Transcript Document

Lecture 6 (Dec. 13):
Chapter 6 (end)
and Chapter 7:
“Entropic Forces at Work”
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We zagen voor de kansverdeling voor de
toestanden van een klein systeem a,
in contact met macroreservoir B (op T):
Dit is het aantal toestanden van het totale systeem gegeven
dat a in een bepaalde toestand zit, ongeacht B.
In evenwicht zijn alle macrotoestanden even waarschijnlijk
(constante P0), dus is de kans op een zekere toestand voor a
Omdat Ea << EB, geldt een Taylorbenadering:
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Gebruik de definitie van temperatuur, T=(dS/dE)-1:
Dit is de Boltzmann verdeling voor klein systeem a
in contact met warmtereservoir op temperatuur T.
a wordt dus alleen door de temperatuur van B bepaald.
Interpretatie van T: “beschikbaarheid van energie van
systeem B”.
Als T groot is, is de kans dus groter dat Ea groter is.
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Systeem met 2 energietoestanden
Stel een systeem verkeert in 2 mogelijke toestanden met
energie E1 en E2
Dan is de kans P1 versus P2 om in toestand S1 versus S2
te verkeren:
Combineren met normeringsconditie, P1+P2=1:
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Vrije energie voor microscopisch
(sub)systeem
Voor een macroscopisch open systeem hadden we de
vrije energie van systeem a geïntroduceerd, die alleen
via de temperatuur T van B afhing.
Voor een microscopisch systeem kunnen we nu hetzelfde
doen:
De energie is hier vervangen door een gemiddelde energie.
Er zijn immers grote fluctuaties mogelijk als het systeem
klein is.
hier is
E a º å Pj × E j
j
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De Boltzmann verdeling minimaliseert deze vrije energie als
we nemen
Sa = -kB å Pj log Pj
(Shannon)
j
Pj º p(E j ) µ e
-E j / kB T
in YT 6G)
Het minimum van de vrije energie wordt dan gegeven door
Z = åe
- E j / k BT
met
j
Z heet de partitie functie
dit volgt uit : Pj =
e
-E j / kB T
Z
en Fa = E a - T× Sa
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voorbeeld: Complexe 2-state systemen
(e.g. macromoleculen)
In een complex systeem van vele toestanden die kunnen worden
onderverdeeld in 2 subklassen, is het mogelijk dat er zelden
overgangen van de ene naar de andere toestand plaatsvinden.
In dat geval is het zinnig om de vrije energie van beide klassen
toestanden afzonderlijk te beschouwen:
Fa , I = á Ea ñ I - TS a, I
Fa , II = á Ea ñ II - TS a, II
We vinden dan:
PI
- DF / k BT
=e
PII
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RNA-folding
Een streng RNA heeft
een zogenaamde
haarspeld-configuratie, of
een open configuratie.
Door middel van het uitoefenen van een kracht
wordt de vrije energie van
de afzonderlijke toestandsklassen van het molecuul
gemanipuleerd, waardoor
de overgangsfrequentie manipuleerbaar wordt. Verrassend hierbij
is het feit dat een dergelijk complex systeem goed voldoet aan de
beschrijving als 2-state complex systeem.
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Analogous to the concept of a mechanical force
on objects in a force field (potential energy, U), for which:
dU pot
f =dx
we introduced in Ch. 6 the entropic force for statistical systems:
dFa
fa = dx
Chapter 7 reviews several 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
• hydrofobia
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Entropic force 1: two examples of the ideal gas
The entropic force is the pressure. Why is that?
I: Volume fixed: L3
N identical particles
Energy: kinetic energy + (fixed)
internal energy
L
L
L
What is the Free Energy of the gas?
We compute this most conveniently
from the partition function, Z
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Partition sum:
Z = åe
-E j /(kB T )
(= normalization factor over
all allowed states j)
j
With N particles in fixed volume L x L x L:
Ej =
p 2j
2m
+ eint
State of particle j is determined by its position, rj, and momentum, pj
Macrostate for the entire (canonical) ensemble of N particles is then:
[r1, p1],[r2, p2 ],[r3, p3 ],[r4 , p4 ], ,[rN , pN ]
Z@
Partition sum is:
å
åe
-E j /(kB T )
Joint probabilities!
all positions all energies j
Z =C×
+¥
L
òd ròd
3
1
0
-¥
3
p1 × e
- p12
+¥
L
/(2kB T )
òd r òd
3
N
0
3
pN × e
- p N2 /(2kB T )
-¥
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Z =C×
Z =C×
+¥
L
òd ròd
3
3
1
0
-¥
L
+¥
òd ròd
3
1
L
Z = C × [ ò d r1 ×
3
/(2kB T )
3
3
-¥
3
-( p12 + + p N2 )/(2kB T )
ò d r ]ò d
N
pN × e
-¥
+¥
3
pN × e
- p N2 /(2kB T )
0
N
L
3
N
+¥
0
0
òd r òd
òd r òd
p1 ×
+¥
L
L
-¥
0
0
3
p1 × e
- p12
+¥
3
p1
-¥
òd
3
pN × e
-( p12 + + p N2 )/(2kB T )
-¥
Z = Z(L) = C'×L3N
The Free Energy is then computed by:
Fa = -kB T × lnZ = C - kB T × ln L
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Þ Fa (L) = C - kB T × N lnL = E a - T × S
3
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The entropic force is then found by:
dFa
kB TN
fa = =
º pa
3
d(L )
V
and the entropy of the gas is ~
Gas law!
