Thermodynamic of Phase Diagram

Thermodynamic of Phase Diagram
by Dr. Srimala

Content 1.0 Introduction 2.0 Thermodynamically stable phase
3.0 Unary Heterogeneous Systems 3.1 P - T Diagram -Unary, Single Component Phase Diagram 3.2 logP – 1/T Diagram -Unary, Single Component Phase Diagram 3.3 Conclusion-Unary p - T Diagrams 3.4 G-T Phase Diagrams G - T Diagram - Unary, Single Component Phase Diagram – V T Diagram - Unary, Single Component Phase Diagram - L, V T Diagram - Unary, Single Component Phase Diagram-a, L,V T Diagram - Single Component Phase Diagram -a, b, L, V

3.5 Metastability 3.6 Chemical Potential and Gibbs Free Energy of Single Component Phases 3.7 Enthalpy & Entropy of Transformation 3.8 Compute Phase Equilibria from Free Energy Relations 4.0 Binary System 4.1 Binary liquid system 4.2 Binary solutions with total solid solubility 4.3 Binary systems without solid solution 5.0 Free Energy-Composition (G-X) Diagram 5.1 Free energy diagrams of total solubility systems 5.2 Free energy diagram for binary solutions with a miscibility gap 5.3 Free energy diagram of binary systems without solid solution (eutectic system) 6.0 Phase boundary Calculations

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.

2.0 Thermodynamically Stable Phases
Thermodynamics can be used to predict weather the system is in equilibrium and to analyze the phase stability and phase transformations. 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

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.

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

Cont… Many element exhibit more than one phase form in solid state-called Allotropes Example : 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. 1 P - T Diagram -Unary, Single Component Phase Diagram
P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + L a + V Show regions of stability of phases in terms of the state variables T & P. a + L + V

Cont.. a, solid crystal L, liquid a + L V + L a + L + V V, vapor a + V
Consider phase stability and phase transformation at a constant pressure, Po, eg. one-atm. P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + L a + V a + L + V Po TM TV

Single phase (a, L, V) stability regions are 2-D, an area.
Cont….. Can change temperature and pressure independently and remain in region. P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + L a + V a + L + V Single phase (a, L, V) stability regions are 2-D, an area.

Two phase (a+L, L+V, a+V) stability regions are 1-D, a line.
Cont… Can pick temperature or pressure (not both) and the other is fixed by the phase boundary. P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + L a + V a + L + V Two phase (a+L, L+V, a+V) stability regions are 1-D, a line.

Cont….. L, liquid a, solid crystal a + L V + L a + L + V V, vapor
Cannot pick temperature or pressure. Both are fixed by the triple point. P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + V a + L a + L + V Three phase (a+L+V) stability regions are 0-D, a point. Note: More than three phases cannot exist at equilibrium for a unary system.

Cont…. L, liquid a + L a, solid crystal V + L a + L + V V, vapor a + V
Metastable extensions of two-phase stability lines extrapolate into opposite single phase regions. P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + L a + V a + L + V

Cont…. C.P. X L, liquid a + L a, solid crystal V + L a + L + V
Note the critical point where the properties of the liquid and vapor phases merge. P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + L a + V a + L + V C.P. X

3.2 logP – 1/T Diagram -Unary, Single Component Phase Diagram
1/T (K-1) P (atm) T (K) a, solid crystal V, vapor L, liquid V + L a + L a + V a + L + V On a plot of logP vs. 1/T the lines for two-phase equilibrium become (approx.) straight.

Cont…. a + L L, liquid V + L a, solid crystal a + L + V V, vapor a + V
Again, consider stability regions for constant pressure conditions. T (K) a + L L, liquid a, solid crystal V + L Log P TMa TV P (atm) 1 atm a + L + V V, vapor a + V 1/T (K-1)

Cont… L, liquid a + L V + L a, solid crystal a + L + V V, vapor a + V
Again, note metastable extensions of two-phase equilibrium boundaries. T (K) L, liquid a + L V + L a, solid crystal P (atm) Log P 1 atm a + L + V V, vapor a + V 1/T (K-1)

Cont…. L, liquid a + L V + L a, solid crystal a + L + V V, vapor a + V
Slopes of lines for two-phase equilibria are inversely proportional to magnitude of volume expansion on heating. T (K) L, liquid a + L V + L a, solid crystal P (atm) Log P 1 atm a + L + V V, vapor a + V 1/T (K-1)

Recall Gibbs Phase Rule
For a system at equilibrium, F = C - P + 2 C = the number of components (1 so far in this chapter) P = the number of phases present F = 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)

Cont …… For a one-component system the phase rule becomes F = 3 - P Components Phases Degrees of Freedom 1 2 2 1 2 1 1 2 3

3.3 Conclusion-Unary p - T Diagrams
Single phase regions are represented by an area: 2-D & 2 degrees of freedom. Two phase equilibria are represented by a line: 1-D & 1 degree of freedom. 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 at equilibrium. Properties of the liquid and vapor phases merge at the critical point.

