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Lecture 6 (Dec. 13): Chapter 6 (end) and Chapter 7: “ Entropic Forces at Work ”

Lecture 6 (Dec. 13): Chapter 6 (end) and Chapter 7: “ Entropic Forces at Work ”. 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

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Lecture 6 (Dec. 13): Chapter 6 (end) and Chapter 7: “ Entropic Forces at Work ”

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  1. Lecture 6 (Dec. 13): Chapter 6 (end) and Chapter 7: “Entropic Forces at Work”

  2. 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:

  3. 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.

  4. 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:

  5. 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.

  6. De Boltzmann verdeling minimaliseert deze vrije energie als we nemen in YT 6G) Het minimum van de vrije energie wordt dan gegeven door met Z heet de partitie functie

  7. 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: We vinden dan:

  8. RNA-folding Een streng RNA heeft een zogenaamde haarspeld-configuratie, of een open configuratie. Door middel van het uit- oefenen van een kracht wordt de vrije energie van de afzonderlijke toestands- klassen 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.

  9. Analogous to the concept of a mechanical force on objects in a force field (potential energy, U), for which: we introduced in Ch. 6 the entropic force for statistical systems: 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

  10. I: Volume fixed: L3 N identical particles Energy: kinetic energy + (fixed) internal energy What is the Free Energy of the gas? L L L Entropic force 1: two examples of the ideal gas The entropic force is the pressure. Why is that? We compute this most conveniently from the partition function, Z

  11. (= normalization factor over all allowed states j) Partition sum: With N particles in fixed volume L x L x L: Macrostate for the entire (canonical) ensemble of N particles is then: Partition sum is: State of particle j is determined by its position, rj, and momentum, pj Joint probabilities!

  12. The Free Energy is then computed by:

  13. Gas law! and the entropy of the gas is ~ Note that: T A f p V=A•L L The entropic force is then found by: II. A similar approach is followed when p=constant, V changes

  14. And since P(L) is a probability: work on the gas

  15. That is: (again the Gas Law!) Giving: Also note that: (with Z(p) the partition function of gas+piston) MAKE YOUR TURN 7A

  16. with c=N/Vsol. (Van het Hoff) 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.

  17. Height zf? In this case there is also the atmospheric pressure. Pressure difference:

  18. 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! => regulationof c is needed! What is the maximal work for the osmotic machine? MAKE YOUR TURN 7B Is osmosis relevant/significant for cells?

  19. Cells are very crowded!

  20. In cells molecules of very different sizes and shapes are active: This introduces an entropic force, due to depletion! Fig. 2.4

  21. Depletion zone for the small particles large particle small particle In (a): effective volume for the small particles: In (b): effective volume for the small particles: (met <1) What is the depletion effect? Depletion is an entropic effect caused by combining particles with different sizes and shapes. I.e.: Veff increases => Seff increases => F decreases!

  22. 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!

  23. p0 z0 Hydrostatic pressure in a constant external force (g). z is external force/volume Osmotic force results from ‘rectification’ of Brownian motion! How is that? 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:

  24. In x-direction: In z-direction: In y-direction: We look at the total force balance on this volume element:

  25. we get: Note that when the force=constant, with 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: but p(z) and c(z) are both unknown!

  26. But the force on a particle also obeys: and Boltzmann tells us: Combine: Note: 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 this force. How??

  27. 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

  28. H20 flow leftward (reverse osmosis) H2O flow rightward (osmotic machine) 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 (---). now there is a pressure difference => flow through the pore • Driven osmosis: due to pressure difference ∆p • Spontaneous osmosis: due to a concentration difference ∆c (VhHoff)

  29. Flow through the pore: Poiseuille flow (Ch. 5): Q=[m3/s] Combine driven+spontaneous osmosis: (Lp is the ‘filtration coefficient’) Flux: flow per area per second: If ∆c=0: jv= -Lp∆p If ∆p=0: jv=Lp∆c • kBT

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