Chapter 7 Lecture

Biological PhysicsNelson

Updated 1st Edition

Slide 1-1

Entropic Forces at Work

Slide 1-2

Last time

• We need Chapter 6 on ENTROPY, visit here:

• Main points:

• And the partition function Z

• -kBT ln(Z) minimizes the free energy!

• Entropic forces f = - dFa/dx

Slide 1-3

A great Sci-fi quote

• “The struggle between life elements is the

struggle for the free energy of a system.”

• DUNE, Frank Herbert, 1967

Slide 1-4

Outline

• 7.1 Microscopic view of entropic forces

– Fixed volume and pressure approaches

• 7.2 Osmotic pressure

– From ideal gas law and depletion forces

• 7.3 Beyond Equilibrium: Osmotic flow

– Rectification of Brownian motion

• 7.4 A repulsive interlude

– Electrostatics, Gauss’ Law and more

• 7.5 Properties of water (extra reading)

– Water generates an entropic attraction

between nonpolar objects

Slide 1-5

Comment & Focus questions

• F = E – TS, macroscopic systems tend to lower

their free energy even if:-

– The change actually increases the energy

(but increases TS more)

– The change actually decreases entropy (but

decreases E/T more)

• What keeps cells full of fluid?

• How can a membrane push fluid against a

pressure gradient?

• Osmotic pressure is a simple example of an

entropic force

Slide 1-6

• Constant volume case leads to (pressure case,

extra-reading). Derive p = kBNT/V from partition

function where in this case Z is

• And Fa=-kBTln(Z(L)) where many of the terms

above are just constants →

• So Z is a constant × L3N →

F(L) ＝ A - kBTNlnL3

• Using entropic force then recovers the ideal gas

law:

7.1 Microscopic view of entropic forces

Slide 1-7

Osmosis and entropic forces?

Slide 1-8

7.2.1 Equilibrium osmotic pressure from

ideal gas law

• Consider Fig. 1.3 (previous slide). For a dilute

concentration we apply same methods as for an

ideal gas to find the free energy

• This time V is RHS of chamber in Fig. 1.3 and

derivative of ln Z is again N/V and (c = N/V)

• How about when there is air (or atmospheric

pressure)? In this case we get Δp=zf ρm g

• See figure 7.1 and the thistle-tube experiment

– How does this work? Try Your Turn 7B

Slide 1-9

Osmotic pressure experiment

thistle-tube experiment

Slide 1-10

7.2.1 Cell safety: Laplace’s formula

• Estimates

– Homework/whiteboard for now …

– Result is

• Surface tension, Σ, estimated ~1.5×10-3 N m-1

– Enough to rupture a eukaryotic cell

membrane

• Summary

– Osmotic pressure is significant for cells …

Slide 1-11©1993. Used by permission of Springer-Verlag.

7.2.2 Depletion force for large molecules

Slide 1-12

7.2.2 Osmotic pressure creates a depletion

force between large molecules

• The hierarchy in Fig. 7.2 leads to surprising

“entropic effects”

– Depletion interaction or molecular crowding

• Consider two LARGE “sheep” in a bath

containing smaller “sheepdogs” with number

density c. The sheepdogs herd the larger sheep

together. How (see Fig. 7.3)

• In order to decrease Fa depletion zone is short

range of order 2a (a size of sheepdog)

• Pressure is change of free energy per change of

volume →

Slide 1-13

The depletion zone (interactions)

Slide 1-14©1998, American Chemical Society. Used by permission.

Experimental data (depletion is real!)

Slide 1-15

7.3.1 Beyond Equilibrium: Osmotic Flow

• Where does osmotic pressure come force from?

• Ultimately, from membrane itself (see Fig.7.5)

• Using & hydrostatic eq:

• Leads to Eq. 7.13: where 𝑓 = −𝑑𝑈

𝑑𝑍

• How do we solve this. We need a result for c(z)?

• Using c=c0 and steps on p. 258 gives

• This is just Δp=c0kBT or more generally:

Slide 1-16

Osmotic flow: Fluid forces

Slide 1-17

7.3.1 Osmotic flow

The equilibrium pressure difference across a

semipermeable membrane equals kBT times the

difference in concentration between the two force-free

regions on either side of the membrane.

• Read 7.3.2 for homework (see section 5.3.4 as

well)

•Generally we have

where 𝐿𝑝 is the filtration constant of the membrane

• Reverse osmosis occurs when Δp≥ΔckBT (see

Fig 7.6d)

Slide 1-18Figure 7.6 (Schematic, sketch graphs.) Caption: See text.

