Last Lecture:• Viscosity and relaxation times increase with

decreasing temperature: Arrhenius and Vogel-Fulcher equations

• First and second-order phase transitions are defined by derivatives of Gibbs’ free energy.

• The glass transition occurs at a temperature where config exp and is dependent on thermal history. In a glass, config > exp .

• Glass structure is described by a radial distribution function.

• The Kauzmann temperature could represent the temperature at which there is a first-order phase transition underlying the glass transition.

HE3 SM

Phase Separation

19 February, 2009

Lecture 5

See Jones’ Soft Condensed Matter, Chapt. 3 and Appendix A

Today’s Question:When are Two Liquids Miscible?

When cooled below a critical temperature, miscible liquids will separate into two phases.

Oil and water

Basic Guiding Principles

• Recall from last week that dG = VdP-TdS.• Since, S increases or stays the same in an isolated

system, at constant P, the condition for the thermodynamic equilibrium of a system is that the Gibbs’ free energy, G, goes to a minimum at equilibrium.

• Helmholtz free energy: F = U - TS, so that in a phase transition at constant T: F = U – TS

• It can also be shown that dF = -PdV - SdT• Therefore at constant V, when a system is in thermal

equilibrium with its surroundings, the Helmholtz free energy, F, also goes to a minimum.

• We see that an increase in S or a decrease in U favours a transition to a more favourable state.

• Whether a transition occurs is thus decided by the balance between U and S.

G + R GR mixture

+

Higher S

Lower S

But what about F?

Unmixed state

Mixed state

Why are some liquids immiscible if a mixture has a higher entropy?

Higher S

In immiscible liquids, U increases upon mixing.

Then assume R + G = 1 (non-compressibility condition).

Let R = Volume of Red

Total Volume

Let G = Volume of Green

Total VolumeVolume fractions

Entropy Calculation from Statistical Thermodynamics

S = k lnThe statistical weight, , represents the number of ways of arranging particles (microstate) in a particular energy state (macrostate).

Boltzmann’s tomb

Meaning of the Statistical Weight

For a given “macrostate” of a system (i.e. a certain volume, pressure, temperature and average composition), there are microstates.

That is, there are ways of arranging the particles in the system to achieve that macrostate.

If all of the microstates are equally likely, then the probability of a particular microstate is p = 1/ , and the Boltzmann equation can be written as

S = k ln = - k ln -1 = - k ln p

(Shannon’s expression)

Change in S on Mixing, Smix

)+(== GRRGunmixmixmix SSSSSS

)ln+ln(ln= GRRGmix kkkS

Let NR be the total number of red molecules and NG be the total number of green ones.

The number of ways of arranging N indistinguishable molecules on N “lattice” sites is N!. Therefore:

)]!ln()!ln())!+[ln((= GRGRmix NNNNkS

But the Stirling approximation tells us that lnN! NlnN-N, for large N. Applying this approximation, we find:

RRRGRGRGRmix NNNNNNNNNkS +ln)+()+ln()+[(=

]+ln GGG NNN

RRRGRGRGRmix NNNNNNNNNkS +ln)+()+ln()+[(=

]+ln GGG NNN

Statistical Interpretation of Smix

Simplifying by grouping NR and NG terms:

)]+

ln(+)+

ln([=G

GRG

R

GRRmix N

NNN

NNN

NkS

If the volumes of red and green molecules are the same, then number fraction and volume fraction are identical:

GR

R

GR

RR NN

NVV

V+

=+

=

Substituting for ln(-1) = - ln():

)]ln(+)ln([= GGRRmix NNkS

(And likewise for G.)

Smix Expressed per Molecule

)]ln(+)ln([= GGRRmix NNkS

Our expression is the entropy change upon mixing all NR+ NG molecules:

Then, Smix per molecule can be found by dividing by the total number of molecules (NR + NG):

]ln+ln[+

=+

= GGRRGRGR

mixmolmix NN

NNk

NNS

S

Note that we have moved the negative outside the brackets. Recognising R and G:

]ln+ln[= GGRRmol

mixS k

Next we need to consider the change in internal energy, U!

Compare to Shannon’s expression: i

ii ppkS ln

Entropic (S) Contribution to Fmix

]ln+ln[= GGRRmol

mixS k

kSmix

G

Change in U on Mixing, Umix

• Previously, we considered the energy of interaction between pairs of molecules, w(r), for a variety of different interactions, e.g. van der Waals, Coulombic, polar, etc.

• We assumed the interaction energies (w) are additive!• When unmixed, there are interaction energies between

like molecules only: wRR and wGG.

