Thermodynamics of Condensed Phases
2.1 THERMODYNAMICS OF METALS AND ALLOYS
2.1.2 Phase Equilibria in Binary-Component Systems
THERMODYNAMICS OF METALS AND ALLOYS 145
4000
3000
2000
1000
00 200 400 600 800
5000
K
A D
Diamond C
E B
J
Graphite
Liquid
Metallic carbon
P (katm)
T (K)
Figure 2.2 Temperature– Pressure unary phase diagram for carbon. From K. M. Ralls, T. H. Courtney, and J. Wulff, Introduction to Materials Science and Engineering. Copyright
1976 by John Wiley & Sons, Inc. This material is used by permission of John Wiley &
Sons, Inc.
point C. There are no degrees of freedom at this point—any variation in temperature or pressure will result in movement into a distinctly separate phase field, and at least one of the phases must necessarily be lost. In the cases where the three phases in coexistence are solid, liquid, and vapor, thisinvariant point is known as thetriple point.
the metallic mixture and determine whether the species are soluble in one another, whether they form intermetallic compounds, or whether they phase separate into two distinct phases of different composition.
The conditions for equilibrium have not changed, and application of the phase rule is conducted as in the previous section. The difference now is that composition can be counted as an intensive variable. Composition is accounted for through direct introduction into the thermodynamic quantities of enthalpy and entropy. The free energy of a mixture of two pure elements, A and B, is still given by the definition
G=H−TS (2.10)
but the enthalpy and entropy for the mixture are now taken as a combination of the enthalpies and entropies of components A and B, plus parameters that accounts for the interaction of the two species due to differences in crystal structure, electronegativity, valence, and atom radii, known as the enthalpy and entropy of mixing, Hmix and Smix, respectively:
H =XAHA◦+XBHB◦+Hmix (2.26) S =XASA◦ +XBSB◦ +Smix (2.27) Here XA and XB are the mole fractions of component A and B, respectively, and are related by (XA+XB)=1. We have used a superscript circle on the enthalpies and entropies of pure components A and B to indicate that these are standard state enthalpies and entropies of the pure components. The standard state of a component in a condensed system is its stable state at the particular temperature and pressure of interest. So, depending on the temperature and pressure of the system, the standard state could be either a liquid or a solid for either of components A and B.
Substitution of Eqs. (2.26) and (2.27) into Eq. (2.10), and using a similar definition for standard state free energies, leads to the following relationship for the free energy of the mixture
G=XA(HA◦−T SA◦)+XB(HB◦ −T SB◦)+Hmix −T Smix G=XAG◦A+XBG◦B +Gmix
(2.28)
where
Gmix =Hmix −T Smix (2.29)
Though Eq. (2.28) is rigorously correct, it contains a rather ill-defined parameter that has been introduced specifically to account for the mixing of the two species, namely, Gmix. In order to proceed further, we must explore this quantity and its definition [given in Eq. (2.29)] more thoroughly.
The entropy of mixing can be thought of as a measure of the increase in the number of spatial configurations that become available to the system as a result of the mixing process,Sconf, which can be shown with statistical thermodynamic arguments to be Sconf =Smix = −R(XAlnXA+XBlnXB) (2.30) where Ris the gas constant.
THERMODYNAMICS OF METALS AND ALLOYS 147
Equation (2.30) shows that the entropy of mixing in a binary component solution is dependent only on the composition (relative number of moles of components) in the solution and is independent of temperature.
