### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Thermodynamics and Kinetics of Materials MSE 3050

UVA

GPA 3.77

### View Full Document

## 42

## 0

## Popular in Course

## Popular in Materials Science Engineering

This 198 page Class Notes was uploaded by Jamison Kirlin on Monday September 21, 2015. The Class Notes belongs to MSE 3050 at University of Virginia taught by Leonid Zhigilei in Fall. Since its upload, it has received 42 views. For similar materials see /class/209594/mse-3050-university-of-virginia in Materials Science Engineering at University of Virginia.

## Popular in Materials Science Engineering

## Reviews for Thermodynamics and Kinetics of Materials

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/21/15

Binary phase diagrams Binary phase diagrams and Gibbs free energy curves Binary solutions With unlimited solubility Relative proportion of phases tie lines and the lever principle Development of microstructure in isomorphous alloys Binary eutectic systems limited solid solubility Solid state reactions eutectoid peritectoid reactions Binary systems With intermediate phasescompounds The ironcarbon system steel and cast iron Gibbs phase rule Temperature dependence of solubility Threecomponent ternary phase diagrams Reading Chapters 151 157 of Porter and Easterling Chapter 10 of Gaskell MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary phase diagram and Gibbs free energy A binary phase diagram is a temperature composition map which indicates the equilibrium phases present at a given temperature and composition The equilibrium state can be found om the Gibbs free energy dependence on temperature and composition We have discussed the G dependence of G of a one component system on T i 3 Slag15 T We have also discussed the G1 dependence of the Gibbs ee energy om composition at a given T GE GXAGA XBGB AHniX TA Sm G 0 XE 1 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary solutions with unlimited solubility 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 Let s consider Gibbs free energy curves for the two phases at different T gt T1 is above the equilibrium melting temperatures of both pure components T1 gt TmA gt TmB gt the liquid phase Will be the stable phase for any composition solid G B solid GA G l1qu1d liquid B G A Gid XAGA XBGB RTXAlnXA XBlnXB MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary solutions with unlimited solubility II Decreasing the temperature below T1 will have two effects 3 Ggqmd and GgqUid will increase more rapidly than Gflid and GE Why 3 The curvature of the GXB curves will decrease Why gt Eventually we will reach T2 melting point of pure component A where Ggqmd G351 solid G B liquid liquid solid GB GA GA MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary solutions with unlimited solubility 111 gt For even lower temperature T3 lt T2 TmA the Gibbs ee energy curves for the liquid and solid phases will cross solid G liquid G B A lid Gso 1 d A 1qu1 G B X1 0 XBXZ 1 As we discussed before the common tangent construction can be used to show that for compositions near crossover of GSOlid and Gliquid the total Gibbs free energy can be minimized by separation into two phases MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary solutions with unlimited solubility IV As temperature decreases below T3 qumd anngqmd continue solid solid to 1ncrease more rap1dly than G A andGB gt Therefore the intersection of the Gibbs free energy curves as well as points X1 and X2 are shifting to the right until at T4 TmB the curves will intersect at X1 X2 1 liquid G A liquid solid G B G B solid GA At T4 and below this temperature the Gibbs ee 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 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary solutions with unlimited solubility V Based on the Gibbs free energy curves we can now construct a phase diagram for a binary isomorphous systems solid G B liquid G A solid GA liquid G B QQ MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary solutions with unlimited solubility VI Example of isomorphous system CuNi the complete solubility occurs because both Cu and Ni have the same crystal structure FCC similar radii electronegativity and valence x I l x 2800 1500 Liquid 1453 c 2500 1400 7 9 39 Q g WM quot9 Solidusline mo g E 1300 E 3 a E i E 3 gt5 1200 e 2200 1100 Solid solution 2000 1085 C 1000 w i I l l i w i w 20 40 60 50 l 0 Cu Composmun w N NI Liquidus line separates liquidfrom liquid solid Solidus line separates solidfrom liquid solid Binary solutions With unlimited solubility VII In onecomponent system melting occurs at a wellde ned melting temperature In multicomponent systems melting occurs over the range of temperatures between the solidus and liquidus lines Solid and liquid phases are in equilibrium in this temperature range E L liquid solution C6 5 cu E a Q O F Solidus 0 0 liquid solution 0 crystallites of solid solution polycrystal 20 4O 6O 80 B solid solution comPOSitiOH Wt MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Interpretation of Phase Diagrams For a given temperature and composition we can use phase diagram to determine 1 The phases that are present 2 Compositions of the phases 3 The relative fractions of the phases Finding the composition in a two phase region 1 Locate composition and temperature in diagram 2 In two phase region draw the tie line or isotherm 3 Note intersection with phase boundaries Read compositions at the intersections The liquid and solid phases have these compositions liquid olid XE XE XE MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Interpretation of Phase Diagrams the Lever Rule Finding the amounts of phases in a two phase region 1 Locate composition and temperature in diagram 2 In two phase region draw the tie line or isotherm 3 Fraction of a phase is determined by taking the length of the tie line to the phase boundary for the other phase and dividing by the total length of tie line a 05 The lever rule is a mechanical analogy to the mass balance calculation The tie line in the two phase region is analogous to a lever a balanced on a fulcrum wa 1 All material must be in one phase or the other Wu WI3 1 Derivation of the lever rule 2 Mass of a component that is present in both phases equal to the mass of the component in one phase mass of the component in the second phase WOLCOL WBCB C 3 Solution of these equations gives us the Lever rule WB C0 COL cot and Wot C0 COL CompositionConcentration weight fraction vs molar fraction Composition can be expressed in Molar fraction XB or atom percent at that is useful when trying to understand the material at the atomic level Atom percent at is a number of moles atoms of a particular element relative to the total number of moles atoms in alloy For twocomponent system concentration of element B in at is t nB Ca ZB mAXlOO CatZXBXlOO Hm Hm Where nmA and nt are numbers of moles of elements A and B in the system Weight percent C wt that is useful when making the solution Weight percent is the weight of a particular component relative to the total alloy weight For twocomponent system concentration of element B in wt is wt mg C X 100 ITIB m A where mA and mB are the weights of the components in the system nA mA nB mB where AA and AB are atom1c m A A m AB weights of elements A and B MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Composition Conversions o 0 o 0 at Welght A to Atomlc A CB WX 100 CB AA CAAB CMA Cit gtlt100 CB AACA AB Atomic toWeight CVBW at CgABat X100 C131Ai13CAAA at C2quot CAAA x100 CgAB Cj AA Of course the lever rule can be formulated for any speci cation of composition ML XBoc XB0XBOL XBL Cata Cato Cata CatL Ma 2 X30 39 XBLXBX 39 XBL Z Cato 39 CatL Catoc 39 CatL WL CW CW2 CW CWtL Wu Z Cwio CW1 CW2 CW1 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Phase compositions and amounts An example 1300 7 Damn E Txe we 2 a 0qu g 7 E g u qumd a 1200 EARS 2 30 40 5 CL Co Camposmon WPn NI CD 35 W1 0 CL 315 wt c5425 wt Mass fracu39ons WL S RS Ca CD Ca CL 068 Wm RRs CD CL CB CL 032 MSE sus Phase Diagzmsand musics Lmnirl zhigjla39 Development of microstructure in isomorphous alloys Tempevaluve as 1300 7 1 200 Hrm Equilibrium very slow cooling LKM Ni 2 Compusmun m N Development of microstructure in isomorphous alloys Equilibrium very slow cooling gt Solidification in the solid liquid phase occurs gradually upon cooling from the liquidus line gt The composition of the solid and the liquid change gradually during cooling as can be determined by the tieline method gtNuclei of the solid phase form and they grow to consume all the liquid at the solidus line MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Development of microstructure in isomorphous alloys Temperature 39EC Nonequilibrium cooling 130 C 1200 1100 Cr 46 Ni LEMON H wise rm l L L 35 Ni LL35 Ni r l a 3973 2O Corruposition M94 Ni Development of microstructure in isomorphous alloys Nonequilibrium cooling Compositional changes require diffusion in solid and liquid phases Diffusion in the solid state is very slow gt The new layers that solidify on top of the existing grains have the equilibrium composition at that temperature but once they are solid their composition does not change gt Formation of layered cored grains and the invalidity of the tieline method to determine the composition of the solid phase The tieline method still works for the liquid phase where diffusion is fast Average Ni content of solid grains is higher gt Application of the lever rule gives us a greater proportion of liquid phase as compared to the one for equilibrium cooling at the same T gt Solidus line is shifted to the right higher Ni contents solidi cation is complete at lower T the outer part of the grains are richer in the lowmelting component Cu Upon heating grain boundaries will melt rst This can lead to premature mechanical failure MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei 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 AHmix gt 0 G T1 G T2ltT1 li uid G q At low temperatures there is a region where the solid solution is most stable as a mixture of two phases cal and 0c2 with compositions X1 and X2 This region is called a miscibility gap MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Eutectic phase diagram For an even larger AH the miscibility gap can extend into the miX liquid phase region In this case we have eutectic phase diagram MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Eutectic phase diagram With different crystal structures of pure phases A similar eutectic phase diagram can result if pure A and B have different crystal structures G Tl G MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Temperature dependence of solubility There is limited solid solubility of A in B and B in A in the alloy having eutectic phase diagram shown below T G T1 The closer is the minimum of the Gibbs free energy curve GO XB to the axes XB O the smaller is the maximum possible concentration of B in phase or Therefore to discuss the temperature dependence of solubility let s nd the minimum of G XB Gmg XAGA XBGB QXAXB RTXAlnXA XBlnXB dG dXB O M1n1mum 0f GXB 1 see problem 1 0f hW 4 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Eutectic systems alloys with limited solubility I 1200 I l l 39 l 39 A 1000 L 779 C TE m o 0 Temperature 0C m 8 on 3 7 Copper 7 Silver phase diagram 200 l l 1 l l Cu Composition Wt Ag As Three single phase regions on solid solution of Ag in Cu matrix 3 solid solution of Cu in Ag matrix L liquid Three twophase regions on L 3 L oc HS Solvus line separates one solid solution from a mixture of solid solutions Solvus line shows limit of solubility MSE 305 Phase Diagams and Kinetics Leonid Zhigilei Eutectic systems alloys with limited solubility II Lead 7 Tin phase diagram 7 300 I qilirl U o E 200 a 183 619 978 7 g 7 v E Invariant or eutectic pomt d F 100 a a Eutectic isotherm 0 l l l l l l 20 40 60 80 l D Pb Composition wt Sn 3quot Eutectic 0r invariant point Liquid and two solid phases co exist in equilibrium at the eutectic composition CE and the eutectic temperature TE Eutectic isotherm the horizontal solidus line at TE Eutectic reaction 7 transition between liquid and mixture of two solid phases on 3 at eutectic concentration CE The melting point of the eutectic alloy is lower than that of the components eutectic easy to melt in Greek Eutectic systems alloys with limited solubility III Compositions and relative amounts of phases are determined from the same tie lines and lever rule as for isomorphous alloys Temperature C 0 o N o a a as o m o l 0 PM Composition wt Sn 5 3 For points A B and C calculate the compositions Wt and relative amounts mass fractions of phases present MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Development of microstructure in eutectic alloys 1 Several different types of microstructure can be formed in slow cooling an different compositions Let s consider cooling of liquid lead 7 tin system as an examp e 400 300 Temperature C mo 1W0 5m T ID 1 Composition wt Sn In the case of leadrich alloy 02 wt of tin solidi cation proceeds in the same manner as for isomorphous alloys eg Cu Ni that we discussed earlier L aL a MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Development ofmicrostructure in eutectic alloys II At compositions between the room temperature solubility limit and the maximum solid solubility at the eutectic temperature B phase nucleates as the a solid solubility is exceeded upon crossing the solvus line g c l n Temperature C V 0 g 11 100 7 a 5 39 l l l 10 20 30 40 5 2 Composition wt Sn MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Development of microstructure in eutectic alloys HI Solidification at the eutectic composition No changes above the eutectic temperature TE At TE all the liquid transforms to x and 3 phases eutectic reaction L 461 Wl 7 Sn Temperature C Ca 51 9 Composition Wl Sn L a B MSE 305 Phase Diagrams and Kinetics Leunid Zhigiiei Development of microstructure in eutectic alloys IV Solidi cation at the eutectic composition Compositions of 0L and B phases are very different gt eutectic reaction involves redistribution of Pb and Sn atoms by atomic diffusion we will learn about diffusion in the last part of this course This simultaneous formation of 0L and B phases result in a layered lamellar microstructure that is called eutectic structure Formation of the eutectic structure in the leadtin system In the micrograph the dark layers are leadreach 0L phase the light layers are the tinreach B phase MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Development of microstructure in eutectic alloys V Compositions other than eutectic but within the range of the eutectic isotherm Primary 0L phase is formed in the on L region and the eutectic structure that includes layers of on and Bphases called eutectic a and eutectic B phases is formed upon crossing the eutectic isotherm L a L a B L I wl SH 300 7 U o 0 200 7 5 1 E a e m e a 7 L 519 wo Sn 1 E l a E 1 mo 7 l l Eutectwca 133 Wm 5m l 7 2 l l l l 20 l 50 so lLo Pb 15m 0 40 Composition Wt Sn MSE 305 Phase Diagams and Kinetics Leonid Zhigilei Development of microstructure in eutectic alloys VI Microconstituent element of the rnicrostructure having a distinctive structure In the case described in the previous page microstructure consists of two rnicroconstituents primary 1 phase and the eutectic structure a n Kx EQti 3235 stem Although the eutectic structure consists of two phases it is a microconstituent with distinct lamellar structure and xed ratio of the two phases MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei How to calculate relative amounts of microconstituents Eutectic microconstit39uent forms from liquid having eutectic composition 619 wt Sn We can treat the eutectic as a separate phase and apply the lever rule to nd the relative fractions of primary on phase 183 wt Sn and the eutectic structure 619 wt Sn We P PQ eutectic W Q PQ primary 300 i L a L 3 200 7 a 5 E E n n n a 7 i a i as E H 100 i 0 PD Sn 183 0 4 619 978 Composition wt Sn MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei How to calculate the total amount of a phase both eutectic and primary Fraction of on phase determined by application of the lever rule across the entire or 3 phase eld W QR PQR a phase W393 P PQR B phase 300 i L U a L or 200 i a a L a E 5 7 P Q R E E 100 7 0 PD Sn 183 C A 619 978 Composition wt Sn MSE 305 Phase Diagrams and Kineu39cs Leonid Zhigilei Binary solutions With AHmix lt 0 ordering If AHmix lt 0 bonding becomes stronger upon mixing gt melting point of the mixture will be higher than the ones of the pure components For the solid phase strong interaction between unlike atoms can lead to partial ordering gt AHmiX can become larger than QXAXB and the Gibbs ee energy curve for the solid phase can become steeper than the one for liquid T2 ltT1 li uid G q G liquid 0 XB 1 T3ltT2 li uid G q 0 XB 1 0 At low temperatures strong attraction between 0 O o O o o unlike atoms can lead to the formation of O O O O O O ordered phase or Binary solutions with AHmix lt 0 intermediate phases If attraction between unlike atoms is very strong the ordered phase may extend up to the liquid T In simple eutectic systems discussed above there are only two solid phases 0c and B that exist near the ends of phase diagrams Phases that are separated from the composition extremes 0 and 100 are called intermediate phases They can have crystal structure different om structures of components A and B MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei AHmiXltO tendency to form highmelting point intermediate phase T liquid Increasing negative AHmix Phase diagrams with intermediate phases example Example of intermediate solid solution phases in CuZn 0c and 11 are terminal solid solutions 5 B y 5 a are intermediate solid solutions lUU 1000 Liquid 800 600 Temperature 400 7 l 2 0 0 Cu Composition VV39tQ tr Z11 Z11 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Phase diagrams for systems containing compounds For some systems instead of an intermediate phase an intermetallic compound of speci c composition forms Compound is represented on the phase diagram as a vertical line since the composition is a speci c value When using the lever rule compound is treated like any