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Last Lecture

Last Lecture

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Last Lecture

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  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 wheretconfigtexp , and its value is dependent on the “thermal history”. In a glass, tconfig>texp . • 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 T0 – but the glass is likely to crystallise at very slow cooling rates, so it is never reached.

  2. Temperature-Dependence of Polymer Viscosity Viscosity above Tg: 1 Pa-s = 10 Poise T0Tg – 50 C Tg G. Strobl, The Physics of Polymers, 2nd Ed. (1997) Springer, p. 229

  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? Two miscible liquids Immiscible oil and water When cooled below a critical temperature, miscible liquids will separate into two phases.

  5. Phase Separated Polymer Structure Phase-separated poly(styrene) and poly(2-vinyl pyridine) Applications in lithography, sensors, membranes/filters, etc. From Cui et al., Polymer45 (2004) pp. 8139–8146

  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 DU and DS.

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

  8. Higher S LetfR = Volume of Red Total Volume LetfG= 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 and unfavourably raises F. Then assumefR + fG = 1 (non-compressibility condition).

  9. 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).

  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. 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: Change in S on Mixing,DSmix Entropy change in mixing R and G particles: Let NR be the total number of red molecules and NG be the total number of green ones.

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

  13. 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. RecognisingfRandfG: 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!

  14. Entropic (S) Contribution toDFmix fG

  15. Change in U on Mixing, DUmix • 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 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.

  16. 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

  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 simplyfR. • 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 green Probability that the reference molecule is red The mean-field approach assumes that a molecule on a given site will havezfRred neighbours andzfGgreen neighbours.

  20. Multiplying through: From before: Factor outf 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

  21. We now define a unitless interaction parameter,c, to characterise the change in the energy of interaction after the swap: After Before 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”.

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

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

  24. Entropic (S) Contribution to DFmix REMINDER:

  25. At constant temperature: Using our previous expression forDSmixmol: In general, where f is the fraction of one phase: Factor out kT: Free Energy of Mixing,DFmix

  26. Dependence ofDFmixonc Mixing not favoured Mixing is favoured Regular solution model c= 5 c= 3 c= 2 c= 1 c= 0 c= -1 c= -2

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

  28. Summary of Observations We have assumed (1) non-compressibility, (2) that molecules are on a lattice, and (3) that volume fraction equals number fraction. When c2, there is a single minimum atfR= 0.5 When c>2, there are two minima inDFmixand a maximum atfR = fG = 0.5. Ascincreases, the two compositions at theDFmix minima become more different. How does this dependence ofDFmixonfdetermine the composition of phases in a mixture of liquids?

  29. Initial,fG:f0=0.7 Phase-Separated:f1=0.5andf2=0.8 Example of Phase Separation of Liquids Mixed state Volume of each phase depends on f0, f1 and f2.

  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 offo. • At a certain temperature, this mixture separates into two phases with volume fractions of G off1andf2. • 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 DFmix for the f1 and f2 phases: Fsep can be easily interpreted graphically!

  31. Fsep f1 f2 Free Energy of System when c < 2 DFmix What if the compositionfowas to separate intof1andf2? . Then the free energy would increase from Fo to Fsep. Fo Conclude: Only a singlephase is stable! f0 1 0 fG

  32. Fsep f1 f2 Stable, co-existing compositions found from minima: Free Energy of System whenc >2 DFmix What if the compositionfowas to separate intof1andf2? . Then the free energy would decrease fromFotoFsep. Fo Conclude: Two phases are stable. fo 1 0

  33. Fsep 2* The stable compositions aref1andf2*! F Does Not Always Decrease! DFmix What happens iffoseparates intof1andf2? F Then Foincreases to Fsep which is not favourable;f2is a metastable composition. . Fo f1 fo f2

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

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

  36. Spinodal where: Co-existence where: Constructing a Phase Diagram T1v<T2 <T3 <T4 <T5 T1 T2 T3 T4 T5 T1<T2<T3….

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

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

  39. Interfacial Energy between Immiscible Liquids Imagine an interfacial area exists between two liquids: L F x • 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 and entropy, S. • 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 isdW = Fdx =gLdx = gdA. In this case, gcontributes to the internal energy, U, by determining the work done on it.

  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 v2/3.

  41. 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

  42. Entropic Contribution tog 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 ingS, leading to a strong decrease ing. At the critical point,c= 2 andgU >0. But because of the entropic contribution,g= 0, and so the interface disappears!

  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 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?