Fundamentals of Polymer Engineering Part 10 docx

9 Thermodynamics of Polymer Mixtures 9.1 INTRODUCTION As with low-molecular-weight substances, the solubility of a polymer (i.e., the amount of polymer that can be dissolved in a given liquid) depends on the temperature and pressure of the system. In addition, however, it also depends on the molecular weight. This fact can be used to separate a polydisperse polymer sample into narrow molecular-weight fractions in a conceptually easy, albeit tedious, manner. It is obvious that any help that thermodynamic theory could afford in selecting solvent and defining process conditions would be quite useful for optimizing polymer fractionation. Such polymers having a precise and known molecular weight are needed in small quantities for research purposes. Although today we use gel permeation chromatography for polymer fractionation, a working knowledge of polymer solution thermodynamics is still necessary for several important engineering applications [1]. In the form of solutions, polymers find use in paints and other coating materials. They are also used in lubricants (such as multigrade motor oils), where they temper the reduction in viscosity with increasing temperature. In addition, aqueous polymer solutions are pumped into oil reservoirs for promoting tertiary oil recovery. In these applications, the polymer may witness a range of temperatures, pressures, and shear rates, and this variation can induce phase separation. Such a situation is to be avoided, and it can be, with the aid of 374 Copyright © 2003 Marcel Dekker, Inc. thermodynamics. Other situations in which such theory may be usefully applied are devolatilization of polymers and product separation in polymerization reactors. There are also instances in which we want no polymer–solvent interactions at all, especially in cases where certain liquids come into regular contact with polymeric surfaces. In addition, polymer thermodynamics is very important in the growing and commercially important area of selecting components for polymer–polymer blends. There are several reasons for blending polymers: 1. Because new polymers with desired properties are not synthesized on a routine basis, blending offers the opportunity to develop improved materials that might even show a degree of synergism. For engineering applications, it is generally desirable to develop easily processible polymers that are dimensionally stable, can be used at high tempera- tures, and resist attack by solvents or by the environment. 2. By varying the composition of a blend, the engineer hopes to obtain a gradation in properties that might be tailored for specific applications. This is true for miscible polymer pairs such as polyphenylene oxide and polystyrene that appear and behave as single-component polymers. 3. If one of the components is a commodity polymer, its use can reduce the cost or, equivalently, improve the profit margin for the more expensive blended product. Although it is possible to blend two polymers by either melt-mixing in an extruder or dissolving in a common solvent and removing the solvent, the procedure does not ensure that the two polymers will mix on a microscopic level. In fact, most polymer blends are immiscible or incompatible. This means that the mixture does not behave as a single-phase material. It will, for example, have two different glass transition temperatures, which are representative of the two constituents, rather than a single T g . Such incompatible blends can be homogenized somewhat by using copolymers and graft polymers or by adding surface-active agents. These measures can lead to materials having high impact strength and toughness. In this chapter, we presen t