Last Lecture:

1. 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 tconfigtexp and is dependent on thermal history. In a glass, tconfig>texp . • Glass structure is described by a radial distribution function. • Liquid crystals have order between that of liquids and crystals.

2. 3SM Phase Separation 14 February, 2008 Lecture 5 See Jones’ Soft Condensed Matter, Chapt. 3 and Appendix A

3. Today’s Question:When are Two Liquids Miscible? Oil and water When cooled below a critical temperature, miscible liquids will separate into two phases.

4. 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: DF = DU – TDS (at const. T) • Likewise at constant V, the Helmholtz free energy, F, also goes to a minimum at equilibrium. • We also see that an increase in S or a decrease in U favours a transition. • Whether a transition occurs is thus decided by the balance between DU and DS.

5. G + R GR mixture Unmixed state + Lower S Mixed state  Higher S But what about F?

6. Higher S Let fR = Volume of Red Total Volume Let fG = Volume of Green Total Volume Volume fractions Why are some liquids immiscible if a mixture has a higher entropy? In immiscible liquids, U increases upon mixing. Then assume fR + fG = 1 (non-compressibility condition).

7. Entropy Calculation from Statistical Thermodynamics Boltzmann’s tomb S = k ln The statistical weight,, represents the number of ways of arranging particles (microstate) in a particular energy state (macrostate).

8. 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)

9. The number of ways of arranging N distinguishable molecules on N “lattice” sites is N!. Therefore: But the Stirling approximation tells us that lnN! NlnN-N, for largeN. Applying this approximation, we find: Change in S on Mixing, DSmix Let NR be the total number of red molecules and NG be the total number of green ones.

10. Simplifying by grouping NR and NG terms: If the volumes of red and green molecules are the same, then number fraction and volume fraction are identical: Substituting for ln(f -1) = - ln(f): Statistical Interpretation of DSmix (And likewise for fG.)

11. Then, DSmixper molecule can be found by dividing by the total number of molecules (NR + NG): Note that we have moved the negative outside the brackets. Recognising fR and fG: Compare to Shannon’s expression: D Smix Expressed per Molecule Our expression is the entropy change upon mixing allNR+ NG molecules: Next we need to consider the change in internal energy, U!

12. Entropic (S) Contribution toDFmix fG

13. Change in U on Mixing, DUmix • 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 likemolecules 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, DUmix = WR+G - (WRR + WGG), which is the difference between the mixed and the unmixed states.

14. Summary Charge-charge Coulombic Dipole-charge Dipole-dipole Keesom Charge-nonpolar Dipole-nonpolar Debye Nonpolar-nonpolar Dispersive Type of InteractionInteraction Energy, w(r) In vacuum:e=1

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

16. Energy of the Unmixed State • Each molecule only “owns” 1/2 of the pair interaction energy. • For each individual molecule:

17. Energy of the Mixed State Probability that a neighbour is red Probability that a neighbour is green Probability that the reference molecule is green Probability that the reference molecule is red The mean-field approach assumes that a molecule on a given site will have zfR red neighbours and zfG green neighbours.

18. Multiplying through: From before: Factor out f terms: Energy of Mixing, DUmix, per Molecule NB: As we did with entropy, we will consider the change in Uper molecule. DUmix = Umix - Uunmix But,fG + fR= 1, so thatfR – 1= -fG and fG -1 = -fR

19. We now define a unitless interaction parameter, c, to characterise the the change in the energy of interaction after the swap: The Interaction Parameter, c Imaginethat a red molecule in a pure red phase is swapped with a green molecule in a pure green phase: Two “sets” of interactions between R and G are gained, but interactions between R & R and between G & G are lost! We see thatccharacterises the strength of R-G interactions relative to their “self-interactions”.

20. and Substituting for c we now find: Internal Energy of Mixing, DUmix We saw previously that: A simple expression for how the internal energy changes when two liquids are mixed. Depends on values of T and c.

21. Energetic (U) Contribution toDFmix Regular solution model c= 5 c <0 favours mixing! c= 3 c= 2 c= 1 c= 0 c= -1 c= -2

22. Entropic (S) Contribution toDFmix

23. At constant temperature: Using our previous expression for DSmixmol: Factor out kT: Free Energy of Mixing, DFmix

24. Dependence ofDFmix on c Mixing not favoured Mixing is favoured Regular solution model c= 5 c= 3 c= 2 c= 1 c= 0 c= -1 c= -2

25. Predictions of Phase Separation Regular solution model c= 3 c= 2.75 c= 2.5 c= 2.25 c= 2

26. Summary of Observations We have assumed non-compressibility, that molecules are on a lattice, and that volume fraction equals number fraction. When c < 2, there is a single minimum atfR = 0.5 When c2, there are two minima in DFmix and a maximum at fR = fG = 0.5. As c increases, the two compositions at the DFmix minima become more different. How does this dependence ofDFmix on f determine the composition of phases in a mixture of liquids?

27. Initial:fG=0.7 Phase-Separated:fG=0.5 and fG=0.8 Phase Separation of Liquids

28. 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 fo. • At a certain temperature, this mixture separates into two phases with volume fractions of G of f1 and f2. • 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!

29. Fsep f1 f2 Free Energy of System with Lowc DFmix What if the composition fo was to separate into f1 and f2? . Then the free energy would increase from Fo to Fsep. Fo Conclude: Only a singlephase is stable! fo 1 0

30. Fsep f1 f2 Stable, co-existing compositions found from minima: Free Energy of System with Highc DFmix What if the composition fo was to separate into f1 and f2? . Then the free energy would decrease from Fo to Fsep. Fo Conclude: Two phases are stable. fo 1 0

31. F Does Not Always Decrease! Fsep 2* The stable compositions are f1 and f2*! DFmix What happens if fo separates into f1 and f2? F Then Foincreases to Fsep which is not favourable; f1 and f2 are metastable. . Fo f1 fo f2

32. Negative curvature Positive curvature Spinodal point Defining the Spinodal Point F Two phases stable: “Spinodal region” Metastable f

33. 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 c1/T. When c >2, two phases are stable; the mixture is unstable. Whenc<2, two phases are unstable; the mixture is stable. When 0 <c<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 c = 2.

34. Constructing a Phase Diagram Spinodal where: Co-existence where: T1<T2<T3<T4<T5 T1 T2 T3 T4 T5 T1<T2<T3….

35. Phase Diagram for Two Liquids Described by the Regular Solution Model Low T Immiscible Miscible High T fA

36. Interfacial Energy between Immiscible Liquids Imagine an interfacial area exists between two liquids: L F x • By moving the barrier a distance dx, we increase the interfacial area by Ldx. The force to move the barrier is F = gL, so that the work done is dW = Fdx = gLdx = gdA. • 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 (enthalpy) and entropy.

37. 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.

38. The net energetic (U) cost of broadening the interface is thus: Thus, we can write: “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

39. Entropic Contribution to g As a result of thermal motion, a liquid interface is never smooth at the molecular level. As the temperature increases, the interface broadens. There is an increase in gS, leading to a strong decrease ing. At the critical point, c = 2 and gU >0. But because of entropic contribution,g= 0, and so the interface disappears!

40. Problem Set 3 1.The phase behaviour of a liquid mixture can be described by the regular solution model. The interaction parameter depends on temperature as c = 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 c 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?