the right column (P1, P2) do influence z1
and z2, because usually the vapor density is much lower than the
density of the liquids.
At the solution B on the right side we have a higher ground z2 for
the vapor, due to the osmosis.
And due to
Raoult's law the vapor pressure
at the surface of B is the partial pressure of the solvent:
Ps = Pspure *xs =
P0 xs = P2
Where xs, is the mole fraction of the solvent in the solution.
One result of the classical theory, which can be found e. g. in the book of
Octave Levenspiel, is that at height z2 the pressure is the same on both
sides, because the height difference z2-z1 is proportional to the
osmotic pressure, but the Raoult's law reduces the partial pressure
also proportional and finally approximating the pressure
by using the first two summands of the taylor series of the exponential function (in the
barometric formula) the pressure
above the pure solvent decreases linear with the height.
Putting it all together and setting the presssure at z2 on the left
side equal to the pressure on the right side we get
P0 * (1 -m*g*(z2-z1)/(k*T)) = P0 *xs
The height difference z2-z1 must be substitued via the formula
for the hydrostatic pressure difference
ΔPhydro = g *ρ * (z2-z1)
which is equal to the osmotic pressure n' *k *T/V.
So we get
n' *k *T/V = g *ρ * (z2-z1)
=>
z2-z1=n' *k *T/(V *g *ρ)
=> (Subtitution into the equation for the pressure at z2, after
reducing P0)
1 -m *g *(n' *k *T/(V *g *ρ))/(k*T) = xs
= 1 -m *n'/(V *ρ) = xs
for the mole fraction of the solvent.
Because the density ρ is defined as m*n/V, we get
xs = 1 -m *n'/(V *m*n/V) = 1 - n'/n = n/n -n'/n = (n-n')/n
This is not exact xs because it is defined as n/(n+n'), but the
difference to xs is only a second order term:
delta = n/(n+n') -(n-n')/n
=>
n*delta = n2/(n+n') -(n-n')
=>
(n+n')*n*delta = n2 -(n+n')*(n-n') = n2 -n2 -n*n' +n*n' -n'2 = -n'2
=>
delta = -n'2/((n+n')*n)
Because we have only low solute concentrations, we have n' << n and therefore delta is close to
-(n'/(n+n'))2, which is the square of the (small) mole fraction of the
solute; Q.E.D..
So above the surface of the solution there is the same vapor pressure on the
left and the right side and therefore no vapor flow.
This classical theory is consistent, but only when the solute has the same
density as the solvent, because otherwise we would have to apply the
barometric formula for the solute in the solvent and therefore different
concentrations or mole fractions at the membrane M and at the surface at
z2 !
So the classical theory can NOT be applied to e. g. glass microspheres or big
polystyrene microspheres in water. In these cases we have a higher partial
pressure on the right side at z2, while the left side is unchanged. This gives
a vapor flow from the right to the left side, at thermal equilibrium!
To be accurate, we have to take into account that in this case z2 is
reduced a little
due to the higher density of the solution, but when we use heavy microspheres,
we can use a half-density-height of several mm at a hight difference of
several m and with a solution surface that is more than 99,99 % pure solvent at
the surface (z2).
So the difference of the average density of the solution and the solvent is
less than 0.05 % and therefore negligible.
Another problem of the classical theory is that the vapor is treated as ideal gas,
so that close to the
critical point
the barometric formula can not be applied.
Due to the fact that the compressibility is lower at high pressure, the pressure gradient in the
vapor is higher simply because the density decreases slower with the upward
decresing pressure.
By using barotropy (the gas density is
higher than the fluid density, see chapter 8.2 in
http://www.knaw.nl/waals/pdf/fluids_complete.pdf
and
Kammerlingh Onnes first helium experiment in 1905:
http://www.lt25.nl/abstracts/LT3177.pdf
and
http://www.knaw.nl/waals/pdf/Onnes03.pdf
), the osmosis sketch gets changed: Bottom and
top get switched; we have osmosis upside down.
A simple example is a gas with a density that is two times the
density of the liquid: The buoyancy of the liquid in the gas is two times
the gravity force so the resulting force for the liquid (l), buoyancy force - gravity force =
-2*Ml*g +Ml*g, is the same as the gravity force but with another sign.
An example (with a lower density difference) is a mixture of hydrogen and helium at
a temperature of about 25 K and microspheres as solute.
At room temperature xenon and water can be
used, because the xenon gas density at the critical point is 1100
kg/m3 (=1.1 times the water density) at 16.6°C and 5840 kPa:
http://www.uigi.com/rare_gases.html.
Instead of xenon other dense liquids/gases like
bromine
can be used, because the critical point of bromine is at 588 K, 10.34 MPa
and 1184 kg/m3 and below the critical point of
water, which is at 647 K, 22.064 MPa and 328 kg/m3.
As soluted particles glass or plastic microspheres can be used.
Because at room temperature the vapor pressure of water is low, it would be not
efficient. Alternative liquids
like
pentane
have a higher vapor pressure, lower viscosity and a significant lower density:
Pentane has a density of only 626 kg/m3, which is a little less than 2/3 the
density of water, and a vapor pressure of 562 hPa at 293 K, which is 24 times
the vapor pressure of water.
The critical point of pentane is at 469.8 K, 3.37 MPa and 232 kg/m3.
So with alternative liquids in critical xenon we are close to the example where
the buoyancy of the liquid in the gas is two times the gravity force.
Barotropy does change neither the osmotic pressure (van't Hoff law) nor the partial
pressure (Raoult's law) or the barometric formula, but it changes the sign of the height difference
z2-z1 ! This means that the pressure difference due to the barometric formula
has (at first approximation) the same absolute value, but with the other sign!
So the pressure below the pure solvent increases from the surface at z1 to the
height z2, where we have only the partial pressure
Ps = Pspure *xs < Pspure
on the right side, at the surface of the solution.
So at thermal equilibrium at z2 we have a higher pressure on the left side as
on the right side and therefore a solvent vapor flow from the pure solvent to
the solution, driven by the barometric formula on the left side and the
Raoult's law on the right side (and ambient heat).
An important point is that the height differnce z1-z2 is inverse proportional to the
buoyancy of the solvent in the gas and not simply proprtional to the density,
and that the buoyancy can be adjusted via the pressure (e. g. by reducing or
increasing the volume of the container).
Last Change: 2008-11-01