the classical Flory–Huggins theory, which can explain a large number of observations regarding the phase behavior of concen- trated polymer solutions. The agreement between theory and experiment is, however, not always quantitative. Additionally, the theory cannot explain the phenomenon of phase separation brought about by an increase in temperature. It is also not very useful for describing polymer–polymer miscibility. For these reasons, the Flory–Huggins theory has been modified and alternate theories have been advanced, which are also discussed. Thermodynamics of Polymer Mixtures 375 Copyright © 2003 Marcel Dekker, Inc. 9.2 CRITERIA FOR POLYMER SOLUBILITY A polymer dissolves in a solvent if, at constant temperature and pressure, the total Gibbs free energy can be decreased by the polymer going into solution. Therefore, it is necessary that the following hold: DG M ¼ DH mix À T DS mix < 0 ð9:2:1Þ For most polymers, the enthalpy change on mixing is positive. This necessitates that the change in entropy be sufficiently positive if mixing is to occur. These changes in enthalpy and entropy can be calculated using simple models; these calculations are done in the next section. Here, we merely note that Eq. (9.2.1) is only a necessary condition for solubility and not a sufficient condition. It is possible, after all, to envisage an equilibrium state in which the free energy is still lower than that corresponding to a single-phase homogeneous solution. The single-phase solution may, for example, separate into two liquid phases having different compositions. To understand which situation might prevail, we need to review some elements of the thermodynamics of mixtures. A partial molar quantity is the derivative of an extensive quantity M with respect to the number of moles n i of one of the components, keeping the temperature, the pressure, and the number of moles of all the other components fixed. Thus,  MM i ¼ @M @n i  T;P;n j ð9:2:2Þ It is easy to show [2] that the mixture property M can be represented in terms of the partial molar quantities as follows: M ¼ P i  MM i n i ð9:2:3Þ For an open system at constant temperature and pressure, however, dM ¼ P i  MM i dn i ð9:2:4Þ but Eq. (9.2.3) gives dM ¼ P i  MM i dn i þ P i n i d  MM i ð9:2:5Þ so that P i n i d  MM i ¼ 0 ð9:2:6Þ which is known as the Gibbs–Duhem equation. 376 Chapter 9 Copyright © 2003 Marcel Dekker, Inc. Let us identify M with the Gibbs free energy G and consider the mixing of n 1 moles of pure component 1 with n 2 moles of pure component 2. Before mixing, the free energy of both components taken together, G comp ,is G comp ¼ P 2 i¼1 g i n i ð9:2:7Þ where g i is the molar free energy of component i. After mixing, the free energy of the mixture, using Eq. (9.2.3), is as follows: G mixture ¼ P 2 i¼1  GG i n i ð9:2:8Þ Consequently, the change in free energy on mixing is DG M ¼ P 2 i¼1 ð  GG i À g i Þn i ð9:2:9Þ and dividing both sides by the total number of moles, n 1 þ n 2 ,yieldsthe corresponding result for 1 mol of mixture, Dg m ¼ P 2 i¼1 ð  GG i À g i Þx i ð9:2:10Þ where x i denotes mole fraction. It is common practice to call the partial molar Gibbs free energy  GG i the chemical potential and write it as m i . Clearly, g i is the partial molar Gibbs free energy for the pure component. Representing it as m 0 i , we can derive from Eq. (9.2.10) the following: Dg m ¼ x 1 Dm 1 þ x 2 Dm 2 ð9:2:11Þ where Dm 1 ¼ m 1 À m 0 1 and Dm 2 ¼ m 2 À m 0 2 . Because x 1 þ x 2 equals unity, Eq. (9.2.11) can be written Dg m ¼ Dm 1 þ x 2 ðDm 2 À Dm 1 Þð9:2:12Þ Differentiating this result with respect to x 2 gives dDg m dx 2 ¼ dm 1 dx 2 þðDm 2 À Dm 1 Þþx 2 dm 2 dx 2 À dm 1 dx 2  ð9:2:13Þ ¼ðDm 2 À Dm 1 Þþx 2 dm 2 dx 2 þ x 1 dm 1 