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

Equilibria7. First-Order Phase© C. Rose-Petruck, Brown University, 19987.1 Classification of Phase TransitionsWe 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 dependent7. 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. Phasetransitions 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 figure7. 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)vap7. 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 volumesof 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 yields7. 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 with7. 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 ofv(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∫MOM7. 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... µii7. 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 degreesof freedom, pressure and temperature. With two phases, C = 1, R = 0, P = 2, we have one degree offreedom, 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. Thus27. 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.2222127. 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 from7. 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 curvatureof 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 ofboth 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, alloysPhase 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. At7. 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

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