The phase diagram derivedfrom these DSC curves is indicated by the dotted line using the same temperature axisand a horizontal concentration axis [1].. TheDSC runs D, E, and F reveal a s
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To review a general phase diagram in the condensed state with two components,
a series of DSC heating-traces, labeled A–F, are shown in Fig 7.1 with a verticaltemperature axis and a horizontal heat-flow-rate axis The phase diagram derivedfrom these DSC curves is indicated by the dotted line using the same temperature axisand a horizontal concentration axis [1] The DSC trace A is for a sample with aconcentration somewhat beyond the pure component, x1 The single, broad, meltingpeak suggests a solution of the two components The beginning and end of meltingindicate the positions of the solidus and liquidus, as represented by the dotted lines.Both change with variations in the overall concentration When analyzing run B, asecond endotherm can be seen, a peritectic transition It is best identified togetherwith run C at a higher mole fraction of component 2 At the peritectic temperature,
a second solid solution that exists at larger x2turns unstable This becomes obviouswhen completing the full phase diagram, as is shown in area 6 of Fig 7.2, below TheDSC runs D, E, and F reveal a simple eutectic phase diagram of crystals of component
2 with an intermediate compound (point 5 in Fig 7.2) The phase diagram isinterpreted in Fig 7.2 Proof of the assumed phases usually needs a detailed X-raystructure analysis of all the indicated phase areas
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Fig 7.2
Only the eutectic portion on the far right of Fig 7.2 fulfills the conditions ofcomplete solubility in the liquid phase and complete separation in the crystals, as wasassumed in Fig 2.27 The pure crystals, marked by the concentration x1= 1, and thecompound “5,” participating in the eutectic “3,” have a small solubility in theircrystals, the areas “9” and “8,” respectively All the solid solutions indicated in thediagram need additional specification of the concentration-dependence of the chemicalpotential of the components in the single phase areas (I) of limited solubility (“8,” “6,”and “9”) The solid solution “6” decomposes above the peritectic temperature “7” intothe solid solution “9” and the liquid solution All two-phase areas (II) consist of thetwo phase-separated compositions given at any chosen temperature by the points ofintersections of a horizontal with the boundaries of the phase-areas At equilibrium,the possible numbers of phases are governed by the phase rule: P + F = C + 2 (seeSect 2.5.7) In Fig 7.2, three of the condensed phases are in contact at the eutectictemperatures “3” and “4,” and the peritectic temperature “7.” Adding the gas phase,omitted in the present discussion, leads to four phases (P) and no degrees of freedom(F) These quadruple points are thus the points of sharp transitions in the two-component phase diagram of Fig 7.2 (C = 2)
The next step towards the description of phase diagrams that include molecules is to change from the just discussed ideal solution to the real solutions,mentioned in Sect 2.2.5 For this purpose one can look at the phase equilibriumbetween a solution and the corresponding vapor The simplest case has a negligiblevapor pressure for the second component, 2 The chemical potentials for the firstcomponent, 1, in solution and in the pure gas phase are written in Fig 7.3 asEqs (1–3), following the discussion given in Fig 2.26 For component 1, thechemical potential of the solvent, 1, is defined in Eq (1) for its pure state, x1= 1.0,but at its vapor pressure, po
macro-, not at atmospheric pressure It must then be written as
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Fig 7.3
Eq (2) for the gas phase, but with its chemical potential 1 ' defined at atmosphericpressure p = 1.0 atm Equation (3) expresses then the chemical potential of the (pure)solvent vapor over the solution Note, that in these expressions one always uses atm
as unit of pressure, a non-SI unit of pressure (see Fig 2.3, for conversion factors seeSect 4.5.1) At equilibrium, the two logarithmic terms in Eqs (1) and (3) must beequal, 1
s= 1, and produce Raoult’s law Raoult’s law is indicated by the diagonal
in the graph of Fig 7.3 At concentrations approaching the pure solvent, x1 1.0, thisequation must always hold