dx 2 From Eq. (9.2.6), however, P 2 1 x i dm i equals 0. Therefore, Eq. (9.2.13) becomes dDg m dx 2 ¼ Dm 2 À Dm 1 ð9:2:14Þ Thermodynamics of Polymer Mixtures 377 Copyright © 2003 Marcel Dekker, Inc. and solving Eq. (9.2.14) simultaneously with Eq. (9.2.11) yields Dm 1 ¼ Dg m À x 2 dDg m dx 2 ð9:2:15Þ Dm 2 ¼ Dg m þ x 1 dDg m dx 2 ð9:2:16Þ Thus, if Dg m can be obtained by some means as a function of composition, the chemical potentials can be computed using Eqs. (9.2.15) and (9.2.16). The chemical potentials are, in turn, needed for phase equilibrium calculations. Let us now return to the question of whether a single-phase solution or two liquid phases will be formed if the DG M of a two-component system is negative. This question can be answered by examining Figure 9.1, which shows two possible Dg m versus x 2 curves; these two curves may correspond to different temperatures. It can be reasoned from Eqs. (9.2.15) and (9.2.16) that the chemical potentials at any composition x 2 can be determined simply by drawing a tangent to the Dg m curve at x 2 and extending it until it intersects with the x 2 ¼ 0 and x 2 ¼ 1 axes. The intercept with x 2 ¼ 0 gives Dm 1 , whereas that with x 2 ¼ 1givesDm 2 . Following this reasoning, it is seen that the curve labeled T 1 has a one-to- one correspondence between Dm 1 and x 2 or, for that matter, between Dm 2 and x 2 . This happens because the entire curve is concave upward. Thus, there are no two composition values that yield the same value of the chemical potential. This implies that equilibrium is not possible between two liquid phases of differing compositions; instead, there is complete miscibility. At a lower temperature T 2 , however, the chemical potential at x 0 2 equals the chemical potential at x 00 2 . Solutions of these two compositions can, therefore, coexist in equilibrium. The points x 0 2 and x 00 2 are called binodal points, and any single-phase system having a composition between these two points can split into these two phases with relative FIGURE 9.1 Free-energy change of mixing per mole of a binary mixture as a function of mixture composition. 378 Chapter 9 Copyright © 2003 Marcel Dekker, Inc. amountsofeachphasedeterminedbyamassbalance.Phaseseparationoccurs becausethefreeenergyofthetwo-phasemixturedenotedbythepointmarkedDg islessthanthefreeenergyDg*ofthesingle-phasesolutionofthesameaverage composition.PointsS 0 andS 00 areinflectionpointscalledspinodalpoints,and betweenthesetwopointstheDg m curveisconcavedownward.Asolutionhaving acompositionbetweenthesetwopointsisunstabletoeventhesmallest disturbanceandcanloweritsfreeenergybyphaseseparation.Betweeneach spinodalpointandthecorrespondingbinodalpoint,however,Dg m isconcave upwardand,therefore,stabletosmalldisturbances.Thisiscalledametastable region;here,itispossibletoobserveasingle-phasesolution—butonlyfora limitedperiodoftime. Thepresenceofthetwo-phaseregiondependsontemperature.Forsome solutions,atahighenoughtemperaturecalledtheuppercriticalsolution temperature,thespinodalandbinodalpointscometogetherandonlysingle- phasemixturesoccurabovethistemperature.ThissituationisdepictedinFigure 9.2onatemperature–compositiondiagram.Here,thelocusofthebinodalpoints iscalledthebinodalcurveorthecloudpointcurve,whereasthelocusofthe spinodalpointsiscalledthespinodalcurve.Next,wedirectourattentionto determiningthefree-energychangeonmixingapolymerwithalow-molecular- weightsolvent. 