S = kB N lnV
II. A similar approach is followed when p=constant, V changes
Note that:
T
A
p
V=A•L
L
f
L ?
L º
L max
ò L × P(L) × dL
0
P(L)?
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P(L) = P(r1
rN , p1
pN , pPiston ,L)
N
åp
2
j
work on
the gas
p
f ×L
= C × exp()
2mkB T 2MkB T k B T
2
Piston
j=1
L Max
ò P(L) × dL = 1
And since P(L) is a probability:
0
L Max
ò L × P(L, r
1
L =
rN , )d r1 d pN d pPiston dL
3
3
3
0
L Max
ò P(L, r1
0
rN , )d 3 r1 d 3 pN d 3 pPiston dL
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L Max
That is:
L =
- fL / kB T
N
e
×
L
× L ×dL
ò
0
L Max
- fL / kB T
N
e
×
L
×dL
ò
MAKE YOUR TURN 7A
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Giving:
(N + 1)kB T NkB T
L =
»
f
f
Also note that:
(again the Gas Law!)
d(-kB T ln Z( p)) dF( p)
L =
º
dp
dp
(with Z(p) the partition function of gas+piston)
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Osmotic pressure as an example of Free Energy transformation:
The ‘osmotic machine’
The inverse
‘osmotic machine’
Concentration in the solution is low: ‘ideal gas’ approximation.
pequilibrium = c × kB T
with c=N/Vsol. (Van het Hoff)
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Height zf?
In this case there is also the atmospheric pressure.
Pressure difference:
kB T
zf =
×c
rm g17
What is the maximal work for the osmotic machine?
MAKE YOUR TURN 7B
Is osmosis relevant/significant for cells?
Some numbers regarding cells:
±30% cell volume = protein molecules,
‘spheres with radius about 10 nm’
=> c(protein) = 7•1022/m3=0.12mM/liter
Osmotic pressure:
20m
This excess pressure suffices to rupture the
cell membrane! => regulation of c is needed!
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Cells are very crowded!
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In cells molecules of very different sizes and shapes are active:
This
introduces
an entropic
force, due to
depletion!
Fig. 2.4
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What is the depletion effect?
large
particle
Depletion zone
for the small
particles
small
particle
Depletion is an entropic effect caused by combining particles
with different sizes and shapes.
In (a): effective volume for the small particles:
In (b): effective volume for the small particles:
(met
I.e.: Veff increases => Seff increases => F decreases! <1)
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Experimental support for the depletion effect (Dinsmore, 1998):
The depletion effect is particularly large when the big molecules
precisely fit in each other! In that case: ΔV=A•(2R)
n.b. The type of molecules/particles is not relevant! 22
Osmotic force results from ‘rectification’ of Brownian motion!
How is that?
p0
p(z) = p0 + rm g× (z0 - z)
z0
Hydrostatic pressure in a constant external
force (g).
z
Now consider the case where:
- the force on a volume element depends on position, and is
therefore not constant, like gravity
(this force originates from the semipermeable membrane)
- the force is directed along +z:
dV = dx × dy × dz
F(z)
is external force/volume
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We look at the total force balance on this volume element:
In x-direction:
p(z)(x- 1 dx ) dydz = p(z)(x + 1 dx ) dydz
In y-direction:
p(z)(y- 1 dy ) dxdz = p(z)(y + 1 dx ) dxdz
2
2
In z-direction:
2
2
F(z)dxdydz = p(z + 12 dz)dxdy - p(z - 12 dz)dxdy
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F(z)dxdydz - p(z + 12 dz)dxdy + p(z - 12 dz)dxdy = 0
p(z + 12 dz) - p(z - 12 dz) dp
F(z) =
=
dz
dz
Note that when the force=constant,
p(z) =
with
F = -r m g
ò Fdz = C - r
we get:
m
g× z
p(0) = rm g× z0 + p0
Consider particles in a suspension, density c(z), an that on each particle
works a force f(z) in the z-direction (z=distance from membrane).
Low Reynolds-number case: in equilibrium f(z) is compensated by
a viscous counter force. So:
dp
F(z) = c(z) × f (z) =
dz
but p(z) and c(z)
are both unknown!
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But the force on a particle also obeys:
dU(z)
f (z) = dz
and Boltzmann tells us:
Combine:
c(z) = c 0e
F(z) = c(z) × f (z) = -c 0e
-
U (z)
kB T
-
U (z)
kB T
dU(z)
×
dz
d(e-U (z )/ kB T ) )
dc dp
Note: F(z) = c 0 k B T
= kB T
=
dz
dz dz
Note: the osmotic pressure arises through a force!
(it is not just due to the presence of a solvent).
Concentration gradients themselves do NOT cause pressure!
The semipermeable membrane is essential in generating
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this force. How??
Semipermeable membrane: colloidal particles
carry momentum => membrane delivers an
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.
Note: without a pressure
gradient across the pore =>
no net flow through
the pore
MAKE YOUR TURN 7C
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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)
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Flow through the pore: Poiseuille flow (Ch. 5):
Q=[m3/s]
Z hydr
8hL
1
= 4 º
pR
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:
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