1 chapter 7 continued: entropic forces at work (jan. 10, 2011) mechanisms that produce entropic...
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3 General case: spontaneous and driven osmosis combined Two situations in which an active pump keeps the pressure gradient across the membrane the same ( __ ) or even larger (---). Driven osmosis: due to pressure difference ∆p Spontaneous osmosis: due to a concentration difference ∆c (VhHoff) now there is a pressure difference => flow through the pore H 2 O flow rightward (osmotic machine) H 2 0 flow leftward (reverse osmosis)TRANSCRIPT
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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
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Semipermeable membrane: colloidal particles carry (letward) momentum => membrane delivers an equal and opposite counter force, effectively directed rightward, dragging water molecules along!
Pressure difference due to the transfer ofmomentum is given by the Van het Hoffrelation (‘Gas law’): ∆p = kBT∆c.
MAKE YOUR TURN 7C
Note: without a pressure gradient across the pore =>
no net flow throughthe pore
Wat veroorzaakt osmotische druk?
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General case: spontaneous and driven osmosis combined
Two situations in which an active pumpkeeps the pressure gradient across the
membrane the same (__) or even larger (---).
• Driven osmosis: due to pressure difference ∆p• Spontaneous osmosis: due to a concentration difference ∆c (VhHoff)
now there is a pressuredifference => flow through
the pore
H2O flow rightward (osmotic machine)
H20 flow leftward (reverse osmosis)
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Flow through the pore: Poiseuille flow (Ch. 5): Q=[m3/s]
Flux: flow per area per second:
If ∆c=0: jv= -Lp∆p
If ∆p=0: jv= Lp∆c • kBT
(Lp is the ‘filtration coefficient’)
Combine driven+spontaneous osmosis:
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Entropic force by electrostatic interactions:
(now we account for the interactions between the molecules)
Macroscopic bodies are electrically neutral. Why is that?
2 mm
Water drop How much work is needed to ionize 1% of the water molecules?
+ +
++++
++ +
++++ +
Ions will start to form a positively charged shell around the drop
Number of charges:
!!!
Make YOUR TURN 7D
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In the cell:Thermal motion can make large neutral macromoleculesto split into charged subunits!
Take an acid macromolecule like DNA A negatively charged macro-ion is left behind + a surrounding cloud of positive cations.
DNA-- -
- +
+
+
+
+
+ - -
Equilibrium betweenthe urge to increase
entropy, and the cost by having to overcome
the electric field (potential energy)
Looked at from far away theE-field of the macromolecule is neutral,
but from nearby (nm scale) it’s not!
‘diffuse charge layer/electric double layer’
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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 andstick together.
What are the properties of the surrounding cation-cloud?
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negative surface charge density (C/m2)
positive volume charge density (C/m3)
Gauss law: (close to the surface)
(away from the surface)
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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:
But what is V(x)?
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From: and:
We also had the Boltzmann charge concentration:
we get the Poisson equation: (I)
and note that for the charge density:
(II)
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To solve (I) and (II) we first introduce the Bjerrum length in water:
“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):
The Poisson-Boltzmann equation
How do we solve this P-B equation? Boundary conditions!
Your Turn 7E
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Boundary conditions for V(x):
(close to the surface)
We further adopt:
yielding: Moreover:(no charge, pure H2O)
YOUR TURN 7F
The solution is a function of the form V(x) = B ln (1 + x/x0) (see Book, page 268), which gives:
with
The “Gouy-Chapman layer”
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An interesting case: two negative macro ions withdiffuse charge clouds, approaching each other:
When D<2x0 the ions start to sense each other. The cations areforced to stay between the ions (neutrality), which creates an osmotic pressure!
We can again use the Poisseuille-Boltzmann eqn., but now withdifferent boundary conditions!
(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
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Second boundary condition:
(close to the surface, x=±D)
Question 7-ii: Verify that now
So that:
Grafic solutionfor
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The osmotic force between the macro ions is:kBT times the concentration difference:
Comparison of theory with experiment
Voor de durf-al:YOUR TURN 7G
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Werkcollege van 11-1-2011:
Your Turns 7C, 7D, 7E en 7FOpgave 7.4