9.3THEFLORY^HUGGINSTHEORY TheclassicalFlory–Hugginstheoryassumesattheoutsetthatthereisneithera changeinvolumenorachangeinenthalpyonmixingapolymerwithalow- molecular-weightsolvent[3–5];theinfluenceofnon-athermal(DH mixing 6¼0) behaviorisaccountedforatalaterstage.Thus,thecalculationofthefree-energy FIGURE9.2Temperature–compositiondiagramcorrespondingtoFigure9.1. Thermodynamics of Polymer Mixtures 379 Copyright © 2003 Marcel Dekker, Inc. change on mixing at a constant temperature and pressure reduces to a calculation of the change in entropy on mixing. This latter quantity is determined with the help of a lattice model using formulas from statistical thermodynamics. We assume the existence of a two-dimensional lattice with each lattice site having z nearest neighbors, where z is the coordination number of the lattice; an example is shown in Figure 9.3. Each lattice site can accommodate a single solvent molecule or a polymer segment having a volume equal to a solvent molecule. Polymer molecules are taken to be monodisperse, flexible, initially disordered, and composed of a series of segments the size of a solvent molecule. The number of segments in each polymer molecule is m, which equals V 2 =V 1 ,the ratio of the molar volume of the polymer to the molar volume of the solvent. Note that m is not the degree of polymerization. We begin with an empty lattice and calculate the number of ways, O,of arranging n 1 solvent molecules and n 2 polymer molecules in the n 0 ¼ n 1 þ mn 2 lattice sites. Because the heat of mixing has been taken to be zero, each arrangement has the same energy and is equally likely to occur. The only restriction imposed is by the connectivity of polymer chain segments. It must be ensured that two segments connected to each other lie on the nearest neighboring lattice sites. Once O is known, the entropy of the mixture is given by k ln O; where k is Boltzmann’s constant. 9.3.1 Entropy Change on Mixing In order to calculate the entropy of the mixture, we first arrange all of the polymer molecules on the lattice. The identical solvent molecules are placed thereafter. If j FIGURE 9.3 Schematic diagram of a polymer molecule on a two-dimensional lattice. 380 Chapter 9 Copyright © 2003 Marcel Dekker, Inc. polymer molecules have already been placed, the number of lattice sites still available number n 0 À jm. Thus, the first segment of the ð j þ 1Þst molecule can be arranged in n 0 À jm ways. The second segment is connected to the first one and so can be placed only in one of the z neighboring sites. All of these may, however, not be vacant. If the polymer solution is relatively concentrated so that chain overlap occurs, we would expect that, on average, the fraction of neighboring sites occupied ð f Þ would equal the overall fraction of sites occupied. Thus, f ¼ jm=n 0 . As a result, the second segment of the ð j þ 1Þst molecule can be placed in zð1 Àf Þ ways. Clearly, the third segment can be placed in ðz À 1 Þð1 À f Þ ways, and similarly for subseq uent segments. Therefore, the total number of ways O jþ1 in which the ð j þ 1Þst polymer molecule can be arranged is the product of the number of ways of placing the first segment with the number of ways of placing the second segment and the number of ways of placing each subsequent segment. Thus, O jþ1 ¼ðn 0 À jmÞzð1 Àf Þ Q m 3 ðz À 1Þð1 À f Þð9:3:1Þ where the symbol Q denotes product. As a consequence, O jþ1 ¼ðn 0 À jmÞzðz À1Þ mÀ2 ð1 À f Þ mÀ1 ffiðn 0 À jmÞðz À1Þ mÀ1 ð1 À f Þ mÀ1 ¼ðn 0 À jmÞðz À1Þ mÀ1 1 À jm n 0  mÀ1 ¼ðn 0 À jmÞ m z À 1 n 0  mÀ1 ð9:3:2Þ The total number of ways of arranging all of the n 2 polymer molecules, O p ,isthe product of the number of ways of arranging each of the n 2 molecules in sequence. This fact and Eq. (9.3.2) yield O p ¼ Q n 2 À1 j¼0 ðn 0 À jmÞ m z À 1 n 0  mÀ1 "# ð9:3:3Þ where the index only goes up to n 2 À 1 because j ¼ 0 corresponds to the first polymer molecule. The development so far assumes that all of the polymer molecules are different. They are, however, identical to each other. This