Lecture 3: Solutions: Activities and. Phase Diagrams

Transcription

1 Lecture 3: Solutions: Activities and Lecture plan: Phase Diagrams Gibbs phase rule vapour composition two-component phase diagrams phase diagrams in material science: microstructures in isomorphous binary systems microstructures in eutectic alloys liquid crystals problems

2 Phase diagrams what is the composition (number of phases and their amount and composition) at equilibrium at a given temperature; what happens to the system when is cools down/heats up we can predict the structure and the properties of the system at low temperature. we can understand development and preservation of non-equilibrium structures design materials of required properties iron-carbon diagram

3 Phase diagrams water-surfactant-oil That s the base of all modern engineering from swiss knife to food and cosmetics! iron-carbon diagram

4 Phase diagrams Constituent a chemical species that is present Component a chemically independent constituent of the system (i.e. not connected by a chemical reaction) CaCO3() s CaO() s + CO2( g) Phase1 Phase2 Phase3 C = 2 Variance the number of intensive variables that can be changed independently without disturbing the number of phases at equilibrium. Phase rule (J.W. Gibbs): F=C-P+2 variance number of phases number of components Indeed: number of variables would be: number of equations: P*(C-1)+2 C*(P-1)

5 One component diagrams C=1 therefore F=C-P+2=3-P

6 One component diagrams Detection of phase transitions and building a phase diagram is based on calorimetry measurements

7 Two-components diagrams C=2 therefore F=4-P. We have to reduce degree of freedom e.g. by fixing T=const Vapour pressure diagrams Raoult s Law p = x p p = x p * * A A A B B B p = p + p = p + x ( p p ) * * * A B B A A B

8 Two-components diagrams The composition of vapour pa From Dalton s law: ya = ; yb = p From Raoult s law: p p p = x p *; p = x p * A A A B B B B pa * y = ; y = 1 y p * + ( p * p *) x A B A B A B A p p * A * B

9 Two components diagrams

10 Two components diagrams

11 Two components diagrams Relative amount and the composition of phases in equilibrium can be found on the phase diagram The lever rule nl = nl β β

12 Two-components diagrams Temperature-composition diagrams Fractional distillation Distillation of mixtures number of theoretical plates

13 Two-components diagrams Temperature-composition diagrams Azeotropes Azeotrope, evaporation w/o change in composition nl A-B interacation stabilize the mixture A-B interacation destabilize the mixture

14 Two components diagrams Immiscible liquids Will boil at lower temperature! boiling condition p = p + p = 1atm A B can be used for steam distillation of heat sensitive components

15 Two components diagrams Liquid-liquid phase diagrams: partially miscible liquids hexane nitrobenzene

16 Two components diagrams Δ G = nrt( κ lnκ + κ ln κ + βκ κ ) mix A A B B A B G κ Δ mix = 0 ln + β(1 2 κ) = 0 κ 1 κ Upper critical solution T

17 Two components diagrams Lower critical temperature is usually caused by breaking a weak complex of two components

18 Two components diagrams Upper critical temperature is less than the boiling point Boiling occur before liquids are fully miscible

19 Eutectic composition Liquid-solid phase diagrams

20 Liquid-solid phase diagrams Eutectic halt

21 Liquid-solid phase diagrams Reacting systems peritectic line: 3 phases are in equilibrium Incongruent melting: compounds melts into components

22 Phase diagrams and Microstructure

23 Binary phase diagrams Phase diagram with total solubility in both liquid and solid state: isomorphous system T( C) homogeneous liquid solution of Cu and Ni. L (liquid) liquidus L + (FCC solid solution) solidus phases: L (liquid) (FCC solid solution) 3 phase fields: L L + homogeneous solid solution of Cu and Ni. wt% Ni Cu-Ni phase diagram

24 C 0 =35 wt% Ni Cu-Ni phase diagram at T A : Only liquid, composition of liquid is given by the overall composition (C 0 =35 wt% Ni) at T D : Only liquid, composition of liquid is given by the overall composition (C 0 =35 wt% Ni) at T B : Both L and are present Composition at T B : Liquid phase (L) of 32% Ni Solid phase () of 43% Ni Weight ratio: Information we can extract from the diagram: the phases present; composition of the phases percentage of fraction of the phases T( C) TA 1300 TB 1200 TD WL S R (43 35) = ; W = = = 73% W R S + R (43 32) 20 L (liquid) L + R A B S D 3235 CL C o tie line liquidus L + solidus (solid) 43 C wt% Ni

25 Development of microstructure in a Cu-Ni alloy Equilibrium case (very slow cooling)

26 Development of microstructure in a Cu-Ni alloy Non-Equilibrium case (real) Fast cooling: Cored structure Slow cooling: Equilibrium structure First to solidfy: 46wt%Ni Last to solidfy: < 35wt%Ni Uniform C: 35wt%Ni How we can prevent coring and get equilibrium structure?

