Lecture 3: Solutions: Activities and Phase Diagrams 21-09-2010 • Lecture plan: – 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 Phase diagrams ¾what is the composition (number of phases and their amount and composition) at equilibrium at a given temperature; ¾what happens to the system when is cools down/heats up ¾we can predict the structure and the properties of the system at low temperature. ¾we can understand development and preservation of non-equilibrium structures ¾design materials of required properties iron-carbon diagram Phase diagrams water-surfactant-oil That’s the base of all modern engineering from swiss knife to food and cosmetics! iron-carbon diagram 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) ⎯⎯→ CaCO32() s←⎯⎯ CaO () s+ CO (g ) C = 2 Phase1 Phase2 Phase3 • 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 number of phases variance number of components Indeed: number of variables would be: P*(C-1)+2 number of equations: C*(P-1) One component diagrams C=1 therefore F=C-P+2=3-P One component diagrams Detection of phase transitions and building a phase diagram is based on calorimetry measurements 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 ** pA ==xpAA p B xp BB *** p =+=pppAB B + xpp AAB() − Two-components diagrams • The composition of vapour pp yy==AB; From Dalton’s law: ABpp From Raoult’s law: pxppxpAAA==*; BBB * pA * * yyyA = ;1BA=− pA pppxBABA*(+− * *) * pB Two components diagrams Two components diagrams Two components diagrams Relative amount and the composition of phases in equilibrium can be found on the phase diagram The lever rule nlα αββ= nl Two-components diagrams Temperature-composition diagrams Fractional distillation • Distillation of mixtures number of theoretical plates Two-components diagrams Temperature-composition diagrams •Azeotropes Azeotrope, evaporation w/o change in composition nlα α A-B interacation stabilize the mixture A-B interacation destabilize the mixture Two components diagrams • Immiscible liquids Will boil at lower temperature! boiling condition p =+=pAB p1 atm • can be used for steam distillation of heat sensitive components Two components diagrams • Liquid-liquid phase diagrams: partially miscible liquids hexane nitrobenzene Two components diagrams ΔmixGnRT=++(lnκ Aκκ A B ln κβκκ B A B ) ∂ κ Δ G =+−=0lnβκ (12)0 ∂κ mix 1−κ Upper critical solution T Two components diagrams Lower critical temperature is usually caused by breaking a weak complex of two components Two components diagrams Upper critical temperature is less Boiling occur before liquids are fully than the boiling point miscible Liquid-solid phase diagrams Eutectic composition Liquid-solid phase diagrams Eutectic halt Liquid-solid phase diagrams • Reacting systems peritectic line: 3 phases are in equilibrium Incongruent melting: compounds melts into components Phase diagrams and Microstructure 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. 1600 • 2 phases: L (liquid) 1500 L (liquid) α (FCC solid solution) 1400 • 3 phase fields: s du L ui α 1300 liq + us L + α L id ol 1200 s α α 1100 (FCC solid solution) homogeneous solid solution of Cu and Ni. 1000 0 20 40 60 80 100 wt% Ni Cu-Ni phase diagram Cu-Ni phase diagram Information we can extract from the diagram: ¾the phases present; ¾composition of the phases ¾percentage of fraction of the phases C =35 wt% Ni 0 T(°C) at TA: Only liquid, composition of liquid A T tie line s is given by the overall composition A idu liqu (C0=35 wt% Ni) 1300 L (liquid) at T : Only liquid, composition of liquid D + α is given by the overall composition L s B du (C =35 wt% Ni) oli 0 TB s at TB: Both L and α are present R S α Composition at T : L + α B 1200 D • Liquid phase (L) of 32% Ni TD (solid) • Solid phase (α) of 43% Ni • Weight ratio: 20 303235 4043 50 W SR(43− 35) L C C C ===;73%Wα = L o α WRα SR+−(43 32) wt% Ni Development of microstructure in a Cu-Ni alloy Equilibrium case (very slow cooling) Development of microstructure in a Cu-Ni alloy