1

1

Thermodynamic of

Phase Diagram

by Dr. Srimala

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Content 1.0 Introduction

2.0 Thermodynamically stable phase

3.0 Unary Heterogeneous Systems3.1 P - T Diagram -Unary, Single Component Phase Diagram3.2 logP – 1/T Diagram -Unary, Single Component Phase Diagram3.3 Conclusion-Unary p - T Diagrams3.4 G-T Phase Diagrams

3.4.1 G - T Diagram - Unary, Single Component Phase Diagram – V3.4.2 - T Diagram - Unary, Single Component Phase Diagram - L, V3.4.3 - T Diagram - Unary, Single Component Phase Diagram-, L,V3.4.4 - T Diagram - Single Component Phase Diagram -, L, V

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3.5 Metastability3.6 Chemical Potential and Gibbs Free Energy of Single Component Phases3.7 Enthalpy & Entropy of Transformation3.8 Compute Phase Equilibria from Free Energy Relations

4.0 Binary System4.1 Binary liquid system4.2 Binary solutions with total solid solubility4.3 Binary systems without solid solution

5.0 Free Energy-Composition (G-X) Diagram5.1 Free energy diagrams of total solubility systems5.2 Free energy diagram for binary solutions with a miscibility gap5.3 Free energy diagram of binary systems without solid solution

(eutectic system)

6.0 Phase boundary Calculations

4

1.0 Introduction

• A phase diagram is a graphical representation of all the equilibrium phases as a function of temperature, pressure, and composition.

• Phase diagrams arise from minimizing free energies for each phase. They can be used to describe gas - liquid – solid transitions, polymorphic solid-to-solid transitions, stable phases in alloys of different composition, etc.

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5

For example, for transformation occurring at constant T and P the relative stability of the phases in a system is determined by their Gibbs free energies, ∆G = Gfinal - Ginitial = ∆ H - T ∆ S

∆G < 0 => process is allowed∆G > 0 => process is forbidden∆G = 0 => equilibrium

2.0 Thermodynamically Stable Phases

Thermodynamics can be used to predict weather the system isin equilibrium and to analyze the phase stability and phase transformations.

6

Cont….

• Usually, only one phase of a given substance is stable at any given temperature and pressure.

• At some conditions of temperature and pressure, two or more phases may exist in equilibrium.

• A slight change in temperature or pressure will favor one phase over others. The conversion of one phase to another is a phase transition.

• Phase transitions occur with a decrease (spont.) or no change (equil.) in Gibbs energy.

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3.0 Unary Heterogeneous Systems

A system is considered to be unary if it consists of a single chemical component for the range of states under study.

Example CO2 or H2O

Homogeneous system - consists of a single phase

Heterogeneous system - consists of more than one phase

All the elements may exists in at least 3 distinct states of matter or phase- solid, liquid and gas

8

Cont…

Many element exhibit more than one phase form in solid state-called AllotropesExample : BCC and FCC structures are allotropic forms of iron

The BCC phase change to FCC iron at 910oC, 1 atm pressure-called allotropic transformation

In general, Phase Transformation- any change in the phase form of a system

Example : •melting or fusion-which change solid to liquid•boiling or vaporization-which change from liquid to vapour

3

9T (K)

+ L + V

, solid crystal

V, vapor

L, liquid Show regions of stability of phases in terms of the state variables T & P.

3. 1 P - T Diagram -Unary, Single Component Phase Diagram

10

Consider phase stability and phase transformation at a constant pressure, Po, eg. one-atm.

TM TVT (K)

, solid crystal

V, vapor

L, liquid

+ L + V

Cont..

11

Cont…..

Single phase

(, L, V) stability regions are 2-D, an area.

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V

Can change temperature and pressure independently and remain in region.

12

Cont…

Two phase (L, L+V, +V)

stability regions are 1-D, a line.

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V

Can pick temperature or pressure (not both) and the other is fixed by the phase boundary.