other phase except they appear not as a wide region but as a vertical line l l l l l l l l intermetallic compound Temperature we a MgZPb 9 l l l l O 20 40 so a Mg Compositlon WVa Pb Pb This diagram can be thought of as two joined eutectic diagrams for MgMgsz and MgszPb In this case compound Mgsz 19 wt Mg and 81 wt Pb can be considered as a component A sharp drop in the Gibbs free energy at the compound composition should be added to Gibbs free energy curves for the existing phases in the system Stoichiometric and nonstoichiometric compounds Compounds which have a single wellde ned composition are called stoichiometric compounds typically denoted by their chemical formula Compound With composition that can vary over a nite range are called nonstoichiometric compounds or intermediate phase typically denoted by Greek letters T T nonstoichiometric stoichiometric Common stoichiometric compounds compound AB A B5 ASB4 A4B3 A3B2 ASB3 c0mp0siti0nat 500 455 444 429 400 375 compound AZB ASB2 A3B A4B ASB A6B c0mp0siti0nat 333 286 250 200 167 143 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Eutectoid Reactions The eutectoid eutecticlike in Greek reaction is similar to the eutectic reaction but occurs from one solid phase to two new solid phases Eutectoid structures are similar to eutectic structures but are much ner in scale diffusion is much slower in the solid state Upon cooling a solid phase transforms into two other solid phases 5 7 7 2 in the example below Looks as V on top of a horizontal tie line eutectoid isotherm in the phase diagram 700 7 a I 500 7 7 8 E 9 Eutectold 500 7 l o 90 Compusmon Wm zm Eutectic and Eutectoid Reactions Temperature Z 1 6 Eutectic y 39 temperature 05 y 3 Eutectoid a 3 temperature a 6 Eutectoid Eutectic composition composition Composition The above phase diagram contains both an eutectic reaction and its solidstate analog an eutectoid reaction MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Peritectic Reactions A peritectic reaction solid phase and liquid phase will together form a second solid phase at a particular temperature and composition upon cooling eg L 0c lt gt B These reactions are rather slow as the product phase will form at the boundary between the two reacting phases thus separating them and slowing down any further reaction a liquid liquid Temperature liquid Peritectoid is a threephase reaction similar to peritectic but occurs om two solid phases to one new solid phase on B y MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Example The Iron Iron Carbide Fe Fe3C Phase Diagram In their simplest form steels are alloys of Iron Fe and Carbon C The FeC phase diagram is a fairly complex one but we will only consider the steel part of the diagram up to around 7 Carbon 153B C 1200 f E e y Austemte 2 14 43930 a E E 1000 a a g 912 C 3 Fe3C 800 quot 727cc 39Y 0 75 0022 600 u Ferme u Fe3C Cemenme Feat A i i i i i i 5 6 67 3 4 Fe Composmon weu c MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Phases in Fe Fe3C Phase Diagram aferrite solid solution of C in BCC Fe Stable form of iron at room temperature 0 The maximum solubility of C is 0022 wt Transforms t0 FCC yaustenite at 912 C y austenite solid solution of C in FCC Fe 0 The maximum solubility of C is 214 wt Transforms to BCC 5ferrite at 1395 C Is not stable below the eutectic temperature 727 C unless cooled rapidly 8ferrite solid solution of C in BCC Fe The same structure as ocferrite Stable only at high T above 1394 C Melts at 1538 C Fe3C iron carbide or cementite This intermetallic compound is metastable it remains as a compound inde nitely at room T but decomposes very slowly within several years into ocFe and C graphite at 650 700 C FeC liquid solution A few comments on Fe Fe3C system C is an interstitial impurity in Fe It forms a solid solution with on y 5 phases of iron Maximum solubility in BCC ocferrite is limited max 0022 wt at 727 C BCC has relatively small interstitial positions Maximum solubility in FCC austenite is 214 wt at 1147 C FCC has larger interstitial positions Mechanical properties Cementite is very hard and brittle can strengthen steels Mechanical properties also depend on the microstructure that is how ferrite and cementite are mixed Magnetic properties 0c ferrite is magnetic below 768 C austenite is nonmagnetic Classi cation Three types of ferrous alloys Iron less than 0008 wt C in oc ferrite at room T Steels 0008 214 wt C usually lt 1 wt ocferrite Fe3C at room T Cast iron 214 67 wt usually lt 45 wt MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Temperature 96 Eutectic and eutectoid reactions in Fe Fe3C Eutectic 430 wt C 1147 0C L y Fe3C 1200 y Austem te 1000 y Fe3C 800 727 C 0022 600 a Ferme a F630 Cemenme Peach A 1 2 3 4 5 6 67 Fet Composit on wt C Eutectoid 076 wtC 727 OC y076 wtoo C or 0022 wt C Fe3C Eutectic and eutectoid reactions are very important in heat treatment of steels Development of Microstructure in Iron 7 Carbon alloys Microstructure depends on composition carbon content and heat treatment In the discussion below We consider slow cooling in which equilibiimn is maintained Microstructure of eutectoid steel 1 1i 1000 7 y y Fe3C 900 7 V Y 7 800 a 39y 7 Tempevamre c1 700 500 7 a 4 FESC 1 l Cum DGSIUDH wt A C Microstructure of eutectoid steel 11 When alloy of eutectoid composition 076 wt C is cooled slowly it forms perlite a lamellar or layered structure of two phases xferrite and cementite Fe3C The layers of alternating phases in pearlite are formed for the same reason as layered structure of eutectic structures redistribution C atoms between ferrite 0022 wt and cementite 67 wt by atomic diffusion Mechanically pearlite has properties intermediate to soft ductile ferrite and hard brittle cementite r39 awquot quot 3939quotZquot l A In the micrograph the dark areas are 6 9 4 Fe3C layers the light phase is 0c n 39 Austemte gram boundary rgv 3596 5 c r 4K Agilkv 95 Ferrite a Kg 2 Growth direction of pearlite V Carbon diffusion Microstructure of hypoeutectoid steel 1 Compositions to the le of eutectoid 0022 076 Wt C hypoeutectoid less than eutectoid Greek alloys 7 17 aFe3C 1000 Yam peramre C7 quot jgtPeamte 600 page Prneulecmld u Eutecimd a 500 7 a rage 3quot w 10 20 Co Compusman mm c Microstructure of hypoeutectoid steel 11 Hypoeutectoid alloys contain proeutectoid fern39te formed above the eutectoid temperature plus the eutectoid perlite that contain eutectoid fern39te and cementite Pearlite F830 Proeutectoill a Eutectoid a Microstructure of hypereutectoid steel 1 Compositions to the right of eutectoid 076 214 Wt C hypereutectoid more than eutectoid Greek alloys lure 8 lempeva 7 7Fe3C aFe3C P 39y Page 1000 7 y w 7 g 900 7 regc 7 500 A 7 v a y 700 0 7 quot Pea lite son a 7 Proeutectold Fesc Emectmd FeEC 500 7 a Fejc 2 n L Composmon wl c Microstructure of hypereutectoid steel 1ow 1 1ow 1 9 1200 P 1200 939 9 5 5 a a g 1 g 1 1 800 1 1 800 1 1 1 mm 1 1 1 1 1 mm 1 1 1 1 1 2 4 6 2 4 o e Compoamorw WI 2 F630 e Compos1t1on WIDa 2 F630 1ow w wow w P 1200 P 1200 9 9 5 5 a a E w E U 1 800 1 1 800 1 1 1 mm 1 1 1 1 1 mm 1 1 1 1 1 2 4 6 2 4 6 e Compoamorw WI 2 F630 e Composmom WI 2 F630 1ow 1 1ow 1 P 1200 P 1200 9 9 5 5 a a g 1 g 1 1 800 1 800 a 1 1 o 1 1 mm 1 1 1 1 1 mm 1 1 1 1 1 2 4 6 2 4 6 e Compoamorw WI 2 F630 e Composmom WI 2 F630 Microstructure of hypereutectoid steel 11 Hypereutectoid alloys contain proeutectoid cementite formed above the eutectoid temperature plus perlite that contain eutectoid ferrite and cementite How to calculate the relative amounts of proeutectoid phase a or Fe3C and pearlite Application of the lever rule With tie line that extends from the eutectoid composition 075 Wt C to x 7 a Fe3C boundary 0022 Wt C for hypoeutectoid alloys and to a Fe3C 7 Fe3C boundary 67 Wt C for hipereutectoid alloys 1 F530 Temperature Composmnn mm C Fraction of 1 phase is determined by application of the lever rule across the entire 0 Fe3C phase eld Example for hypereutectoid alloy With composition C1 Fraction of pearlite W1 X VX 67 C1 67 076 Fraction of proeutectoid cementite Wm V VX C1 076 67 076 y FeSC Temperature Composmun mm C The Most Important Phase Diagram in the History of Civilization 2300 E 6 H aarmmn ALPHAJRON mu 79 LIQUID L 7 Y L w AUSTENITE GAMMA mom 2200 utec 39 L id sSC 391 Pt int F130 IRON 130quot CARBIDE swamurn a y 39f F sc 09 ut to39d 39 quot 39 1333 0F o nint 1000 a F93C 83 200 A 430 667 gt Euro HYPER vaoeurscnc 1A HYPEREUTEchc M11213031 EUquotEG on STEE cAsnnoN vx b v pearlite naturally formed composite hard amp brittle ceramic F e3C soft BCC iron ferrite MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei The Gibbs phase rule 1 Let s consider a simple onecomponent system solid In the areas Where only one phase is stable both pressure and temperature can be independently varied Without upsetting the equilibrium gt there are 2 degrees of freedom T Along the lines Where two phases coexist in equilibrium only one variable can be independently varied Without upsetting the twophase equilibrium P and T are related by the Clapeyron equation gt there is only one degree of freedom At the triple point Where solid liquid and vapor coexist any change in P or T would upset the threephase equilibrium gt there are no degrees of freedom In general the number of degrees of freedom F in a system that contains C components and can have Ph phases is given by the Gibbs phase rule FC Ph2 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei The Gibbs phase rule 11 Let s now consider a multicomponent system containing C components and having Ph phases A thermodynamic state of each phase can be described by pressure P temperature T and C l composition variables The state of the system can be then described by PhgtltCl2 variables But how many of them are independent The condition for Ph phases to be at equilibrium are Ta TB TY Phl equations Pa PB PY Phl equations u 2 MBA 2 NA Phl equations C sets of equations LL01 2 pi M3 Phl equations Therefore we have Ph lgtltC 2 equations that connect the variables in the system The number of degrees of eedom is the difference between the total number of variables in the system and the minimum number of equations among these variables that have to be satis ed in order to maintain the equilibrium FPhC1 Ph 1C2C Ph2 F C Ph 2 Gibbs phase rule MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei The Gibbs phase rule example an eutectic systems FC Ph2 Pconst 139 Q E FC Ph1 C22 1 F3 Ph Inn mi miliun gt In onephase regions of the phase diagram T and XB can be changed independently gt In twophase regions F 1 If the temperature is chosen independently the compositions of both phases are xed gt Three phases L on B are in equilibrium only at the eutectic point in this two component system F 0 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Multicomponent systems I The approach used for analysis of binary systems can be extended to multicomponent systems w 40 A A 60 50 50 T VAVAVAVAVA VVVV so 20 90 m n 0 m 20 30 4a 50 51 70 so 90 100 3 7 wt Representation of the composition in a ternary system the Gibbs triangle The total length of the red lines is 100 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Multicomponem systems 11 The Glbbs free energy sur ng instead of curves for a binary system can be plotted for all the possible phases and for diifenent tennpeiatures A B The chemical potentials of A B and c of any phase in this ystem are given by the points where the angemfzal plane to the i s free energy surfaces intersects the A B and c ax s MSE sns use Diagram and Kinzljcs Lmnitl zhigjlsi Multicom ponent systems 111 A threephase equilibrium in the ternary system for a given temperature can be derived by means of the tangential plane construction For two phases to be in equilibrium the chemical potentials should be equal that is the compositions of the two phases in equilibrium must be given by points connected by a common tangential plane eg l and m The relative amounts of phases are given by the lever rule eg using tieline lm A three phase triangle can result from a common tangential plane simultaneously touching the Gibbs free energies of three phases eg points x y and z Eutectic point fourphase equilibrium between on B y and liquid An example of ternary system Stainless 1 Steel D I w Ternary39Phase Diagram B Etalnless statel l YFeH39i39 Fe Weight Fernant Nickel r A A A h A h A n a 10 20 30 40 EU ED T D 8D 9U The ternary diagram of NiCrFe It includes Stainless Steel wt of Cr gt 115 wt of Fe gt 50 and Inconeltm Nickel based super alloys Inconel have very good corrosion resistance but are more expensive and therefore used in corrosive environments Where Stainless Steels are not suf cient Piping on Nuclear Reactors or Steam Generators MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Another example of ternary phase diagram oil water surfactant system Inverse Missile surfactant Hexagonal Elcemlnuous Cubic Uni or Mulli Lamalmr Valle Radlike Mlcelte 1 Micalla W t EIimn nuaus 3 er Micmemulsion Drawing by Carlos Co University of Cincinnati Surfactants are surfaceactive molecules that can form interfaces between immiscible uids such as oil and water A large number of structurally different phases can be formed such as droplet rodlike and bicontinuous microemulsions along With hexagonal lamellar and cubic liquid crystalline phases MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Review of classical thermodynamics Fundamental Laws Properties and Processes 2 Entropy and the Second Law Concepts of equilibrium Reversible and irreversible processes The direction of spontaneous change Entropy and spontaneousirreversible processes Calculation of entropy in isochoric and isobaric processes Calculation of entropy in reversible and irreversible processes Reading Chapters 31 35 314 317 ofGaskell or the same material in any other textbook on thermodynamics MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei 211d Law of Thermodynamics A system left to itself will either 1 Remains in the same state inde nitely system is in equilibrium The system Will move from the equilibrium state only if acted on by some external impact 2 Will move of its own accord to some other state The system in this case is in a nonequilibrium state and Will move towards equilibrium state in a natural or spontaneous process Spontaneous transition from nonequilibrium state to the equilibrium state cannot be reversed Without application of an external force it is an irreversible process Examples of spontaneousirreversible processes Mixing of different gases Heat ow from hot to cold objects 0 O o o 39 0 o 39 T1 ltT2 o 0 o o o o T1 T 7 Spontaneous processes are not necessarily instantaneous processes kinetics MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei How to predict which process will proceed spontaneously lSt law U and H are state functions if the system is going from A to B AUA gtB AUB gtA or AHA gtB AHB gtA But st law does not tell us which reaction forward or reverse is natural or spontaneous one Intuition The energetic driving force the tendency to fall to a lower potential energy eg release heat in a reaction Indeed often spontaneous changes are exothermic reaction produces heat eg H2 gas 12 02 gas gt H20 liquid AH 286 kJmol391 But reactions can also occur spontaneously Which are endothermic eg H20 liquid105 C gt H20 gas105 C AH 44 kJmol391 A negative sign of AH favors but does not guarantee spontaneity MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Quanti cation of irreversibility It is desirable to nd some common measure of the tendency of a system to change spontaneously This measure should be gt A thermodynamic property state function gt It should change in a characteristic manner eg always increase when a process proceeds spontaneously Such function entropy from TpOTET transformation in Greek has been introduced by Clausius in 1850 Second Law of Thermodynamics can be formulated in different ways One possible fomiulation is There exist a state function the entropy S which for all reversible processes is de ned by dS dqreVT and for all irreversible processes is such that dS gt dqT or in general dS 2 dqT Entropy is a state function whose change is de ned for a reversible process at T where Q is the heat absorbed AS QT For an isolated system dq 0 gt dS 2 O Entropy is maximized in any spontaneous process This will be the basis for de nition of the equilibrium conditions MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Why qT For qualitative understanding of Why the quantity qT is used as a measure of the degree of irreversibility let s consider an example from Gaskell 33 Let s consider two irreversible processes 1 Conversion of work to heat 2 Flow of heat down a T gradient Three processes with different degree of irreversibility D T2gtT1 D T2 T2 T1 qgt 33 G Q qT2ltqT1 CD is less q irreversible T T1 3 than 1 gt All three processes are irreversible The 3 is just a sum of CD and should be the most irreversible Both heat production and temperatures are important in de ning a scale of irreversibility MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Summary on Entropy gt Entropy is a state function gt When the weightheat reservoir system discussed above undergoes a spontaneous process which causes the adsorption of heat q at a constant temperature T the entropy produced by the system AS qT The increase in entropy caused by the process is thus a measure of the degree of irreversibility gt The increase in entropy due to the occurrence of an irreversible process arises om the degradation of energy potentially available for useful work into heat gt In a reversible process the driving force is in nitesimal and the process proceeds at an in nitesimal rate the system moves through a continuum of equilibrium states and the entropy is not created can only be transferred from one part of the system to another For more on entropy in reversible processes see Gaskell 34 39 gt The entropy of an adiabatic system cannot decrease It increases in an irreversible process and remains constant during a reversible process MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei The Second Law Again We can reformulate the second law in the following way For every thermodynamic system there exist an extensive state function called