Slide 1-19

Figure 7.6 Caption

Figure 7.6: (a) A literal model of a semipermeable membrane, consisting of a

perforated wall with channels too small for suspended particles to pass. (b)

The force along z exerted by the membrane on approaching particles is

−dU/dz, where U is the potential energy of one particle. (c) In equilibrium, the

pressure p is constant inside the channel (between dashed lines), but p falls in

the zone where the particle concentration is decreasing. (d) Solid curve: If the

pressure on both sides is maintained at the same value, osmotic flow through

the channel proceeds at a rate such that the pressure drop across the channel

(from viscous drag) cancels the osmotic pressure jump. Dashed curve: In

reverse osmosis, an external force maintains a pressure gradient even greater

than the equilibrium value. The fluid flows oppositely to the case of ordinary

osmotic flow, as seen from the reversed slope of the pressure profile inside the

channel.

Slide 1-20

4.6.1 Recall Membrane Diffusion*

• Imagine a long thin membrane/tube of Length L,

with one end in ink C(0)=c0 and in water C(L)=0

• This leads to a quasi-steady state so we set

dc/dt =0 and hence d2c/dx2=0

• This means that c is constant and js=-DΔc/L

where Δc0=cL-c0 and subscript s means the flux

of solute not water

• Now define js=-PsΔc where Ps is the permeability

of the membrane.

Slide 1-21

• In simple cases Ps roughly relates to the width of

the pore and thickness of the membrane (length

of pore)

• Using dN/dt=-Ajs leads to

Slide 1-23

Last week’s homework: Your Turn 4C

Generalizing Fig. 7.6 to concentrations 𝑐1 & 𝑐2• a) Δ𝑝 = 𝑝1 − 𝑝2 < 𝑐1 − 𝑐2 𝑘𝐵𝑇

• b) Δ𝑝 = 𝑝1 − 𝑝2 > 𝑐1 − 𝑐2 𝑘𝐵𝑇 (reverse osmosis)

Slide 1-24

A REPULSIVE INTERLUDE

Section 7.4

Slide 1-25

7.4.1 Bulk electroneutrality?

• Example: For a raindrop with R=1 mm, how

much work is needed to remove 1 electron from

1% of the water molecules in the drip?

• If we take away charge then we form a shell with

electrostatic potential energy

Eelectrostatic =1

2𝑞𝑉 𝑅 =

𝑞2

8𝜋𝜀0𝑅where

𝑒2

4𝜋𝜀0= 2.3 ⋅ 10−28 Jm.

• This implies

𝑞1%

𝑒

2=

103𝑘𝑔

𝑚3

6⋅1023

0.0018 𝑘𝑔×

4𝜋

310−3𝑚 3 × 0.01

2

×

2.3⋅10−28

2𝑅= 2 ⋅ 1011 𝐽 ≫ 𝑘𝐵𝑇𝑟 = 4.1 ⋅ 10−21𝐽 .

• So electrically neutral!

Slide 1-26

7.4.1 Electrostatic interactions are critical

for cell functioning

• From Your Turn 7D, if R=1 nm we see that

separation energy ≈ 0.7𝑘𝐵𝑇𝑟.

• Hence on nanoscales DNA is a macroion which

is nearly fully dissociated (cf. Chapter 8)

• Even though the electrostatic interaction is of

long range in vacuum, in solution the screening

effect of counterions reduces its effective range,

typically to a nanometer or so.

• Macroions will not feel each other until they’re

nearby. Once nearby, detailed surface pattern of

+ve and –ve residues on a protein can be felt by

its neighbor, not just the overall charge.

Slide 1-27

Continued ...

• While each individual electrostatic interaction

between matching charges is rather weak

(compared to kBTr), still the combined effect of

many such interactions can lead to strong

binding of two molecules—if their shapes and

orientations match precisely.

• Molecular recognition implies we understand the

counterion cloud around a charged surface

• This implies Gauss’s Law

• This is not enough as we also need to worry

about orientation (stereospecificity)

Slide 1-28

7.4.2 The Gauss Law (for later use)

• You should have already studied the Gauss law:

which for a flat charged surface gives

• Considering Fig. 7 (next slide) we find that a

solid element satisfies

implying

Slide 1-29

7.4.2 The Gauss Law

Slide 1-30

7.4.3 Counterions on charged surfaces

• Example: You might want to coax DNA into a

cell for gene therapy, but both are negatively

charged

– We want to know by how much?

• Consider the slightly simpler problem depicted

in Figure 7.8a

– Assume monovalent counterions e.g. Na+

• Problem is that we are not given the charge

distribution?