4

13

Cont…..

Three phase

(+L+V) stability regions are 0-D, a point.

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V

Cannot pick temperature or pressure. Both are fixed by the triple point.

Note: More than three phases cannot exist at equilibrium for a unary system. 14

Cont….Metastable extensions of two-phase stability lines extrapolate into opposite single phase regions.

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V

15

Cont….Note the critical point where the properties of the liquid and vapor phases merge.

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V

C.P.X

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1/T (K-1)

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V

On a plot of logP vs. 1/T the lines for two-phase equilibrium become (approx.) straight.

3.2 logP – 1/T Diagram -Unary, Single Component Phase Diagram

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17

Cont….

1/T (K-1)

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V1 atm

TM

TV

Again, consider stability regions for constant pressure conditions.

18

Cont…

1/T (K-1)

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V1 atm

Again, note metastable extensions of two-phase equilibrium boundaries.

19

Cont….

1/T (K-1)

T (K)

, solid crystal

V, vapor

L, liquid

+ L + V1 atm

Slopes of lines for two-phase equilibria are inversely proportional to magnitude of volume expansion on heating.

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Recall Gibbs Phase Rule

For a system at equilibrium,

F = C F = C -- P + 2P + 2

C = the number of components (1 so far in this chapter)P = the number of phases presentF = the number of degrees of freedom (the number of intensive

variables such as temp, pres, or mol frac that can be changed without disturbing the number of phases in equilibrium)

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21

Cont ……

For a one-component system the phase rule becomesF = 3 - P

Phases Degrees of Freedom Components

21 1 21 22 11 23 0

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3.3 Conclusion-Unary p - T Diagrams

Single phase regions are represented by an area: 2-D & 2 degrees offreedom.

Two phase equilibria are represented by a line: 1-D & 1 degree offreedom.

Three phase equilibria are represented by a point (the triple point): 0-D & no degrees of freedom.

No regions on a P - T diagram show more than 3 phases coexisting atequilibrium.

Properties of the liquid and vapor phases merge at the critical point.

23

Cont…

Two phase equilibrium boundaries extend into stable single phaseregions & indicate potential metastable equilibria.

Slopes of two-phase equilibrium lines increase as the volumeexpansion on heating decreases.

Slopes of two-phase equilibrium lines are positive for volumeexpansions on heating and slopes are negative for volumecontractions.

Two-phase equilibrium boundaries are approximately straight onlog P vs. I/T diagrams.

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3.4 G-T Phase Diagrams

Gib

bs e

nerg

y, G

Temperature

G = H - TS, so the slope of the lines at left

= (dG/dT) = -S

This assumes that H and S are constant with temp.

Since S is positive for all phases of all substances, the slopes are all negative.

Note that the gas phase has the steepest slope; the solid phase, the least steep.

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Change in Entropy

Ssteam > Sliquid water > Sice

Relative Entropy Example:

Third Law Entropies:

All crystals become increasingly ordered as absolute zero is approached (0K = -273.15°C) and at 0K all atoms are fixed in space so that entropy is zero.

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T (K)

Fixed PV

V

-SV

3.4.1 G - T Diagram - Unary, Single Component Phase Diagram - V

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3.4.2 G - T Diagram - Unary, Single Component Phase Diagram - L, V

T (K)

L

L

Fixed P

-SV

-SL

V

V

L

28

Cont….

T (K)

L/

L/

Fixed PV/

VTV

L

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3.4.3 G - T Diagram - Unary, Single Component Phase Diagram - , L, V

T (K)

Fixed P

L

-SV

-SL

-S

V

30

Cont…

T (K)

L/

Fixed PV/

VTV

L

/

/

L/

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Cont…

T (K)

L/

Fixed PV/

V

TV

L

/

/

L/

MT

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3.4.4 G - T Diagram - Single Component Phase Diagram -, L, V

T (K)

L/

Fixed PV/

VTV

L

/

/

L/MT

/

/

MT /T

ST

ST

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T (K)

Fixed P

VTV

L

MT /T

Equilibrium Phases and Transformations, Only

Cont…

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Sketch curves representing the variation of the molar Gibbs free energy with temperature at the pressure corresponding to triple point for an element. Repeat this sketch for a pressure slightly above and below the triple point.