the entropy which can be calculated by a reversible path from an arbitrary chosen reference state by integrating the heat absorbed by the system divided by the absolute temperature The entropy of a system plus its surroundings together forming the universe an isolated system never decreases and increases in any irreversible process The 3rd law of thermodynamics The third law of thermodynamics states that if one could reach absolute zero temperature all the thermal motion of atoms could be removed and a complete internal equilibrium all bodies would have the same entropy In other words a body at absolute zero could exist in only one possible state which would possess a definite energy called the zeropoint energy This state is de ned as having zero entropy The 3rd law has been rst formulated by Walter Nernst and also known as the Nernst heat theorem The 3rd law allows us to de ne absolute values of entropy at a given T T T T c dSJJFPdT thus STSOIPdT where 800 0 0 0 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Combined statement of the 1St and 211d laws For a closed system and a reversible process dU Sq 5w Sq TdS and assuming that work due to the volume change is the only form of work performed by the system SW PdV Combining these equations together we get dU TdS PdV U USV gt dU 6U6SVdS 6U6VSdV Comparing the equations we see that T BUBSV P BU6VS S and V are referred to as the natural choice of independent variables to describe the internal energy If the function USV is known we have all information about the system ie U can be calculated directly and T and P can be calculated om the derivatives of the function Thus U USV is a fundamental equation In a system with U const and V const equilibrium occurs when the entropy is maximized d8 0 the definition of S through the second law of thermodynamics allows to predict the direction of spontaneous change in a system that initially is not in equilibrium MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Entropy calculation Constant volume process 5C1 Sq dS F cV Igt 5q chT Tf c dSZEEE JdSASJ4 T T Ti T Constant pressure process 5C1 Sq d8 cP 1 ij Igt 8q cPdT Tf c dSZEEE deASjJaT T Ti T Example Calculate the difference in entropy between 3 moles of 02 gas at 800 K and at 300 K Pressure is 1 atrn in each case Cp3Ot418x10317l7leVP2hnolWCq 800 800 4 deAS3j aT3j E9418xm4 mfo T T m0 T T 300 330h1 99 418x103B00 3m 85x104 12131 300 800 300 94 J K391 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Entropy production reversible phase transformation Example One mole of liquid lead at its equilibrium freezing temperature 600 K freezes slowlyreversibly at l atrn pressure to solid state Calculate the entropy production Cpliquid 324 3lgtltlO393 T Jmol39lK39l Cpsolid 236 975 gtlt10393 T Jmol39lK39l AHm 4810 Jmol391 latent heat of melting Entropy createdproduced AS S ahd Sthres zuid sthres ASPb Asthres nal mitial 4810 ASPquot q W negative heat is removed qthres Asthres T 600 positive heat ows into reservoir AS Aspb Asthres 0 There is change in entropy in both lead and the thermal reservoir The sum however is zero in reversible process MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Entropy production irreversible process 1 Example One mole of supercooled liquid lead spontaneously freezes at 590 K and l atrn pressure to the solid state Calculate the entropy production Cpliquid 324 3lgtltlO393 T Jmol39lK39l Cpsolid 236 975 gtlt10393 T Jmol39lK39l AHm 4810 Jmol391 latent heat of melting Entropy createdproduced AS SPb soli thres Pb thres d Sim1 AS AS d Sthres SPb nal liqui First let s calculate change in entropy for Pb Liquid Pb Solid Pb The entropy changes along 6000K I the actual path is unknown II the process is irreversible But since entropy is a state I In function we can calculate l the entropy change along a reversible path 5900K Actual path ASHES may MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Entropy production irreversible process 2 I isobaric reversible heating 600 6 600 c 600 4 A51 01qu T j PdTj 39 31x10 3 T 590 T 590 T 590 T 324ln 00031600 5900514JK II isothermal isobaric process of freezing A811 E 8017 JK T 111 isobaric reversible cooling 590 590 590 A81 015qu T idT j 23396 975gtlt103 T 600 T 600 T 600 T 2361n 000975590 600 0494 JK ASPb ASI ASH ASIII 0514 8017 0494 7997 JK MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Entropy production irreversible process 3 There is also a change in entropy in the surroundings This relates to the heat transfer to the constant temperature reservoir at constant pressure AsthreS 3 Achres T T Achres AHPb AHI AHH AHHI 600 600 AHI jcPdT j32431x10 3T1T 306 J 590 590 AH AHm 4810 J AHHI Sjoc dT 53023 6 9 75gtltIO3T1T 294 J P 600 600 AH AHI AH AHIH 306 4810 294 4798 J thres As hres AH 8134 JK T 590 Entropy produced AS 2 ASPquot As hres 7997 JK 8134 JK 0137 JK Can the freezing at 590 K be carried out reversibly MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Entropy and phase transformations The example considered above shows that the entropy change in the universe is positive if a lowtemperature phase converts into a high temperature phase at a temperature above the equilibrium transition temperature Or if the highT phase converts to a lowT one at T below the transition temperature Otherwise it is negative and the transition is prohibited by the 2nd law of thermodynamics will not go spontaneously MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei The Statistical Interpretation of Entropy Physical meaning of entropy Microstates and macrostates Statistical interpretation of entropy and Boltzmann equation Con gurational entropy and thermal entropy Calculation of the equilibrium vacancy concentration Reading Chapter 4 of Gaskell Optional reading Chapter 158 of Porter and Easterling MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei What is the physical meaning of entropy Entropy is introduced in phenomenological thermodynamics based on the analysis of possible and impossible processes We know that heat ows from a hot region to a cold region of a system and that work can be irreversibly transferred into heat To describe the observations the entropy and the 2nd law stating that entropy is increasing in an isolated system have been introduced The problem with phenomenological thermodynamics is that it only tells us how to describe the empirical observations but does not tell us why the 2nd law works and what is the physical interpretation of entropy In statistical thermodynamics entropy is de ned as a measure of randomness or disorder Intuitive consideration In a crystal atoms are vibrating about their regularly arranged lattice sites in a liquid atomic arrangement is more random Sliquid gt Ssohd Atomic disorder in gaseous state is greater than in a liquid state Sgas gt Sliquid o o 390 o 39 o a 55 0 39 o o 0 o 55 o o 39 39 o o o o o o L 39 o 39 G o o 0 Does this agrees with phenomenological thermodynamics Melting at constant pressure requires absorption of the latent heat of melting q AHm therefore ASm AHIn TIn the increase in the entropy upon melting correlates with the increase in disorder MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Microstates and Macrostates A macroscopic state of a system can be described in terms of a tates eg particlesWe can specify coordinates and velocities of all atoms The 2m1 law can be stated as follows The equilibrium state of an isolated system is the one in which the number of possible microscopic states is the largest Example for making it intuitive rolling dice g Macrostate 7 the total ofthe dice ach die have 6 microstates the system of 2 dices has 6X636 microstates a system ofN dice has 6 microstates For two dice there are 6 mysmicrostates to get macrostate 7 but only one microstate that correspond to 2 or 12 The most likely macrostate is abig number N of dice the macrostate for which the number of possible microstates is amaximum is 3 5 Ifyou shake a1arge bag ofdice and roll them it is likely that you get the total close to 35xN for which the number of my to make it from individual dice is maximum An isolated thermodynamic system is simi1ar a thermal uctuations do s 39ng the macrostate corresponds to the largest number or microslales Actually the system ofdice is closer to aquantum system with discrete states In the classical case the states f an e have to replace the sum over states by integrals over phase space MXE 3051711332 Dizng and Kinetics Lennid Zhigiiei Statistical interpretation of entropy If we combine two systems the number of microstates multiply remember 6636 for two dice At the same time we know that entropy is an extensive quantity 8MB SA SE and if we want to relate enthalpy to the number of microstates we have to make sure that this equation is satisfied If we take logarithm of the number of microstates the logarithms adds when we put systems together The quantity maximized by the second law can be defined then by equation written on Ludwig Boltzmann s tombstone in Vienna SkBan m where Q is the number of microstates k is the Boltzmann constant it is the same constant that relates kinetic energy to temperature but it was first introduced in this equation and S is the entropy The 2 1 law can be restated again An isolated system tent toward an equilibrium macrostate with maximum entropy because then the number of microstates is the largest The entropy is related to the number of ways the microstate can rearrange itself without affecting the macrostate NISE 305 Phase Diagrams and Kinetics Leonid Zhigilci Heat ow and production of entropy Phenomenological thermodynamics transfer of energy from hot to cold is an irreversible process that leads to the production of entropy Consideration of probabilities of microstates can lead to the same conclusion TA UA 2A TB UB QB When the thermal contact is made between the two systems the number of microstates QAQB is not in the maximum and heat starts to ow increasing the value of QAQB The heat ows until the increase in QA caused by the increase in UA is greater than the decrease in QB caused by the decrease in UB The heat ow stops when QAQB reaches its maximum value 51nQ AQB 0 If we only have a heat exchange no work 5q 5qA 5qB 51m QA Sq 51noB q BA BB smvoB 51noA 51nQB 3 ii kB TA TB SanA QB 0 only when TA TB reversible heat transfer is only possible after temperatures are equal MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Con gurational entropy and thermal entropy Thermal entropy In the example on the heat transfer the number of microstates Q can be thought as the number of ways in which the thermal energy can be divided between the atoms The thermal entropy of a material gt Increases as the temperature increases gt Decreases as the cohesive energy increases The thermal entropy plays important role in many polymorphic transitions With increasing T the polymorphic transition is from a phase with lower entropy to the one with higher entropy Con gurational entropy Entropy can be also considered in terms of the number of ways in which particles themselves can be distributed in space Mixing of elements in two crystals placed in physical contact or gases in two containers mass transport leads to the increase of Sconf and is similar to the heat transfer case when Sth is increasing sthT u 0 u39 s T conf MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Con gurational entropy equilibrium vacancy concentration The configurational entropy of a crystal refers to the distinguishable ways the atoms can be arranged on the lattice sites For monoatomic crystals the only origin of configurational entropy is the presence of crystal defects such as vacancies 7 lattice sites without atoms The rem oval of an atom from its lattice site gt breaks bonds and increases the internal energy of the material by 8f gt Increases the randomness of the atomic configuration and thus S conf39 Let s first calculate the configurational entropy Consider a crystal lattice with N lattice sites and Nl atoms one vacancy Assume that N is large enough so that we can neglect the surface in our consideration and assume that all lattice sites are equivalent In the system of N lattice sites we can have a vacancy in any of the N sites 7 N different configurations are possible and SIconf kB lnN If there are two vacancies for each location of the first vacation we have Nl locations of the second one Therefore Szconf kB In 2 N Nl gtgt Slconf 2 comes from the fact that vacancies are identical and the states where one vacancy is at the site i and the second one at j are identical to the state ji 53cm kB In 16 N N1N2 gtgt S2 confgt MSE 305 Phase Diagrams and Kinajcs Leonid Zhigilei Equilibrium vacancy concentration 2 The number of distinct con gurations for n vacancies is l N n l n N Sconf kB In The equilibrium concentration of vacancies is the one that corresponds to the minimum of free energy A U TS Neglecting a small change in the thermal entropy due to the change in the vibrational frequencies of atoms around a vacancy formation entropy and ignoring the vacancyvacancy interactions we can write N A neg kBTln n N n If N is large enough introduction of the rst vacancy Will reduce the ee energy for any nite T A1 2 8i kBT lnN lt 0 The equilibrium concentration of vacancies can be found from 6A an 0 nneq MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Equilibrium vacancy concentration 3 68 Let s rst consider inf using Stirling formula for big an numbers 1n N z N In N N 68 f a N a k 1 k 1N 1 1 N 6n BaninnKN n i Bann nn n n kB NlnN N nlnnn N n1nN nN n n kB NlnN nlnn N n1nN n n hhnr4h N w cv nR Dk3mNnj N n n 1 1 1 1 k 1n z k In B N n B N For the equilibrium concentration of vacancies n 6A 2 ineg T820Ilf 2 8i kBT 1n eq 0 an nzn an nzneq N 1 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Equilibrium vacancy concentration 4 f V kBT neq NeXp For metals svf z 9kBTm We can estimate that at room temperature in copper there is one vacancy pet 1015 lattice atoms Whereas just below the melting point there is one vacancy for every 10000 atoms very strong temperature dependence A The equilibrium concentration U of the vacancies is defined by the balance between the internal potential energy of n vacancy formation and the eq 1ncrease of the configuratlonal n entropy TS kBT In The formation energy svf in crystals that we used in the derivation is practically indistinguishable om the formation enthalpy va Which has to be used if the pressure and not the volume is kept constant MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Demonstration of the effect of con gurational entropy Rapid stretching or unstretching of a wide rubber band is nearly isentropicadiabatic process there is almost no transfer of heat to or from the rubber band during this fast processes The stretching decreases the configurational entropy Sconf molecular structure becomes more ordered Since the stretching is isentropic then the thermal entropy Sth must increase to compensate for the decrease of Scmf Since S11 is a function of T only therefore the T should increase as a result of the stretching Similarly T should decrease as a result of rapid unstretching You can carry out this experiment and measure the T by your lip MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Summary I What is entropy Entropy is the measure of the quotprobabilityquot of a given macrostate or essentially the same thing the number of microstates possible for the given macrostate What is a microstate and a macrostate Any particular arrangement of atoms where we look only on average quantities is a macrostate Any individual arrangement de ning the properties e g positions and momentary velocity of all the atoms for a given macrostate is a microstate For a microstate it matters what individual particles do for the macrostate it does not What is the probability of a macrostate The probability for a macrostate the number of possible waysmicrostates to generate the same macrostate divided by the number of all microstates all possible combinations of the dice which is a constant Remember examples on playing dice and on equilibrium concentration of vacancies Example with vacancies we just have to nd how many ways of microstates are there to arrange n vacancies macrostate in a crystal of N lattice sites After we nd the probability we can use the Boltzmann equation to calculate the entropy and we can use the equilibrium condition to select the most likely macrostate the number of vacancies in equilibrium MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Summary 11 Just knowing the internal energy U of a system with a constant volume and temperature is not enough to determine what the equilibrium con guration is There could be many macrostates with the same U That39s why just minimizing U or H is not good enough we have to minimize A U TS or G H TS to nd the equilibrium con guration of the system Of all the macrostates possible for a given U or H the one with the largest entropy at the given temperature will be the one that the system will adopt Equilibrium is a state in which the best possible balance between a small energy and a large entropy is achieved High entropies often mean high energies and vice verse both quantities are opposed to each other The entropy part in A or G becomes more important at high temperatures The entropy of a certain macrostate can be calculated by the statistical de nition of the entropy S using the Boltzmann entropy equation SkBan MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Binary Solutions gt Composition as a thermodynamic variable gt Gibbs free energy of binary solutions gt Entropy of formation and Gibbs free energy of an ideal solution gt Chemical potential of an ideal solution gt Regular solutions Heat of formation of a solution gt Activity of a component Henry s and Raoult s laws gt Real solutions interstitial solid solutions ordered phases intermediate phases compounds gt Equilibrium in heterogeneous systems Reading Chapter 13 of Porter and Easterling Chapters 95 96 99 910 ofGaskell Mmm v w w39l39 Solid Solutions terminology Solid solutions are made of a host the solvent or matrix which dissolves the minor component solute The ability to dissolve is called solubility gt Solvent in an alloy the element or compound present in