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7. First-Order Phase Equilibria 7.1 Classification of Phase Transitions

**Equilibria****7.** **First**-**Order** **Phase**© C. Rose-Petruck, Brown University, 1998**7.**1 **Classification** **of** **Phase** **Transitions**We know phase transitions from daily experience, water evaporates, water freezes, and materials boil orevaporate. In general, all materials exhibit various forms that are characterized by different physicalproperties such as density, viscosity, or molecular structure. A system under constant pressure andtemperature conditions, where virtually all phase transitions occur, is best described by the Gibbs freeenergy. Two phases in equilibrium under such conditions have equal Gibbs free energy. However, ifsome system's variable changes other parameters might not necessarily adapt to these changes and thesystem might not always immediately assume the point **of** lowest possible Gibbs free energy. Instead thesystem might "get stuck" in some local minimum **of** the Gibbs potential. Such a Gibbs potential is depictedin the figure.**7.** **First**-**Order** **Phase** **Equilibria** - 1 © C. Rose-Petruck, Brown University, 1999

The left minimum corresponds to a phase with higher density then the right minimum. For example, iceswims which means that its density is smaller than that **of** liquid water. If the water is cooled slowly belowthe freezing point it does not immediately freeze and settles into the left minimum. Superheated watermight suddenly bubble in a beaker unless we take precautions. However, the system parametersfluctuate. If the barrier between the minima is small the system will eventually "jump" into the globalminimum on the right. We could induce this transition by tapping against the container with thesupercooled water. In general, the positions **of** the local and global minima are temperature dependent**7.** **First**-**Order** **Phase** **Equilibria** - 2 © C. Rose-Petruck, Brown University, 1999

The minima correspond to the liquid and the solid phase, respectively. The system in equilibrium willalways assume the state **of** lowest Gibbs free energy. Both phases have different molar volumes. It isinstructive to plot the position **of** the minima as a function **of** temperature as in the next figure.**7.** **First**-**Order** **Phase** **Equilibria** - 3 © C. Rose-Petruck, Brown University, 1999

These curves should be extended into the pressure direction. Projection **of** the resulting surfaces into thep-T-plane give us a phase diagram. Two typical phase diagrams are shown in the next figure. While theleft one in typical for most substances, the right depicts the behavior **of** water.**7.** **First**-**Order** **Phase** **Equilibria** - 4 © C. Rose-Petruck, Brown University, 1999

The figure shows that at some particular temperature and some particular pressure the solid, the fluid,and the gas phase are in equilibrium. This point is called triple point. This point uniquely defines atemperature for each material. The triple point **of** water at 273.15K and 0.61kPa defines the Kelvintemperature scale.Systems do not necessarily maintain the number **of** minima when the temperature changes. In the nextfigure the phase diagram and the potential surface **of** a system that can undergo second order phasetransitions is shown.**7.** **First**-**Order** **Phase** **Equilibria** - 5 © C. Rose-Petruck, Brown University, 1999

While the phase diagram looks like that **of** a typical substance, the potential surface clearly shows that theisolated minima a low temperature merge at higher temperature. As the temperature increases from T 1 toT 6 the system moves along the liquid-gas coexistence curve. The difference **of** the molar volumes **of** thetwo phases gradually disappears. At T 6 the two phases cannot be distinguished any more. This point inthe p-T-diagram is called the critical point. The distinction between gas and liquid cannot be made anymore. From the critical point on we call both phases together the liquid phase in contrast to the solidphase.In general, there are many types **of** phase transitions: Melting and crystallization, evaporation andcondensation, but also solid-solid, conducting-superconducting, or fluid-superfluid transitions. **Phase**transitions are generally classified according to the Ehrenfest classification. The order **of** a phasetransition is defined to be the order **of** the lowest-order derivative, which changes discontinuously at thephase boundary. The first three orders are given in the figure**7.** **First**-**Order** **Phase** **Equilibria** - 6 © C. Rose-Petruck, Brown University, 1999

