Transcript Document

Chapter 7 Continued:
Entropic Forces
at Work
(Jan. 10, 2011)
Mechanisms that produce entropic forces:
• in a gas: the pressure
• in a cell: the osmotic pressure
• in macromolecular solutions: depletion
• in ionic interactions: electrostatic forces between macromolecules
• hydrofobia
1
Wat veroorzaakt osmotische druk?
Semipermeable membrane: colloidal particles
carry (letward) momentum => membrane delivers a
equal and opposite counter force, effectively
directed rightward, dragging water molecules
along!
Pressure difference due to the transfer of
momentum is given by the Van het Hoff
relation (‘Gas law’): ∆p = kBT∆c.
Note: without a pressure
gradient across the pore =>
no net flow through
the pore
MAKE YOUR TURN 7C
2
General case: spontaneous and driven osmosis combined
H20 flow leftward (reverse osmosis)
Two situations in which an active pump
keeps the pressure gradient across the
membrane the same (
now there is a pressure
difference =>
flow through
the pore
__) or even larger (---).
H2O flow rightward (osmotic machine)
• Driven osmosis: due to pressure difference ∆p
• Spontaneous osmosis: due to a concentration difference ∆c (VhHoff)
3
Flow through the pore: Poiseuille flow (Ch. 5):
Q=[m3/s]
8hd
1
Z hydr =
º
4
p R A × Lp
(Lp is the ‘filtration coefficient’)
Flux: flow per area per second:
If ∆c=0: jv= -Lp∆p
If ∆p=0: jv= Lp∆c • kBT
Combine driven+spontaneous osmosis:
4
Entropic force by electrostatic interactions:
(now we account for the interactions between the molecules)
Macroscopic bodies are electrically neutral. Why is that?
Water drop
2 mm
++
++
+
+
+
+
+
+
+
+
+ +
How much work is needed to ionize 1% of
the water molecules?
Ions will start to form a
positively charged
shell around the drop
E pot (R) =
q2
8pe0 R
Number of charges:
N = NA ×
mdruppel
rV
= 6 ×10 23 × druppel ´1% =1.4 ×1018
mmol
0.018
2
e
1
E pot (R) = N 2 ×
×
» 2.1011 J !!!
4 pe0 2R
Make YOUR
TURN 7D
5
In the cell:
Thermal motion can make large neutral macromolecules
to split into charged subunits!
Take an acid macromolecule like DNA
 A negatively charged macro-ion is left behind
 + a surrounding cloud of positive cations.
+
+
-
+
Equilibrium between
the urge to increase
entropy, and the cost by
having to overcome
the electric field
(potential energy)
-
+
DNA
-
-
+
+ ‘diffuse charge layer/electric double layer’
Looked at from far away the
E-field of the macromolecule is neutral,
but from nearby (nm scale) it’s not!
6
Two types of forces work on macromolecules:
• depletion drives them together
• electrostatic repulsion moves them apart
Therefore the macromolecules are not all closely packed together
Moreover:
• these forces act at very short range (order of nm)
• these forces have a strong geometric dependency
(stereo-specificity)
This ensures that certain molecules can find each other and
stick together.
What are the properties of the surrounding cation-cloud?
7
-s q
rq (x)
Gauss law:
negative surface charge density (C/m2)
positive volume charge density (C/m3)
sq
(close to the surface)
E opp = e
dE rq (x) (away from the surface)
=
dx
e
8
When charge distributions have a complex geometry: use
‘Mean Field Approximation’:
each ion senses on average the same electric field at x
Question: what is c+(x)? (the cation distribution in average potential V(x))
- Far away from the surface: c+(x) -> 0
- Ions move independent of each other through the field V(x):
Boltzmann distribution:
-eV (x )/ kB T
+
0
c (x) = c × e
But what is V(x)?
9
From:
dE rq
=
dx
e
dV (x)
and: E = dx
we get the Poisson equation:
rq
d 2V
=2
dx
e
(I)
We also had the Boltzmann charge concentration:
c + (x) = c 0 × e-eV (x )/ kB T
and note that for the charge density:
rq (x) º e × c + (x) = e × c 0 × e
-eV (x)/ kB T
(II)
10
To solve (I) and (II) we first introduce
the Bjerrum length in water:
B
e2
º
4pe × kB T
“How closely can two equal charges
be brought together with an
energy of kBT”
At room temp, monovalent charges in H2O: 0.71 nm
and the normalized potential:
Rewriting in these
new variables yields
from (I) and (II):
eV (x)
V (x) º
kB T
d 2V
-V
= -4p B c 0e
2
dx
Your Turn 7E
The Poisson-Boltzmann equation
How do we solve this P-B equation? Boundary conditions!
11
Boundary conditions for V(x):
E opp
sq
dV
=- =e
dx
yielding:
dV
dx
opp
We further adopt:
(close to the surface)
sq
= 4p B
e
Moreover:
dV
E¥ =dx
(no
= 0 charge,
pure
¥
H2O)
V (0) = 0
The solution is a function of the form V(x) = B ln (1 + x/x0)
(see Book, page 268), which gives:
kB T
x
V (x) = 2
ln(1+ )
e
x0
with
x0 º
e
2p Bs q
The “Gouy-Chapman layer”
YOUR TURN 7F
12
An interesting case: two negative macro ions with
diffuse charge clouds, approaching each other:
When D<2x0 the ions start to sense each other. The cations are
forced to stay between the ions (neutrality), which creates an
osmotic pressure!
We can again use the Poisseuille-Boltzmann eqn., but now with
different boundary conditions!
V midden º V(0) = 0
(V(x) symmetric re. centre)
Now try a solution of the form: V(x)=A•ln cos(x):
Question 7-i: verify that A=2 and
b = 2p B c 0
13
Second boundary condition:
E opp
sq
dV
=- =e
dx
(close to the surface, x=±D)
Question 7-ii: Verify that now
So that:
2b tan(bD) = 4p B (s q /e)
c0
V (x) = 2lncos(bx) c + (x) =
(cos bx) 2
Grafic solution
for 
14
The osmotic force between the macro ions is:
kBT times the concentration difference:
b2
=
kB T
2p B
Comparison of theory with experiment
Voor de
durf-al:
YOUR TURN 7G
15
Werkcollege van 11-1-2011:
Your Turns 7C, 7D, 7E en 7F
Opgave 7.4
16