Cont… Two phase equilibrium boundaries extend into stable single phase regions & indicate potential metastable equilibria. Slopes of two-phase equilibrium lines increase as the volume expansion on heating decreases. Slopes of two-phase equilibrium lines are positive for volume expansions on heating and slopes are negative for volume contractions. Two-phase equilibrium boundaries are approximately straight on log P vs. I/T diagrams.

3.4 G-T Phase Diagrams ! 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. Gibbs energy, G ! Temperature

Change in Entropy Relative Entropy Example:
Ssteam > Sliquid water > Sice Third Law Entropies: All crystals become increasingly ordered as absolute zero is approached (0K = °C) and at 0K all atoms are fixed in space so that entropy is zero.

3.4.1 G - T Diagram - Unary, Single Component Phase Diagram - V
Fixed P -SV G (T) - GO (J/mole) V T (K)

3.4.2 G - T Diagram - Unary, Single Component Phase Diagram - L, V
Fixed P -SV G (T) - GO (J/mole) L -SL L L V T (K)

Cont…. V/ Fixed P G (T) - GO (J/mole) L/ L L/ V TV T (K)

3.4.3 G - T Diagram - Unary, Single Component Phase Diagram - a, L, V
Fixed P -SV V a G (T) - GO (J/mole) -Sa L -SL T (K)

Cont… V/ Fixed P G (T) - GO (J/mole) L/ a a/ a/ L L/ V TV T (K)

Cont… V/ Fixed P G (T) - GO (J/mole) L/ a a/ a/ L L/ V T (K) TV

3.4.4 G - T Diagram - Single Component Phase Diagram - a, b, L, V
Fixed P G (T) - GO (J/mole) L/ b/ a/ a b a/ L b/ L/ V TV T (K)

Cont… Fixed P Equilibrium Phases and Transformations, Only
G (T) - GO (J/mole) a b L V TV T (K)

Exercise 1.1 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.

Solution 1.1 L, liquid a + L a, solid crystal V + L a + L + V V, vapor
P - T Diagram based on the question L, liquid a + L a, solid crystal V + L P (atm) P>Ptriple point P=Ptriple point P<Ptriple point a + L + V V, vapor a + V T (K)

Solution 1.1 P (triple pt.) V G (T) - GO (J/mole) Triple Point: a+L+V
T (K)

Solution 1.1 P > P (triple pt.) V Melting Point: a+L
Boiling Point: L+V G (T) - GO (J/mole) L a a L a L V T (K)

Solution 1.1 P < P (triple pt.) V G (T) - GO (J/mole)
Sublimation Point: a + V G (T) - GO (J/mole) L a a L V T (K)

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.

Exercise 1.2 The standard Gibbs energy of formation of metallic white tin (a-tin) is 0 at 25 oC and that of nonmetallic gray tin (b-tin) is 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 a- (white) tin at 25 oC.

Exercise 1.3 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

Solution 1.3: L/ G (T) - GO (J/mole) / Tt a a/ a/  / L Tm T (K)

Entropy of  > Entropy of 
Exercise 1.4 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 

Solution 1.4  a G 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 a phase. In order for  phase to appear at higher temperatures the free energy of  should fall rapidly than the a as the temperature is raised . In other words, the free energy curve for  is steeper than the a. this means the entropy of  should be larger than a  a G T (K)

3. 6. Chemical Potential and Gibbs Free
3.6 Chemical Potential and Gibbs Free Energy of Single Component Phases Combined 1st & 2nd law with extensive functions: By definition (recall gibbs free energy function):

Extensive enthalpy: Substitute for dU/ in enthalpy expression: Extensive Gibbs free energy: Substitute for dH/ in free energy expression:

For constant T & P (set dT=0, dP=0, & rearrange):
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 Yielding: In a unary system the chemical potential of a phase is equal to its molar Gibbs free energy at that T & P.

3.7 Enthalpy & Entropy of Transformation
At equilibrium --- G=H -TS and Thus,

3.8 Compute Phase Equilibria from Free Energy Relations

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 T

Similar argument for other phase in the system For liquid,
Fig 7.3b Fig 7.3c T Similar argument for other phase in the system For 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 the solid 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.