Similarly, it can be shown that the enthalpy of mixing can be approximated by the internal energy of mixing [see Eq. (2.4)] for condensed phases, which in turn can be related to the mole fractions of the two components, and an interaction energy, α, which also has units of joules:
Hmix =αXAXB (2.31)
Ifα <0, exothermic mixing occurs (Hmix <0) due to the fact that an A–B bond is stronger (more negative) than either the A–A or B–B bonds and unlike nearest neigh- bors are favored in the system. Endothermic mixing (Hmix >0) occurs when like nearest neighbors are favored, and A–A and B–B bonds are stronger (more negative) than A–B bonds, resulting inα >0. For an ideal solution, the bond energies of A–A, B–B, and A–B bonds are identical, there is no preference for either atom as a nearest neighbor in solution, and both the interaction energy and the heat of mixing are zero:
αideal =Hmixideal =0 (2.32)
The free energy of mixing is found by substituting (2.31) and (2.30) into (2.29) to obtain
Gmix =αXAXB+RT (XAlnXA+XBlnXB) (2.33) which is then substituted into (2.28) forGmix to obtain
Greg=XAG◦A+XBG◦B +αXAXB+RT (XAlnXA+XBlnXB) (2.34) Solutions whose free energies follow Eq. (2.34) are said to be regular, to distinguish them from irregular solutions, wherein the entropy of mixing is governed by a rela- tionship other than Eq. (2.30). The free energy for a mixture that behaves as anideal solution (α=0) reduces to
Gideal =XAG◦A+XBG◦B+RT (XAlnXA+XBlnXB) (2.35) Keep in mind that at this point, the mixtures can be either solid or liquid, depending on the temperature, and that both solid and liquid mixtures may coexist at certain temperatures and compositions. Thus, when performing actual free energy calculations using Eqs. (2.34) or (2.35), the standard state free energies for both components must be carefully selected. This process is illustrated in Example Problem 2.1.
The conditions for equilibrium between two phases in coexistence are still the same as for the case of a single component; that is, Eqs. (2.19)–(2.21) still apply. The phase rule is also still applicable. Before applying these principles to a binary system, let us first examine how a binary phase diagram can be constructed from the free energy equations.
2.1.2.1 Binary Phase Diagrams. Just as in the case of the unary phase diagram in the previous section, we can construct graphical representations of the equilibrium phases present in a binary component system by first plotting, with the assistance of Eq. (2.33), free energy of mixing versus composition for a system consisting of two
Example Problem 2.1
Cesium, Cs, and rubidium, Rb, form ideal solutions in the liquid phase, and regular solutions in the solid phase. Their standard state Gibbs free energy changes of melting as a function of temperature,G◦m,Cs=(G◦Cs,L−G◦Cs,S) and◦m,Rb=(G◦Rb,L−G◦Rb,S)respectively, are given by the following empirical relationships (inJ)
G◦m,Cs =2100−6.95T G◦m,Rb=2200−76.05T 1. Determine the melting temperatures for Cs and Rb.
2. Determine the relationship for Gibbs free energy of mixing as a function of composi- tion at 9.7◦C (282.7 K) for both liquid and solid solutions.
Solution
1. The melting temperatures are found from the free energy expressions by setting them equal to zero, since for each pure component, the liquid and solid free energies must be equal for both phases to co-exist at the melting temperature. For Cs
G◦m,Cs=2100−6.95Tm,Cs =0 Tm,Cs=302 K=29.0◦C Similarly,Tm,Rb=312 K (38.9◦C).
2. First, we note that the temperature under consideration (282.7 K) is below the melting point of both elements, so that the solid-state free energies of both components should be selected as the reference state for our free energy calculations. For the liquid solution, which is ideal, the free energy change of mixing relative to the solid components is found by subtracting the weighted average (in terms of mole fraction) of the standard state free energies of the components from the free energy of the ideal mixture, as given by Eq. (2.35)
GidealL =[XCsG◦Cs,L−XRbG◦Rb,L+RT (XCslnXCs+XRblnXRb)]
−[XCsG◦Cs,S−XRbG◦Rb,S]
which, upon combining terms and using the definitions of free energy changes of melting, simplifies to
G◦L=XCsG◦m,Cs+XRbG◦m,Rb+RT (XCslnXCs+XRblnXRb)
Substitution of the empirical free energy relationships provided for each component, simpli- fication usingXCs+XRb=1, and evaluation atT =282.7 K gives
GidealL =205XRb+133(1−XRb)+(8.314)(282.7) [(1−XRb)ln(1−XRb)+XRblnXRb]
This is the free energy change of mixing for the liquid as a function of composition at 282.7 K.
