Phase Transformation of Materials

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1 2009 fall Phase Transformation of Materials Eun Soo Park Office: Telephone: Office hours: by an appointment 1

2 Contents for previous class Chapter 1 - Equilibrium Thermondynamics and Phase Diagrams dg = 0 Lowest possible value of G o desire to change ad infinitum - Phase Transformation Δ G = G2 G1 < 0 - Single component system Gibbs Free Energy as a Function of Temp. and Pressure G T P = S, G P T = V ( dp dt ) eq ΔH = T ΔV eq Clausius-Clapeyron Relation : - Driving force for solidification ΔG = LΔT T m - Classification of phase transition First order transition: CDD/Second order transition: CCD 2

3 Contents for today s class - inary System - Gibbs Free Energy in inary System Ideal solution and Regular solution - Chemical potential and ctivity 3

4 * Single component system One element (l, Fe), One type of molecule (H 2 O) : 평형상태압력과온도에의해결정됨 * inary System (two component), : 평형상태압력과온도이외에도조성의변화를고려 - Mixture ;, ; 각각의성질유지, boundary 는존재, 섞이지않고기계적혼합 - Solution ; ; atomic scale 로섞여있다. Random distribution Solid solution : substitutional or interstitial - compound ; ;, 의위치가정해짐, Ordered state 4

5 - Mixture ;, ; 각각의성질유지, boundary 는존재, 섞이지않고기계적혼합 5

6 - Solution ; ; atomic scale 로섞여있다. Random distribution Solid solution : substitutional or interstitial 6

7 - Compound ; ;, 의위치가정해짐, Ordered state MgCi3 7

8 lloying: atoms mixed on a lattice partial or complete solid solution Solid Solution vs. Intermetallic Compounds Crystal structure Pt 0.5 Ru 0.5 Pt structure (fcc) Surface PtPb is structure ruthenium 루테늄 백금류의금속원소 ; 기호 Ru, 번호 44 8

9 Solid Solution vs. Intermetallic Compounds Pt 0.5 Ru 0.5 Pt structure (fcc) PbPt is structure 9

10 Solid solution: Crystalline solid Multicomponent yet homogeneous Impurities are randomly distributed throughout the lattice Factors favoring solubility of in (Hume-Rothery Rules) Similar atomic size: r/r 15% Same crystal structure for and Similar electronegativities: χ χ 0.6 (preferably 0.4) Similar valence If all four criteria are met: complete solid solution If any criterion is not met: limited solid solution 10

11 complete solid solution limited solid solution inary Solutions : 합금에대한열역학기본개념도입위해 2원고용체, 1기압의고정된압력조건11고려

12 Gibbs Free Energy of inary Solutions * Composition in mole fraction, + = 1 1. bring together mole of pure and mole of pure 2. allow the and atoms to mix together to make a homogeneous solid solution. 12

13 Gibbs Free Energy of The System In Step 1 - The molar free energies of pure and pure pure ; pure ; G ( T, P) ( T, P) G G1 = G + G J/mol ;, (mole fraction) Free energy of mixture 13

14 Gibbs Free Energy of The System In Step 2 G 2 = G 1 + ΔG mix J/mol Since G 1 = H 1 TS 1 and G 2 = H 2 TS 2 nd putting H mix = H 2 - H 1 S mix = S 2 -S 1 G mix = H mix -T S mix H mix : Heat of Solution i.e. heat absorbed or evolved during step 2 S mix : difference in entropy between the mixed and unmixed state. How can you estimate ΔH mix and ΔS mix? 14

15 - Ideal solution 가정1 ; H mix =0 : Mixing free energy ΔG mix ; 와 가 complete solid solution (, ; same crystal structure) ; no volume change Entropy can be computed from randomness by oltzmann equation, i.e., G mix = H mix -T S mix ΔG mix = -TΔS mix J/mol S = k ln w w : degree of randomness, k: oltzman constant thermal; vibration ( no volume change ) Configuration; atom 의배열방법수 ( distinguishable ) S = S + S th config 15

16 Ideal solution Excess mixing Entropy If there is no volume change or heat change, w w config config = 1 before _ solution _( pure _ pure) ( = + )!!! after _ solution _(, ) umber of distinguishable way of atomic arrangement ΔS mix = S after S before = k ln ( +! )! k! ln1 = 0, = using Stirling s approximation and 0, + = ln! ln R = k 0 0 이므로 = k[( o ln o o ) ( o ln o o ) ( o ln o o )] 16