Exercise 1.1

35T (K)

, solid crystal

V, vapor

L, liquid

+ L + VP>Ptriple point

P=Ptriple point

P<Ptriple point

P - T Diagram based on the question

Solution 1.1

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Solution 1.1

T (K)

L

L

V

V

Triple Point: a+L+V

P (triple pt.)

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Solution 1.1

T (K)

L

L

V

V

Melting Point: a+L

L

P > P (triple pt.)

Boiling Point: L+V

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Solution 1.1

T (K)

L

L

V

V

Sublimation Point: + V

P < P (triple pt.)

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3.5 Metastability

In systems at constant T & P Gibbs free energy is a minimum at equilibrium.

The phase with the lowest Gibbs free energy is the most stable.

When phases with higher Gibbs free energies form, they are metastable.

The greater the deviation of their free energies from the stable phase, the lower the stability of the metastable phase.

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Exercise 1.2

• The standard Gibbs energy of formation of metallic white tin (-tin) is 0 at 25 oC and that of nonmetallic gray tin (-tin) is +0.13 kJ mol-1 at the same temperature. Which is the thermo-dynamically stable phase at 25 oC?

Solution 1.2:• The thermodynamically stable phase is the one with the lower

Gibbs energy, which would be the - (white) tin at 25 oC.

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Compute G-T diagram to show the stable region of , and L based on the information given.

G <G at T<Tt and hence is more stable

G <G at T>Tt and hence is more stable

Gliq <G at T>Tm and hence liquid is more stable

Tt <Tm

Exercise 1.3

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T (K)

/

L/

LTm

/

/

/Tt

Solution 1.3:

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and are 2 solid phases which are possible to appear in the substance A. At low temperature is more stable. Prove thermodynamically requirements for the appearance of allotropic transformation form to at constant pressure is

Entropy of > Entropy of

Exercise 1.4

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Solution 1.4

T (K)

Recall :(dG/dT) = -S

= the slope of the free energy curve

Thus the slope of the curve is the negative entropy of the phase. In order for phase to appear at higher temperatures the free energy of should fall rapidly than the as the temperature is raised . In other words, the free energy curve for is steeper than the this means the entropy of should be larger than

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3.6 Chemical Potential and Gibbs Free Energy of Single Component Phases

Combined 1st & 2nd law with extensive functions:dn VPd - STd Ud

By definition (recall gibbs free energy function): VSn

U,

/

46

dn dPV dTS- Gd

Substitute for dU/ in enthalpy expression:

Extensive enthalpy: dP VPd Ud Hd V

dn dPV STd Hd

Substitute for dH/ in free energy expression:

Extensive Gibbs free energy: dT STd Hd Gd S

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dn dPV dTS- Gd

For constant T & P (set dT=0, dP=0, & rearrange):

PTPT n

nGn

G,,

/

PTnG

,

/

Note G’ is the gibbs free energy of the system and G is the gibbs free energy per mole of the system (molar Gibbs free energy). Therefore

GnnGn

nGn

GPTPTPT

,,,

/

G Yielding:

In a unary system the chemical potential of a phase is equal to its molar Gibbs free energy at that T & P.

nGG

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3.7 Enthalpy & Entropy of Transformation

At equilibrium --- GG and

TSHTSH SSTHH

Thus, /// STH

/

//

THS

V

VLVL

THS

//

M

LL

THS

//

G=H -TS

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3.8 Compute Phase Equilibria from Free Energy Relations

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Chemical Potential Surface and the structure of the Unary Phase Diagram

Because of in a unary system

(7.4)The coefficients are the molar entropy and molar volume of the phase.example for phase,

(7.5)

The molar entropy and molar volume for phase can be computed as function of temperature and pressure from heat capacity, expansion and compressibility data.