At higher solute concentrations (x2= 1 x1), however, deviations occur for realsolutions and must be evaluated in detail The deviations from the ideal solution areoften treated by introduction of an activity a1replacing the concentration x1 A well-known interpretation of the activity for the strong interactions, as are found for ionsdissolved in water, is given by the Debye-Hückel theory [2] Clusters of water arebound to the solute and reduce p1 Repulsions, on the other hand, reject the solventfrom the solution into the pure gas phase and increase p1, as is shown in the example
of Fig 7.3 At high solute concentration x2, the indicated tangent can approximate thevapor pressure and is called Henry’s law
For macromolecules the vapor pressure of a the low molar solvent is decreased somuch that the Henry’s law activity, a1, remains practically zero for a wide concentra-tion range Only close to x1= 1.0 does the vapor pressure approach Raoult’s law, i.e.,
a large mole fraction of low-molar-mass solvent dissolves in a polymer melt withoutproducing a noticeable vapor pressure Perhaps, this should not be surprising if oneremembers that x1must be large compared to x2, the mole fraction of the polymericsolute, to achieve comparable masses for the two components The Flory-Hugginsequation, to be described next, will resolve this problem and describes the vaporpressure in Fig.7.3 for cases such as natural rubber dissolved in benzene
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as before The RHS is identical to the prior case and written on the right in Eq (2).One needs to remember that Hfrefers to one mole of large molecules, i.e., it has amuch bigger effect than for small molecules, perhaps by a factor of 1,000! The left-hand side, the mixing expression, will be given by the Flory-Huggins equation
The free enthalpy of mixing, Gmix, will first be treated by considering a new idealentropy of mixing that depends on molecular size Then, it will be expanded to a realsolution by adding the interaction term, G*, seen in Eq (3) Easiest is theHildebrand derivation [3] One assumes that ideal liquids have a universal freevolume fraction,1
vf On mixing, the molecules expand into the available free volume,
as in an ideal gas They change from the free volumes in the pure states, Vf1= V1vfand Vf2= V2vf, to the total free volume Vf1,2= Vtotalvfwith a change in the entropy of
Sideal=n1R ln (Vf1,2/Vf1) n2R ln (Vf1,2/Vf2), where n1and n2are the moles of therespective molecules Since the free volume fraction is assumed to be constant duringthe mixing, this leads to Eq (4) in Fig 7.4 with the definition of the volume fractions
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Fig 7.5
written as Eq (5) and Eq (6) An identical expression can be derived by placing thesolution on a lattice where the solvent molecules occupy one unit cell each, and themacromolecules use x unit cells each (x = V2/V1) [4,5] Note that with x = 1, i.e., forequal sizes of both components, the equation describing the mixing reverts back to theresult for the ideal gas in Fig 2.24 (v1= x1and v2= x2)
A simple calculation clarifies the effect of molecular size Assuming two types ofmolecules mixed to a fixed volume V with fractions v1= v2= 0.5 Allowing large andsmall molecules of ratio 1,000:1, Eq (4) yields a S = 693 R when both molecules aresmall (x = 1.0, and n1= n2= 500), when one molecule is large and the other one small,
x = 1,000, and now S = 347 R (n1= 500 and n2= 0.5), and when both molecules arelarge, x is again 1.0, but this time S = 0.7 R (n1= n2= 0.5) One concludes that forsmall molecules the positive entropy strongly drives the dissolution and mayovercome the endothermic heat of solution which usually exists in the absence ofspecific dipole or H-bond interactions between solvent and solute Large moleculesstill may dissolve in a small molecular solvent for entropic reasons Macromolecules,
in contrast, will not dissolve other macromolecules, unless the overall free enthalpy
is negative, i.e., the heat-of-dissolution is exothermic (see Sect 2.2.3)
The last step in the derivation involves the evaluation of G*
, the interaction term,which may help the dissolution of a macromolecule by becoming exothermic.Equation (7) in Fig 7.4 and the sketch in Fig 7.5 show how one can assess G*
fromthe number of contacts a macromolecule can make with its low-molar-mass neighbors
If the interaction parameter, ... v1, the heat offusion of the macromolecule in Eq (15) of Fig 7.6 can be replaced by Hf/x, the heat
of fusion of the macromolecule per reference volume of the low-molar-masscomponent... case of two components being part of a single molecule istreated with the example of a block copolymer as described in Fig 1.19 and Sect 3.4.The treatment of shape effects and the consideration of. .. of the molecularsegment, N (ordinate), and the fraction of A in the molecule, f (abscissa) Of specialinterest is the limit of phase separation for small values of3 N Depending on thestrength of