27 Binary Eutectic Systems: Sn-Pb Sn-Pb system: limited solubility in solid state 3 single phase regions (L, a, b); T E =183 0 C, no liquid below T E. Eutectic composition 61.9% At the eutectic temperature: LC ( ) ( C ) + β ( C ) E E βe T( C) L R L (liquid) + β L+β 183 C S β 0 Pb For a 40wt%Sn-60wt%Pb alloy at 150C, find... --the compositions of the phases: Ca = 11wt%Sn Cb = 99wt%Sn Co Sn Co, wt% Sn W = = 67wt % W β = = 33 wt %

28 Microstructures in binary systems C o < 2wt%Sn Result: --polycrystal of grains. T( C) L: Cowt%Sn L L 200 TE : Cowt%Sn L + (Pb-Sn System) β Co 2 (room T solubility limit) 30 Co, wt% Sn

29 Microstructures in binary systems 400 T( C) L: Cowt%Sn 2wt%Sn < C o < 18.3wt%Sn Result: -- polycrystal with fine β crystals TE 100 L L + + β : Cowt%Sn β L Co (sol. limit at Troom) 18.3 (sol. limit at TE) 30 Co, wt% Sn

30 Microstructures in binary systems Eutectic composition

31 Microstructures in binary systems: eutectic and around T( C) 300 L 200 TE 100 L + + β L + β β (Pb-Sn System) hypoeutectic: Co=50wt%Sn Co Co hypoeutectic hypereutectic 60 eutectic Co, wt% Sn hypereutectic: (illustration only) 175μm eutectic: Co=61.9wt%Sn 160μm β β β β β β eutectic micro-constituent From: W.D. Callister, Materials Science and Engineering: An Introduction, 6e.

32 IRON-CARBON (Fe-C) PHASE DIAGRAM 2 important points -Eutectic (A): L γ + Fe 3 C -Eutectoid (B): γ +Fe 3 C 1600 δ T( C) γ γ+l (austenite) +γ R B γ γ γ γ 1148 C R L 727 C = Teutectoid A γ+fe3c S +Fe3C L+Fe3C S Fe3C (cementite) 120μm Result: Pearlite = alternating layers of and Fe3C phases. (Adapted from Fig. 9.24, Callister 6e. (Fig from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) (Fe) Co, wt% C Fe3C (cementite-hard) (ferrite-soft) Ceutectoid Adapted from Fig. 9.21,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) 21

33 HYPOEUTECTOID STEEL γ γ γ γ γ γ γ γ γ γ γ γ w =s/(r+s) wγ =(1-w) 1600 δ w =S/(R+S) w Fe3C =(1-w) T( C) 600 γ γ+l (austenite) r s R S Co 0.77 pearlite w pearlite = wγ 727 C L 1148 C γ+fe3c +Fe3C L+Fe3C Fe3C (cementite) Co, wt% C 100μm (Fe-C System) Hypoeutectoid steel Adapted from Figs and 9.26,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) Adapted from Fig. 9.27,Callister 6e. (Fig courtesy Republic Steel Corporation.) 22

34 Fe3C γ γ γ γ γ γ γ γ γ γ γ γ 1600 δ T( C) γ γ+l (austenite) R w Fe3C =r/(r+s) 600 wγ =(1-w Fe3C ) +Fe3C w =S/(R+S) w Fe3C =(1-w) HYPEREUTECTOID STEEL 0.77 pearlite w pearlite = wγ r Co s S L 1148 C γ+fe3c L+Fe3C Adapted from Fig. 9.30,Callister 6e. (Fig copyright 1971 by United States Steel Corporation.) Fe3C (cementite) Co, wt% C (Fe-C System) 60μm Hypereutectoid steel Adapted from Figs and 9.29,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) 23

35 Liquid crystals Mesophase an intermedediate phase between solid and liquid. Example: liquid crystal Liquid crystal substance having a liquid-like imperfect order in at least one direction and longrange positional or orientational order in at least one another direction Nematic Smectic calamitic (rod-like) discotic Cholesteric

36 Nematic crystals in LCD

37 Problems (to solve in the class) 6.1a: At 90ºC the vapour pressure of methylbenzene is 53.3kPa and that of 1.2-dimethylbenzene is 20kPa. What is the composition of a liquid mixture that boils at 90ºC when the pressure is 0.5 atm. What is the composition of the vapour produced. down 6.9b: sketch the phase diagram of the system NH 3 /N 2 H 4 given that the two substances do not form a compound and NH 3 freezes at -78C, N 2 H 4 freezes at +2C, eutectic formed with mole fraction of N 2 H and melts at -80C. 6.10b Describe the diagram and what is observed when a and b are cooled down