We shall discuss first-order transition in the next section. These transitions are, e.g., characterized bychanges in enthalpy or specific volume. An example for a second order transition is the conductingsuperconductingtransition in metals at low temperatures. Other types **of** second order transitions aresolid-solid (structural) transition in crystals.Latent HeatAs discussed above, systems undergo phase transitions in order to assume the lowest Gibbs free energyas the temperature is varied. We know thatG ≡ H − TS . (6.60)In comparison to an ideal gas under equivalent conditions, a substance in the solid phase has a largelynegative enthalpy because **of** the strong binding forces between the molecules. The entropy is rathersmall because the molecules are well ordered. Thus the solid phase has an overall smaller Gibbs freeenergy and is, therefore, the stable phase at low temperatures. At high temperature the gas phase hasvery little enthalpy but large entropy. Thus, the gas phase is the stable phase at high temperatures.This is depicted in a little more detail in the next figure that is similar one **of** the figures above.**7.** **First**-**Order** **Phase** **Equilibria** - 7 © C. Rose-Petruck, Brown University, 1999

We calculate the slope **of** the red curve⎛ ∂G⎞⎜ ⎟⎝ ∂p⎠T , N= V. (6.50)The slopes **of** the curves that correspond to different phases are not necessarily identical. In general,while the Gibbs free energy is equal for phases in contact, the entropy **of** a system will changesdiscontinuously upon phase transformation. With ∆G = 0 we obtain∆ H = T∆S. (**7.**1)This implies that the change **of** entropy is associated with a heat exchange between the phases incontact. For the solid-liquid transition this energy is called heat **of** fusion∆ H = T ∆S. (**7.**2)For the liquid-gas transition this energy is called heat **of** vaporizationThe pressure dependence **of** the Gibbs potential isfusvapfusvapfus∆ H = T ∆S. (**7.**3)vap**7.** **First**-**Order** **Phase** **Equilibria** - 8 © C. Rose-Petruck, Brown University, 1999

⎛ ∂G⎞⎜ ⎟⎝ ∂p⎠T , N= V. (6.50)Since the specific volume **of** a substance doesn't changes less for the solid-liquid transition than for theliquid-gas transition, the first one is less pressure sensitive than the later one.**7.**2 **Phase** BoundariesWe now apply the consideration **of** the previous section in a quantitative manner. In the next figure acoexistence curve, i.e., a curve in the p-T-diagram is drawn.The slope **of** the curve is directly determined by the properties **of** the properties **of** the phases. We shallderive now an expression that relates the slope **of** the curve to the latent heat and change **of** the specificvolume.The Clapeyron EquationThe chemical potentials **of** the states A, A' as well as B, B' are in equilibrium. We assume that thepressure and temperature differences between state A and state B are infinitesimally small. Thus,dp = p B− p Aand dT = T B− TA. The slope **of** the curve is then dp dT .**7.** **First**-**Order** **Phase** **Equilibria** - 9 © C. Rose-Petruck, Brown University, 1999

The chemical potentials on both sides **of** the phase boundary are equal.Thus,Both sides can be written asµA= µ A', (**7.**4)µB= µ B'. (**7.**5)µ −BµA= µB'µA'− . (**7.**6)µ − µ = dG = pdv − sdT , (**7.**7)BAµ − µ = dG'= p'dv − s'dT . (**7.**8)B 'We rearrange the equation and getA'dpdTs'−s=v ' −v(**7.**9)ordpdT∆s∆v= . (**7.**10)The change in entropy and volume are the discontinuities due the latent heat and different molar volumes**of** the two phases. Using equation (**7.**1) we get the Clapeyron equationdpdT∆hT∆v= . (**7.**11)This equation is exact and applies to equilibrium between any two phases, that is the melting andvaporization processes. For example, the molar heat **of** fusion **of** water is 6.0kJ/mol. This means, whenfreezing (at 273.15K) the enthalpy **of** one mole **of** water changes by -6.0kJ. The specific volume increasesby 1.6x10 -6 m 3 /mol. We get thendpdT− 6.0kJ273.15K⋅1.6⋅10NKm7= = −1.37287⋅10. (**7.**12)− 6 32In contrast to other materials, the slope for water is negative as expected.mWe can use the result to calculate the melting temperature **of** ice at a pressure **of** 1000 bar = 10 8 Pa = 10 8N/m 2 :From (**7.**11) followsdTT∆vdp∆h= . (**7.**13)Inserting the result **of** (**7.**12) and the pressure change yields**7.** **First**-**Order** **Phase** **Equilibria** - 10 © C. Rose-Petruck, Brown University, 1999