Non-Equilibrium case (real) • Fast cooling: • Slow cooling: Cored structure Equilibrium structure Uniform C : First α to solidfy: α 46wt%Ni 35wt%Ni Last α to solidfy: < 35wt%Ni How we can prevent coring and get equilibrium structure? Binary Eutectic Systems: Sn-Pb T(°C) Sn-Pb system: 300 ¾limited solubility in solid state L (liquid) ¾3 single phase regions (L, a, b); ¾T =183 0C, no liquid below T . L + α E E 200 α 183°C L+β β ¾Eutectic composition 61.9% 18.3 61.9 97.8 150 At the eutectic temperature: R S 100 α + β LC()EEER α ( Cαβ )+ β ( C ) 0 11 20 40 60 80 99100 PbC Sn o Co, wt% Sn • For a 40wt%Sn-60wt%Pb alloy at 150C, find... --the compositions of 59 the phases: Wα = = 67 wt % Ca = 11wt%Sn 88 29 Cb = 99wt%Sn W = = 33wt % β 88 Microstructures in binary systems T(°C) L: Cowt%Sn 400 • Co < 2wt%Sn L • Result: α --polycrystal of α grains. 300 L L + α α 200 (Pb-Sn TE α: C wt%Sn o System) 100 α + β 0 10 20 30 Co Co, wt% Sn 2 (room T solubility limit) Microstructures in binary systems T(°C) L: Cowt%Sn 400 L • 2wt%Sn < Co < 18.3wt%Sn L • Result: 300 α --α polycrystal with fine L + α β crystals. α: Cowt%Sn α 200 TE α β 100 α + β 0 10 20 30 2 Co Co, wt% Sn (sol. limit at Troom) 18.3 (sol. limit at TE) Microstructures in binary systems • Eutectic composition Microstructures in binary systems: eutectic and around T(°C) 300 L L + α 200 L + β TE α β (Pb-Sn α + β System) 100 Co Co hypoeutectic hypereutectic 0 0 20 40 60 80 100 Co, wt% Sn 18.3 eutectic 97.8 61.9 hypoeutectic: Co=50wt%Sn hypereutectic: (illustration only) eutectic: Co=61.9wt%Sn α β α β α α β β α β α β 175μm 160μm eutectic micro-constituent From: W.D. Callister, “Materials Science and Engineering: An Introduction”, 6e. IRON-CARBON (Fe-C) PHASE DIAGRAM • 2 important T(°C) 1600 points δ -Eutectic (A): 1400 L L ⇒ γ + Fe C 3 γ γ+L A 1200 1148°C L+Fe3C -Eutectoid (B): (austenite) R S γ ⇒α+Fe3C 1000 γ γ γ+Fe3C α γ γ + 800 γ α B 727°C = Teutectoid R S C (cementite) 600 3 α+Fe3C Fe 400 0 1234566.7 (Fe) 0.77 4.30 120μm Co, wt% C Result: Pearlite = Fe3C (cementite-hard) alternating layers of α (ferrite-soft) α and Fe3C phases. (Adapted from Fig. 9.24, Callister 6e. eutectoid Adapted from Fig. 9.21,Callister 6e. (Fig. 9.21 (Fig. 9.24 from Metals Handbook, 9th ed., C adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 9, Metallography and Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM Microstructures, American Society for International, Materials Park, OH, 1990.) Metals, Materials Park, OH, 1985.) 21 HYPOEUTECTOID STEEL T(°C) 1600 δ L 1400 (Fe-C γ γ γ γ+L System) 1200 L+Fe C γ γ (austenite) 1148°C 3 γ γ 1000 Adapted from Figs. γ γ γ+Fe3C 9.21 and 9.26,Callister 6e. (Fig. 9.21 adapted α 800 r from Binary Alloy αγ γ s 727°C α Phase Diagrams, 2nd α ed., Vol. 1, T.B. γ γ R S C (cementite) wα =s/(r+s) 600 3 Massalski (Ed.-in- α+Fe3C Chief), ASM wγ =(1-wα) Fe International, Materials 400 Park, OH, 1990.) α 0 Co 1234566.7 α pearlite 0.77 C , wt% C α o wpearlite = wγ Hypoeutectoid wα =S/(R+S) 100μm steel wFe3C =(1-wα) Adapted from Fig. 9.27,Callister 6e. (Fig. 9.27 courtesy Republic Steel Corporation.) 22 HYPEREUTECTOID STEEL T(°C) 1600 δ L 1400 (Fe-C γ γ γ γ+L System) 1200 L+Fe C γ γ (austenite) 1148°C 3 γ γ 1000 γ γ γ+Fe3C Adapted from Figs. Fe C 9.21 and 9.29,Callister 3 6e. (Fig. 9.21 adapted 800 γ γ r s from Binary Alloy γ γ α Phase Diagrams, 2nd R S C (cementite) ed., Vol. 1, T.B. 600 3 wFe3C =r/(r+s) α+Fe C Massalski (Ed.-in- 3 Chief), ASM wγ =(1-wFe3C) Fe International, Materials 400 Park, OH, 1990.) 0 1234566.7Co pearlite 0.77 Co, wt% C wpearlite = wγ wα =S/(R+S) 60μm Hypereutectoid steel wFe3C =(1-wα) Adapted from Fig. 9.30,Callister 6e. (Fig. 9.30 copyright 1971 by United States Steel Corporation.) 23 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 long- range positional or orientational order in at least one another direction Nematic Smectic calamitic discotic (rod-like) Cholesteric Nematic crystals in LCD 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.