reduces the total number of possible arrangements by a factor of n 2 !, and it is therefore necessary to divide the right-hand side of Eq. (9.3.3) by n 2 !. Having arranged all of the polymer molecules, the number of ways of fitting all of the indistinguishable solvent molecules into the remaining lattice Thermodynamics of Polymer Mixtures 381 Copyright © 2003 Marcel Dekker, Inc. sites is exactly one. As a result, O p equals O, the total number of ways of placing all the polymer and solvent molecules on to the lattice. Finally, then, S mixture ¼ k ln O ð9:3:4Þ and using Eq. (9.3.3) properly divided by n 2 !: S mixture k ¼Àlnðn 2 !Þþm P n 2 À1 j¼0 lnðn 0 À jmÞþðm À 1 Þ P n 2 À1 j¼0 ln z À 1 n 0  ð9:3:5Þ Because j does not appear in the last term on the right-hand side of Eq. (9.3.5), that term adds up to ðm À 1Þn 2 ln½ðz À 1Þ=n 0 . Also, the first term can be replaced by Stirling’s approximation: ðn 2 !Þ¼n 2 ln n 2 À n 2 ð9:3:6Þ Now, consider the summation in the second term: P n 2 À1 j¼0 lnðn 0 À jmÞ¼ P n 2 À1 j¼0 ln  m  n 0 m À j  ¼ n 2 ln m þ P n 2 À1 j¼0 ln  n 0 m À j  ð9:3:7Þ Furthermore, P n 2 À1 j¼0 ln  n 0 m À j  ¼ ln  n 0 m  þ ln  n 0 m À 1  þÁÁÁþln  n 0 m À n 2 þ 1  ¼ ln n 0 m  n 0 m À 1  n 0 m À 2  ÁÁÁ n 0 m À n 2 þ 1 hi ¼ ln n 0 m  n 0 m À 1  ÁÁÁ n 0 m À n 2 þ 1  n 0 m À n 2  ÁÁÁ1 n 0 m À n 2  ÁÁÁ1 8 > < > : 9 > = > ; ¼ ln ðn 0 =mÞ! ðn 0 =m À n 2 Þ!  ð9:3:8Þ 382 Chapter 9 Copyright © 2003 Marcel Dekker, Inc. Combining all of these fragments and again using Stirling’s approximation in Eq. (9.3.8) yields S mixture k ¼Àn 2 ln n 2 þ n 2 þ m  n 2 ln m þ  n 0 m  ln  n 0 m  À n 0 m À  n 0 m À n 2  ln  n 0 m À n 2  þ  n 0 m À n 2  þðm À 1 Þn 2 ln z À 1 n 0  ð9:3:9Þ which, without additional tricks, can be simplified to the following: S mixture k ¼Àn 2 ln  n 2 n 0  þ n 2 À mn 2 À n 1 ln n 1 n 0  ð9:3:10Þ þðm À 1 Þ½n 2 lnðz À 1Þ Adding to and subtracting n 2 ln m from the right-hand side of Eq. (9.3.10) gives the result S mixture k ¼Àn 2 ln mn 2 n 0  À n 1 ln n 1 n 0  þ n 2 ½ðm À 1Þlnðz À 1Þþð1 ÀmÞþln m ð9:3:11Þ The entropy of the pure polymer S 2 can be obtained by letting n 1 be zero and n 0 be mn 2 in Eq. (9.3.11): S 2 k ¼ n 2 ½ðm À 1Þlnðz À 1Þþð1 À mÞþln mð9:3:12Þ Similarly, the entropy of the pure solvent S 1 is obtained by setting n 2 equal to zero and n 1 equal to n 0 : S 1 k ¼ 0 ð9:3:13Þ Using Eqs. (9.3.11)–(9.3.13), DS mixing ¼ DS mixture À S 1 À S 2 ¼Àkn 1 ln n 1 n 0  þ n 2 ln mn 2 n 0  ð9:3:14Þ From the way that m and n 0 have been defined, it is evident that n 1 =n 0 equals f 1 , the volume fraction of the solvent, and mn 2 =n 0 equals f 2 , the volume fraction of the polymer. As a result, DS ¼Àk½n 1 ln f 1 þ n 2 ln f 2 ð9:3:15Þ Thermodynamics of Polymer Mixtures 383 Copyright © 2003 Marcel Dekker, Inc. [...]... side of Eq (9.6.2) are negligible compared to the third term As a consequence, DGM ffi DHM and miscibility depends entirely on the energetics of intermolecular interactions In other words, a negative value of De or, equivalently, of the interaction parameter is needed to assure polymer polymer miscibility Example 9.4: If the 1 g of polymer of Example 9.1 is dissolved in 9 g of a different polymer of molecular... viscosity of each of the two solvents is the same Use literature values of the molar energy of vaporization and the molar volume to estimate the solubility parameter of water at 25 C The glass transition temperature of poly(ether ether ketone) (PEEK), a semicrystalline polymer, is 145 C, whereas that of poly(ether imide) (PEI), an amorphous polymer miscible with PEEK, is 215 C What will be the Tg of a... from each of the two equations yields the following:   1 1 À 1À ð9:3:34Þ wc ¼ ð2f2c ÞÀ1 2f2c ð1 À f2c Þ m 1 wc ¼ ð9:3:35Þ 2ð1 À f2c Þ2 