Integration of Equation 7.5 yield a function

can be visualized graphically in Fig 7.3a. Fig 7.3a

G

VdPSdTdGd

dPVdTSdGd

),( PTT

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Similar argument for other phase in the systemFor liquid,

can be visualized graphically in Fig 7.3b

The chemical potential of these 2 phase can be compared at any given (T,P) combination if and only if the reference state used in their computation is the same.

The reference state is chosen to be thesolid phase at (To, Po) and now it is possible to construct surfaces for both surface on one graph Fig 7.3 c

The two surfaces intersect along a space curve AB-at any point on this curve, T, P, of the two phases are identical.

),( LLLL PT

Fig 7.3b

Fig 7.3cT

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Ts = TL Ps= PL s= L

The two surfaces intersect along a space curve AB-at any point on this curve, T, P, of the two phases are identical.

This is the conditions that must be met in order for the solid and liquid phase to coexist in equilibrium

The projection of this line onto (P, T) plane is the phase boundary that defines the (s+L) two phase equilibrium and the limit of stability of the solid and liquid phases.

Fig 7.3cT

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53

Intersection of solid and gas (COD)-represents the (s+G) two phase equilibriumC’O’D’on the (P, T) plane- sublimation curve

Intersection of liquid and gas (EOF)-represents the (L+G) equilibriumE’O’F’on the (P, T) plane- vaporization curve

All three surface intersect at a single point O. At this point T, P, of all the 3 phases are the same and all 3 phases coexist in equilibrium. The projection onto (T,P) plane, O’ is the triple point for the 3 phases (s+L+G)

Fig 7.4 shows the construction when the chemical potential surface for the gas phase is added.

Fig 7.4

4.0 Binary systems

• Binary liquid system• Binary system without solid solution• Binary system with solid solution

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4.1 Binary liquid system

When two liquid are brought together, they may

• Totally dissolve in one another in all proportions (total miscibility),

• Partially dissolve in one another or• Be completely immiscible

Species A

Species BImmiscible

Partially Immiscible

Total miscibility

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Immiscible Mixtures

A (droplets) + (continuous)in separate phases

A and B in gas

Pure APure B

PAsat

PBsat

These are equivalent

=GasWe may regard this

mixture as

The liquid A is in equilibrium with its vapour at the vapour pressure of PA and The liquid B is in equilibrium with its vapour at the vapour pressure of PBTotal pressure= PA + PB

Partially-Miscible Liquid Systems

• The hexane-nitrobenzene system is seen at left. This system has an upper critical temperature.

• The region of miscibility is outside the dome-shaped area. When the components are miscible there is only one phase present.

• For any given temp inside the dome shaped area, forms two liquid solution

• The relative amounts of the two phases present will be given by the lever rule.

Fully

miscible

Immiscible

Other Partially-Miscible Systems

• Two other types of partially-miscible systems are seen at left.

• The triethylamine-water system has a lower critical temperature. That is, the components get more miscible as the temperature gets lower.

• The nicotine-water system has both an upper and a lower critical temperature. Only between 61 and 210 oC are there two phases present.

For binary system in which two components are mutually soluble in all proportion in both liquid and solid state, the possible phase diagram are as shown below

4.2 Binary solutions with total solid solubility

Azeotropes, low-boiling

Azeotropes, high-boiling

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T-x Diagrams• For an ideal solution the T-x diagram looks

like that at left.

• Here the liquid phase lies below the vapor.

• As the temp is raised from a starting point at a1, boiling occurs at a2 , with the vapor comp’n being at a2. The distillate is richer in the more volatile component.

• Successive distillations will separate the pure components.

