PETER PAZMANY 

SEMMELWEIS UNIVERSITYCATHOLIC UNIVERSITY 



Development of Complex Curricula for Molecular Bionics and Infobionics Programs within a consortial* framework** 
Consortium leader 


PETER PAZMANY CATHOLIC UNIVERSITY 

Consortium members 


SEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER 

The Project has been realised with the support of the European Union and has been co-financed by the European Social Fund *** 
**Molekuláris bionika és Infobionika Szakok tananyagának komplex fejlesztése konzorciumi keretben 
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INTRODUCTION TO BIOPHYSICS 

(Bevezetés a biofizikába) 
COLLIGATIVE PROPERTIES 

(Kolligatív sajátságok) 
GYÖRFFY DÁNIEL, ZÁVODSZKY PÉTER 



. 
Colligative properties are properties of a dilutesolution that depend only on the number ofparticles in the solution but do not depend onthe properties of them, like mass or size etc. 
. 
Colligative properties are: 
– 
Osmotic pressure 

– 
Lowering of vapour pressure 

– 
Elevation of boiling point 

– 
Depression of freezing point 




Osmotic pressure 

. 
Osmotic pressure is one of the colligative
properties of a solution 

. 
To formalize it, let us set out from the known fundamental law: 
– In a heterogenous – consisting of more than onephase –, closed system at equilibrium, the chemicalpotential of an uncharged substance is the same in all phases between which it can pass freely 


. 
Consider a closed box containing water in
equilibrium with its vapour 

. 
Suppose some vapour goes into the liquid
phase (or vice versa) 

. 
We would like to know what the free energychange of the system is 
. G . G
dG system =..dnl ...dnv =.l dnl ..v dnv
. nl . nv 


Liquid-vapour system 




. 
Since the system is closed, i.e. only energyexchange is possible between the system andits environment, we know that 
nl .nv =constant 

so 
dnl .dnv =0 
and 
dnl =-dnv 



. 
Thus the free energy change is 
dG=.l dnl -.v dnv=..l -.v .dnl 
based on which 
dG 

=.l -.v
dnl 



. 
Let us plot the free energy, G, as a function of the number of moles in the liquid phase, nl 
. 
We can see that there pure liquid or pure
vapour are both unstable conditions 

. 
Therefore the free energy decreases movingaway from either extreme 
. 
Somewhere there is a minimum to this 
function 

. 
By definition this is the equilibrium position 
. 
At this point the slope of the curve equals zero 


G vs. nl 




. 
Since at equilibrium the slope of the curve is 



.
.G .

=0
. nl .l -.v =0 
and 

.l =.v 


. 

Now we will apply this fundamental law toosmotic pressure 
. 

Let us consider the following figure 
. 

In chamber A there are water and some polysaccharide to which the membrane isimpermeable 
. 

In chamber B there is pure water 


Demonstration of the osmotic pressure 




. 

Water will flow from side B to side A generating a pressure difference proportional to h, the difference in water levels in the standpipes at equilibrium 
. 

The excess pressure on side A is the osmotic pressure and is signed by . 


. 

As we know, the pressure is a function of depth 
..Pa .=h.m .·.. kg / m3 .·g . m/sec2 . 
Pa =N /m2 

. 

. can be converted to atm by noting 
1atm=101325 Pa 



. 
For our case 

.H 2 O =1000 kg/ m3 
and 

g =9.80 m /sec2 


. 

From the fundamental law discussed earlier, the chemical potential of water must be thesame on both sides of the membrane 
. 

This is not true for the polysaccharide whichcan not pass the membrane 


. 

µ is a function of the temperature, thepressure and the composition of the solution 
.1, A.T,p.. ,x1, A.=.1,B .T ,p,x1, B. 

. 

Since temperature is constant on both sides, we can consider µ to be a function of only thepressure and the composition 
.= f . p,x. 



. 