Copyright © 2003 Marcel Dekker, Inc Thermodynamics of Polymer Mixtures 389 F IGURE 9.4 Solvent chemical potential as a function of polymer volume fraction for m ¼ 100 0 The value of w1 is indicated on each curve (Reprinted from Paul J Flory, Principles of Polymer Chemistry... ð9:3:52Þ Thermodynamics of Polymer Mixtures TABLE 9.1 393 UCST Data for Solutions of PS in DOP Molecular weight ( 10 5 ) UCST ( C) Molar volume ratio ( 10 3 ) 2.00 2.80 3.35 4.70 9.00 18.00 5.9 7.4 8.0 8.8 9.9 12.0 0.456 0.639 0.770 1.072 2.069 4.131 Source: Ref 10 pffiffiffiffi so that a plot of 1=Tc versus ½ð1=2mÞ þ ð1= mފ should be a straight line with a slope of 1=yc and an intercept of 1=y These are, in... ð9:5:6Þ 2Vp where jepp j is the energy of interaction between two polymer segments and Vp is the volume of 1 mol of polymer segments Because polymers generally decompose on heating, dp cannot be obtained using data on the energy of vaporization, and an indirect method is needed An examination of Eq (9.5.5) shows that DHM vanishes when the solubility parameters of the two components equal each other... Mixtures 385 energy of interaction (a negative quantity) between two polymer segments is represented by e22 and that between a polymer segment and a solvent molecule by e12 , the total energy of interaction for the single polymer segment is zf2 e22 þ zf1 e12 Because the total number of polymer segments in the lattice is n0 f2 , the interaction energy associated with all of the polymer segments is z... Rangel-Nafaile, C., A B Metzner, and K F Wissbrun, Analysis of Stress-Induced Phase Separations in Polymer Solutions, Macromolecules, 17, 1187–1195, 1984 PROBLEMS 9.1 What is the value of m, the ratio of the molar volume of polymer to the molar volume of solvent, for polystyrene of 250,000 molecular weight dissolved in toluene? Thus, determine wc and the polymer volume fraction corresponding to the upper critical... 9.3.4 Phase Behavior of Monodisperse Polymers If we mix n1 moles of solvent with n2 moles of polymer having a known molar volume or molecular weight (i.e., a known value of m), the chemical potential of the solvent in solution is given by Eq (9.3.30) If we fix w1 , we can easily plot ðm1 À m0 Þ=RT as a function of f2 By changing w1 and repeating the procedure, 1 we get a family of curves at different... be the Tg of a blend of these two materials containing 10% by weight PEI? Speculate on what the presence of the PEI in the blend might do to the rate of crystallization of PEEK Would a polymer that hydrogen-bonds with itself be more likely or less likely to form miscible blends with other polymers compared with a polymer that does not hydrogen-bond with itself? Use the results of Example 9.2 and the... Critical Points in Polymer Solutions, Polymer, 1, 20–26, 1960 19 Krigbaum, W R., and P J Flory, Statistical Mechanics of Dilute Polymer Solutions: IV Variation of the Osmotic Second Coefficient with Molecular Weight, J Am Chem Soc., 75, 1775–1784, 1953 20 Flory, P J., Phase Equilibria in Solutions of Rod-Like Particles, Proc Roy Soc London, A234, 73–89, 1956 21 Patterson, D., Free Volume and Polymer Solubility: . and composed of a series of segments the size of a solvent molecule. The number of segments in each polymer molecule is m, which equals V 2 =V 1 ,the ratio of the molar volume of the polymer to. 1 n 0  mÀ1 ð9:3:2Þ The total number of ways of arranging all of the n 2 polymer molecules, O p ,isthe product of the number of ways of arranging each of the n 2 molecules in sequence. This. the total number of ways O jþ1 in which the ð j þ 1Þst polymer molecule can be arranged is the product of the number of ways of placing the first segment with the number of ways of placing the second