Fractional Distillations• The horizontal lines in the two-phase region, joining

liquid and vapor phases in equilibrium but of different composition, are called tie lines.

• These are labeled 1, 2, 3, etc. at left.

• A fractional distillation consists of starting with liquid, heating to boiling, condensing the distillate, and repeating.

• The efficiency is quantified by the number of the theoretical plates, the number of effective vaporisation and condensation steps that are required. The closer together the liquid and vapor curves are, the more theoretical plates are needed to achieve a given degree of separation of components.

heat

ing

boiling

condensation

High efficient

less efficient

Azeotropes, high-boiling

• Unfortunately, not all solutions are ideal. Often there is a maximum or minimum in the boiling-point curve.

• Note that the liquid and vapor curves converge at the maximum (composition b)

• Once this composition is reached, no further separation is possible by simple distillation.

• Composition b is called an azeotrope.

• The system HCl-H2O exhibits this behavior.

Azeotropes, low-boiling• In the high-boiling azeotrope, a pure

component could be separated in the distillate, and the residue approached the azeotropic composition.

• There are also low-boiling azeotropes, in which pure components can be separated in the residue but the distillate approaches the azeotropic composition.

• A well-known system with a low-boiling azeotrope is EtOH- H2O. The azeotropic comp’n is 95% EtOH.

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4.3 Binary systems without solid solution

Eutectic System

• It is commonly found that many materials are highly miscible in the liquid state, but have very limited mutual miscibility in the solid state. Thus much of the phase diagram at low temperatures is dominated by a 2-phase field of two different solid structures-one that is highly enriched in component A (the α phase) and one that is highly enriched in component B (the β phase). These binary systems, with unlimited liquid state miscibility and low or negligible solid state miscibility, are referred to as eutectic systems.

• The behavior just described, where the two components are completely miscible at high temperatures in the liquid state and phase-separated into two solids at low temperatures would be represented by a phase diagram as follows:

• The solid forms of A and B are essentially immiscible.

• Their liquids are complete miscible.• No solid-state compounds are formed between A and B.

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5.0 Free Energy-Composition (G-X Diagram)

• For a binary alloy consisting of components A and B, we fix the pressure (typically P = 1 atm).

T

TmA

XB

T2

T1

T4

T3tie line

solid

liquid

0 1Xl Xs

=TmB

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BXB

T1

solid

liquid

0 1

G

A

Xl

At T1 the free energy of the liquid is everywhere below that of the solid. The graphical construction yields the values for the liquid phase chemical potentials.

T

TmA

XB

T2

T1

T4

T3tie line

solid

liquid

0 1Xl Xs

=TmB

18

69

At T2 the free energy of the liquid coincides with that of the solid at the single point X=1. The graphical construction yields the values for the solid-phase chemical potentials.

T2

solid

liquid

0 1

G

XB

B

=TmB

T

TmA

XB

T2

T1

T4

T3tie line

solid

liquid

0 1Xl Xs

=TmB

70

A

B

GT3

solid

liquid

0 1XsXl

liquid + solid

At T3 the “common tangent” construction shows that a mixed-phase region exists between Xl and Xs, between which the chemical potentials of A and B in each phase are equal. Single phase regions—either solid or liquid—exist outside those composition limits.

T

TmA

=TmB

XB

T2

T1

T4

T3tie line

solid

liquid

0 1Xl Xs

• Common tangents to free energy curves define composition regions where phase separation (two phase equilibria) occurs.

• Phase separation lowers the overall free energy by splitting the homogenous system into a weighted mix of two separate phases, which each have lower free energy than the starting homogeneous phase.

Compositions of the phases in two-phase regions are given by the tangent points, and the amount of each phase is determined by the lever rule.

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XB

T4

solid

liquid

0 1

G

At T4, the solid phase everywhere has a lower free energy than does the liquid, and is the stable phase at all compositions.