THERMODYNAMICS OF METALS AND ALLOYS 149
A similar evaluation for the solid solution starts with Eq. (2.34), since the solid solution is regular and not ideal. This equation simplifies rapidly, since the standard state free energies of the components are in the solid phase in both the separate and mixed states, such that
GregS =αXAXB+RT (XAlnXA+XBlnXB) which, when evaluated at 282.7 K, becomes
GregS =αXRb(1−XRb)+(8.314)(282.7)[XRblnXRb+(1−XRb)ln(1−XRb)]
Cooperative Learning Exercise 2.1
Using the data in Example Problem 2.1 and assuming thatα=668 J, determine the com- position of a mixture of Cs and Rb that will meltcongruently at 282.7 K — that is, will go from a solid solution to a liquid solution without changing composition (going through phase separation).
Person 1: Calculate the Gibbs free energy of mixing for the liquid solution fromXRb=0 to 1, in increments of 0.1. (Note: You will have to use approximations forXRb=0.0 and 1.0. Why?)
Person 2: Calculate the Gibbs free energy of mixing for the solid solution fromXRb=0 to 1, in increments of 0.1. (Note: You will have to use approximations forXRb=0.0 and 1.0. Why?)
Plot both of your values on one graph, and estimate the value ofXRbat which the free energies of the liquid and solution solutions are equivalent.
Answer : X
≈ Rb
0.45.This isalso easilyaccomplished
using as preadsheetand
equations olver.
pure elements A and B. Such a plot is shown in Figure 2.3. It should be noted that the free energy of mixing is selected as the ordinate in these plots, since we are concerned only with the change in free energy that results when the two components are taken from separate, standard states into the mixture. The mixture may be either a liquid or a solid.
Figure 2.3 requires careful consideration. Figures 2.3a through 2.3e are free energy of mixing curves as a function of concentration (here plotted asXB, whereXB =1.0 at pure B andXB =0 at pure A) at decreasing temperatures (T5> T4 > T3 > T2 > T1).
Temperatures T1 and T2 are selected such that they are at or below the melting point (TB) of the lowest melting point species, B. Temperatures T4 and T5 are selected such that they are at or above the melting point (TA) of the higher melting species, A.
TemperatureT3is between the melting points of the individual components. Figure 2.3f results from application of the equilibrium criteria at all temperatures, including T1
throughT5. Notice that the binary-component phase diagram is a plot of temperature versus composition, and not pressure versus temperature as was the case for a unary phase diagram. This has to do with the relative importance of our intensive variables.
O
A B
T5 > TA
S
L
Xb (a)
∆Gmix
T4 = TA
S L
A Xb B
(b ) O
∆Gmix
TA > T3 > TB
S
XS XL L
A Xb B
(c ) O
∆Gmix
T2 = TB
S L
A Xb B
(d ) O
∆Gmix
T1 < TB
S L
A Xb B
(e) O
∆Gmix
A Xb B
XS XL T4 T5
T3 T2
T1
(f ) Liquid
Solid
L + S
T
Figure 2.3 Free energy of mixing curves for solid and liquid phases at various temperatures (a– e) and resulting temperature– composition phase diagram for a completely soluble binary component system (f). From O. F. Devereux,Topics in Metallurgical Thermodynamics. Copy- right1983 by John Wiley & Sons, Inc. This material is used by permission of John Wiley &
Sons, Inc.
For the unary diagram, we only had one component, so that composition was fixed.
For the binary diagram, we have three intensive variables (temperature, pressure, and composition), so to make anx–ydiagram, we must fix one of the variables. Pressure is normally selected as the fixed variable. Moreover, pressure is typically fixed at 1 atm.