Above the critical temperature and in certain p-T-ranges the van der Waals equationRT ap = − . (6.103)2( v − b) vis in agreement with the stability criteria derived in chapter 6 stable. One **of** the criteria was2∂ U2∂V≥ 0(6.221){2∂ U2∂V∂ ⎛ ∂U⎞= ⎜ ⎟∂V⎝ ∂V⎠T , N⎛ ∂p⎞= −⎜⎟⎝ ∂V⎠T , N≥ 0(**7.**22){⎛ ∂p⎞⎜ ⎟⎝ ∂V⎠T , N≤ 0. (**7.**23)**7.** **First**-**Order** **Phase** **Equilibria** - 14 © C. Rose-Petruck, Brown University, 1999

The criterion is violated in the range MLKJF in the next figure.At this time it is interesting to note that, in contrast to the van der Waals fluid, the ideal gas satisfies allstability criteria in the entire parameter space. For instance, we derived the pressure dependence **of** the ideal gasg = G0⎛ p ⎞+ RT ln⎜⎟⎝1bar⎠. (6.140)In agreement with equation (6.228) this Gibbs potential is always a concave function **of** the intensive variable, thepressure.Because **of** this stability violation this portion if the phase diagram cannot describe a real physical system.Other processes must supersede the isotherm in this region. We shall see shortly that these processesare phase transitions between the vapor and liquid phase. The form **of** the isotherm determines the molarGibbs free energy, i.e., the chemical potential. The Gibbs-Duhem relationd = −sdT+ vdpµ (5.18)can be integrated in order to obtain the chemical potential( p) dp φ( T )µ = ∫ v + . (**7.**24)Here φ ( T ) denotes the integration constant. In the right half **of** the figure the isotherm is displayed with**7.** **First**-**Order** **Phase** **Equilibria** - 15 © C. Rose-Petruck, Brown University, 1999

the volume as a function **of** the pressure. The difference between the molar Gibbs energy at point A andat point B is thenB( p)µB− µA= ∫ v dp . (**7.**25)ALet us ignore for a moment that only parts **of** the van der Waals isotherm represent a real physicalsystem. We simply integrate the isotherm from point A along the path **of** v(p). The resulting molar Gibbspotential at each temperature as a function **of** pressure is shown in the next figure. The labels in thisfigure refer to equivalent positions in the figure correspond to equivalent labels in the figure above.The potential increases when integrating the volume from A to F. Further integration along the path **of**v(p) to point M reduces the potential as the pressure drops. From point M to point S the potential ismonotonically increasing again. It is important to realize that the slope **of** this branch is different than thebranch from A to F. A real system in equilibrium will always assume the state **of** lowest potential. Thus thesystem will always proceed to point D which is identical to point O. Further increase **of** the pressure movethe system along the path OQRS. Thus, at the intersection, points D - O, the slope changesdiscontinuously. This is a signature **of** a first-order phase transition.φ is temperature dependent we do not know the exact shape **of** theGibbs potential surface as a function id pressure and temperature. But we do have a rather good mentalpicture **of** its shape as depicted in the next figure. We can observe the transition for a region **of** first-orderphase transitions to a region **of** second-order phase transitions.However, since the constant ( T )**7.** **First**-**Order** **Phase** **Equilibria** - 16 © C. Rose-Petruck, Brown University, 1999

What are the consequences **of** the particular shape **of** the molar Gibbs potential. A real physical systemwill never assume the states E,F,J,L,M, and N. It will instead simply proceed along the lower potentialbranch. As the volume **of** the system is reduced phase transitions will keep the pressure constant point Ois reached. The real isotherm has the following form.**7.** **First**-**Order** **Phase** **Equilibria** - 17 © C. Rose-Petruck, Brown University, 1999