T

TmA

XB

T2

T1

T4

T3tie line

solid

liquid

0 1Xl Xs

=TmB

19

Based on the Gibbs free energy curves we can construct a phase diagram for a binary isomorphous systems

Let’s construct a binary phase diagram for the simplest case: A and B components are mutually soluble in any amounts in both solid (isomorphous system) and liquid phases, and form ideal solutions.

.

Example :

5.1 Free energy diagrams of total solubility systemsWe have 2 phases – liquid and solid. Let’s consider Gibbs free energy curves for the two phases at different T

T1 is above the equilibrium melting temperatures of both pure components: T1 > Tm(A) > Tm(B) > the liquid phase will be the stable phase for any composition

Decreasing the temperature below T1 will have two effects:

GAliquid and GB

liquid will increase more rapidly than GAsolid and GB

solid

The curvature of the G(XB) curves will decrease.

Eventually we will reach T2 – melting point of pure component A, where GA

liquid = GBsolid

For even lower temperature T3 < T2=Tm(A) the Gibbs free energy curves for the liquid and solid phases will cross

As temperature decreases below T3 GAliquid and GB

liquid continue to increase more rapidly than GA

solid and GBsolid

Therefore, the intersection of the Gibbs free energy curves, as well as points X1 and X2 are shifting to the right, until, at T4 = Tm(B) the curves will intersect at X1 = X2 = 1

20

At T4 and below this temperature the Gibbs free energy of the solid phase is lower than the G of the liquid phase in the whole range of compositions – the solid phase is the only stable phase.

Construction of the phase diagram

5.2 Free energy diagram for binary solutions with a miscibility gap

Let’s consider a system in which the liquid phase is approximately ideal, but for the solid phase we have .Hmix > 0

At low temperatures, there is a region where the solid solution is most stable as a mixture of two phases α1 and α2 with compositions X1 and X2.

This region is called a miscibility gap.

21

• It is commonly found that many materials are highly miscible in the liquid state, but have very limited mutual miscibility in the solid state. Thus much of the phase diagram at low temperatures is dominated by a 2-phase field of two different solid structures-one that is highly enriched in component A (the α phase) and one that is highly enriched in component B (the β phase). These binary systems, with unlimited liquid state miscibility and low or negligible solid state miscibility, are referred to as eutectic systems.

• The behavior just described, where the two components are completely miscible at high temperatures in the liquid state and phase-separated into two solids at low temperatures would be represented by a phase diagram as follows:

5.3 Free energy diagram of binary systems without solid solution (Eutectic System)

Eutectic phase diagram with different crystal structures of pure phases

Cont… Cont…

22

Cont… Cont…

Eutectic phase diagram with same crystal structures of pure phases

Cont…

23

Cont…

90

6.0 Phase boundary Calculations

Tm (A)

A BXB

XAL

XAs

XBS

For the phase equilibrium of and liquid as shown in the figure, we have with composition XA

S in equilibrium with liquid of composition XA

L .

L

91

Cont…

)(SoAG

At equilibrium, partial molal Gibbs free energy for each component is the same in both the phase

Note :Gi=Gio + RTln ai

represent free energy change in melting

(fusion).