• When mixed, there is then a new interaction energy between unlike molecules: wRG.

• At a constant T, the kinetic energy does not change with mixing; only the potential energies change.

• So,Umix = WR+G - (WRR + WGG), which is the difference between the mixed and the unmixed states.

SummaryType of Interaction Interaction Energy, w(r)

Charge-charge rQQ

o421 Coulombic

Nonpolar-nonpolar 62

2

443

r

hrw

o

o

)(_=)(

Dispersive

Charge-nonpolar 42

2

42 rQ

o )(_

Dipole-charge24 r

Qu

ocos_

42

22

46 kTruQ

o )(_

Dipole-dipole

62

22

21

43 kTruu

o )(_

Keesom

321

22

21

4 rfuu

o ),,(_

Dipole-nonpolar

62

2

4 ru

o )(_

Debye

62

22

4231

ru

o )()cos+(_

In vacuum: =1

Mean-Field Approach

• Describes the molecules as being on a 3-D lattice.

• Assumes random mixing, i.e. no preference for a particular lattice site.

• Then the probability that a site is occupied by a red molecule is simply R.

• We will only consider interaction energies (w) between each molecule and its z closest neighbours - neglecting longer range interactions.

Energy of the Unmixed State

• Each molecule only “owns” 1/2 of the pair interaction energy.

• For each individual molecule:

1G

RRRRR WzU 2=GGGGG WzU 2=

)+(2

=+= RRRGGGRRGGunmix wwz

UUU

1R

Energy of the Mixed State

)]+(+)+([= GGGRGRGRGGRRRR wwzwwzU 21

Probability that a neighbour is red

Probability that a neighbour is green

Probability that the reference molecule is red

Probability that the reference molecule is green

The mean-field approach assumes that a molecule on a given site will have zR red neighbours and zG green neighbours.

Energy of Mixing, Umix, per Molecule

)]+(+)+([= GGGRGRGRGGRRRRmix wwzwwzU 21

Umix = Umix - Uunmix

])(++)[(= GGGGRGGRRRRRmix wwwzU 2222

But, G + R = 1, so that R – 1= -G and G -1 = -R

]+[= GGRGRGGRRRGRmix wwwzU 22

NB: As we did with entropy, we will consider the change in U per molecule.

)+(=+= RRRGGGGGRRunmix WWzUUU 2From before:

)+(= GGRGRRGRmix wwwzU 22 Factor out terms:

]++[= GGGRGGRRRRmix wwwzU 22 22 Multiplying through:

The Interaction Parameter,

Two “sets” of interactions between R and G are gained, but interactions between R & R and between G & G are lost!

We now define a unitless interaction parameter, , to characterise the change in the energy of interaction after the swap:

)(= GGRRRG wwwkTz

22

We see that characterises the strength of R-G interactions relative to their “self-interactions”.

Imagine that a red molecule in a pure red phase is swapped with a green molecule in a pure green phase:

)+(= GGRGRRGRmix wwwzU 22

Internal Energy of Mixing, Umix

We saw previously that:

Substituting for we now find:

GRmix kTU =

A simple expression for how the internal energy changes when two liquids are mixed. Depends on values of T and .

)(= GGRRRG wwwkTz

22

and

Energetic (U) Contribution to Fmix

= 5

= 3

= 2

= 1

= 0

= -1 = -2

Regular solution model

kTUmix

GRmix kTU =

<0 favours mixing!

Entropic (S) Contribution to Fmix

]ln+ln[= GGRRmol

mixS kk

Smix

Free Energy of Mixing, Fmix

GRmix kTU =

At constant temperature:

mixmixmix STUF =

Using our previous expression for Smixmol:

))ln+ln((= GGRRGRmix kTkTF

)ln+ln+(= GGRRGRmix kTF Factor out kT:

Dependence of Fmix on

Regular solution model = 5

= 3

= 2

= 1

= 0

= -1

= -2

mixmixmix STUF =

kTFmix

Mixing is favoured

Mixing not favoured

Predictions of Phase Separation

Regular solution model

= 3

= 2.75

= 2.5

= 2.25

= 2

kTFmix

Summary of Observations

When 2, there are two minima in Fmix and a maximum at R = G = 0.5.

When 2, there is a single minimum at R = 0.5

How does this dependence of Fmix on determine the composition of phases in a mixture of liquids?

As increases, the two compositions at the Fmix minima become more different.

We have assumed non-compressibility, that molecules are on a lattice, and that volume fraction equals number fraction.

