Fundamentals of Polymer Engineering Part 10 docx

9.2 CRITERIA FOR POLYMER SOLUBILITYA polymer dissolves in a solvent if, at constant temperature and pressure, thetotal Gibbs free energy can be decreased by the polymer going into soluti

In the form of solutions, polymers find use in paints and other coatingmaterials They are also used in lubricants (such as multigrade motor oils), wherethey temper the reduction in viscosity with increasing temperature In addition,aqueous polymer solutions are pumped into oil reservoirs for promoting tertiaryoil recovery In these applications, the polymer may witness a range oftemperatures, pressures, and shear rates, and this variation can induce phaseseparation Such a situation is to be avoided, and it can be, with the aid of

thermodynamics Other situations in which such theory may be usefully appliedare devolatilization of polymers and product separation in polymerizationreactors There are also instances in which we want no polymer–solventinteractions at all, especially in cases where certain liquids come into regularcontact with polymeric surfaces.

In addition, polymer thermodynamics is very important in the growing andcommercially important area of selecting components for polymer–polymerblends There are several reasons for blending polymers:

1 Because new polymers with desired properties are not synthesized on aroutine basis, blending offers the opportunity to develop improvedmaterials that might even show a degree of synergism For engineeringapplications, it is generally desirable to develop easily processiblepolymers 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 agradation in properties that might be tailored for specific applications.This is true for miscible polymer pairs such as polyphenylene oxideand polystyrene that appear and behave as single-component polymers

3 If one of the components is a commodity polymer, its use can reducethe cost or, equivalently, improve the profit margin for the moreexpensive blended product

Although it is possible to blend two polymers by either melt-mixing in anextruder or dissolving in a common solvent and removing the solvent, theprocedure does not ensure that the two polymers will mix on a microscopiclevel In fact, most polymer blends are immiscible or incompatible This meansthat the mixture does not behave as a single-phase material It will, for example,have two different glass transition temperatures, which are representative of thetwo constituents, rather than a single Tg Such incompatible blends can behomogenized somewhat by using copolymers and graft polymers or by addingsurface-active agents These measures can lead to materials having high impactstrength and toughness

In this chapter, we present the classical Flory–Huggins theory, which canexplain 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 thephenomenon of phase separation brought about by an increase in temperature It

is also not very useful for describing polymer–polymer miscibility For thesereasons, the Flory–Huggins theory has been modified and alternate theories havebeen advanced, which are also discussed

9.2 CRITERIA FOR POLYMER SOLUBILITY

A polymer dissolves in a solvent if, at constant temperature and pressure, thetotal Gibbs free energy can be decreased by the polymer going into solution.Therefore, it is necessary that the following hold:

DGM¼ DHmix
TDSmix< 0 ð9:2:1ÞFor most polymers, the enthalpy change on mixing is positive This necessitatesthat the change in entropy be sufficiently positive if mixing is to occur Thesechanges in enthalpy and entropy can be calculated using simple models; thesecalculations are done in the next section Here, we merely note that Eq (9.2.1) isonly a necessary condition for solubility and not a sufficient condition It ispossible, after all, to envisage an equilibrium state in which the free energy is stilllower than that corresponding to a single-phase homogeneous solution Thesingle-phase solution may, for example, separate into two liquid phases havingdifferent compositions To understand which situation might prevail, we need toreview some elements of the thermodynamics of mixtures

A partial molar quantity is the derivative of an extensive quantity M withrespect to the number of moles ni of one of the components, keeping thetemperature, the pressure, and the number of moles of all the other componentsfixed Thus,

Let us identify M with the Gibbs free energy G and consider the mixing of

n1 moles of pure component 1 with n2 moles of pure component 2 Beforemixing, the free energy of both components taken together, Gcomp, is

Dgm¼P2

i¼1

ð
GGi
giÞxi ð9:2:10Þ

where xi denotes mole fraction

It is common practice to call the partial molar Gibbs free energy
Gi thechemical potential and write it as mi Clearly, gi is the partial molar Gibbs freeenergy for the pure component Representing it as m0

i, we can derive from

Eq (9.2.10) the following:

Dgm¼ x1Dm1þ x2Dm2 ð9:2:11Þwhere Dm1¼ m1
m0

1 and Dm2¼ m2
m0

2 Because x1þ x2 equals unity,

Eq (9.2.11) can be written

Dgm¼ Dm1þ x2ðDm2
Dm1Þ ð9:2:12ÞDifferentiating this result with respect to x2 gives

1 xidmiequals 0 Therefore, Eq (9.2.13) becomes

dDgm

and solving Eq (9.2.14) simultaneously with Eq (9.2.11) yields

FIGURE9.1 Free-energy change of mixing per mole of a binary mixture as a function

of mixture composition

amounts of each phase determined by a mass balance Phase separation occursbecause the free energy of the two-phase mixture denoted by the point markedDg

is less than the free energyDg* of the single-phase solution of the same averagecomposition Points S0 and S00 are inflection points called spinodal points, andbetween these two points theDgmcurve is concave downward A solution having

a composition between these two points is unstable to even the smallestdisturbance and can lower its free energy by phase separation Between eachspinodal point and the corresponding binodal point, however, Dgm is concaveupward and, therefore, stable to small disturbances This is called a metastableregion; here, it is possible to observe a single-phase solution—but only for alimited period of time

The presence of the two-phase region depends on temperature For somesolutions, at a high enough temperature called the upper critical solutiontemperature, the spinodal and binodal points come together and only single-phase mixtures occur above this temperature This situation is depicted in Figure9.2 on a temperature–composition diagram Here, the locus of the binodal points

is called the binodal curve or the cloud point curve, whereas the locus of thespinodal points is called the spinodal curve Next, we direct our attention todetermining the free-energy change on mixing a polymer with a low-molecular-weight solvent

The classical Flory–Huggins theory assumes at the outset that there is neither achange in volume nor a change in enthalpy on mixing a polymer with a low-molecular-weight solvent [3–5]; the influence of non-athermal (DHmixing 6¼ 0)behavior is accounted for at a later stage Thus, the calculation of the free-energy

FIGURE9.2 Temperature–composition diagram corresponding toFigure 9.1

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 thehelp of a lattice model using formulas from statistical thermodynamics

We assume the existence of a two-dimensional lattice with each lattice sitehaving z nearest neighbors, where z is the coordination number of the lattice; anexample is shown in Figure 9.3 Each lattice site can accommodate a singlesolvent molecule or a polymer segment having a volume equal to a solventmolecule Polymer molecules are taken to be monodisperse, flexible, initiallydisordered, 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 V2=V1, theratio of the molar volume of the polymer to the molar volume of the solvent Notethat m is not the degree of polymerization

We begin with an empty lattice and calculate the number of ways, O, ofarranging n1 solvent molecules and n2polymer molecules in the n0¼ n1þ mn2lattice sites Because the heat of mixing has been taken to be zero, eacharrangement has the same energy and is equally likely to occur The onlyrestriction imposed is by the connectivity of polymer chain segments It must

be ensured that two segments connected to each other lie on the nearestneighboring lattice sites Once O is known, the entropy of the mixture is given

by k lnO; where k is Boltzmann’s constant

In order to calculate the entropy of the mixture, we first arrange all of the polymermolecules on the lattice The identical solvent molecules are placed thereafter If j

FIGURE9.3 Schematic diagram of a polymer molecule on a two-dimensional lattice

polymer molecules have already been placed, the number of lattice sites stillavailable number n0
jm Thus, the first segment of the ð j þ 1Þst molecule can

be arranged in n0
jm ways The second segment is connected to the first oneand 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 thatchain overlap occurs, we would expect that, on average, the fraction ofneighboring sites occupied ð f Þ would equal the overall fraction of sites occupied.Thus, f ¼ jm=n0 As a result, the second segment of the ð j þ 1Þst molecule can

be placed in zð1
f Þ ways Clearly, the third segment can be placed in

ðz
1Þð1
f Þ ways, and similarly for subsequent segments Therefore, thetotal number of ways Ojþ1 in which the ð j þ 1Þst polymer molecule can bearranged is the product of the number of ways of placing the first segment withthe number of ways of placing the second segment and the number of ways ofplacing each subsequent segment Thus,