with ref to pure solid A

with ref to pure liquid A

Free energy axis

)(LoAG

) (

)()()()(

lnln

)(

LSoLA

SA

SoA

LoA

LoA

LA

SoA

SA

LA

SA

GaRTaRT

GGGGGG

GG

)(SoAG

)( S

AGFree energy change on

fusion at ToK

)( LAG

LAaRT ln

SAaRT ln

92

LA

SA

AMM

LA

SA

AM

AMAM

LA

SAL

ASA

LSOA

AM

AMAM

LSOA

LSOA

LSOA

aaRT

TTH

aaRT

TH

TH

aaRTaRTaRTG

TH

THG

STHG

G

ln1

ln

lnlnln

K Tat fusion of changeenergy free

)(

)(

)()(

)(

)(

)()(

)(

)(

o)(

S H/T

Using this we have

24

93

And if the solutions are assumed ideal

Similarly for the component B

)(

)(

)(

)(

)(

)(

11ln

11ln

11ln

BM

BMLB

SB

AM

AMLA

SA

AM

AMLA

SA

TTRH

XX

TTRH

XX

TTRH

aa

Defination of ideal & non idealSpecies A Species B Ideal Mixture: = 1

+

Non-Ideal Mixture of A and B: > 1

A-B Repulsive InteractionsExaggerated

95

2)(

)(

)(

2)(

)(

)(

ln

11ln

ln

11ln

M

AMLB

AM

AMLB

M

AMLA

AM

AMLA

TT

RH

X

TTRH

X

TT

RH

X

TTRH

X

Further if the diagram indicates no solid solubility, we have

In several cases, for the region close to the melting point, one can state as an approximation,

Also noting that ln(1-x)-x when x<<1, we can say

96

Tm (A)

A BXB

XAL

XAs

XBL

Tm (B)

Phase diagram of full solid solubility is shown below. Calculation of the solidus and liquidus of this type of diagram can be made as follows, using the assumption of ideal solution for solid and liquid state.

As per equations derived before,

)(

)( 11lnAM

AMLA

SA

TTRH

X

X

)(

)( 11lnBM

BMLB

SB

TTRH

X

X

25

97

cont….

1 and 1

and 21

LB

LA

SB

SA

LB

SB

LA

SA

XXXX

X

Xk

X

Xk

SA

SB

LA

LB

LA

SA

LA

XXXX

XkX

kkkX

11

.

1

1

12

2

Let

Also

Solving algebraically,

Effects of relative values of entropy of fusion of the components

on the shape of the phase diagram with full solid and liquid solubility

)1(

)1(o1

m

mTH

S

99

a) Pure iron has melting point 1539oC and on addition of 1% (weight) of silicon lowers it to 1527oC. Find the silicon content in solid iron in equilibrium with liquid at 1527oC. Heat of fusion of iron is 15490 J/mol

b) At 1492oC, delta iron contains a maximum of 0.10% (by weight) of carbon. What is the carbon content of the liquid iron in equilibrium with this solid?

Exercise 1.1 Solution (1.1a)

18121

18001

314.815490

9802.0ln

11ln

9802.00198.01

0198.0)56/99()28/1(

28/1

)(

)(

SFe

FeM

FeMLFe

SFe

LFe

LSi

X

TTRH

XX

X

X

Tm (pure Fe) =1539+273=1812K, Tm (with Si)=1527+273=1800K

26

x100%

% 65.0)56)(9870.0()28)(013.0(

)28)(013.0( %wt

013.09870.01

0.98700.9802 x 0069.1

01 x 855.69802.0

ln 3

SSi

SFe

SFe

X

X

X

18121

17651

314.8154909954.0ln

11ln

9954.001 x 65.41

01 x 65.4)56/999.0()12/001.0(

12/001.0

)(

)(

3

3

LFe

FeM

FeMLFe

SFe

SFe

SC

SC

X

TTRH

XX

X

X

X

Solution (1.1b)

x100%

% 692.0)56)(9685.0()12)(0315.0(

)12)(0315.0( %wt

0315.09685.01

9685.00278.19954.0

0278.19954.0

0274.09954.0ln

LC

LFe

LFe

LFe

X

X

X

X

104

Copper and nickel are fully soluble in each other in solid and liquid states. Assuming ideal solutions, calculate the solidus and liquidus curves of the Cu-Ni diagram using following data : (calculate at 1400, 1500, 1600 and 1700K).