Phase Separation of Liquids

Phase-Separated: 1=0.5 and

2=0.8

Initial, G: 0=0.7

Mixed state

Free Energy of a System of Two Liquids

• A system of two mixed liquids (G and R) will have a certain initial volume fraction of liquid G of o.

• At a certain temperature, this mixture separates into two phases with volume fractions of G of 1 and 2.

• The total volume of the system is conserved when there is phase separation.

• The free energy of the phase-separated system can be shown to be:

Fsep can be easily interpreted graphically!

)(--

)(--

212

101

12

02

mixmixsep FFF

Free Energy of System with Low

10

Fmix

0

.Fo

Fsep

1 2

What if the composition o was to separate into

1 and 2?

Then the free energy would

increase from Fo to Fsep.

Conclude: Only a single

phase is stable!G

Free Energy of System with High

10 o

Fo

Fsep

1 2

What if the composition o was to separate into

1 and 2?

Then the free energy would

decrease from Fo to Fsep.

Conclude: Two phases are

stable.

.

Fmix

Stable, co-existing compositions found from minima: 0d

dF

F Does Not Always Decrease!

F

o

Fo

Fsep

1 2

.

What happens if o separates into 1 and 2?

Then Fo increases to Fsep which is not favourable; 2 is a metastable composition.

2*

The stable compositions are 1 and 2*!

Fmix

F

Two phases stable:

“Spinodal region”

Negative curvature

Positive curvature

02

2

>d

Fd

02

2

<d

Fd

Defining the Spinodal Point

Spinodal point

02

2

=d

Fd

Metastable

)(= GGRRRG wwwkTz

22

Determining a Phase Diagram for Liquids: Regular Solution Model

Recall that:

As the interaction energies are only weakly-dependent on T, we can say that 1/T.

When >2, two phases are stable; the mixture is unstable.When <2, two phases are unstable; the mixture is stable.

When 0 < <2, mixing is not favoured by internal energy (U). But since mixing increases the entropic contribution to F, a mixture is favoured.

A phase transition occurs at the critical point which is the temperature where = 2.

Constructing a Phase Diagram

T1

T2

T3

T4

T5

kTFmix

T1<T2<T3….

Co-existence where:

0=d

dF

Spinodal where:

02

2

=d

Fd

T1<T2<T3<T4<T5

Phase Diagram for Two Liquids Described by the Regular Solution Model

A

Immiscible

Miscible

T1~

Low T

High T

Interfacial Energy between Immiscible Liquids

Imagine an interfacial area exists between two liquids:

• By moving the barrier a distance dx, we increase the interfacial area by Ldx. The force to move the barrier is F = L, so that the work done is dW = Fdx = Ldx = dA. In this case, contributes to the internal energy, U, by determining the work done on it.

• The interfacial tension (N/m) is equivalent to the energy to increase

the interfacial area (J/m2). • The interfacial energy is a FREE energy consisting of contributions from internal energy, U (enthalpy) and entropy, S.

L F

x

sU T =

U or “Energetic” Contribution to Interfacial Energy

At the molecular level, interfacial energy can be modelled as the energy (or U) “cost” per unit area of exchanging two dissimilar molecules across an interface.

For a spherical molecule of volume v, its interfacial area is approximately v2/3.

“Energetic” Contribution to Interfacial Energy

Two new RG contacts are made: +2wRG, but at the same time, a GG contact and an RR contact are lost: - wGG - wRR

The net energetic (U) cost of broadening the interface is thus: )( RRGGRG www2

21

Thus, we can write: 3/23/2 =)2(2

1=

zv

kTwww

v RRGGRGU

Entropic Contribution to

As a result of thermal motion, a liquid interface is never smooth at the molecular level.

As the temperature increases, the interface broadens.

At the critical point, = 2 and U >0. But because of the entropic contribution, = 0, and so the interface disappears!

There is an increase in S, leading to a strong decrease in .

Problem Set 31. The phase behaviour of a liquid mixture can be described by the regular solution model. The interaction parameter depends on temperature as = 600/T, with T in degrees Kelvin.

(a) Calculate the temperature of the critical point.(b) At a temperature of 273 K, what is the composition (volume fractions) of the co-existing phases?(c) At the same temperature, what are the volume fractions of the phases on the spinodal line?

2. Octane and water are immiscible at room temperature, and their interfacial energy is measured to be about 30 mJm-2. The molecular volume of octane and water can be approximated as 2.4 x 10-29 m3. (a) Estimate the parameter for octane and water.(b) What can you conclude about the difference between the interaction energy of octane and water and the “self-interaction” energy of the two liquids?