This allows us to plot the most commonly manipulated variables in a binary component system: temperature and composition.
THERMODYNAMICS OF METALS AND ALLOYS 151
In Figure 2.3a, at a temperatureT5, the liquid phase free energy of mixing is lower than the solid phase free energy of mixing at all compositions. This means that the liquid is more stable at this temperature for all compositions, and indeed this is the case in Figure 2.3f. Any composition at temperatureT5 will be located in the single- phase, liquid region of the phase diagram. A similar situation applies to temperatureT4, shown in Figure 2.3b. Any changes in composition of the system at these temperatures e.g., adding more B to the mixture) will be reflected in a commensurate change in composition of the liquid phase. At temperatures T2 and T1, Figures 2.3d and 2.3e, respectively, the solid solution is the most stable at all compositions. Since A and B mix completely at all compositions, any change in composition at these temperatures e.g., adding more A to the mixture) will result in a corresponding change in the solid alloy.
The situation shown in Figure 2.3c is a little more complex. Here, the solid phase has the lowest free energy at A-rich compositions (XB < XS), and the liquid phase has the lowest free energy at B-rich compositions (XB > XL). In between the two minima of the solid and liquid free energy curves, XS < XB < XL, both the liquid and solid phases can coexist. Why is this? Recall that the stability criteria are as follows for a two-phase system:
Tα =Tβ (2.14)
Pα =Pβ (2.15)
µα =µβ (2.16)
Pressure has already been fixed, as has temperature for the diagram in Figure 2.3c.
The only criterion of concern here is that the chemical potentials of the two phases be equal. Recall that the chemical potential is defined as the derivative of the free energy with respect to composition
µi = ∂G
∂ni
T ,P ,nj
(2.13) or, for the two phases under consideration here,
µS=
∂Gmix,S
∂XB
T ,P ,nL
=µL=
∂Gmix,L
∂XB
T ,P ,nS
(2.36) Graphically, this condition can only occur when the two free energy of mixing curves share the same tangent, which represents the derivative of the curve at that point. This tangent is shown in Figure 2.3c. This is the situation that exists at all temperatures between T4 and T2, so there exists a two-phase region at all compositions between these two temperatures, as reflected in Figure 2.3f. Recall that the minima of the two free energy of mixing curves at any given temperature are located by taking the second derivative and setting it equal to zero:
∂2Gmix,S
∂X2B
=0,
∂2Gmix,L
∂X2B
=0 (2.37)
The locus of points generated by these conditions at various temperatures constructs the phase boundary lines between the solid and liquid phases shown in Figure 2.3f.
That is, there is a different solid and liquid composition, XS andXL, resulting from the tangent between the free energy of mixing curves, at all temperatures betweenT2
and T4, as illustrated in Figure 2.3c. We will return to the conditions stipulated by Eq. (2.37) in subsequent sections.
So, for this binary solution of components A and B, which mix perfectly at all compositions, there is a two-phase region at which both solid and liquid phases can coexist. The uppermost boundary between the liquid and liquid+solid phase regions in Figure 2.3f is known as theliquidus, or the point at which solid first begins to form when a melt of constant composition is cooled under equilibrium conditions. Similarly, the lower phase boundary between the solid and liquid+solid phase regions is known as thesolidus, or the point at which solidification is complete upon further equilibrium cooling at a fixed composition.
Let us examine a real alloy phase diagram. Figure 2.4 is the binary phase diagram for the system Cu–Ni. First of all, note that the composition is indicated in terms of weight fraction of one of the components, in this case Ni, and not mole fraction. The conversion between the two is straightforward if the molecular weights are known, and weight fraction is used only because it is more useful experimentally than mole fraction.
In terms of the phase rule, we must utilize Eq. (2.24) instead of (2.23), since we have only one noncompositional intensive variable,T (pressure is fixed), so N=1.
