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

2. Phase Separation 3SCMP 16 February, 2005 Lecture 5 See Jones’ Soft Condensed Matter, Chapt. 3

3. Today’s Question: When are Two Liquids Miscible? At 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. Likewise at constant V, the Helmholtz free energy, F, also goes to a minimum at equilibrium. F = U - TS, so that in a phase transition at constant T:  F =  U - T  S. We 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  U and  S.

5. G + R  GR mixture +  Higher S Lower S

6. Why are some liquids immiscible if a mixture has a higher entropy? Higher S In immiscible liquids, U increases upon mixing. Let  R = Volume of Red Total Volume Let  G = Volume of Green Total Volume Then assume  R +  G = 1 (non-compressibility condition). Assume red and green molecules have the same volume.

7. Entropy Calculation from Statistical Thermodynamics S = k ln  The thermodynamic probability, , represents the number of ways of arranging particles in a particular energy state.

8. Change in S on Mixing,  S mix Let N R be the total number of red molecules and N G be the total number of green ones. 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 large N. Applying this approximation, we find:

9. Statistical Interpretation of  S mix Simplifying by grouping N R and N G terms: If the volumes of red and green molecules are the same, then number fraction and volume fraction are identical: Substituting for ln(  -1 ) = - ln(  ): (And likewise for  G.)

10.  S mix Expressed per Molecule Our expression is the entropy change upon mixing all N R + N G molecules: Then,  S mix per molecule can found by dividing by the total number of molecules (N R + N G ): Note that we have moved the negative outside the brackets. Recognising  R and  G : Next we need to consider the change in internal energy, U!

11. Entropic (S) Contribution to  F mix

12. Change in U on Mixing,  U mix 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 assume the interaction energies (w) are additive! When unmixed, there are interaction energies between like molecules only: w RR and w GG. When mixed, there is then a new interaction energy between unlike molecules: w RG. At a constant T, the kinetic energy does not change with mixing; only the potential energies change. So,  U mix = W R+G - (W RR + W GG ), which is the difference between the mixed and the unmixed states.

13. Mean-Field Approach Describe the molecules as being on a lattice. Assume random mixing. Then the probability that a site is occupied by a red molecule is simply  red. We will only consider interaction energies (w) between each molecule and its z closest neighbours - neglecting longer range interactions.

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

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

16. Energy of the Mixed State 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 z  R red neighbours and z  G green neighbours.

17. Energy of Mixing,  U mix, per Molecule  U mix = U mix - U unmix But,  G +  R = 1, so that -  G =  R - 1 and  G -1 = -  R NB: As we did with entropy, we will consider the change in U per molecule. From before: Factor out  terms:

18. The Interaction Parameter,  We now define a unitless interaction parameter, , to characterise the the change in the energy of interaction when a red molecule in a pure red phase is swapped with a green molecule in a pure green phase: Interactions between R and G are gained, but interactions between R & R and between G & G are lost! We see that  characterises the strength of R-G interactions relative to their “self-interactions”.

19. Free Energy of Mixing,  F mix We saw previously that: Substituting for  we now find: A simple expression for how the internal energy changes when two liquids are mixed. Depends on values of T and . and

20. Energetic (U) Contribution to  F mix  = 5  = 3  = 2  = 1  = 0  = -1  = -2 Regular solution model  <0 favours mixing!

21. Entropic (S) Contribution to  F mix

22. Free Energy of Mixing,  F mix At constant temperature: Using our previous expression for  S mix mol : Factor out kT:

23. Dependence of  F mix on  Regular solution model  = 5  = 3  = 2  = 1  = 0  = -1  = -2 Mixing is favoured Mixing not favoured

24. Predictions of Phase Separation Regular solution model  = 3  = 2.75  = 2.5  = 2.25  = 2

25. Summary of Observations When 2  , there are two minima in  F mix and a maximum at  R =  G = 0.5. When  < 2, there is a single minimum at  R = 0.5 How does this dependence of  F mix on  determine the composition of phases in a mixture of liquids? As  increases, the two compositions at the  F mix minima become more different. We have assumed non-compressibility, that molecules are on a lattice, and that volume fraction and number fraction are equal.

26. Phase Separation of Liquids Phase- Separated:  G =0.5 and  G =0.8 Initial:  G =0.7

27. 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: F sep can be easily interpreted graphically!

28. Free Energy of System with Low  1 0  F mix oo. FoFo F sep 11 22 What if the composition  o was to separate into  1 and  2 ? Then the free energy would increase from F o to F sep. Conclude: Only a single phase is stable!

29. Free Energy of System with High  1 0 oo FoFo F sep 11 22 What if the composition  o was to separate into  1 and  2 ? Then the free energy would decrease from F o to F sep. Conclude: Two phases are stable.  F mix

30. F Does Not Always Decrease! F oo FoFo F sep 11 22. What happens if  o separates into  1 and  2 ? Then F o increases to F sep which is not favourable;  1 and  2 are metastable. 2*2* The stable compositions are  1 and  2 *!

31. F  Two phases stable Spinodal region Negative curvature Positive curvature Defining the Spinodal Point Spinodal Metastable

32. 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 enthalpy (or 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.

33. Constructing a Phase Diagram T1T1 T2T2 T3T3 T4T4 T5T5 T 1 <T 2 <T 3 …. Co-existence where: Spinodal where: T 1 <T 2 <T 3 <T 4 <T 5

34. Phase Diagram for Two Liquids Described by the Regular Solution Model AA Immiscible Miscible Low T High T

35. Interfacial Energy between Immiscible Liquids Imagine an interfacial area exists between two liquids: By moving the barrier a distance x, we increase the interfacial area by Lx. The force to move the barrier is F =  L, so that the work done is Fx =  Lx =  A. The interfacial tension (N/m) can also be associated with the energy to increase the interfacial area (J/m 2 ). The interfacial energy is a FREE energy consisting of contributions from enthalpy and entropy. L F x

36. 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 v 2/3.

37. “Energetic” Contribution to Interfacial Energy Two new RG contacts are made: +2w RG, but at the same time, a GG contact and an RR contact are lost: - w GG - w RR The net energetic (U) cost of broadening the interface is thus: Thus we can write:

38. 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,  = 0, and so the interface disappears! There is an increase in  S, leading to a strong decrease in .

39. 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  = 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 m 3. (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?