1. Last Lecture Viscosity and relaxation times increase strongly with decreasing temperature: Arrhenius and Vogel-Fulcher equations First and second-order phase transitions are defined by the derivatives of Gibbs’ free energy with respect to the state variables (P, V and T). The glass transition occurs at a temperature where  config   exp, and its value is dependent on the “thermal history”. In a glass,  config >  exp. Glass structure is described by a radial distribution function with short-range order but no long-range order. The Kauzmann temperature could represent the temperature at which there is a first-order phase transition underlying the glass transition – possibly at a temperature of T 0 – but the glass is likely to crystallise at very slow cooling rates, so it is never reached.

2. Temperature-Dependence of Polymer Viscosity G. Strobl, The Physics of Polymers, 2 nd Ed. (1997) Springer, p. 229 Viscosity above T g : T 0  T g – 50  C TgTg 1 Pa-s = 10 Poise

3. PH3-SM (PHY3032) Soft Matter Lecture 5 Phase Separation 1 November, 2011 See Jones’ Soft Condensed Matter, Chapt. 3 and Appendix A

4. Today’s Question: When and Why are Two Liquids Miscible? When cooled below a critical temperature, miscible liquids will separate into two phases. Immiscible oil and waterTwo miscible liquids

5. Phase Separated Polymer Structure From Cui et al., Polymer 45 (2004) pp. 8139–8146 Phase-separated poly(styrene) and poly(2-vinyl pyridine) Applications in lithography, sensors, membranes/filters, etc.

6. Basic Guiding Principles Helmholtz free energy: F = U – TS (where U is the internal energy), so that: dF = dU – TdS – SdT But the First Law tells us: dU = TdS – PdV for a 3-D, compressible system. Thus, it can 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 (i.e. decreased F). Whether a transition occurs is thus decided by the balance between  U and  S.

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

8. Why are some liquids immiscible if a mixture has a higher entropy? Higher S In immiscible liquids, U increases upon mixing and unfavourably raises F. Then assume  R +  G = 1 ( non-compressibility condition ). Let  R = Volume of Red Total Volume Let  G = Volume of Green Total Volume Volume fractions

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

10. 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)

11. 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 indistinguishable 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: Entropy change in mixing R and G particles:

12. 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 the number fraction and volume fraction are identical: Substituting for ln(  -1 ) = - ln(  ): (And likewise for  G. )

13.  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 be 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! Compare to Shannon’s expression:

14. Entropic (S) Contribution to  F mix GG

15. Change in U on Mixing,  U mix Previously, we considered the potential 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: 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.

16. 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

17. 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.

18. Energy of the Unmixed State Each molecule only “owns” 1/2 of the pair interaction energy. For each individual molecule, consider “like” interactions:

19. 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.

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

21. 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: 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: AfterBefore

22. Internal Energy of Mixing,  U mix We saw previously that: Substituting for  we now find: This is a simple expression for how the internal energy changes when two liquids are mixed. It depends on values of T and . and

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

24. Entropic (S) Contribution to  F mix REMINDER:

25. Free Energy of Mixing,  F mix At constant temperature: Using our previous expression for  S mix mol : Factor out kT: In general, where  is the fraction of one phase:

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

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

28. 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 (1) non-compressibility, (2) that molecules are on a lattice, and (3) that volume fraction equals number fraction.

29. Example of Phase Separation of Liquids Phase- Separated:  1 = 0.5 and  2 = 0.8 Initial,  G :  0 = 0.7 Mixed state Volume of each phase depends on  0,  1 and  2.

30. 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 must be conserved when there is phase separation. The free energy of the phase-separated system can be shown to be a function of  F mix for the  1 and  2 phases: F sep can be easily interpreted graphically!

31. Free Energy of System when  < 2 1 0  F mix 00. 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! GG

32. Free Energy of System when  >2 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 Stable, co-existing compositions found from minima:

33. 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;  2 is a metastable composition. 2*2* The stable compositions are  1 and  2 *!  F mix

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

35. 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.

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

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

38. Unstable region: http://youtu.be/NSpOX9mfX3g http://youtu.be/__gGYCJz3-c Visualisation of Phase Separation Note that when a liquid mixture phase separates, interfaces between the two phases are created. The interface between the phases has an interfacial energy, .  1 2 2 2

39. 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/m 2 ). The interfacial energy is a FREE energy consisting of contributions from internal energy, U and entropy, S. L F x

40. U or “Energetic” Contribution to Interfacial Energy At the molecular level, interfacial energy can be modelled as the energy (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.

41. “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:

42. 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 .

43. 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?