The area I equals area II. This can easily be show by splitting the integral from point D to point O into foursections.OFK∫ vdp = ∫vdp+ ∫vdp+ ∫vdp+ ∫0 =vdp (**7.**26)D∫D∫F∫⇓vdp − vdp = vdp − vdp . (**7.**27)The left side equals area I, the right side equals area II. ThusFDFKKMarea I = area II . (**7.**28)Between the points D and O the molar Gibbs potential does not change, even though work is done on thesystem. During the course **of** such and isobaric and isothermal compression the natural variables **of** theGibbs potential remain constant. Consequently, the Gibbs potential remains constant as well.Within the two-phase region any given point denotes a mixture **of** the two phases. The ratio **of** the liquidand vapor phase can be calculated by the lever rule. Let's denote the molar volumes **of** the liquid and thevapor phase with v l and v g . The molar volume **of** the mixed system is denoted by v = V/N. if x l and x g areMKO∫MOM**7.** **First**-**Order** **Phase** **Equilibria** - 18 © C. Rose-Petruck, Brown University, 1999

the mole fractions **of** the two phases we writeThis impliesV = Nv = Nx v + Nx v . (**7.**29)llggxlvg= (**7.**30)vg− v− vlandxgv − vv − vl= . (**7.**31)glThus, the mole fraction **of**, e.g., the liquid is the length **of** the lever from point D to the actual volume **of** thesystem divided by the lever OD.**7.**4 Multicomponent Systems: Gibbs **Phase** RuleWe already know that the fundamental equation for a multicomponent system is( S V , N , N ),1 2U U ,...,= (3.33)N ror in molar form( s v,x , x ),1 2u u ,...,= (**7.**32)x rwith x i = N i /N. Since the sum **of** the mole fractions equals one, only r-1 variables are independent. Atequilibrium all potentials, the internal energy, the enthalpy, the Gibbs free energy, and the Helmholtz freeenergy are convex functions **of** the mole fractions. If, however, the stability criteria are not satisfied for amulticomponent system phase transitions may occur. The chemical composition, the enthalpy, or themolar volume can differ in each phase.Let us consider a container with a mixture **of** components that do not chemically react with each other.Suppose that we observe the system to have different phases. Since in equilibrium the Gibbs potentialdoesn't change dg = 0 for any change **of** variables **of** the Gibbs potential. For instance, the potentialdoesn't change if some fraction **of** the some chemical component changes from one phase into anotherphase. It certainly cannot change into the phase **of** another component since we excluded chemicalreactions. We thus writewhich implies0 = dg = µ dx − µ dx (**7.**33)( B)i( A)( B)iµii( A)iµ = . (**7.**34)This argument applies for any two phases for each chemical component i. Therefore, we can writeThis is equivalent toµ = == for all i. (**7.**35)( A)( B)( P)iµi... µii**7.** **First**-**Order** **Phase** **Equilibria** - 19 © C. Rose-Petruck, Brown University, 1999

g g = ... = g( A)i= for all i. (**7.**36)( B)i( P)iThe result for a single component system that we discussed at the beginning **of** this chapter is certainlyincluded in these results.Now we consider possible chemical reactions among the components. We know that for chemicalreactions⎛ dG ⎞⎜ ⎟⎝ dξ⎠T , p=∑iµ ν = 0ii. (6.149)From these results we can derive an expression for the number **of** degrees **of** freedom **of** a system as afunction **of** the number **of** chemical components, the number **of** reactions, and the number **of** phases.For C chemical components, we have C equations (**7.**35), which contain C(P-1) conditions. Eachchemical reaction introduces a further condition **of** the form (6.149). Suppose that the are R chemicalreactions. The number **of** degrees **of** freedom, which remain to the system, aref total= + CP − C(P −1)− R = 2 + C − R2 . (**7.**37)Usually, we are not concerned with the amount **of** each phase present but only with their composition andnumber and only with the number **of** intensive degrees **of** freedom. The extend **of** each phasecorresponds to one degree **of** freedom. Thus, we have to subtract P from f total .f = ftotal − P = 2 + C − R − Pint(**7.**38)This result is known as the Gibbs phase rule. With one phase, C = 1, R = 0, P = 1, we have two degrees**of** freedom, pressure and temperature. With two phases, C = 1, R = 0, P = 2, we have one degree **of**freedom, pressure or temperature. This is the Clapeyron equation. Three phases can only exist in a welldefinedpoint, the triple point. We have no freedom to choose the temperature or the pressure. Thesevariables are determined by the system's properties.Solubility gaps in binary systemsWe shall now discuss binary, non-ideal mixtures. Depending on the mole fraction **of** each component themixture can be homogenous or heterogeneous. For instance, at room temperature the maximumsolubility **of** carbon disulphide in methanol is 50% by weight. The maximum solubility **of** methanol incarbon disulphide is 2.5% by weight. Any other mole fraction in between will result in phase separationswith one phase containing 2.5% methanol and the other 50% carbon disulphide.A homogeneous mixture will separate into two phases if the Gibbs free energy **of** the heterogeneousmixture is lower than **of** the homogeneous mixture.Let's first consider two components that are not mixed with molar fractions x and (x - 1). The total molarGibbs potential isg ( 1−x)µ + xµ= . (**7.**39)1This molar potential is that **of** an ideal solution in which the chemical components do not interact with onanother. If we now mix the components that molar Gibbs potential might change due to interactionbetween the components, entropy and volume changes. Thus2**7.** **First**-**Order** **Phase** **Equilibria** - 20 © C. Rose-Petruck, Brown University, 1999