Cu (A) Ni (B)

Melting point, K 1356 1728

Heat of fusion (J/mole) 12,790 17,154

Exercise 1.2

27

105

)(

)(

)(

)(

11ln

11ln

BM

BMLB

SB

AM

AMLA

SA

TTRH

XX

TTRH

XX

Solution 1.2

106

Temp

1400

1500

1600

1700

)(

)( 11lnBM

BMLB

SB

TTRH

XX

)(

)( 11lnAM

AMLA

SA

TTRH

XX

107

Temp

1400

1500

1600

1700

)(

)( 11lnBM

BMLB

SB

TTRH

XX

)(

)( 11lnAM

AMLA

SA

TTRH

XX

0357.01356

11400

1314.8

12790ln

LA

SA

X

X

1089.01356

11500

1314.8

12790ln

LA

SA

X

X

2296.01356

11700

1314.8

12790ln

LA

SA

X

X

2797.01728

11400

1314.8

17154ln

LB

SB

XX

1730.01356

11600

1314.8

12790ln

LA

SA

X

X

0197.01728

11700

1314.8

17154ln

LB

SB

XX

1815.01728

11500

1314.8

17154ln

LB

SB

XX

0955.01728

11600

1314.8

17154ln

LB

SB

XX

108

Temp

1400

1500

1600

1700

SA

SB XX 1 1 L

A

SA

XXk 1

12

2

kkkX L

A

2 LB

SB

XXk .1

LA

SA XkX L

ALB XX 1

28

109

Temp

1400 0.965 1.323 0.870 0.902 0.130 0.098

1500 0.897 1.199 0.591 0.659 0.409 0.341

1600 0.841 1.100 0.325 0.386 0.675 0.614

1700 0.795 1.020 0.071 0.089 0.929 0.911

SA

SB XX 1 1 L

A

SA

XXk 1

12

2

kkkX L

A

2 LB

SB

XXk .1

LA

SA XkX L

ALB XX 1

110

Temperature/K

1728 K

Ni composition

solid

liquid

Cu Ni0.2 0.4 0.6 0.8 1.0

1300

1400

1500

1600

1700

1200

1800

1358 K

111

Exercise 1.3

The compound Ca2B2O5-CaSiO3 form a simple eutectic system with no solid solubility. Using the following data and assuming the liquid solutions to be ideal, calculate the phase boundaries.

Ca2B2O5 CaSiO3

Melting point, K 1583 1813

Heat of fusion (J/mole) 100,834 56,066

(A) (B)

112

)(

)(

)(

)(

11ln

11ln

BM

BMLB

AM

AMLA

TTRH

X

TTRH

X

Solution 1.3:

29

113

Temp Temp

1300 1400

1350 1500

1400 1600

1450 1700

1500 1800

1550

)(

)( 11lnAM

AMLA TTR

HX

)(

)( 11lnBM

BMLB TTR

HX

114

Temp Temp

1300 1400

1350 1500

1400 1600

1450 1700

1500 1800

1550

)(

)( 11lnAM

AMLA TTR

HX

189.01583

11300

1314.8

100834ln

L

AX

267.01583

11350

1314.8

100834ln

L

AX

495.01583

11450

1314.8

100834ln

L

AX

367.01583

11400

1314.8

100834ln

L

AX

)(

)( 11lnBM

BMLB TTR

HX

334.01813

11400

1314.8

56066ln

L

BX

654.01583

11500

1314.8

100834ln

L

AX

849.01583

11550

1314.8

100834ln

L

AX

460.01813

11500

1314.8

56066ln

L

BX

609.01813

11600

1314.8

56066ln

L

BX

781.01813

11700

1314.8

56066ln

L

BX

973.01813

11800

1314.8

56066ln

L

BX

115

Temperature /K

1813 K

CaSiO3 compositionCa2B2O5 CaSiO3

0.2 0.4 0.6 0.8 1.0

1300

1400

1500

1600

1700

1200

1800

1583 K

1900

The phase boundary exist at 1480K with Ca2B2O5 = 58% and CaSiO3= 42%

Below this point no liquid phase can exist