17 Ideal solution Excess mixing Entropy mix ΔS = R( ln + ln ) ΔG mix = -TΔS mix mix ΔG = RT ( ln + ln ) 17

18 Ideal solution Since ΔH = 0 for ideal solution, G 2 = G 1 + ΔG mix Compare G solution between high and low Temp. G 1 ΔG mix 18

19 1) Ideal solution G = H-TS = E+PV-TS Chemical potential The increase of the total free energy of the system by the increase of very small quantity of, dn, will be proportional to dn. 소량첨가에의한내부에너지변화계산 dg' =μ dn (T, P, n : constant ) μ : partial molar free energy of or chemical potential of G' μ = n T, P, n G' μ = n T, P, n For - binary solution, dg' =μ dn +μdn For variable T and P dg' = SdT + VdP +μ dn +μdn 19

20 Chemical potential 과 Free E 와의관계 1) Ideal solution For 1 mole of the solution (T, P: constant ) G =μ +μ Jmol dg = μ d +μd 1 dg d = μ μ μ =μ dg d dg G= μ +μ dg =μ +μ d d dg =μ d dg =μ d μ = G+ ( 1 ) dg d 20

21 Chemical potential = ( G + RT ln ) + ( G + RT ln ) μ μ 21

22 Regular Solutions Ideal solution : ΔH mix = 0 Quasi-chemical model assumes that heat of mixing, ΔH mix, is only due to the bond energies between adjacent atoms. Structure model of a binary solution G mix = H mix -T S mix 22

23 Regular Solutions ond energy umber of bond - ε P - ε P - ε P Internal energy of the solution E= Pε + Pε + Pε 1 Δ Hmix = Pε where ε = ε ( ε +ε) 2 23

24 Δ Hmix = 0 Completely random arrangement ideal solution P = az bonds per mole a : vogadro's number z : number of bonds per atom Regular Solutions 1 ε = 0 ε = ( ε + ε ) 2 ε < 0 P ε 0 H mix = Ω where ε > 0 P Δ H = P ε mix Ω = a zε Ω >0 인경우 24

25 Regular Solutions G 2 = G 1 + ΔG mix G = G + G + Ω + RT ( ln + ln ) Reference state 0 0 Pure metal G G = 0 = H mix -T S mix G mix = H mix -T S mix 25

26 Phase separation in metallic glasses 26

27 Chemical potential 과 Free E 와의관계 2) regular solution For 1 mole of the solution (T, P: constant ) G = E+PV-TS G = H-TS G=μ +μ Jmol G = G + G + Ω + RT ( ln + ln ) = = μ ( G + Ω(1 ) + RT ln ) + μ ( G + Ω(1 ) + RT ln ) μ μ = = G G + + Ω Ω (1 (1 ) ) RT RT ln ln μ = G + RTln μ = G + RTln Ideal solution 복잡해졌네 --;; 27

28 ctivity, a : mass action 을위해 effective concentration ideal solution regular solution μ = G + RTln μ = G + RTln μ = G + RTlna μ = G + RTlna 2 μ = G + Ω (1 ) + RT ln μ = G a ln( + Ω (1 ) 2 Ω ) = (1 RT a γ = + RT 2 ) ln 28

29 Solution 에서 a 와 와의관계 조성따른 activity 변화 a a Line 1 : (a) a =, (b) a = ideal solution Rault s law Line 2 : (a) a <, (b) a < ΔH mix <0 Line 3 : (a) a >, (b) a > ΔH mix >0 For a dilute solution of in ( 0) a γ = constant a γ = (Henry's Law) 1 (Rault's Law) 29

30 degree of non-ideality? a a =γ, a =γ γ : activity coefficient a Ω ln = (1 ) RT μ = G + RTlna ctivity 는 solution 의상태를나타내는 조성과 Chemical potential 과상관관계가짐. Chemical Equilibrium (μ, a) multiphase and multicomponent (μ iα = μ iβ = μ iγ = ), (a iα = a iβ = a iγ = ) 30

31 Contents for today s class - inary System mixture/ solution / compound - Gibbs Free Energy in inary System G 1 = G + G J/mol G 2 = G 1 + ΔG mix J/mol Ideal solution ( H mix =0) Regular solution mix ΔG = RT ( ln + ln ) 1 Δ Hmix = Pε where ε = ε ( ε +ε) 2 G = G + G + Ω + RT( ln + ln ) - Chemical potential and ctivity G' μ = n T, P, n μ = G + RTlna 31

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