( u − Ts + pv) = ∆u− T∆s+ p∆v∆ g = ∆. (**7.**40)The volume change can usually be neglected. There is, however, always an entropy change because themixture is less ordered than the separate components. The entropy change can be expressed as themixing entropy **of** two gases∆s− = x1R( x) + (1 − x)ln( − x)ln . (**7.**41)This symmetrical function is zero for x = 0 and x = 1. It reaches it maximum at x = 0.5. The entropycontributes negatively to the chemical potential in equation (**7.**42).The mixing energy depends on the interaction potential between the molecules **of** equal and differentcomponents. If each atom has z neighbors the initial molar energy, i.e., the molar energy before mixingthe components, isu12[( 1−x)V xV ]initalzNA11+= . (**7.**43)After mixing the components the probability that a molecule has a nearest neighbor **of** the 1 st componentis (1-x) and **of** the 2 nd component is x. The final molar energy is thenThusufinal1zN2A2 2[( 1−x) V + x V + 2x( 1−x)V ]= . (**7.**44)111∆ u = ufinal− uinitial= zNAx( 1−x)( 2V12−V11−V22). (**7.**45)2The mixing energy is zero for all x if V 12 is the mean **of** V 11 and V 22 otherwise it is parabolic. The entropyalways favors a homogeneous mixture. A negative mixing energy favors a homogeneous mixturewhereas positive mixing energy hinders mixing. If the mixing energy is zero the mixture is ideal.The various contributions to the chemical potential are depicted in the next figure.222212**7.** **First**-**Order** **Phase** **Equilibria** - 21 © C. Rose-Petruck, Brown University, 1999

While the mixing energy is essentially independent **of** temperature, the contribution **of** the mixing entropyis proportional temperature in equation (**7.**44). Thus the shape **of** the chemical potential changes as afunction **of** temperature. The next figure depicts the chemical potential for temperatures increasing fromT 1 to T 5 .**7.** **First**-**Order** **Phase** **Equilibria** - 22 © C. Rose-Petruck, Brown University, 1999

Depending on the particular graph **of** the chemical potential, the mixture might be heterogeneous. In thefollowing we examine the condition for phase separations. We consider a chemical potential as shown inthe next figure.**7.** **First**-**Order** **Phase** **Equilibria** - 23 © C. Rose-Petruck, Brown University, 1999

In the range from a to b the red graph does not depict a stable system. Instead the system will assumethe state **of** lowest chemical potential. Thus, the stable graph is a line "bridging" the unstable region froma to b. This conclusion relies on the fact that the stable functional form **of** the chemical potential, like allother potentials, is a convex function **of** the extensive parameters and a concave function **of** the intensiveparameters. Well, is the stable potential a convex function **of** mole fraction?Let's assume that a system has two subsystems, denoted 1 and 2. The total Gibbs potential for a systemisG −ST+ pV + µ N= . (**7.**46)The mole number is an extensive parameters. A stable functional form **of** the Gibbs potential is then aconvex function **of** the mole number. For instance, the total Gibbs potential is the sum **of** the Gibbspotential **of** the subsystemsG Gµ= . (**7.**47)1+ G2= −ST+ pV + µ1N1+2N2{Gµ = −sT+ pv + µ + µN= . (**7.**48)1x12x2Thus the chemical potential is a convex function **of** the mole fraction. This is obvious from**7.** **First**-**Order** **Phase** **Equilibria** - 24 © C. Rose-Petruck, Brown University, 1999