greater amount V Solute in an alloy the element or compound present in lesser amount V Solid Solution homogeneous maintains crystal structure contains randomly dispersed impurities substitutional or interstitial gt Second Phase as solute atoms are added new compounds structures are formed or solute forms local precipitath gt Solubility Limit of a component in a phase is the maximum amount of the component tha can be dissolved in it eg alcohol has unlimited solubility in water sugar has a limited solubility oil is virtually insoluble The same concepm apply to solid hases Cu and Ni are mutually soluble in any amount unlimited solid solubility while C has a limited solubility in Fe Whether the addition of impurities results in formation of solid solution or second phase depends the nature of the impurities their concentration and temperature pressure v 39dn39 1 Composition as a thermodynamic variable Real materials are almost always mixtures of different elements ra er than pure substances in addition to T and P composition is also a variable To understand conditions for equilibrium and phase diagrams like the one below we have to understand how the Gibbs free energy ofa given phase depends on composition temperature and pressure Since many real experimenm are performed at 1 atrn we will focus on GT composition WW Ma Ma y Ausveme W a Fen V2 2N w CM Cemenme m z 1 s om won Deglns x gt 3 m mummy m Alulul ue la 11 m W Y mum q 1 up 5 Armmun Sam solullon N55 5 m a umun m gamma hon y W w W n 4 mmquot g3 awn cm m Yvru am 2 A 39 t l l A l v mm r Mm I I and 3 7 i v 39 myquot I my quot mm 1 todatmnla 39 quot390 wan n n I Gammacmmm Conwnrumpeanlm 5 mmmm Nun mm terms l Muunmr mm a me l Hwerrautectama Casl lmn Gibbs free energy of a binary solution 1 any proportions form a so id solu ion with the same crystal structure unlimited solid solubility Example Cu and Ni 1 mol of homogeneous solid solution contains XA mol ofA and XE mol ofB XA and XE are the mole fractions ofA and B in the alloy XA XE 1 Let s consider two steps of mixing 1 Bring together XA mol ofpure A and XE mol ofpure B 2 Mix A and B to make a homogeneous solution After step 1 the free energy ofthe system is O O O O 0 0 Gstepl XAGA XBGB g 8 O O G XAGA XE GE G epl GE GA Gibbs free energy per mole before mixing 0 XE l M r 1n n a 7 u Gibbs free energy of a binary solution 11 After step 2 the free energy of the system is G G AGrmx stepZ stepl Where AGrmx is the change of the Gibbs free energy caused by the mixing AGm AHrmx 7 TA Srmx AH H H heat of mixing of the components mix smpz my heat of formation of a solution difference in entropy between A Smix Sstepz Sstepl mixed and unmixed states entropy of formation of a solution Let s first consider an ideal solution 7 interactions between atoms AA B and AB are identical an A m T e free ener han e upon mixing is only due to the change in configurational entropy AG TA Smix A Sm SstepZ Sstepl SLepl kBlnl0 there is only oneway the atoms can be arranged before mixing Therefore A SmX Smpz Entropy of formation of an ideal solution 1 Statistical definition of entropy s kB In a Statistical mechanics if we have NM objects and N of them are special or different the number of ways the objects can be arranged number ofmicrostates is Remember for vacancies we had Q 7 NMV NW N7 number oflaltice sites 7 N N MW r1 7 number ofvacariczes For mixing ofNA particles of type A with NE atoms oftype B N N W Asm 7ktm7kEm NVNE a A N xNA Q Using Stirling formula for big numbers lanm N n N 7 N ASm kElnNA NE lnNAlnNE kENANElINA NENANENAlnNANA NEh11g NE k NAm7NAIjANE N m7NAIjENE MMquot l W MF In l 39d7hquotl39 Entropy offarrnation of an ideal solution 11 N A N NANE x x E A E NANE If the total amount of material is 1 mol NA NB Avogadro s number Na and NA KANE NE XENE NakE R Therefore ASmX 7kE NAln L NEln L NANE NANE 7RXAlnXA x inx Ang 7TA Sm RTXAlnXA XBlnXB For the total Gibbs free energy of an ideal solution G d G Gstepl AGEX Gstepl XAGA XEGE stepz For a nonideal solution we have to take into account heat of formation AHmX AGmx Ame TA SmX MMquot l W Gibbs free energy of an ideal solution AG 7TA Sm RTXAlnXA XBlnXB G 1 XAGA XBGB RTXAlnXA XBlnXB 0 l AGax T2 gt T1 T2 XE 1 Ga GalTl Decrease of GA and GE GAT with T is due to the Giasz temperature dependence ofthe thermal entropy GAIL E is aT P Chemical potential of an ideal solution Remember de nition of the chemical potential ofthe species i i M E dGrSdTVdPZ do an PT F an PTn For a binary solution at constant P and T dG lLAdnA p dnE Addition of XA mol of atoms A and XE mol of atoms B i e dnAdnE XAXE will increase the size of the stem by 1 mol without changing composition and therefore without changing A and 115 The Gibbs free energy of an ideal solution will increase in this case by the molar Gibbs free energy G G AXA HBXB Jmol aG SinceXAXE 1 GllA llB iiiAXB R la llA E Also Gm XAGA XEGB RTXAlnXA XElnXE Gm GE Therefore for an ideal solution 7 RTlnXE uA GA RTlnXA GA ya uE GE RTlnXE A ptA and E depend on the 0 XE 1 composition ofthe phase in mos i i n i Regular solutions Heat offormation of a solution I For a nonideal solution we have to take into account heat of formation AHmlx AGmx AHm X iTA Sm x AH X gt 0 mixing is endothermic heat absorbed AH X lt 0 mixing is exothermic heat released Let s account for the heat offorrnation Ame 0 using a simple model called statistical or quasichemical model In this model the heat of mixing is only related to the bond energies between adjacent atoms The assumption is that the interatomic distances and bond energies are independent of composition 3 types ofbonds A 7A bond energy EAA A 7 B bond energy EAE B r B bond energy EBB Ifthere are FAA PEE PAB bonds of each type the internal energy ofthe solution is E PAAEAA PBBEBB PABEAB Regular solutions Heat of formation ofa solution 11 Let s calculate the internal energy ofthe solution E PAAEAA PEEEEE PAEEAE Ifz is the coordination number of an atom in a crystal then the number of atoms NA and NE are related to the number of bonds as NAz2PAAPAE NEz2PEEPAE Nz P Nz P p Aii p Ei AA 2 2 oo 2 2 Uslng these expresslons for PM and PEB ln the expresslon for energy we have E N5 NAZ Nz E E E EE P E 7 EH 2 2 4 2 energy ofunmixed componenm Therefore the energy ofmixing heat of formation AHmQ is AHmiX Hstepz T E E Hmp m AEmx PAE EAE 7 Mmm quotl 39rl7hquotl39 Regular solutions Heat offormation of a solution 111 statistical or quasichemical model 1 EAA EEE 2 nergy of mixing heat of formation Ame calculated withir s Itquot EAE the solution is ideal AHmo If EAE gtm AHmx gt o 7 atoms will tend to be 2 surrounded by atoms of the same type E E If EAE lt AHm lt 7 atoms will tend to be 0 surrounded by atoms of different type But for small differences between EAB and EAAEBE2 and for high T we can still consider a random arrangement of atoms in a solution such solutions are called regular solutions Then PAB ZNMXAXB and AHm X QXAXB where E E QZN EAE Ame Qgt0 Regular solutions Gibbs free energy I AG X AHmx 7TA Srmx QXAXB RTXAlnXA XBIHXB Qlt0 ForQlt0AH lt07exothermic AH solution 7 mixing is favorable at all T TAS For high 9 and low T PAB 7 max an ordered alloy could be formed 7 AG the assumption of random mixing is m not valid solution is not regular AH QX X 1 mix A a XE Q gt0 highT Q gt0 lowT AH m AH m 0 0 AG m AGW TASm TAsmx 1 XE XE For Q gt 0 AHmx gt 0 7 mixing formation of AB pairs is avoided at low T At high T entropy helps to mix At low T 39 39 occur7 nlutinn i not re ular Regular solutions Gibbs free energy 11 AGm Ame 7 TA Srmx QXAXB RTXAlnXA X3111XB Q gt 0 low T Addition of a small amount of solute always leads to the decrease AH mix ofAGmlx and G Why AG TASm XE G XAGA XBGB QXAXB RTXAlnXA XBlnXB Or using de nition of chemical potential G AKA BXB where for regular solution pA GA 7917 XA2 RTlnXA pa 7 GB 917 XE 2 RTlnXE Regular solutions Gibbs free energy 111 G XAGA XBGB QXAXB RTXAlnXA XBlnXB G HAXA PBX HA 7 GA 217XA2 RTlnXA pg 7 GE 917 XE2 RTlnXE Or we can introduce activities aA and aE of components A and B so that expressions for the chemical potentials would have the same form as for ideal solution A GA RTln M G GE uE GE RTln aE 7RTln aB GA 3 For ideal solutions a XA an aE E For regular solutions aA and aE are related to XA and XE by a Q 1 A 7 17X quotXA RT A a Q 2 1 E 7 17X quotXE RT E A Mmm v 39d7hquotl39 Activity Henry s law Raoult s law 1 a Q 2 In A 17X AH gt0 XA RT A N a Q a w In E 17X 3 3 5 g9 XE RT E w 0 9 AHmX lt0 0 For an ideal solution aA XA and aE XE For a regular solution with AHmx lt 0 activity ofthe components is less than in an ideal solution for AHmlx gt 0 7 higher than in an ideal solution The ratio aAXA is called the activity coef cient 7A of A For a dilute solution ofB in A XB 7 0 we have 3A X l Raoult s law m a is com Henry s law YA XE A Activity and chemical potential are measures of the tendency of an atom to leave a solution If the activity or chemical potential is low the atoms are reluctant to leave the solution MMquot I 171 Real solutions For regular solutions we assumed a random arrangement of atoms in a solution For many real materials this is not a valid assumption If 0 lt 0 the internal energy is minimized by increasing the number of AB bonds 7 can lead to ordered solution at low T if O gt 0 the internal energy is minimized by increasing the number of AA and BB bonds 7 can lead to clustering at low T T e arrangement of atoms is a result ofcompromise between the lowest internal energy and highest entropy Degree of clustering or ordering decreases with increasing T since the entropy contribution TS to Gibbs free energy becomes more important random ordered Mmm r urn1 Substitutional Solid Solutions Max solute concentration 50 at e g CuNi unlimited solid solubility Factors for high solubility gt Atomic size factor atoms need to fit gt solute and solvent atomic radii should be within 7 15 gt Crystal structures of solute and solvent should be the same gt Electronegativities of solute and solvent should be comparable otherwise new intermediate phases are encouraged gt Generally more solute goes into solution when it has higher valency than solvent Interstitial Solid Solutions Nonnally max solute concentration g 10 e g go 1 atofC in oFe 13cc Factors for high solubility gt For fcc bcc hcp structures the voids or interstices between the host atoms are relatively small gt atomic radius of solute should be signi cantly less than solvent Mmm r urn1 leads charge transfer and formation of strong ionic bonding eg M n 2Sn Real solutions If atoms A and B have different sizes the statistical or quasi chemical model will underestimate A X The ener of the elastic strain fields due to the mismatch in atomic sizes should be taken into account If the size difference is large the contribution of the strain energy term could as important as the chemical bonding term If the size difference between the atoms is very large then interstitial solid solutions are energetically favorable In systems with strong chemical bonding between the atoms ere is a tendency for formation of intermediate phases The intermediate phases can have a different crystal structure and may be highly ordered Intermediate phases The intermediate phases can have a crystal structure that is different from the one ofthe pure components and an additional Gibbs free energy curve for the intermediate phase should be considered G G GA GA 0 XE l 0 XE 1 If an intermediate phase have a specific composition and small deviations from the ideal composition cause a rapid rise in G the phase is called compoun ompounds typically have a stoichiometric composition Aan where n and m are integers The structure of intermediate phases is determined by Relative atomic size RARE N 1 16 7 Laves phases eg Manz MgNiz RA gtgt RE 7 interstitial compounds eg Fe C Valency stability of phases depends on the of valence ll electrons per unit ce Electronegativity different electronegativity of components Equilibrium in Heterogeneous Systems 1 Systems with two or more phases are heterogeneous systems If pure components A and B have di erent equilibrium crystal structures phase a and phase 3 we have to consider two Gibbs ree energy curves one for each phase homogeneous ophase 9 homogeneous 3 phase For compositions near crossover of GBL and GB the total Gibbs free energy can be minimized by separation into two phases Equilibrium in Heterogeneous Systems 111 For composition XE the lowest Gibbs free energy is Gan for a homogeneous system The total Gibbs free energy can be additionally decreased by separation into two phases for example with compositions X g and Equilibrium in Heterogeneous Systems 11 Let s consider a systems with two phases a and 3 that have compositions XEBL and XEB and Gibbs free energies GBL and GB If the average composition of the system is XE the molar free energy GM is given by the point on a straight line that connects GBL and GB and the relative number of moles of phases a and 3 can be found by the lever rule Ma X13a X130 pr XE Mg X130 XBWQB X13 Molar free energy of the phase mixture with molar fractions of phases a and 3 L3L2 MGETW Gw MBL and MB MMquot l W Equilibrium in Heterogeneous Systems IV The total Gibbs free energy is at minimum if phases a and 3 have compositions X and B These are the equilibrium compositions ofthe phases In equilibrium tangenm to GBL and GB curves are the same lie on a c mmon lin Therefore the condition for heterogeneous equilibrium is equity of the chemical potentials and activities of each component in the two phases LE GE RTln 21E 1 A G A RTln a Binary phase diagrams Binary phase diagrams and Gibbs free energy curves Binary solutions with unlimited solubility Relative proportion of phases tie lines and the lever principle Development ofmicrostructure in isomorphous alloys Binary eutectic systems limited solid solubility Solid state reactions eutectoid peritectoid reactions Binary systems with intermediate phasescompounds The ironcarbon system steel and cast iron Gibbs phase rule Temperature dependence of solubility Threecomponent ternary phase diagrams Reading Chapters 151 7157 ofPorter and Easterling Chapter 10 of Gaskell Mmm v urn1 Binary phase diagram and Gibbs free energy A binary phase diagram is a temperature composition map which indicates the equilibrium phases present at a given temperature and composition The equilibrium state can be found from the Gibbs free energy dependence on temperature and composition We have discussed the G dependence ofG ofa one component system on T We have also discussed the dependence ofthe Gibbs free energy from composition at a given T GXAGAXEGE AHmrTASm G Binary solutions with unlimited solubility 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 7 liquid and solid Let s consider Gibbs free energy curves for the two phases at different T gt T1 is above the equilibrium melting temperatures of both pure components Tl gt Tm A gt TmB 7 the liquid phase will be the stable phase for any composition G sum 3 G 2mm T G gqurd G 2 Gsulid 0 XB l G XAGA XEGE RTXAlnXA XElnXE Mmm v urn1 Binary solutions with unlimited solubility II I 39 T1 Will 0 2 Gym and Glgqmd will increase more rapidly than GT and G Why 2 The curvature ofthe GXE curves will decrease Why gt Eventually we will reach T2 7 melting point of pure component A where G G G SEEIle T 2 Gliquid Gliquid Gsulid E A A Gsulid 0 XB 1 Binary solutions with unlimited solubility 111 gt For even lower temperature T3 lt T2 TmA the Gibbs free energy curves for the liquid and solid phases will cross liquid G A solid GA 39 39 solidi quotgt SOME liquid 5 liquid As we discussed before the common tangent construction can be used to show that for compositions near crossover of Gsul d and Gl qmd the total Gibbs free energy can be minimized by separation into two plumes Mmm quotl 39rl7hquotl39 Binary solutions with unlimited solubility V Based on the Gibbs free energy curves we can now construct a phase diagram for a binary isomorphous systems 1 d solid G 3 T3 G a Gsulid Grim Gliiquid T1 solid solid liquid liquid T T1 T2 T2 T3 T4 T4 T5 Mmm quotl 39d7hquotl39 Binary solutions with unlimited solubility IV As temperature decreases below T3 G1 andGl qh d continue suiia suiia to increase more rapidly than GA 811ng gt Therefore the intersection ofthe Gibbs free energy curves as well as points X1 and X2 are shifting to the right until at T4 TmB the curves will intersect at Xl X2 1 G liquid T4 A thmd liquid solid G E G E solid GA 0 XB 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 ithe solid phase is the only stable phase Binary solutions with unlimited solubility VI Example of isomorphous system CuNi the complete solubility occurs because both Cu and Ni have the same crystal structure FCC similar radii electronegativity and Valence 7 2800 Liquid 1453 C7 2600 moo 7 O i g quuldus llne Solidus m 2m 15 1300 E 3 5 E 7 c gt2 5 200 7 7 2200 1 Solid solution 7 2000 lDBSC moo cm CompuSiiion M9 N0 Ni Liquidus line separates liquid from liquid solid Solidus line separates nlid from limiid solid Binary solutions with unlimited solubility VII I onecomponent system melting occurs at a welldefined melting temperature In multicomponent systems melting occurs over the range of temperatures between the solidus and liquidus lines Solid and liquid phases are in equilibrium in this temperature range liquid solution Temperature liquid solution crystallites of solid solution A 0 40 60 80 B polycrystal solid solution COmPOSilion Wt M mm d 71 Interpretation ol Phase Diagrams For a given temperature and composition we can use phase diagram to determine 1 The phases that are present 2 Compositions of the phases 3 The relative fractions ofthe phases Finding the composition in a two phase region 1 Locate composition and temperature in diagram 2 In two phase region draw the tie line or isotherm 3 Note intersection with phase boundaries Read compositions at the inters ections The liquid and solid phases have these compositions Interpretation of Phase Diagrams the Lever Rule Finding the amounts of phases in a two phase region 1 Locate composition and temperature in diagram 2 In two phase region draw the tie line or isotherm 3 Fraction of a phase is determined by taking the length ofthe tie line to the phase boundary or the other phase and dividing by the total length oftie line