– We need to find it, but an electrostatic force is

long range so we need to know positions of all

the other charges

Slide 1-31

7.4.3 Counterions near surfaces

Slide 1-32

7.4.3 Charged surfaces are surrounded by

neutralizing ion clouds

• We need to evaluate the effect of many ions all

moving around and repelling each other?

• This is too complicated so we make a very

common assumption in physics:

• The Mean Field Approximation (MFA)

– For a large no. of ions we assume that all the

other ions contribute to an average charge

density ＜ρ＞ and can ignore individual

charges ρi between ions

– We’ll use this to solve the Poisson-Boltzmann

equation but first we need to derive it ...

Slide 1-33

7.4.3 Poisson-Boltzmann Equation

• The Boltzmann part comes from the MFA

assuming c+(x→∞)→0 with

c+(x)=c0exp[-eV(x)/kBT]

and V(0)=0 fixes c0=c+(x=0) is the no. of counter-

ions at the surface

• The Poisson relation comes from bulk Gauss

law: dE/dx = ρｑ /ε with Ε=-dV/dx gives

d2V/dx2=-ρｑ /ε

Slide 1-34

7.4.3 Poisson-Boltzmann Equation*

• Equating the charge now leads to

d2V/dx2=-ec+(x)/ε=-ec0/ε exp[-eV(x)/kBT]

• Defining

and dimensionless variables

gives

• Problem: How to solve this equation?

Slide 1-35

Mean Field Approximation

Slide 1-36

7.4.3 Boundary conditions

• Like any differential equation we need to find

boundary or initial conditions: d2z/dt2=-g does

NOT tell you how high a rock will rise into the air

– Need to know how hard you threw it or it’s

initial speed and height.

• Solution for Poisson-Boltzmann case?

• Use boundary Gauss Law -dV/dx|surface = - σｑ /ε

or in terms of new variables

NB. Remember σｑ is a positive number

Slide 1-37

Solutions of the PB equation: Case 7.8a

• We have the BC at surface (previous slide) plus

another solution at infinity: dṼ/dx= 0 and Ṽ (0)=0

• PB eqn is non-linear and generally very difficult

to solve; however the exponent implies Ṽ(x)=ln x

might be a solution, but this blows up at x=0:

• Therefore try

• We need to find B and x0? (class exercise) →

• This is the “Gouy-Chapman” layer with length x0

Slide 1-38

PB solution continued: Case 7.8a

• A diffuse layer forms with thickness x0: The

counterions are willing to pay some electrostatic

energy to gain entropy

• Note, turning off thermal motion (T→0) implies

an infinite Bjerrum lenth and hence x0→0.

• How much energy? Think of two sheets of

charge with capacitive energy E=qtot2/(2C) where

C = εA/x0 . Combining the preceding formulae

we find (Case 7.8c?):𝐸

𝑎𝑟𝑒𝑎≈ 𝑘𝐵𝑇

𝜎𝑞

𝑒

for no added salt

Slide 1-39

7.4.4 Repulsion of like charged surfaces

• Solution of the PB equation for two negatively

charged surfaces, Case 7.8b

• We now have a symmetric solution with 𝑉 0 =

0: ⇒ 𝑉 = 2 ln 𝛽𝑥, where 𝛽 = 2𝜋ℓ𝐵𝑐0 and

𝑐+ 𝑥 = 𝑐0 cos 𝛽𝑥 −2

• Finally, applying surface charge b.c. leads to

4𝜋𝑙𝐵𝜎𝑞

𝑒= 2𝛽tan 𝐷𝛽

• This is a great example of solving a problem

graphically (used a lot in quantum mechanics)

• The important point is that this has been

observed experimentally (see next next slide)

Slide 1-40

Graphical solution to Poisson-Boltzmann

• Two oppositely charged surfaces (Fig. 7.8b)

Slide 1-41

Comparison with experiment: it works!

Slide 1-42

The Big Picture: Conclusion

• What is the big picture from Chapter 7?

• Entropic forces ubiquitous in the cellular world

– Depletion Forces

– Osmosis

– Reverse osmosis

• Neutral objects can be charged on nanoscales

Slide 1-43

Homework I

• Read Section 7.5

– Special properties of water

• Figure out Figure 7.6 and try

– “Your Turn 7A, 7B and 7C and Poisson

Relation”

– Think about electric neutrality and whether or

not this is true for either macroscopic or

nanoscopic systems?

– Try 7.4.3:- Track 2 on page 284 (book); pg.

250 of ebook. Including a salt solution

Slide 1-44

NEXT TIME …

Chapter 8: Chemical Forces & Self Assembly