Application of Eq. (2.24) to pointsAandCin Figure 2.4 indicates that the liquid and solid regions (φ=1;C=2) correspond toF =2, which means that both temperature and composition can be altered independently in these regions. PointBis in the liquid+ solid two-phase region (φ=2;C=2), so that F =1. Temperature and composition are not independent in this region; any change in temperature necessarily results in a
1500
1400
1300
1200
1100
1000
0 20 40 60 80 100
(Cu) (Ni)
Wt. % Ni
T(°C)
1300°C D
E F L
B A a
L + a CL= 47 wt. % Ni
CS= 63 wt. % Ni
C0= 53 wt. % Ni C
Figure 2.4 The Cu – Ni phase diagram, illustrating the use of the lever rule. From K. M. Ralls, T. H. Courtney, and J. Wulff,Introduction to Materials Science and Engineering. Copyright 1976 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc.
THERMODYNAMICS OF METALS AND ALLOYS 153
change in composition of both the liquid and the solid phases. The liquidus and solidus phase boundaries are included in this two-phase region.
More involved phase diagrams result when the two elements, A and B, are only slightly soluble in each other, or when more than one solid alloy phase can be formed from the two components. The principles of generating the phase diagram from free energy of mixing versus composition diagrams still apply, however, as do the appli- cation of the stability requirements represented by Eq. (2.36). For example, when A and B are insoluble in the solid phase, a phase diagram such as that shown in Figure 2.5f can result. As in the previous example, the phase boundaries are con- structed from the free energy of mixing curves as a function of composition at various temperatures. The primary difference here is that the solid free energy curve is a straight line instead of a curve with a minimum, such that at intermediate temper- atures, as in Figures 2.3c and 2.3d, tangents can be drawn between the free energy curves of the pure solids (which occur at the pure component free energies) and the liquid solution, sometimes involving both pure components, as in Figure 2.3d.
When the free energy of the pure solids segment is the tangent to the liquids solution (located exactly at the minimum of the liquid solution free energy curve), an inter- esting phenomenon occurs. It is at this temperature–composition combination, called the eutectic point, at which the liquid, solid A, and solid B coexist. According to the phase rule, there should be zero degrees of freedom, and indeed this is the case (φ=3, C=2, N =1). We will learn more about the eutectic and other three-phase transformations in section 2.1.2.3.
Another typical binary phase diagram that results when two different solid solutions of A and B can form, here labeledαandβ, is shown in Figure 2.6f. Again, the phase diagram is constructed from the free energy of mixing as a function of concentration curves. In this case, it is always possible to draw a tangent between the free energy curves of theαandβphases at low temperatures, and in some cases, one tangent will be sufficient to touch all three curves for theα,βand liquid phases, as in Figures 2.6d and 2.6e. Note that unlike the binary component system in Figure 2.5 in which A and B were never soluble in the solid phase, in this system, A and B are always soluble in each other to some extent. Thus, the composition of the solidαphase will vary, as will that of theβphase. This can be thought of as each major component having an ability to dissolve small amounts of the impurity phase, either as interstitial or substitutional impurities (see Section 1.1.3). This phase diagram also has a eutectic point, found at the intersection of theXL composition line and theT1 isotherm.
Close examination of a variety of binary component phase diagrams allows us to draw a number of generalizations regarding the spatial relationship between phases in a diagram. These are summarized in Table 2.2.
2.1.2.2 The Lever Rule. The phase rule tells us that there is only one degree of freedom in a two-phase region of a binary component phase diagram—for example, the solid + liquid region of the Cu–Ni phase diagram in Figure 2.4. If we change temperature in this region, the compositions of the two phases must also change. This means that the relative amounts of the two phases must, in turn, be adjusted. If there is an equal amount (in terms of weight) of the solid and liquid phase, regardless of their compositions, removing some of component B from the liquid and moving it to the solid reduces the weight of the liquid phase and increases the weight of the solid phase. The relationship between composition of the two phases and the relative amount