Ojþ1¼ ðn0
jmÞzð1
f Þ Qm

3

ðz
1Þð1
f Þ ð9:3:1Þwhere the symbolQ

denotes product As a consequence,

Having arranged all of the polymer molecules, the number of ways offitting all of the indistinguishable solvent molecules into the remaining lattice

sites is exactly one As a result,OpequalsO, the total number of ways of placingall the polymer and solvent molecules on to the lattice Finally, then,

Smixture¼ k ln O ð9:3:4Þand using Eq (9.3.3) properly divided by n2!:

by Stirling’s approximation:

ðn2!Þ ¼ n2ln n2
n2 ð9:3:6ÞNow, consider the summation in the second term:

Combining all of these fragments and again using Stirling’s approximation in

ln



n0m



n0

m
n2

þ

be mn2 in Eq (9.3.11):

S2

k ¼ n2½ðm
1Þ lnðz
1Þ þ ð1
mÞ þ ln m ð9:3:12ÞSimilarly, the entropy of the pure solvent S1is obtained by setting n2equal to zeroand n1 equal to n0:

DS ¼
k½n1ln f1þ n2ln f2 ð9:3:15Þ

which is independent of the lattice coordination number z The change in entropy

on mixing n1moles of solvent with n2moles of polymer will exceed by a factor

of Avogadro’s number the change in entropy given by Eq (9.3.15); multiplyingthe right-hand side of this equation by Avogadro’s number gives

DS ¼
R½n1ln f1þ n2ln f2 ð9:3:16Þwhere R is the universal gas constant and n1 and n2 now represent numbers ofmoles Note that if m were to equal unity, f1 and f2 would equal the molefractions and Eq (9.3.16) would become identical to the equation for the change

in entropy of mixing ideal molecules [2] Note also that Eq (9.3.16) does notapply to dilute solutions because of the assumption that f equals jm=n0 and isindependent of position within the lattice

Example 9.1: One gram of polymer having molecular weight 40,000 and density

1 g=cm3 is dissolved in 9 g of solvent of molecular weight 78 and density0.9 g=cm3

.(a) What is the entropy change on mixing?

(b) How would the answer change if a monomer of molecular weight 100were dissolved in place of the polymer?

Solution:

(a) n1¼ 9=78 ¼ 0:115; n2¼ 2:5  10
5; f1¼ ð9=0:9Þ=½ð9=0:9Þ þ 1 ¼0:909; f2 ¼ 0:091 Therefore, DS ¼
R½0:115 ln 0:909 þ 2:5

10
5ln 0:091 ¼ 0:011R

(b) In this case,DS ¼
R½n1ln x1þ n2ln x2, with n2¼ 0:01; x1¼ 0:92,and x2 ¼ 0:08 so that DS ¼ 0:035R

If polymer solutions were truly athermal,DG of mixing would equal
TDS, and,based on Eq (9.3.16), this would always be a negative quantity The fact thatpolymers do not dissolve very easily suggests that mixing is an endothermicprocess andDH > 0 If the change in volume on mixing is again taken to be zero,

DH equals DU, the internal energy change on mixing This latter change arisesdue to interactions between polymer and solvent molecules Because intermole-cular forces drop off rapidly with increasing distance, we need to consider onlynearest neighbors in evaluating DU Consequently, we can again use the latticemodel employed previously

Let us examine the filled lattice and pick a polymer segment at random It issurrounded by z neighbors Of these, zf2are polymeric and zf1are solvent If the

energy of interaction (a negative quantity) between two polymer segments isrepresented by e22and that between a polymer segment and a solvent molecule by

e12, the total energy of interaction for the single polymer segment is

where the factor of 1

2has been added to prevent everything from being countedtwice

Again, by similar reasoning, the total energy of interaction for a singlesolvent molecule picked at random is

n0f2ze22

2For pure solvent, the corresponding quantity is

n0f1ze11

2From all of these equations, the change in energy on mixing, DU, is thedifference between the sum of the interaction energy associated with the polymerand solvent in solution and the sum of the interaction energy of the purecomponents Thus,

where De ¼ ð1=2Þð2e12
e11
e22Þ, and the result is found to depend on theunknown coordination number z Because z is not known, it makes sense to lump