The lever rule is a mechanical analogy to the s balance calculation The tie line in the two phase region is analogous to a lever balanced on a fulcrum Derivation of the lever rule 1 All material must be in one phase or the other Wa WB 2 Mass of a component that is present in both phases equal to the mass of the component in one phase mass 0 component in the second phase WaCa B B D 3 Solution of these equations gives us the Lever rule WB CU Cu CB Cu and Wm C CU CB Cu Comp ositionConcentr ation weight fraction vs molar fraction Composition can be expressed in trying to under tand the ma erial at the atomic level Atom percent at is a number of moles atoms of a particular element relative to the total number of moles atoms in alloy For twocomponent system concentration of element B in at is Molar fraction 3 or atom percent at that is useful when s t n Ca E quot Agtlt100 C XEgtlt100 nm nm Where nmA and nt are numbers of moles of elements A and B in the system Weight percent C wt that is useful when making the solution Weight percent is the weight of a particular component relative to the total alloy weight For twocomponent system concentration of element B in wt is m V Egtlt 100 InB InA where mA and mE are the weights of the components in the system A mA E ma where AA and AE are atomic AA AE weightsofelementsAandB 1 H 1 H Comp ositjon Conver sions CWA Weight to Atomic C x 100 cE AA cA AE CWA Ci gtlt100 cE AA cA AE o o it Atomic o toWelght 0 CEquot At CEAE X100 CEAE CAAA it 3 X100 CE AE CAAA Of course the lever rule can be formulated for any speci cation of composition ML 7 09 agoX XBL 7 ca a Caz ca a Caz Ma XBU XBLXBOL XBL Cato call catat call wt 7 Cm cm cwt cwt W th CM cm th Mpquot l w Phase compositions and amounts An example 1300 7 Liquid 7 Q We iiiie g u Liquid g 7 7 E e ti Liquid a 1200 7 i 2 so 40 5 LL Ca Campasitian Wi Ni Cu 35 wt CL 315 wt 0 05425 wt Mass fractions WL S RS CuL Ca Cm CL 068 Wm RRSCD cgCm CL032 allquot v 39dh39w39 Development of microstructure in isomorphous alloys Equilibrium very slow cooling i i ti 46 Ni 1300 7 L 32 Ni 15 Ni 8 t L 39u 43 ms Ni 2 L 24 Ni N E i 3 E 2 1200 i mr i i i 2 30 no 50 Development of microstructure in isomorphous alloys Equilibrium very slow cooling gtSolidi cation in the solid liquid phase occurs gradually upon cooling from the liquidus line gt The composition of the solid and the liquid change gradu during cooling as can be determined by the tieline method lt gtNuclei of the solid phase form and they grow to consume all the liquid at the solidus line Development oimicrostructure in isomorphous alloys Nonsequilibrium cooling 1300 7 quot40 Ni luuw N g o 2 200 7 Hm Binary solutions with a miscibility gap e onsider system in which the liquid phase is approximately ideal but for the solid phase we have Anni gt o G Tl G liquid G solid 7 T3 aim XE At lowtemperatures there is aregion where the solid solution i most stable as a mixture of two phases an wi compositions x and X This region is called amiscibility gap MFM Development ofmicrostructurein isomorphous alloys Nonsequilibrium cooling Compositional changes require diffusion in solid and liquid phases Diffusion in the solid state is very slow 2 The new layers grains and the invalidity of the tieline method to determine the composition ofthe solid phase e tieline method still Works for the liquid phase where dus line is shifted to the right higher Ni contents solidification is complete at lower T the outer part of the grains are richer in the lowmelting component Cu Upon heating grain boundaries will melt first This can lead to premature mechanical ai ure mp u Eutectic phase diagram For an even larger Ame the miscibility gap can extend into the liquid phase region In this case we have eutectic phase diagram G l ll l 53 7 3 t 9 3 s Eutectic phase diagramwilh di 39erem crystal uuctures or pure phases lmllar euteeae phase diagram ean result if pure A andB have A s different erystal struetures G Temperature dqiendence ofsolubility e s limited snlid snluhility ofA tn B andB tn A tn the alloy having euteeae phase diagam shoWn below liquid T a t W Xx The eloser is the rnrnrrnurn of the Gibbs free energy curve to the axes XE 0 the smaller is the maximum possible eoneentraaon of B in phase a T er re to drseuss the ternperature dependenee of solubrhty let39s find the rnrnrrnurn of Gian GquotS XAGA XEGE QXAXE RTXAlnXA XElnXE dc dXs 39 M mmm f 305 gt See problem 1 of th L39 id x quot1 5 liquid OLB Copper 7 Silver phase diagram CumpusmunwlquotaAg sliquid a Eutectic systems 7 alloys with limited solubility II t t Lead 7 Tm phase diagram Tnnpuature c tsn W Cumpustttun wtvo Sn u eetie or invariant paint quid and two solid phases cor enrst equilibrium at the euteetre eornposrtron c2 and the euteetreternperature E Eutectic isntlnsrm the horizontal sohdus line at TE sohdphases ot 13 at euteeae eoneentraaon cE Th rneltrng pornt of the euteeae alloy is lower than that of the eornponents Eutectic easy m melt in Greek Eutectic systems e alloys with limited solubility 111 Composltlons and relatlve amounts of phases are oletermlneol from the same he llnes anollever rule as forlsomorphous alloys Tnnpnature c l 1 l l l l l l n n An 5 vn mu Cumpuslttun vth Sn tan For polnts A B and c calculate the composltlons vvt and relatlve amounts mass fractlons of phases present Development or miu39ostructure in eutectic alloys I Several different types of mlcrostructure can be formed ln slovv coollng an different composltaon Let39s consloler coollng of llqutolleaoletln system as an exam le 1n the case ofleadrnch alloy 02 vvt of tln sollolflcatlon proceeols tn the same manner as for lsomorphous alloys ue Nlt atvve discussed earller Tempelature c L 4aL Cumpusuunv o Sn Development or microsuucture in euteotic alloys 11 At composltlons between the room temperature solublllty llmlt d e maximum solid solublllty at the cute phase nucleates as the 1 solid solublllty ls exceeded upon crosslng the solvus llne Tnnpnature c H lt rs pa 1Lltnlt Cumpuslttun vth Sn Tnnpelature c Development or microstructure in eutectic alloys 11 Solidi cation at the eutectic composition No changes above the eutectlc temperature TE At TE all the llquldtxansforms to ctanol bphases Eutectic reaction Development of microstructure in eutedic alloys IV Solidi cation at the eutectic composition Composltlohs of a and 5 phases are very dlfferent gt euteeue reaetloh lnvolves reds tnb uoh of Pb and Sr atoms by atomle altrusloh we wlll le f s u am about altrusloh m the last part o tlus eourse Thls slmultaneous formatlon of a and phases result m a layered lamellar mlerostrueture that ls ealleol Eutectic Structure Formauoh ofthe euteeue saueture m the leadrtm system In the rmerograph the darklayers are leadrreaeh aphase the llght layers are the unrreach Bphase Development or microstructure in eutectic alloys V Compositions other than eutectic but within the range of th 39 therm e eutectic lso Primary aphase ls formed m the 1 L regloh and the euteetle and euteeu39e p phases ls formed upon erosslhg the euteetle lsothel39m L gt a L gt a p 3 Tempelature c Compusltlun wtvs sh Development of microstructure in euteou39e alloys VI Mcrncnns tuent 7 element of the mlcrostxucture havlng a though the euteeue strueture conslsts of two es lt ls a Al phas mleroeohsutueht wlth alstlhet lamellar saueture and xed rauo e two phases How to calculate relative amounts ofmicroconstituents Euteeue mleroeohstltueht forms from llquld havlng euteetl eomposluoh 619 wt Sn We ear treat the euteetle as a separate phase and apply the leye rule to flnd the relauye fractlons of pnmary a phase 18 3 wtquot Sn and the euteeue structure619 wt So we P PQ eutectic Wa Q 90 yummy me I utL L1 2007 r M E quot E E mi n lm I I 39m was r my 9 a Compusltlun wtvs sh How to calculate the total amount of 0 phase both eutectic and primary Fraction of on phase determined by application of the lever rule across the entire or 3 phase eld Wu QR PQR a phase W P PQR B phase 300 7 L a L Li 200 7 a 5 L 5 15 g 7 P Q R o 2 E 100 7 O Pb SH 183 C39A 619 978 Composition wt Sn M wm Vinnir Tnnnid m n Binary solutions with AHmixlt 0 ordering If AHmix lt 0 bonding becomes stronger upon mixing gt melting point of the mixture will be higher than the ones of the pure components For the solid phase strong interaction between unlike atoms can lead to partial ordering gt iAH i can mix become larger than iQXAXBi and the Gibbs free energy curve for the solid phase can become steeper than the one for liquid G 0 XB 1 B O C At low temperatures strong attraction between 0 O O O O O unlike atoms can lead to the formation of 2 8 0 0 O ordered phase on n n n n n Binary solutions with AHmix lt 0 intermediate phases If attraction between unlike atoms is very strong the ordered phase may extend up to the liquid In simple eutectic systems discussed above there are only two solid phases on and 3 that exist near the ends of phase diagrams Phases that are separated from the composition extremes 0 and 100 are called intermediate phases They can have crystal structure different from structures of componenw A and B MSE 30517hzse Dizg zms and Kinetics Leonid Zhigilei AHmmlt0 tendency to form highmelting point intermediate phase T liquid 0 5 Increasing negative AHmix Phase diagrams with intermediate phases example Example of intermediate solid solution phases in CuZn on and n are terminal solid solutions 3 7 5 2 are intermediate solid solutions w pm Miquot i l i i m iii vi mm mm w 7 m M 171quot 39 Vinnir Tnnnirl iii ilwi Phase diagrams for systems containing compounds For some systems instead of an intermediate phase an intermetallic compound of speci c composition forms Compound is represented on the phase diagram as a Vertical line since the composition is a speci c Value using the lever rule compound is treated like any other phase except they appear not as a wide region but as a Vertical line I i i i i i W intermetallic compound in This diagram can be thought ofas twojoined eutectic diagrams for MgMgsz and MgZPbPb In this case compound Mgsz 19 wt Mg and 81 wt Pb can be considered as a component A sh drop in the Gibbs free energy at the compound composition should be added to Gibbs free energy curves for the existing phases in the system and r A Compounds which have a single wellde ned composition are called stoichiometric compounds typically denoted by their chemical formula Compound with composition that can Vary oVer a nite range are called nonstoichiometric compounds or intermediate phase typically denoted by Greek letters nonstoichiometric stoichiometric Common stoichiometric compounds Icompound l AB A613 l ASB4 l A413 l 1413Z l A513 l Icompositionat 500 455 444 429 400 375 Icompound IAZBI ASBZ l A313 l A413 l A513 l A B l Icompositionat 333 286 250 200 167 143 MSE 30517hzse Dizg zms and Kinetics Lennid Zhigilei Eutectoid Reactions The eutectoid eutecticlike in Greek reaction is similar to the eutectic reaction but occurs from one solid phase to two new soli hases Eutectoid structures are similar to eutectic structures but are muc ner in scale diffusion is much slower in the solid state Upon cooling a solid phase transforms into two other soli phases 5 lt gt y 2 in the example be ow Looks as V on top of a horizontal tie line eutectoid isotherm in the phase diagmm 700 7 a g 500 7 7 8 E g Eutectoid 500 7 Peritectic Reactions Eutectic and Eutecuid Reactions form a seeonol sohol phase at a partreular temperature and TEmPe a m l eomposruon upon Cooling e e g L use b e reaeuons are rather slow as the produet phase wlll form at the boundary between the two reaetrng phases thus separating l Emacs them and slowing down any funherreactlon E a temperature Euteetorol P temperature m E n5 E K E H Euteetorol Euteetrr eomposrtron Composition Composruon Th aboye phase diagram eontarns both an euteetre reaetron and e its solxdrstateanalogtaneutentmdreawon Peritectnid is a threerphase reaetaon Similar to penteeue bu oeeurs from two Solid phases to one new Solid phase 01 13 y Example The Irnnelrnn Carbide FieFu C Phase Diagram Films in 1727ic Phase Diagram In their simplest form steels are alloys of Iron Fe and Carbon C The EC phase diagram is a fairly Complex one but we gt guitar snlidsnlutjnn nfCin BCCFe wlll only Consider the steel part of the diagram up to around 7 Stable pm 0pm a mom tempmmm Carbon The maximum Solubility of Cs 0 022 wt Transforms to cc 39yraustemte at 912 c gt 39leISLEniter snlid salutian arc in FCC Fe e maximum Solubility of cls 214 wt Transforms to BCC Memte at 1395 c Is not stable below the euteeue temperature 727 C unless Cooledrapldly gt Merrite snlid salutian arc in BCC Fe E ure as arfemte Stable only at high T above 1394 c Melts at 1538 c gt FeC irnn carbide ur eemenute This rntermetalhe eompounol ls metastable rt remalns as a eompounol inde nitely at room T but olee mposes r a a very slowly within several years into arFe and graphite at 650 e700 c gt FeC liquid Snllltinn A few comments on F8F93C system C is an interstitial impurity in Fe It forms a solid solution with Gt 7 5 phases ofiron Maximum solubility in BCC oLferrite is limited max 0022 wt at 727 C BCC has relatively small interstitial positions Maximum solubility in FCC austenite is 214 wt at 1147 C FCC has larger interstitial positions Mechanical properties Cementite is very hard and brittle can strengthen steels Mechanical properties also depend on the microstructure that is how ferrite and cementite are mixed Magnetic properties on ferrite is magnetic below 768 C austenite is nonmagnetic Classi cation Three types offerrous alloys I Iron less than 0008 wt C in otiferrite at room T I Steels 0008 214 wt C usuallylt 1wt oLferrite Fe3C at room T I Cast iron 214 67 wt usuallylt 45 wt MSE 3051711332 Dizng and Kinetics Lennid Zhigjlei Eutectic and eutectoid reactions in F8F93C Eutectic 430 wt C 1147 C L 9 y Fe3C Eutectoid 076 wtC 727 C y076 wt C 4 a 0022 wt C Fe3C Eutectic and eutectoid reactions are very important in heat reatment of steels Development of Microstructure in Iron 7 Carbon alloys Microstructure depends on composition carbon content and heat treatment In the discussion below we consider slow cooling in which equilibrium is maintained lVIicrostructure of eutectoid steel 1 1 r 1000 y yt FESC 900 7 7 7 as v E 800 u y 7 0 7oo 7 500 500 7 a FE3C 7 ii 1 l Camposltion mm C Microstructure of eutectoid steel 11 When alloy of eutectoid composition 076 wt C is cooled slowly it forms perlite a lamellar or layered structure of two phases oLferrite and cementite Fe3C The layers of alternating phases in pearlite are formed for the same reason as layered structure of eutectic structures C atoms between ferrite 0022 wt and cementite 67 wt by atomic diffusion Mechanically pearlite has properties intermediate to so ductile ferrite and hard brittle cementite easyg za 39 In the micrograph the dark areas are 9 Fe3C layers the light phase is UL 5 n errite hum lal 39 mm u an Mlernstruemre nfhypnzlltectnid steel 1 Composltlons to the left of euteetord 0 022 7 076 wt C hypnzlltectnid 1255 than eemeterd rGreek alloys 1 11 gt uFejc Mcrnmncture hr hypnzlltectnid steel 11 the euteetord temperature plus the euteetord perhte that contam euteetord farms and cemenme Pearlite Fe3c Proeutectaill u Eutectoid a Mcrnmucture nrhypereeteetnlel steel 1 Composltlons to the rrght of euteetord 076 e 214 wt C hyperzlltectnid mete Lhan eeleemrd rGreek alloys 1 1Fejc gt aFejc l nee V V lane 7 m o v l l CV M erhtrhrer t t Kr n cementum W m Microstructure orhyporeutectoid steel 11 39 alloys contain proeutectoid cementite formed above th eutectoid temperature plus perlite that contain eutectoid ferrite and cementite How to calculate the relative amounts ofproeutectoid phase ot or FeZC and pearlite Application of the lever rule with tie line that extends from the eutectoid composition 075 mu c to o a o FeZC boundary 0022 WI c for hypoeutectoid alloys and to o Fejc a FeZC boundary 67wt c for hipereutectoid alloys 7 a Farm 4 ruJu l i bL 0022 CD H n 1 thumbs Lwlquot m Fraction of o phase is determined by application ofthe lever rule across the entire 0Fe3C phase eld Example for hypereutectoid alloy with composition c Fraction ofpearlite WP x VX 67 a 1 67 a 076 Fraction of proeutectoid cementite Wp c V VX CI 7 076 67 r 076 r FClC r n 7 m0 0022 0 u 10 minimum Mr C Dia mm in the iii t i i i Klimt ii 2500 5 A 5 H uFTNRITF tummuuoxl 2600 W quuu 7 7 1 1 nutmm GAMMA mom mm m c L F93C Y 1 im r50 lkon lsoa mum autumn 11 7 193 1400 ut mid lJM F ix Point limo n9 xw 700 430 667 moo wan uvoozumm nvrsuzuvzcnc CARBON l su so on 3 ms 7 CASYIRON pearlite naturally fanned composite hard amp brittle ceramic FeZC soft BCC iron ferrite The Gibbs phase rule 1 Let s consider asimple onecomponent system e areas where only one phase is stable both pressure and temperature can be Independently varied without upsetting the equilibrium gt there are 2 degrees of freedom Along the lines where two phases coexist in equilibrium only one variable can be Independently varie without upsetting the twophase equilibrium p an are related by the clapeyron equation gt there is only one degree offreedom At the triple point where solid liquid and vapor coexist any change in p or T would upset the threephase equilibrium gt there are no degrees offreedom In general the number of degrees of freedom F in a system that contains c components and can have Ph phases is given by the Gibbs phase rule The Gibbs phase rule 7 example an eutectic systems FC7Ph2 P const 39nmpnsintm gt In onephase regions of the phase diagram T and xE can be changed independently gt In tworphase regions F 1 It the temperamre is chosen independently the compositions oiboth phases are lixed gt Three phases L o p are in equilibrium only at the eutectic point in this two component system 0 an v 39dhi39ln39 The Gibbs phase rule 11 e now consider a multicomponent system containing c components and having Ph phases thermodynamic state of each phase can be described by pressure 1 temperature T and c 1 composition variables The state of the system can be then described by PhxC12 variables But how many ofthem are independent The condition for Ph phases to be at equilibrium are Ph1 equations Phl equations Phl equations c sets ofequations Ph1 equations Therefore we have Ph 7 1XC 2 equations that connect