∑N = cons tan t = ∑ N = N x . (**7.**49)iiIt is clear now that the system will separate into two phases 1 and 2. We discussed already the lever rule,equations (**7.**30), (**7.**31). Using the lever rule we can determine the chemical potential that corresponds toa stable equilibrium **of** the two phases:( x'−x1)' = µ1+ ( µ2− µ1)( x − x )µ . (**7.**49)There exist no phase with mole fraction between the points 1and 2. This phenomenon is called thesolubility gap. The condition for the existence **of** the solubility gap is the existence **of** a concave curvature**of** the chemical potential.We saw in the second figure above the principle shape **of** the entropy as a function **of** the mole fraction.Since -TS enters the Gibbs potential, the solubility gap is a function **of** temperature. The gap tends todisappear as the temperature is raised. This is shown in the next figure.2i1iIn this phase diagram the mole fractions x 1 and x 2 are displayed for increasing temperatures. Mixtureswhich composition lies in the yellow region spontaneously separate into two phases with molar fractionsx 1 and x 2 .**7.** **First**-**Order** **Phase** **Equilibria** - 25 © C. Rose-Petruck, Brown University, 1999

**Phase** transitions between liquid and solid phaseA typical feature **of** solidification from a liquid phase is that the chemical composition **of** the solid isdifferent than the composition **of** the liquid. Certainly, the liquid will be homogeneous because **of** themobility **of** the constituents. The solid, in contrast, will usually be non homogeneous as a result **of** thechanging composition **of** the material being laid down. Furthermore, in contrast to solid-liquid phasetransitions **of** pure substances, the process **of** solidification extends over a range **of** temperatures. In thenext figure the chemical potential for the liquid and the solid phase **of** a binary systems is shown.While both chemical potentials have no solubility gap, the mixture will separate spontaneously into twophases at x l and x s . The diagram is constructed for mixtures in which the melting point **of** B is greater thanthat **of** A. The upper part **of** the figure is drawn for a temperature in the middle **of** the solidification range.The tangent construction gives the equilibrium between the liquid and the solid phases **of** compositions x land x s as discussed before. Plotting the positions **of** x l and x s as a function **of** temperature yields the lowerpart **of** the figure. The upper green curve is called liquidus. Liquid is stable at all temperatures above theliquidus. Similarly, the solid phase is stable below the solidus.We now consider a mixture **of** initial composition x 1l . The mixture is cooled from the temperature T 0 abovethe liquidus to the temperature T 1 on the liquidus. This is shown in the next figure.**7.** **First**-**Order** **Phase** **Equilibria** - 26 © C. Rose-Petruck, Brown University, 1999

At temperature between the liquidus and the solidus at the given initial composition x 1l the mixturespontaneously separates into two phases with mole fractions x 1l and x 1s . Liquid and solid phases havedifferent composition. The solid phase is richer in substance B than the liquid. Consequently, the liquidbecomes increasingly depleted **of** substance B. As the temperature drops further, the mole fractions **of**both phases drop as indicated by the red arrows. Liquid and solid have equal composition only at thetemperatures T mA and T mB .Eutectic phase diagrams, alloys**Phase** diagrams can be much more complicated than the one displayed above. Both, liquid and solidphases may have solubility gap. There may be more than on solid phase. For instance, alloys will have ahomogeneous liquid phase because all atoms are mobile. However, several crystal structures may existwhich have different chemical compositions. The Gibbs potential for such a systems is plotted below. At**7.** **First**-**Order** **Phase** **Equilibria** - 27 © C. Rose-Petruck, Brown University, 1999

t high temperatures the chemical potential **of** the liquid is lower than the potential **of** the solid phasessince the liquid is the less ordered than the solid phases. As the temperature drops the chemicalpotentials **of** solid phase B overlap with that **of** the liquid phase. **Phase** separate along the tangent 1. Ateven lower temperatures an additional phase separation can occur along tangent 2. Now, depending onthe initial compositions the liquid and the solid phase A or B are in equilibrium with the liquid. As thetemperature drops further the tangents rotate (tangents 3 and 4) until they merge into tangent 5. Thistemperature defines a unique minimum temperature at which liquid can exist. This is the eutectictemperature. At even lower temperatures only phase transitions between the solid phases can occur.The corresponding eutectic phase diagram is shown below.**7.** **First**-**Order** **Phase** **Equilibria** - 28 © C. Rose-Petruck, Brown University, 1999

**7.** **First**-**Order** **Phase** **Equilibria** - 29 © C. Rose-Petruck, Brown University, 1999