De along with it and define a new unknown quantity w1, called the interactionparameter:

w1¼zDe

whose value is zero only for athermal mixtures For endothermic mixing, w1 ispositive (the more common situation), whereas for exothermic mixing, it isnegative Combining Eqs (9.3.17) and (9.3.18) yields

DGM ¼ kT½n1ln f1þ n2ln f2þ w1n1f2 ð9:3:20ÞBecause volume fractions are always less than unity, the first two terms inbrackets in Eq (9.3.20) are negative The third term depends on the sign of theinteraction parameter, but it is usually positive From Eq (9.3.18), however, w1

decreases with increasing temperature so that DGM should always becomenegative at a sufficiently high temperature It is for this reason that a polymer–solvent mixture is warmed to promote solubility Also note that if one increasesthe polymer molecular weight while keeping n1; f1; f2, and T constant, n2decreases because the volume per polymer molecule increases The consequence

of this fact, from Eq (9.3.20), is thatDGM becomes less negative, which impliesthat a high-molecular-weight fraction is less likely to be soluble than a low-molecular-weight fraction This also means that if a saturated polymer solutioncontaining a polydisperse sample is cooled, the highest-molecular-weight compo-nent will precipitate first In order to quantify these statements, we have to use thethermodynamic phase equilibrium criterion [2]

mAi ¼ mB

where i ¼ 1; 2 and A and B are the two phases that are in equilibrium In writing

Eq (9.3.21), it is assumed that the polymer, component 2, is monodisperse Theeffect of polydispersity will be discussed later

The chemical potentials required in Eq (9.3.21) can be computed using

Eq (9.3.20), the definition of the chemical potential as a partial molar Gibbsfree energy, and the fact that

DGM ¼ Gmixture
G1
G2 ð9:3:22Þ

so that

Gmixture¼ n1g1þ n2g2þ RT½n1ln f1þ n2ln f2þ w1n1f2 ð9:3:23Þwhere n1and n2now denote numbers of moles rather than numbers of molecules,and g1and g2are the molar free energies of the solvent and polymer, respectively.Differentiating Eq (9.3.23) with respect to n1and n2, in turn, gives the following:

f1¼ n1

n1þ mn2 and f2¼ mn2

n1þ mn2gives the following:

Introducing these results into Eqs (9.3.24) and (9.3.25) and simplifying gives

ðm1
m0Þ=RT as a function of f2 By changing w1and repeating the procedure,

we get a family of curves at different temperatures, because there is a one-to-onecorrespondence between w1and temperature Such a plot is shown inFigure 9.4

for m equaling 1000, taken from the work of Flory [3,5] Note that increasing w1

is equivalent to decreasing temperature

By examining Figure 9.4, we find that for values of w1below a critical value

wc, there is a unique relationship between m1and f2 Above wc, however, the plotsare bivalued Because the same value of the chemical potential occurs at twodifferent values of f2, these two values of f2can coexist at equilibrium In otherwords, two phases are formed whenever w1> wc To calculate the value of wc,note that at w1¼ wc, there is an inflection point in the m1versus f2 curve Thus,

we can obtain wcby setting the first two derivatives of m1with respect to f2equal

to zero Using Eq (9.3.30) to carry out these differentiations,

At w1¼ wc and f2¼ f2c; these two derivatives are zero Solving for wc fromeach of the two equations yields the following:

wc¼ 1

2f2cð1
f2cÞ
1

1m

ð2f2cÞ
1 ð9:3:34Þ

wc¼ 1

Equating the right-hand sides of the two previous equations gives

f2c¼ 1

1 þ ffiffiffiffim

which means that

wc¼1

2þ 1ffiffiffiffim

FIGURE9.4 Solvent chemical potential as a function of polymer volume fraction for

m ¼ 1000 The value of w1 is indicated on each curve (Reprinted from Paul J Flory,Principles of Polymer Chemistry Copyright#1953 Cornell University and copyright #

1981 Paul J Flory Used by permission of the Publisher, Cornell University Press.)