the variables in the system The number of degrees of freedom is the difference between the total number ofvariables in the system and the minimum number of equations among these variables that have to be satis ed in order to maintain the equilibrium F PhC1rPh71C 2 CrPh 2 F C 7Ph 2 e Gibbsphaserule Msru v 39dhi39l39 Mul component systems I The approach used for analysis of binary systems can be U 0102030405060706W90t00 WI Representation ofthe composition in atemary system the Gibbs triangle The total length ofthe red lines is 100 xA xE xc 1 Multicumponent systems 11 The szbsfree energy summer instead of curves for a binary system can be plotted for all the possible phases and for different temperatures A B The chemical potentials of A B and c of an hase in this system are given by the points where the angennalplane to the free energy surfaces intersects the A B an c axis Multicumponent systems 111 A threephase equilibrium in the ternary system for a given temperature can be derived by means of the tangential plane construction A r r two phases to be in equilibrium the chemical potentials should be equal that is the compositions of the two phases in equilibrium must be given by points connected by a common tangential plane eg 1 an m lative amounts of phases are given by the lever rule eg using tieline lm A 39 r39 39 39 can result from 39 39 39 simultaneously touching the Gibbs free energies ofthree phases eg points x y an 2 Eutectic point fourphase equilibrium between o p y and liquid An example of ternary system The ternary diagram of NiCrFe It includes Stainless Steel Wt ofCrgt 115 Wt ofFe gt 50 nAand1nconelkn Nickel based super alloys Inconel have very good corrosion resistance but are more expensive and therefore used in corrosive environments where Stainless Steels are not su icient piping on Nuclear Reactors or Steam Generators Mun v 39dhi139 Another example of ternary phase diagram oil 7 water a surfactant system mum mull anpmil Eltnnvmuum chase S rlactant K um or c m wuttsmnlav Vuldu nduk u Blcwn nuws Mchmml nun 6 il Drawing by Carlus Cu Umversity ufcinunnati Surfactants are surfaceactive molecules that can form interfaces between immiscible fluids such as oil and mter A large umber of structurally different phases can be formed such as droplet rodlike and bicontinuous microemulsions along with hexagonal lamellar and cubic liquid crystalline phases Nucleation and growth kinetics Homogeneous nucleation Critical mdius nucleation rate Heterogeneous nucleation Nucleation in melting and boiling Growth me chanisms VVVVVV Rate of a phase tmnsformation Reading Chapters 41and 42 of Porter and Easterling Solidi cation Kinetics in a pure substance I there is a driving force for solidi cation AGv G L G When a liquid is cooled below the melting temperature 5 At temperature Tquot G H5 T s G3 Hi T s AGv AHv T Asv At temperature Tm AG AHV TmASf 0 AH ASvm V T m For small undercooling AT we can assume that AH and ASv are independent of temperature neglect the difference in CF between liquid and solid m m x AH AH AT AGV m AHV 7 T v Tm Tm At any tempemture below Tm there is a driving force for solidi cation Nevertheless the liquid does not always solidify at Tm If energy is addedremoved quickly the system can be signi cantly undercooled or supercooled To answer why does this happen we should consider kinetics of the phase tmnsformation MF In v 39d7hquotl39 Solidi cation Kinetics in a pure substance 11 Tmnsformation to solid phase requires gt Nucleation ofnew p ase gt Growth of new phase Nucleation can be Heterogeneous 7 the new phase appears on the walls of the container at impurity particles etc Homogeneous 7 solid nuclei spontaneously appear within the undercooled phase liquid liquid liquid solid solid supercooled homogeneous heterogeneous liquid nucleation nucleation MN I an Homogeneous Nucleation I The Gibbs free energy of the system 39th a small spherical cluster of a solid hqmd is G2 VSGVS VLGVL ASLySL where VS is the volume of the solid sphere VL is the volume of liquid ASL is SOhd the solidliquid interfacial area GvS and GvL are e energies per unit volume of solid and liqiud respectively 75L is the solidliquid interfacial energy If at the same temperature we would have a supercooled liquid without any solid nuclei the Gibbs free energy of liquid would G1 Vs VLGVL The formation of a solid nucleus leads to a Gibbs free energy change ofAG G2 G1 VS GVL GVS AsLySL m Gquot G AGV M I negative below Tm always positive Homogeneous Nucleation II For a spherical nucleus of radius r AG 7VSAG V ASLYSL n RAG v 47 r2ySL interfacial energy 7 r2 volume energy 7 r3 For nucleus With a mdius r gt r the Gibbs free energy Will decrease if the nucleus grows r is the critical nucleus size Mmm v 39d7hquotl39 Homogeneous Nucleation III Atrr dAdri47rr2AGv87rrySL 0 SL 5L3 r2y AG167Y AGv 3AGV2 AHW a Using AGviT I 7 Ava AT SL 3 2 AG 16 Y T2m 1 2 mm W Both r and Gquot decrease With increasing undercooling Rate of Homogeneous Nucleation There is an energy barrier of AGquotr for formation of a solid nucleus of critical size r The probability of energy uctuation of size AGquotr is given b t e Arrhenius equation and the rate of homogeneous nucleation is AG quot I m Ioexp 7 kT nuclei per m3 per s 16 SL fr 1 Using AG r Y 2m 2 3AH 2quot AT Where A has a relatively Weak e endence on temperature as compared to ATZ l110m loexpi7 AT very strong dependence 1th There is critical undercooling for homogeneous nucleation ATEr gt there are virtually no nuclei until ATEr is reached and there 39 explosive nucleation at AT 0 ATcr AT Mmm v 39d7hquotl39 Heterogeneous Nucleation I Let s consider a simple example of heterogeneous nucleation of liquid sohd a nucleus of t e s ape of a spherical cap on a Wall of a Heleus container Three surface energies 7m 7 liquid container interface 7L5 7 liquidsolid interface 75c 7 solidcontainer interface Balancing the interfacial tensions in the plane of the container Wall gives yLC 75c 7L5 cose and the Wetting angle 9 is de ned by cose ch ySC 7L5 The formation of the nucleus leads to a Gibbs free energy change ofAG VS AGv ASLySL Ascysc Ascch VS 2 r3 2 cose l 7 cose23 ASL 22 r2 l 7 cose and ASC 2 r2 sin2e AGE n RAG v 47rr21SLS9 AG Se 9 2 cos 9X17 cos eZ4 Heterogeneous Nucleation II AG 1 7r r3AG V 47 rziSLS6 AG 1 se se 2 cosOXli cos e24 Atrr dA Gr 47 rZAGV 87 r 73Lse 0 dr ZYSL 167 VSL r Am MWS RWWm AGiie AG Active nucleation AT MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Nucleation ofmelting For solidliquidvapor interfaces often ySolidVapor gt ySolidLiquid yLiquidVapor In this case no superheating is required for nucleation of the liquid and surface melting can take place below Tm Melting starts from free surfaces Melting of small atomic clusters a crosssection through the center of the cluster is shown simulations by I Sethna Cornell University NISE 305 Phase Diagalns and Kinetics Leonid Zhigilei Heterogeneous Nucleation III r AGet S6AG hom if e 10 se 2 cos617 cos e24 moquot AG 2e AG 1 1quotquot 1 ex 7 A I bet om o P kT I 1h AG quot Ihet 1 6X 7 bet 0 P kT ht h I gtgtI m NISE 305 Phase Diagalns and Kinetics Leonid Zhigilei Short pulse laser melting gt conventional heterogeneous melting melting front propagation from the surface Lun f rr r i r superheated crystal gt homogeneous nucleation and growth of liquid regions inside the crystal aoi f Ke0a O 1 superheated crystal D S Ivanov andL V Zhigilei Phys Rev Lett 91 105701 200339 Phys Rev B 68 064114 2003 NISE 305 Phase Diagalns and Kinetics Leonid Zhigilei Growth and the rate of a phase transformation The next step after the nucleation is growth Growth of a new phase is typically diffusion controlled 7 becomes more active with increasing temperature D Doexp7 Total rate of a phase transformation is a product of nucleation rate driving force increases with quot but diffusion needed for atomic rearrangement slows down with T decrease and growth rate diffusion controlled transformation drivjng diffusion rate force MSE 305 Phase Diagrams and Kinetics Lennid Zhigjlei Growth mechanisms Smooth solidliquid interface typically advance by th I lateral growth of ledges Ledges can result fro intersecting the interface Spiral growth on dislocations AFM images ofgrowing crystal of KDP potassium dihydrogen phosphate by De Yoreo and LNL and Malkin and Kuznetsov University of California MSE 30517hzs2 Dizng and Kinetics Lennid Zhigjlei Growth mechanisms directional dependence The shape of a growing crystal can be affected by the fact that different crystal faces have different growth rates Closepacked lowenergy faces tend to grow slower and as a result they are the ones that are mostly present in a growing crystallite 1 For example water ice Ih has hexagonal crystal symmetry th is re ected in the symmetry of snow crystals The growth rate is fast pamllel to the basal 0001 and prism 1010 faces As a result very small snow crystals have shape of hexagonal prisms As they growth instabilities result in more complex shapes of larger snow crystals u Hhhl Wu osmium pr nfuiiuA Growth instabilities dendrites Material and heat dif rsion limits the rate at which a crystal can grow o en greatly affecting the shape of the growing crystals An example is the MullinsSekerka supersaturated instability Consider a at solid surface vapor growing into a supersatumted vapor If a small bump appears on the surface then the bump sticks out farther into the supersaturated medium and hence tends to grow faster than the surrounding surface crystal As a result the at growth is unstable and a crystal tend to grow into more complex shapes eg snow akes Patterns of snow crystals A letter from the sky I Lnddampl39elkhujg ne mgquot Q munquot c J via i no runwmwc Nakaya diagram The shape of snow crystals depends on the temperature and humidity of the atmosphere in which they have grown Vertical axis shows the density of water vapor in excess of saturation with respect to ice The black curve shows the saturation with respect to liquid water as a inction of erature Water dendrites in ice MSE ms Phase Diagrams and Kinetics Leonid zhigjie39 Twodimensional ice dendrites on windows httpwww msm cam ac uk MSE ms Phase Diagrams and Kinetics Leonid zhigjiei Crystal growth and heat flow during solidification Thermal dendrites Bragard et al Inter nal Sczzncz 10 121 2002 httpwww msm cam ac ukphasetzrans Phase Transitions and Phase Diagrams Onecomponent systems Enthalpy and entropy dependence on P and T Gibbs ee energy dependence on P and T Clapeyron equation Understanding phase diagrams for onecomponent systems Polymorphic phase transitions Driving force for a phase transition First order and secondorder phase transitions Reading 12 of Porter and Easterling Chapter 71 74 of Gaskell 67 and 94 of White MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei PVT Surface of a Pure Substance W http wwwengusf educampbellThermoIThermolmodhtml A pure substance is heated at constant pressure MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei H and S as function of T at constant P In a closed onecomponent system equilibrium at temperature T and pressure P corresponds to the state with minimum Gibbs free energy G Therefore in order to predict what phases are stable under different conditions we have to examine the dependence of G on T and P Let s use thermodynamic relations to predict the temperature dependence of H S and G at constant P T For HT we have c HT H298 I CPdT 6T P 298 T For ST we have ST dT CP 0 T K H R Slope Cp Slope CPT O TK 7 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei G as function of T at constant P For G H TS we have dG SdT VdP and for P const a G S f h 1 6113 ortesope 2 a G CP for the curvature 6T2 P 6T P T Slope CP 0 A T K iTS Slope S G GT for a single phase at P const MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei G as function of T at constant P for liquid and solid phases At all temperatures the liquid has a higher internal energy U and enthalpy H as compared to the solid Therefore G1 gt GS at low T The liquid phase however has a higher entropy S than the solid phase at all T Therefore G1 decreases more rapidly with T as compared to GS At TIn G1T crosses GST and both liquid and solid phases can coeXist in equilibrium G1 GS Slgtss at all T AHm AHm TmASm At TIn the heat supplied to the system will not rise its temperature but will be used to supply the latent heat of melting AHIn that is required to convert solid into liquid At Tm the heat capacity CI 6H6TP is in nite addition of heat does not increase T MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei A typical PT phase diagram for a pure material 1 d critical P 1 atm so 1 pomt normal G P freezing liquid point hquld 1 atm solid gas normal boiling triple point point T T gt 4 4 solid is liquid is i vaporis stable stable stable The red lines on the phase diagram show the conditions where different phases coeX1st 1n equ111br1um Gphasel Gphasez MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei G as function of P at constant T for liquid and solid phases As we can see from the fundamental equation dG VdP SdT the free energy of a phase increases with pressure 6G V gt O 6P T If the two phases have different molar volumes their free energies will increase by different amounts when pressure changes at a xed T G 920 V1 lt VS for water hot 7 V1 gt VS for most materials TO C 1 atm 13 How the unusual change of V upon melting of water could be related to iceskating What is the curvature of the GP at constant T T 1 6V isothermal 5P bulk B V V 5P T compress1b111ty av T modulus MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Equilibrium between two phases Clapeyron equation If two phases in equilibrium have different molar volumes their free energies will increase by different amounts when pressure changes at a xed T The equilibrium therefore will be disturbed by the change in pressure The only way to maintain equilibrium at different pressures is to change temperature as well For two phases in equilibrium G1 GS and dG1 dGS for in nitesimal change in T and P so that the system remains in equilibrium dGl VldP 39 SldT VldP SldT deP SSdT dGS deP SSdT d3 ss s1g dT m VS V1 AV At equilibrium AG AH TA S O and AH TA S d T TAV dP AH Therefore the Clapeyron equation cq The Clapeyron equation gives the relationship between the variations of pressure and temperature required for maintaining equilibrium between the two phases MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei G as function of P and T for liquid and solid phases Schematic representation of the equilibrium surfaces of the solid and liquid phases of water in GTP space The planes show the free energies of liquid and solid phases the intersections of the planes correspond to the P T conditions needed for maintaining equilibrium between the phases G1 GS solid 0 C MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Clapeyron equation examples A typical diagram for a pure material For liquid to gas transition AV Vg V1 gtgt O AHHgH1gtO wehaveto add heat to convert liquid to gas gt0 dT TAV T Therefore E 11 AH cq For liquid to solid transition AV VS V1 lt O for most materials AH HS H1 lt O heat is released upon crystallization Therefore dP AH gt0 dT m TAV Tr This is the case for H20 ice oats in liquid water Ga Ge diamond dP For some materials however AV VS V1 gt O and E j lt 0 eq In general it takes heat AH gt O to proceed from a lowtemperature to a hightemperature phase entropy of a hightemperature phase is higher than the entropy of a lowtemperature phase Therefore the slope of the equilibrium lines in a PT phase diagram of a pure material re ects the relative densities of the two phases MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Clapeyron equation more examples Some materials may exist in more than one crystal structure this is called polymorphism If the material is an elemental solid it is called allotropy diagram shows the stable phases for pure iron at varying The temperature and pressure Closepacked FCC yFe has a smaller molar volume than BCC czFe AV Vy Vmlt 0 At the same time AH mgt o 39Iherefore lt 0 dT q TAV o 510152025303540455 PressureGPa The effect of increasing pressure is to increase the area of the phase 39 am over Which the phases of the smaller molar volume higher diag density is stable MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Phase diagrams of ohocompohom materials Anodm example of allotropy is carbon which can exist as diamond graphite and amoiphous carbon a g Diaman amp mm mm mm Temperature K aurkmmmmueme equilibrium conditions Diamon carbon Graphite oi liquid carbon Why do not we see fullerenes and nanotubes on ch phase diagram From ch phase diagram what phase has higher density near twophase 39 d i graphite Diamond oi liquid MSE sns Mass Diagrams and Kinzljcs Lamid Zhigjlzi Example squeezing diamond from graphite What pressure should we apply to transform graphite to diamond at 298 K In a reference book we can nd that at 298 K ngaphite Hdiamond 391900 J Sgraphite Sdiamond pgraphite gCm3 pdiamond gCm3 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei The driving force for the phase transformation If solid and liquid are in equilibrium GS G1 and a slow addition of heat leads to the melting of some part of the solid but do not change the total G of the system G 111 G1 nS GS const where 111 and 11S are the numbers of moles of liquid and solid phases and G1 and GS are the molar Gibbs free energies If energy is addedremoved quickly the system can be brought out of equilibrium overheated or undercooled the meltingfreezing process is spontaneousirreversible and G is decreasing At temperature Tquot G1 H1 TSl AG G Gs Hs T 8 Gs AG 2 AH TAS AT G1 At temperature Tm AG AHTmAS O T TIn AHm AS T III For small undercooling AT we can neglect AH the difference in CI of liquid and solid AGzAHm T T m phases and assume that AH and AS are m independent of temperature The driving force for solidi cation T MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Firstorder and secondorder phase transitions 1 The classi cation of phase transitions proposed by Ehrenfest is based on the behavior of G near the phase transformation gt Firstorder phase transition rst derivatives of G are discontinuous gt Secondorder phase transition rst