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. Ts = TL Ps= PL s= L Fig 7.3c T

Intersection of solid and gas (COD)-represents
Fig 7.4 shows the construction when the chemical potential surface for the gas phase is added. Intersection of solid and gas (COD)-represents the (s+G) two phase equilibrium C’O’D’ on the (P, T) plane- sublimation curve Intersection of liquid and gas (EOF)-represents the (L+G) equilibrium E’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

4.0 Binary systems Binary liquid system
Binary system without solid solution Binary system with solid solution

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 !

Partially Immiscible Species A Species B Immiscible Total miscibility

We may regard this mixture as
Immiscible Mixtures A and B in gas Pure A Pure B These are equivalent = Gas We 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 PB Total pressure= PA + PB A (droplets) + (continuous) in separate phases

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.

4.2 Binary solutions with total solid solubility
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 Azeotropes, high-boiling Azeotropes, low-boiling

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 a2N. The distillate is richer in the more volatile component. Successive distillations will separate the pure components.

Fractional Distillations
High efficient 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. boiling condensation heating 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.

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.

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). T2 T1 T4 T3 tie line solid liquid 1 Xl Xs =TmB T TmA XB

XB T1 solid liquid 1 G A Xl T2 T1 T4 T3 tie line solid liquid 1 Xl Xs =TmB T TmA B XB 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.

T2 T1 T4 T3 tie line solid liquid 1 Xl Xs T2 solid liquid 1 G XB B =TmB =TmB T TmA XB 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.

G T3 solid liquid 1 Xs Xl liquid + solid T2 T1 T4 T3 tie line solid liquid 1 Xl Xs =TmB T B TmA A XB ! 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.

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.

T2 T1 T4 T3 tie line solid liquid 1 Xl Xs XB T4 solid liquid 1 G =TmB T TmA XB At T4, the solid phase everywhere has a lower free energy than does the liquid, and is the stable phase at all compositions.

5.1 Free energy diagrams of total solubility systems
Based on the Gibbs free energy curves we can construct a phase diagram for a binary isomorphous systems Example : 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. .

We have 2 phases – liquid and solid
We 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 GBliquid will increase more rapidly than GAsolid and GBsolid The curvature of the G(XB) curves will decrease. Eventually we will reach T2 – melting point of pure component A, where GAliquid = 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 GBliquid continue to increase more rapidly than GAsolid 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

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.

5.3 Free energy diagram of 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:

Eutectic phase diagram with different crystal structures of pure phases

Cont…

Eutectic phase diagram with same crystal structures of pure phases

Cont…

6.0 Phase boundary Calculations
XAL L Tm (A) XAs  XBS A B XB For the phase equilibrium of  and liquid as shown in the figure, we have  with composition XAS in equilibrium with liquid of composition XAL .

Cont… At equilibrium, partial molar Gibbs free energy for each component is the same in both the phase represent free energy change in melting (fusion). with ref to pure solid A with ref to pure liquid A Free energy change on fusion at ToK Free energy axis Note :Gi=Gio + RTln ai

DS = DH/T Using this we have

And if the solutions are assumed ideal
Similarly for the component B !

Definition of ideal & non ideal
Species A Species B Ideal Mixture: g = 1 + Non-Ideal Mixture of A and B: g > 1 A-B Repulsive Interactions Exaggerated

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

As per equations derived before,
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, Tm (A) A B XB XAL XAs XBL Tm (B)

cont…. 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

Exercise 1.1 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 J/mol 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?

Tm (pure Fe) =1539+273=1812K, Tm (with Si)=1527+273=1800K
Solution (1.1a) Tm (pure Fe) = =1812K, Tm (with Si)= =1800K

Solution (1.1b)

Exercise 1.2 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

Solution 1.2

Temp 1400 1500 1600 1700

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

liquid solid 1800 1700 1600 1500 Temperature/K 1400 1300 1200
Cu Ni Ni composition

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)

Solution 1.3:

Temp 1300 1400 1350 1500 1600 1450 1700 1800 1550

The phase boundary exist at 1480K with Ca2B2O5 = 58% and CaSiO3= 42%
Temperature /K 1813 K CaSiO3 composition Ca2B2O5 CaSiO3 1300 1400 1500 1600 1700 1200 1800 1583 K 1900 Below this point no liquid phase can exist The phase boundary exist at 1480K with Ca2B2O5 = 58% and CaSiO3= 42%

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