Making use of the known relationship, the
chemical potentials are 

.1, A. p.. ,x1, A.=.1, 0 A. p....RT ln .1, Ax1, A 

and 

.1, B. p,x1, B.=.1, 0 B. p ..RT ln .1,Bx1, B=.1, B. p . 

making use of the fact that x1, B =1 and .1, B =1 


. 

According to the fundamental law 
.1, 0 A. p....RT ln .1, Ax1, A=.1, 0 B. p. 

so 

.1, 0 A. p...-.1, 0 B . p.=-RT ln x1, A- RT ln .1, A 



. 
Now, we wonder how µ changes with p 

. 
From basic thermodynamics, let us recall 
=V
...Gp .
p,n 
and 


.
.G .

=. 
. n 

T,p 


. 

Since we can do partial differentiation in anyorder 
.. ..G 
=
..[ ..]

. p . p . n T,p T,n ..G .. G 
=



[..][..]

. p . n T,p T,n . n . p T,nT,p 


.. G . 
=.V .
[. n ..][ ]
. p T,nT,p . n T,p 
.. V
.V .==V

[ ]..
. n . n
T,p T,p 
which is the partial molar volume 


. 
For the standard chemical potential 
..10 0 
=V 1
. p 

where V0 is the molar volume of the solvent 
. 
After rearrangement we get 
..10 =V 10 . p 


. 	Let us integrate .µ10 from p to p+. assuming that water is incompressible, that is V10 is constant 
p .. 	p.. 
. ..10 =. V 10 . p 
p	p 
. 
We get 
.10 . p...-.10 . p.=V 10·. p...-V 10·p 


. 
Finally we get 
V 10 .=-RT ln x1, A- RT ln .1, A 
. 
Since 
x1. x2 =1, 
x1 =1- x2 

and thus 
V 10 .=-RT ln.1-x2 .- RT ln .1 


. 
Let us suppose that the concentration ofpolysaccharide is 
w =10 g /dm3 
and the molecular weight of it is 
M =25000 Da 
so the molar concentration of it is 
10 g /dm3 
c==4·10- 4 mol /dm3 
25000 g / mol 


. 
Thus the mole fraction of the polysaccharide is 
c 4·10-4 mol /dm3 
x== .7·10-6 c.cH 2 O 55.5604 mol /dm3 
a very small number 


. 
For small numbers 
ln .1-...-. 
and 
ln .1..... 
if 
.«1 


. 
Performing this approximation we get 
V 10 .= RT x2- RT ln .1 
. 
We know that 
n2
x2 = 
n1 .n2 
but since n1 << n2 we can write that 
n2
x2 . 
n1 


. 
Thus 
n2
V 10 .= RT -RT ln .1
n1 

from which 
RT n2 RT ln .1
.= 0 - 0n1 V 1 V 1 


. 
Further simplifications can be performed 

n1 V 10.V 1 
and since the polysaccharide takes up only asmall portion of the volume of the solution 
V 1.V 


. 
Substituting these simplified expressions into the main equation we obtain 
n2 RT ln .1 RT ln .1
.=RT -=RT c2-
VV 10 V 10 


. 
In the lab, we would weigh out a number ofgrams of polysaccharide and report its solutionconcentration in g/dm3 
. 	Let w2 denote the weight of polysaccharide per liter 
. 
Thus 
w2c2 = M 2 
where M2 is the molecular weight of the polysaccharide 


. 
Substituting this expression of the molar concentration into the expression of theosmotic pressure we get 
RT w2 RT 
.= - ln .1
M 2 V 10 


. 
Now let us imagine that we are at great dilution 
. 
Now 
.1.1 
and 
ln .1.0 
thus 
RT w2
.= 

M 2 


. 
We can solve this equation for the molecular weight of the polysaccharide 
RT w2M 2 = 
. 
so measuring the osmotic pressure of a solutionof some substance its molecular weight can bedetermined 


. 
Another common form of the equation at infinite dilution 
.=RT c2 
. 
It can reminds us of the gas law 
.V =n2 RT 
which holds for ideal solutions 