derivatives of G are continuous but second derivatives of G are discontinuous Firstorder phase transition G V I S 39n I a g Ttrs T Ttrs T Ttrs T H Cp AGtrS O AH E trs trs Ttrs Tm T Tm T ASm i 0 S V discontinuous AHtrs 75 0 5T P 6P T C dH eg melting boiling sublimation some P dT P polymorphous phase transitions Firstorder and secondorder phase transitions II Secondorder phase transition G V S Ttrs T Ttrs T Ttrs T H Cp J AGES 2 o Astrs 0 E AHtrs O Tm T T T AVtrs 0 dH KG 25 KG V CPZECTT 5T P 5P T P continuous S and V do not jump at transition 52G 52G discontinuous 5T6P 5T P aP T 6P T eg conductingsuperconducting transition in metals at low temperatures MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Review of classical thermodynamics Fundamental Laws Properties and Processes 1 First Law Energy Balance Thermodynamic functions of state Internal energy heat and work Types of paths isobaric isochoric isothermal adiabatic cyclic Enthalpy heat capacity heat of formation phase transformations Calculation of enthalpy as a function of temperature Heats of reactions and the Hess s law Reading Chapters 2 61 64 of Gaskell or the same material in any other textbook on thermodynamics MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Thermodynamic variables What are thermodynamics variables There are two approaches to describe properties and behavior of a material 1 Microscopic approach to describe the material in terms of microscopic variables positions velocities charges etc of all particles in the system But there are too many particles N A 6022gtlt1023 mol39l and this approach in unpractical in most cases 2 Classical continuum thermodynamics to describe the material in terms of average quantities or thermodynamic variables such as temperature internal energy pressure etc Statistical thermodynamics provides the connection between the classical thermodynamics and the behavior of the microscopic constituents of matter atoms and molecules Although in this course we will focus on classical thermodynamics we will also consider a few elements of statistical thermodynamics in particular in our discussion of heat capacity and entropy What are state variables and functions System at equilibrium can be described by a number of thermodynamic variables that are independent of the history of the system Such variables are called state variables or state functions depending on the context We can describe a system by a set of independent state variables and we can express other variables state functions through this set of independent variables For example we can describe ideal gas by P and T and use V RTP to de ne V For different applications we can choose different sets of independent variables that are the most convenient MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Thermodynamic variables Intensive and extensive variables Intensive properties independent of the size of the system e g T P Extensive properties proportional to the quantity of material e g V U C H S G Example if V1 V2 2 T1 T2 P1 P2 then v12 v1 v2 but T12 T1 T2 We can also consider derived intensive variables eg MassVolume or EnergyVolume that do not depend on the size of the system Internal energy heat and work not very rigorous definitions It is impossible to give a rigorous de nition of energy in physics today we have no knowledge of what energy is the Feynman Lectures on Physics Thermodynamics laws do not de ne energy thermodynamics is dealing with transfer of energy In particular the 1st law of thermodynamics postulate the energy conservation In thermodynamics of materials we usually do not consider the kinetic energy of the centerof mass motion of the system or gravitational energy mgh only internal energy intrinsic to the body is considered Internal energy U is a sum of all potential and kinetic energies in the system not only of mechanical origin Thermodynamics is only dealing with change of U The absolute value of U is not de ned by the laws of thermodynamics but an arbitrary zero point is often chosen for convenience MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Energy heat and work continued Heat is the energy being transferred to a system as a result of temperature difference workless transfer of internal energy Work can be defined as the energy being transferred to a system as a result of a generalized force acting over a generalized distance Examples of work Mechanical work done by force F on a body moving from r1 to r2 along a certain trajectory or path gt F2 5wFdr U1gt2L Fdr l path integral Work due to the volume expansion of a uid or gas done against an external pressure P 5W 2 P dV Electric polarization work where the generalized force is the strength of the electric field E and the generalized displacement is the polarization of the medium D 8W E Magnetic work where the generalized force is the strength of the magnetic field H and the generalized displacement is the total magnetic dipole moment B 5w H dB Work and heat are both functions of the path of the process they are not state functions Systems never possess heat and work Heat and work are transient phenomena describe energy being transferred tofrom the system MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei 1St Law conservation of energy in a thermodynamic process A state function called the internal energy exists for any physical system and the change in the internal energy during any process is the sum of the work done onby the system and the heat transferred tofrom the system AU q w or in differential form dU Sq 5w U internal energy all potential and kinetic energies It is a state function depends only on thermodynamic state of the system eg P V amp T for a simple system q energy added into the system as heat Positive when the system gains heat from outside endothermic process negative when heat ows out of the system exothermic process w work done by the system on its surroundings Positive when work is done by the system and negative if work done on the system If body does work it expends energy and the internal energy of the body must decrease Note that in some textbooks you will nd a plus sign in front of SW work done on the body or minus sign in front of Sq heat ow out of the body We use notation adapted in Gaskell Example gas expands mechanical work is done against an external 2 pressure W2 P dVgt O cooling by adiabatic expansion ext MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Equivalence of heat and work Joule experiment In loule39s experiment the mechanical energy can be measured simultaneously with temperature thermal energy Joule found that gt the loss in mechanical energy is proportional to an increase in temperature of the water and the amount of water used gt the constant or proportionality is 44 lg C modern data is 4186 lg C Heat and work can independently produce identical changes in the system T is not a good measure of heat but heat can be measured through work and heat capacity can be determined 1 P Joule On the existence of an equivalent relation between heat and the ordinary forms ofmechanical power Phil Mag 27 205 1845 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Types of paths A simple onecomponent system can be described by T P and V They are connected by equation of state eg VVPT Therefore two independent variables describe the system and de ne the state functions eg U UPT Let s consider processes when one of the two independent variables is xed V const isochoric process No work is done w lPdV O and the 1st law takes form dU Sq or AU q internal energy can be changed only by heat exchange P const isobaric process V2 V2 w IPdV P IdV PV2 V1 and the 1st law takes form V1 V1 U2 U1 qp PV2 V1 or U2 PVz U1 PVI qp qp is heat added at constant pressure H U PV enthalpy state function since UPV are state functions H H AH change in enthalpy equals to heat added to 2 1 qp the system at constant pressure MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Types of paths continued T const isothermal process Example 39 ideal gas dT 0 therefore dU Sq SW 0 internal energy of an ideal gas is a function only of T Work done depends on the path ie how the external pressure is changing during the transformation For example gt Free expansion no external pressure W O gt Reversible isothermal expansion Pext Pgas at all times reversible process system is always at equilibrium Sq SW PdV RTdVV per mole of gas Integration between states 1 and 2 gives q w RT 1nV2V1 RT 1nP1P2 Work done by the system heat absorbed by the system Q 0 adiabatic process AU w no heat exchange the internal energy can be changed only by work Real processes are often complex P V and T all are changing In this case state functions can be calculated by breaking process into a series of reversible isothermal isobaric or isochoric processes that bring system to correct nal state MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei PVT Surface of 21 Pure Substance MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Types of thermodynamic systems Isolated system No energy and no matter can pass through the boundaries of the system Closed system Energy can pass through the boundaries as heat andor work but matter cannot Adiabatic system No heat can pass through the boundary neither can matter that can carry heat ideal thermos Work can be performed on or by the system Open system Both energy and matter may pass through the boundaries surroundings gt work 4 gt heat surroundings gt work 4 surroundings gt work lt gt heat 4 gt matter lt gt Hf surroundings An alternative formulation of the 1st law of thermodynamics The work done on a system during an adiabatic process is a state function and numerically equal to the change in internal energy of the system MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Heat Capacity The heat capacity C of a system is the ratio of the heat added to the system or withdrawn from the system to the resultant change in the temperature C qAT SqdT Jdeg This de nition is only valid in the absence of phase transitions Usually C is given as speci c heat capacity c per gram or per mol gt New state of the system is not de ned by T only need to specify or constrain second variable 561 Cv E j heat capacity at constant volume dT V heat capacity at constant pressure P dT P The fact that Sq is not a state function and depends on the path is re ected in the dependence of the heat capacity on the path cp 7 cV note that small c is used for the derived intensive quantity per mass per volume or per mole versus capital C for the extensive quantity For a system containing 11 moles Cp ncp and CV ch where cp and cV are molar values cV and cP can be measured experimentally isobaric process dH Sq cPdT isochoric process dU Sq chT H and U can be calculated from cP and cV MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei cV vs cp If material is allowed to expand during heating how this affects its heat 8H an av P 6T P 6T P 6T P Q 5T v 6U 6V 6U cP V P 6T P 6T P 6T V since U UVT dU a U dV a U dT av T M V capacity 6quot Ml Differentiation with respect to T at constant P gives therefore 8T P 6V T 8T P 8T V w av w av w av w cP cvz P 2 P avTaTP 6TV aTP 6TV aTP avT P work of expansion at constant P due to the P temperature increase by dT 5V 5U work of expansion against internal cohesive 6T P 6V T forces due to the temperature increase by dT MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Calculation of enthalpy from heat capacity For P const dH cP dT and integration gives H T2 T2 JdeHZ Hl JcPdT H2H1jcPdT H1 T1 T1 Example Let us nd enthalpy for copper at 500K cP z 244 Jmol39lK391 for copper at 1 atm From the 1st law can only calculate the difference AH need a reference enthalpy Enthalpy at 1 atm and 298 K is called enthalpy of formation H298 For pure elements in their equilibrium states H298 O 500 500 H500 H298 jcPdT 0 I244dT 49kJmol 298 298 Enthalpy of substances other than pure elements can also be calculated The enthalpy of a compound at 298 K standard heat of formation of the substance from the elements Example For oxidation of copper at 25 C Cusolid 12 Ozgas Cuosolid AHES H355 HC H02 H 0 1561kJmol 298 298 298 The reaction is exothermic heat andor work are produced MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Calculation of enthalpy from heat capacity continued In general heat capacity is a function of temperature For example for alumina A1203 the temperature dependence can be described by cpo 1175 104gtlt10393 T 37lgtlt105 T392 in the range 2982325 K 500 HAIzOs H 91 Os I cPTdT 16757 kJm01 500 298 500 1175 104x10393 T 371gtlt105 T392dT 298 The standard heat of formation H 91 O3 and heat capacity cpT are measured experimentally and can be found in themochemical tables e g at httpwebbooknistg0Vchemistry or at the end of Gaskell s textbook In most cases thermal treatment of materials is carried out under atmospheric pressure and no work other than that against the atmosphere is produced Enthalpy change is used to describe such processes MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Enthalpy and phase transformations If the system undergoes a phase transformation or a chemical reaction then the enthalpy change due to the phase change AHtranS has to be included into total enthalpy change Also different phases can have different heat capacities c T For example let s nd enthalpy for copper at 2000 K c1S01id 177 28lgtlt10393 T 3l2gtltlO396 T2l40gtlt10399 T3 686gtltlO4 T392 Jmol39lK391 in the solid state for T 298 1358 K this formula for cpT is given at httpwebbooknistgovchemistry Since this time we are considering a wide range of temperatures we should account for the temperature dependence of c1 c1 z 24 Jmol39lK391 at 298K 26 Jmol39lK391 at 500K 33 Jmol39lK391 at l358K cPliquid 328 Jmol39lK391 in the liquid state nearly independent on T AHm 12 kJ mol391 latent heat of melting always positive 1358 2000 H2000 2 H298 jc 11dTdT AHm j cgqmddr 298 1358 0301221 2 63k mol391 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Heat of formation and phase transformations If the temperature of interest is higher than the melting temperatures for both the metal and its oxide the enthalpy change for Mliquid 12 OZgas MOliquid is then TM m 1 AHT AH298 J CMOsolid cMs011d cOZgas P P 2 P 298 TMO m 1 J 02408011d cM11qu1d c2gas TM 2 P T 1 J cgIomquid cIliqu1d c2gas TMO 2 Mm 1 2 029 enthalpy o l I A H298 temperature Gaskell Chapter 6 4 Heats of Reactions Hess s Law Heat absorbed or released in a given chemical reaction is the same weather the process occurs in one or several steps Hess 1840 Example C graphite 02 gas CO2 gas Q 3936 kJ mol391 C graphite 12 02 gas C0 gas Q 1105 kJ mol391 CO gas 12 02 gas CO2 gas Q 2831 kJ mol391 Total 3936 kJ mol391 gt Hess s Law allows one to calculate Q for reactions that is hard to measure gt Presence of catalysts change the activation energy of reaction but not the net heat of reaction The Hess s law is just a consequence of the 1St law of thermodynamics for P const AH Q Since H is a state function total heat is independent of path MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Calcula on ofheat oftransi on from heats ofreac ons Let s nd if an allotropic transition from diamond to graphite under ambient conditions results in the release or absorption ofheat c graphite 02 gas co2 gas AHm 3935 kJ molquot c diamond 02 gas co2 gas ABM 3955 kJ molquot Then foneaeu39on c diamond c graphite AHm 3935 73955 19 kJ molquot Transition ofdiamond to graphite is an exomemie reaction Graphi za on ofa surface region of diamond at elevated temperature 7 Figure from a computer simulation MSE sns Phase Diagrams and Kimu39cs Lennitl zhigjlsi Theoretical calculation of the heat capacity gt Principle of equipartition of energy gt Heat capacity of ideal and real gases gt Heat capacity of solids DulongPetit Einstein Debye models gt Heat capacity of metals electronic contribution Reading Chapter 62 of Gaskell Chapter 61 66 of White MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Degrees of freedom and equipartition of energy For each atom in a solid or gas phase three coordinates have to be speci ed to describe the atom s position a single atom has 3 degrees of freedom for its motion A solid or a molecule composed of N atoms has 3N degrees of freedom We can also think about the number of degrees of freedom as the number of ways to absorb energy The theorem of equipartition of energy classical mechanics states that in thermal equilibrium the same average energy is associated with each independent degree of freedom and that the energy is 12 kBT For the interacting atoms eg liquid or solid for each atom we have 12 kT for kinetic energy and 12 kT for potential energy equality of kinetic and potential energy in harmonic approximation is addressed by the Virial theorem of classical mechanics Based on equipartition principle we can calculate heat capacity of the ideal gas of atoms each atom has 3 degrees of freedom and internal energy of 32kBT The molar internal energy U32NAkT32RT and the molar constant volume heat capacity is cVdUdTV32R In an ideal gas of molecules only internal Vibrational degrees of freedom have potential energy associated with them For example a diatomic molecule has 3 translational 2 rotational l Vibrational 6 total degrees of freedom Potential energy contributes 12 kBT only to the energy of the Vibrational degree of freedom and Umolecule 72kBT if all degrees of eedom are fully excited MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Temperature and velocities of atoms At equilibrium Velocity distribution is MaxwellBoltzmann 3 2 2 2 m A mv v v dNvT exp 7 dvxdvydvZ V2i 3kBTm lf system is not in equilibrium it is o en dif th to separate different contributions to the kinetic energy and to de ne temperature Acoustic emissions in the fracture simulation in 2D model Figure by BL Holian and R Ravelo Phys Rev B51 11275 1995 Atoms are colored by Velocities relative to the le toright local expansion Velocity Which causes the crack to advance from the bottom up NISE 305 Phase Diagrams and Kinetics Leonid Zhigilci Heat capacity of molecules straightforward application of equipartition principle does not work Classical mechanics should be used with caution when dealing with phenomena that are inherently quantized