3. Starting with the definition of osmotic pressure
p
RT

ln so0ln
Vm.l
p1
Derive the semiempirical expression for a dilute real solution
 RT

 bC `
`
M2
C
`
Where C is the concentration expressed in (g solute)/(dm3 solution) and b is known as
the interaction constant.
Osmotic pressure depends on a semi-permeable membrane that allows only one
component of the solution to pass through the wall for the solution in question.
dG = -SdT +VdP + sum over i components of the chemical potentials product with the
change in number of molecules on either side of the membrane ....Uidni
I was thinking I would be able to do this with access to the math engine at home. I can
rewrite this later with better formatting.
-SdT refers to the passage of heat through the barrier, VdP refers to the mechanical
expansion of work, and the summation refers to the changes in chemical content...the
“chemical work”....Free energy is heat, mechanical and chemical work.
van’t Hoff first deduced the relationship between osmotic pressure, temperature, and
solute concentration.
Now a bit about real gasses:
In the case of an ideal gas we have for mechanical work only the relation:
dG = VdP = RT d (lnP)
which on integration gives G – Go = RT ln (P/Po)
I am working with n = 1 with these equations
In terms of chemical potentials this is U-Uo = RTln(P/Po)
G.N. Lewis introduced a function called the fugacity for a real gas such that we have an
analogous relation for a real gas:
dUi = dGi = RT d (ln fi)
In the case of an ideal gas, the fugacity equals the pressure, and since all gasses behave as
ideal gasses in the limit of small pressures, we have the necessary condition:
f/P ===> 1 as P===> 0
Now some more about the physical system:
If both sides are filled with pure solvent , the pressures over both sides will be the same
and the chemical potential across the membrane will be the same. foleft = foright.
The fugacity is proportional to the vapor pressure of the solvent. If solute is dissolved in
the left half, then the vapor pressure of the solvent on the left side will be less then that on
the right side. The colligative particles block the transport of the liquid into the vapor
phase. so f left< f right. Solvent then tends to move from the right side to the left side.
This solvent movement can be arrested by increasing the pressure over the left side. The
movement of solvent will cease when the at that pressure for which f1 = f1o. That is the
pressure at which the chemical potential of the solvent will be the same on both sides of
the membrane.
dT is zero for an isothermal system.
We are interested in dUi = dGi = RT d (ln fi) and specifically d (ln fi)
for component 1 we have variance in Pressure and in component 2 that will characterize
this differential value and we can take partials with respect to these variables holding T
and the adjoint variable constant formally. partial is denoted 6p/6x\T,P etc
so d ln f1 = 6p(ln f1)/6X2\TP dX2 + 6p(ln f1)/6P\TX2 dP
at equilibrium d ln f1 = 0 at constant temperature dG = VdP = RT d ln f
so the differential in pressure term becomes V/RT
Also in a dilute solution in which Henry’s law holds for the solute, then Raults’s law
holds for the solvent . This means that f1 = fo1 X1 = fo1(1-X2)
ln f1 = ln fo1 + ln (1-X2)
dln f1 = d ln(1-X2) = - dX2/(1-X2) and for low solute levels X2<<<<1
dln f1 = - dX2 so 6p(ln f1)/6X2\TP = -1
Then V/RT dP = dX2 In the limit of dilute solutions, the partial molar volume V with
the bar over it (this is not an average!) , is essentially constant and equal to the partial
molar volume of the pure solvent Vo with the bar.
We can integrate dX2 between the limits of zero ( pure solvent) and some small mole
fraction X2 This obtains P- Po = RTX2/Vo again Vo is with a bar for partial molar
volume . The pressure differential P-Po is the osmotic pressure and is represented by the
capital PI. Since in dilute solutions X2 = n2/(n2 +n1) =approximates n2/n1
and Vbar x n1 = V we obtain the relation:
PI = n2 RT/V = cRT where c is the moles of solute per cubic meter of solution.
Starting with the definition of osmotic pressure
p
RT

ln so0ln
Vm.l
p1
Derive the semi empirical expression for a dilute real solution
 RT

 bC `
`
M2
C
and ln (1/a) = - ln (a) this gives us the final form as shown above.
The key is to expand ln x in a Taylor series for x small and with x small the terms
psol/P1o give only two significant terms for the approximation. With C as defined
in the expression.
http://mathworld.wolfram.com/MaclaurinSeries.html
http://mathworld.wolfram.com/TaylorSeries.html
C’ = M2 C C = n2/V
V = n1 Vm1 psol/Po = 1-x2 expand about x2 small
and use x2 =n2/n1: ln (1-x) = -x +1/2 x^2 plus cubic terms
So PI = - RT/Vm1 { -x2 +1/2 X2^2}
factor out the n1 with n1V1 =V
PI = RT/V {n2 –1/2 n2^2/n1} factor out the n2 with C = n2/V
PI = CRT {1 – ½ n2/n1} or PI/C’ = RT/M2 { 1 – ½ n2/n1}
now n2/n1 = (n2/V)/n1/V = V/n1 x C = V/n1 x C’/M2 = V/n1M2 C’
however V n1 and M2 are essentially constant
PI/C’ = RT/M2 + b C’ as was to be shown using the Maclauren expansion of
ln (1+x)
Then for the problem in question plot PI/C on y axis and C on the x axix. The intercept is
RT/M2 so M2 is RT divided by this intercept.
Then use the semiempirical formula to determine the molar mass of inulin (C6H10O5)x
from the following data at 200C.
C`
103.8
(gsolute/dm3soln)
0.86
 (bar)
77.1
50.9
25.2
0.58
0.31
0.14