. 
Let us see more about the nonideality term in the equation for osmotic pressure 
RT w2 RT 
.= - ln .1

M 2 V 10 

. Let us recall that if we plot µ1 as a function of x1 
we get a straight line with slope R·T for the ideal case 
. 
The distance between this line and the real 
curve is -RT ln .1 



. 	
Now, let us plot -RT ln .1 as a function of ln x1 

. 	
The slope of the curve approaches zero as ln x1 approaches zero 


. 
This is because the real and ideal curves 
coincide at 

ln x1 =0 
since 
.1 =1 at x1 =1 


-RT ln .1 vs. ln x1 




. 
We can change the axes by multiplying thevertical axis by 1/(R·T·V10) and the horizontal axis by 55.56·M2 which converts x2 to w2 
n2 n2 c2 c2 w2
x2 = .== = 
n1 .n2 n1 c1 55.56 55.56 M 2 


-1/V10 ln .1 vs. w2 




. 
We can use a Taylor expansion to express thisfunction relating 
1
- ln .1
V 10 
to 
w2 



. 
The Taylor expansion is 
- 10 ln .1 =B0. B1 w2 .B2 w22.... V 1 
where Bi's are constants 
. 	This is an infinite series, but Bi gets very small further out 


. Now we should evaluate what B0 and B1 are 
. 
When 
w2 =0 
then 
1
- 0 ln .1 =B0V 1 
so 
B0 =0 


.ln .1
V 10


. 1 . 

=B1 .2 B2 w2
. w2 
which is also zero 
B1 =0 


. 
The curve starts off as a parabola 
1
- ln .1 =B2 w22 .B3 w32.... 
V 10 
so 
w2
.=RT . RT . B2 w22. B3 w32.....
M 2 


. 
In general we do not know the values of B's 
but this is the form of the equation 
. Now, let us divide through by w2 
. 1 
=RT .B2 w2.' higher order terms' 
w2 .M 2 . 

where the higher order terms can be neglected 
. 
Now the nonideality term is approximated by a constant B2 times w2 


. 
This is the virial expansion 
– 
1/M2 is the first virial coefficient 

– 
B2 is the second virial coefficient 


. 
Experimentally we can not measure . at infinite dilution 
. 
We must measure . at finite concentration 
where nonideality is small but still present 



Introduction to biophysics: Colligative properties 



Introduction to biophysics: Colligative properties 


Introduction to biophysics: Colligative properties 



Introduction to biophysics: Colligative properties 
www.itk.ppke.hu 



. 
Two additional rules: 
– In an osmometer, if side A contains a polysaccharideat one concentration and side B contains a polysaccharide at a different concentration then 

.=RT . c 

– If we put more polysaccharides on the same side ofan osmometer then 
.=RT . ci 

i 


dissociable ions 
. 
Let us consider a container consisting of twochambers separated by a semipermeablemembrane, called osmometer 
. 
Let us put a protein in side A that dissociatesto give one Na+ ion 
. 
There will be then the same number of 
proteins (negatively charged) as Na+ions 



. 
According to this if 
[ P- ]A =a 
where [P-]A is the molar concentration of the protein in chamber A, then 
[ Na.]=a 
. 
Therefore, now 
.=RT 2 a 
because . is a colligative property 


Introduction to biophysics: Colligative properties 

Proteins with bound dissociable Na+ ions 




. 
Na+can diffuse across the membrane but the law of 'electroneutrality' states that therecannot be a finite (large) charge separation insolution 
. 
On microscopic scale, we violate this law allthe time (for example battery or themitochondrial membrane etc.) 


. 	If we measured M2 without taking this into account, it will be too small by a factor of 2 
. 
If the protein dissociates to give 10 Na+, then the M2 would be off by a factor of 10 


. 
Before we understood about polyelectrolytes, people thought proteins had a very smallmolecular weight 
. 
Also the molecular weight appeared to vary alot 
. 
This leads to the colloidal theory of protein 
structure 

. 
We can solve this problem by dumping a lot ofsalt into the system