For example let s try to use equipartition theory to calculate the heat capacity of water Motion Degrees of freedom U cV Translational 3 3 X 12 RT 15R Rotational 3 3 X 12 RT 15R Vibrational 3 6 X 12 RT 3R Total cV 6R But experimental cV is much smaller At T 298 K H20 gas has cV 3038R What is the reason for the large discrepancy Rotation n gt T 1 Vibration ms am MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Heat capacity of molecules straightforward application of equipartition principle does not work continued What is the reason for the large difference between the prediction of classical calculations cV 6R and much smaller experimental cV 3038R at 25 C Rotation n Vibrationx The table shows the Vibrational frequencies of water along with the population of the rst excited state at 600 K Translation v cm39l EXphvkT 3825 10 X 10394 1654 19 X 10392 3936 80 X 10395 For the high frequency OH stretching motions there should be essentially no molecules in the rst Vibrational state even at 600 K For the lower frequency bending motion there will be about 2 of the molecules eXCited Contributions to the heat capacity can be considered classically only if En hv ltlt kBT Energy levels with EH 2 kT contribute little if at all to the heat capacity Only translational and rotational modes are eXCited the contribution from Vibrations is only 003 8R MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Heat capacity of solids Dulong Petit law In 1819 Dulong and Petit found experimentally that for many solids at room temperature cV z 3R 25 JK391mol391 This is consistent with equipartition theory energy added to solids takes the form of atomic Vibrations and both kinetic and potential energy is associated with the three degrees of freedom of each atom Ptgt Ktgt ngT The molar internal energy is then U 3NAkT 3RT and the molar constant volume heat capacity is cV 6U6TV 3R Although cV for many elements at room T are indeed close to 3R lowT measurements found a strong temperature dependence of cV Actually cV gtOasT gtOK 25 20 L 0391 cv joulesmoleK a diamond o l l I l l J l 0 50 100 150 200 250 300 temperature K MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Heat capacity of solids Einstein model The lowT behavior can be explained by quantum theory The rst explanation was proposed by Einstein in 1906 He considered a solid as an ensemble of independent quantum harmonic oscillators vibrating at a frequency 1 Quantum theory gives the energy of ith level of a harmonic quantum oscillator as 8i i 12 ho where i 012 and h is Planck s constant For a quantum harmonic oscillator the EinsteinBose statistics must be applied rather than MaxwellBoltzmann statistics and equipartition of energy for classical oscillators and the statistical distribution of energy in the vibrational states gives average energy h ltUtgt Chmk 1 There are three degrees of freedom per oscillator so the total internal energy per mol is 3N Aho Note you do not need to thkBT I remember all these scary quantum mechanics equations for testsexams but you do h 2 need to understand the basic 0 e huk BT concepts behind them 3N k C A B kBT V 5T V ehukBT 12 The Einstein formula gives a temperature dependent cV that approaches 3R as T gt 0C and approaching 0 as T gt 0 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei The High Temperature Limit of the Einstein Speci c Heat Let s show that Einstein s formula approaches Dulong Petit law at high T For high temperatures a series expansion of the exponential gives h D ehDkBT z kBT The Einstein speci c heat expression then becomes h 2 h 2 h 3NAkB U elmkBT 3NAkB U 1 U k B T k B T k B T c w 2 ho kBT chmkBT 12 3NAkB1 kh jz 3NAkB 3R In the Einstein treatment the appropriate frequency in the expression had to be determined empirically by comparison with experiment for each element Although the general match with experiment was reasonable it was not exact Einstein formula predicts faster decrease of cV as compared with experimental data Debye advanced the treatment by treating the quantum oscillators as collective modes in the solid phonons MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Heat capacity of solids Debye model Debye assumed a continuum of frequencies With a distribution of gv avz up to a maximum frequency VD called the Debye frequency This eads to the following expression for the Debye speci c gv heat capacity T genT x4eX cv QNAkBe I ex 12dx D r 0 Where X hvkT and 9D thk NDebye characteristic temperature 1214Nk T 3 A B 5 an For low temperatures Debye39s model predicts cV good agreement With experimental results m We can see that cV depends Cw JKquot quot101391 on TeD With 9D as the scaling function for different A E materials m 1 9D N vD N strength of the 1 W9 interatomic interaction N D H I i i quot n I I quot 539 i 39 Iain 39 39 25 1atomic weight V i I I I 1 force constant For a harmonic osclllator u 21 reduced mass MSE 305 Phase Diagams and Kinetics Leonid Zhigilei Heat capacity of metals electronic contribution cV 6U6TV therefore as soon as energy of electrons are changing with T they will make contribution to CV To contribute to bulk speci c heat the valence electrons would have to receive energy from the thermal energy kT But the Fermi energy is much greater than kT and the overwhelming majority of the electrons cannot receive such energy since there are no available energy levels within kT of their energy The small fraction of electrons which are within kT of the Fermi level de ned by FermiDirac statistics does make a small contribution to the speci c heat This contribution is proportional to temperature cvel VT and becomes signi cant at very low temperatures when cV yT AT3 for metals only 2 1o Law of Dulehg and Petit 39 I 1 3mm f ppreaches 3 SPEC DulungPetit E1 3339 at high temp x 10 2 i if 39 Low temperature 104 T5l behavior matches Debye model 10 I I I l I I I I I 1411 1D 103 1e 10 T3 K3 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei 4 10 Law of Dulung and Petit Copper specific heat c an1 matches Debye heat contributes 103 7 IDE Departs from Baby model at low temp where electan specific 1 Energy Band Structures Metals Empty states T g 7 j g El Er a e Electrm j t 7 7 exmtauon FiHed states Semiconductors and Insulators E 7 E 7 e i 2 i E E 3 E g 1 i g n 8 g Free r e ectron T a J a Q a g 2 5 E W Electron g exmtanon m y x 1 y 3 n c 3 u lt Hole In 5 c lt2 5 c v I g f 3 y we ence g 3 Q X band g MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Summary 1 Make sure you understand language and concepts Degrees of freedom Equipartition of energy Heat capacity of ideal and real gases Heat capacity of solids DulongPetit law VVVVV Quantum mechanical corrections Einstein and Debye models Heat capacity of gas solid or liquid tends to increase with temperature due to the increasing number of excited degrees of freedom requiring more energy to cause the same temperature rise The theoretical approaches to heat capacities discussed in this lecture are based on rather rough approximations anharmonicity is neglected phonon spectrum is approximated by v2 in Debye model etc In practice cpT in normally measured experimentally and the results are described analytically eg CI A BT CT392 for a certain range of temperatures MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Kinetics and Diffusion Basic concepts in kinetics Kinetics of phase transformations Activation free energy barrier Arrhenius rate equation Diffusion in Solids Phenomenological description Flux steadystate diffusion Fick s first law Nonsteadystate diffusion Fick s second law Driving force for diffusion Diffusion in ideal and real solutions Thermodynamic factor Diffusion against the concentration gradient Spinodal decomposition Solutions to the diffusion equation Numerical integration Analytical solution Applications Chemical homogenization Carburization of steel MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Kinetics basic concepts Thermodynamics can be used to predict what is the equilibrium state for a system and to calculate the driving force for a transformation from a metastable state to a stable equilibrium state How fast the transformation occurs is the question addressed by kinetics Let s consider transition from a metastable to equilibrium state The transformation between the initial and final states involves rearrangement of atoms the system should go through a transformation or reaction path Since the initial and final states are metastable or stable ones the energy of the system increases along any transformation path between them MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Kinetics basic concepts Transition path G G1 G2 Initial state Final state metastable ACtiVated State equilibrium or another metastable G1 and G2 are the Gibbs free energies of the initial and final states of the system AG G2 G1 is the driving force for the transformation AG3 is the activation free energy barrier for the transition the maximum energy along the transformation path relative to the energy of the initial state MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Kinetics basic concepts In order for a system to proceed through the transition path to the equilibrium state it has to obtain the energy that is sufficient to overcome the activation barrier The energy can be obtained from thermal uctuation when the thermal energy is pooled together in a small volume Statistical mechanics can be used to predict the probability that a system gets an energy that exceeds the activation energy This process is called thermal activation The probability of such thermal uctuation or the rate at which a transformation occurs depends exponentially on temperature and can be described by equation that is attributed to Swedish chemist Svante Arrhenius rate exp AGX T exp AHa k T B T B AGa AHa TASa Arrhenius equation can be applied to a wide range of thermally activated processes including diffusion that we consider next Arrhenius equation was first formulated by J J Hood on the basis of his studies of the variation of rate constants of some reactions with temperature Arrhenius demonstrated that it can be applied to any thermally activated process MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei What is diffusion Diffusion is mateIial transpon by atomic motion Atoms of type A Atoms of type E 20 Most kinetic processes in mateIials involve dif ision omogeneous mateIi s can become homogeneous by diffusion compositions of phases can change by dif ision For an active diffusion to occur the tempemture should be high enough to overcome energy baniers to atomic motion MSE sus Phase Diang and Kinetics Leonid Zhig39lei Phenomenological description of diffusion diffusion ux The ux of diffusing atoms J is used to quantify how fast diffusion occurs The ux is de ned as either number of atoms diffusing through unit area and per unit time e g atomsmZsecond or in terms of the mass ux mass of atoms diffusing through unit area per unit time eg kgmZsecond J Let s consider steady state diffusion the diffusion ux does not change With time Example diffusion of gas molecules through a thin metal plate gas at gas at C1 pressure P1 pressure P1 lt P2 C2 concentration pro le MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei SteadyState Diffusion Fick s first law Concentration gradient dCdX Kg m394 is the slope at a particular point on concentration pro le Fick s rst law the diffusion ux along direction X is proportional to the concentration gradient dC J D Where D is the diffusion coef cient dX C X The minus sign in the equation means that diffusion is down the concentration gradient Fick s rst laW applies to steady state ux in a uniform concentration gradient MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei NonsteadyState Diffusion Fick s second law I In most practical cases steadystate conditions are not established ie concentration gradient is not uniform and varies with both distance and time Let s derive the equation that describes nonsteadystate diffusion along the direction X Consider an element of material with dimensions dX dy and dz dV dX dy dz JX dV gt JXdX dAX dy dz X xdX acx t 6H JXD J 2 de 6X XdX X 6X The number of particles that diffuse into the volume dV during time dt is JXdAth from the left and JXdXdAxdt from the right From the balance of coming and going particles JX JW dAth dcx tdV Md And using expressions 8C X2 t Z i D 8C X t for JX and JXdX we have t 8x 8X XJX J XdX MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei NonsteadyState Diffusion Fick s second law 11 ngDm 6t 6x 6X If dependence of D on X and C l is ignored aC x t 2 D 62C x t m3t1 m2t1xm3m2 8t 8X 2 Fick s second law relates the rate of change of composition with time to the curvature of the concentration profile C C C 3 T U u X X X Concentration increases With time in those parts of the system Where concentration profile has a positive curvature And decreases Where curvature is negative MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Driving force for diffusion I The empirical Fick s first law assumes proportionality between the diffusion ux and the concentration gradient But thermodynamics tells us that any spontaneous process should go in the direction of minimization of the free energy As we can see from the examples below atoms can diffuse from regions of high concentration towards the regions of low concentration down the concentration gradient left as well as from the regions of low concentration towards the regions of high concentration up the concentration gradient right Arich B rich 3 J Arich 8 rich G G 0 o i CAO X1 12 XB 1 0 O N US K MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Driving force for diffusion 11 G G o o quot13 CAO l 11 12 B 0 XB 1 0 XE 1 Diffusion occur so that the free energy is minimized and is therefore driven by the gradient of free energy The chemical potential of atoms of type A can be defined as the free energy per mole of A atoms G HAXA HBXB Therefore the free energy gradient can be expressed through the chemical potential gradient J M C an A Where M A is the atomic mobility X A A ax ofA atoms In both cases the A and B atoms are diffusing from the regions Where chemical potential is high to the regions Where chemical potential is lower The driving force for diffusion is gradient of chemical potential Atoms migrate so as to remove differences in chemical potential Diffusion ceases at equilibrium when 0h 0l2 0h 0 2 HB MB HA HA MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Driving force for diffusion 111 Chemical potential gradient is the driving force for diffusion a JX MBCB a where MB is mobility of B atoms JX DB 6C3 8X 6p 6p D M C BM X B 39gt B B BaCB B BaXB if al lB gt 0 then DB gt 0 gradient of chemical potential B is in the same direction as the concentration gradient if al lB lt0 then DB lt 0 diffusion occurs against the B concentration gradient For example we can identify regions with negative auaXB in a system with miscibility gap 5amp4 We will discuss the aXB behavior of homogeneous solution cooled within the miscibility gap later after deriving equations for diffusion ux in ideal and regular solutions L MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Driving force for diffusion IV Lets consider diffusion driven by the chemical potential gradient for ideal and regular solutions JX MBCB 6 6X For an ideal solution B GB RTlnXB auB auB 6X13 6XB 6CB 6X 6X13 6X XB 6X CB 6X aCB aCB D B 6X For a regular solution ME 2 GB 01 XB 2 RTlnXB 6H3 aHB aX13 201 XBXB6XB ax 6XB ax E 1 XB RT 6X 1ZQXAXB aCB D M RTF CB RT 6X B B I l The factor in brackets is termed the thermodynamic factor F It defines how interatomic interaction affects the diffusion of the atoms in the presence of concentration gradient MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Driving force for diffusion V As shown below the thermodynamic factor is the same for both species A and B at a given composition and is related to the curvature of the free energy curve Gmg XAGA XBGB QXAXB RTXA1nXA XBlnXB 1 XBGA XBGB 91 XB XB RT1 XB1n1 XBXBlnXB 6G aXB GA GB Q ZQXB RTln1XB 1XB1nXB B lXB GA GB Q ZQ XB RT 1n1XB1nXB B GB GA Q1 2XBRT1r XE 1 1 2 2 RT B XB XAXB 2 2 2 RT aXB aHB 1 2Q XAXBjaCB ZEF aCB 6X CB RT 6X CB 6X F 1 29 XAXB XAXB 62G RT RT aXBZ MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Driving force for diffusion VI The presence of a strain energy gradient an electric field or a temperature gradient can also affect the diffusion and in particular can induce diffusion of atoms against the concentration gradient For example in the presence of a strain energy gradient the equation for the chemical potential Will include an elastic strain energy term EX For a regular solution we have B GB 91 XB2 RTlnXB E 6 FaCB 6E 6X CB 6X 6X JB 2 MBCB aBB MBRTF aC B MBCBa E 6X 6X 6X ac D C 613 J D B B B D M RTF B B ax RTF 6x B B MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Diffusion against the concentration gradient Spinodal Decomposition When the free energy curvature is negative the thermodynamic factor F is negative and the diffusion is directed against the concentration gradient 62G 82G 2 lt 0 Igt F lt 0 G 6sz 8X B DB MBRTF lt 0 A lt0 J B 2 DB 0 1 8X 0 XB 1 T Chemical spinoda1 Miscibility gapamp 0 2 T1 0 XB 1 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Spinodal Decomposition Homogeneous solution cooled into the miscibility gap will decompose into ocl and d2 so that the total free energy of the system decreases The mechanism of decomposition into ocl and d2 is different within the chemical Spinodal region and outside in the nucleation regions Let s consider small uctuations around the average composition XBO x X F1 spatial coordinate Free energy decreases as a result of an arbitrary infinitesimal uctuation in composition the system is unstable 2 8 G 8X2B lt0 GA x X15 xg X XB XB ltlt 12 0 XB z XB z XB uctuations are small Free energy increases as a result of an infinitesimal uctuation in composition the system is stable with respect to small uctuations G X15 X03 XE B MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Spinodal Decomposition Although the system within the miscibility gap but outside the Spinodal region is stable metastable with respect to small uctuations it is unstable to the separation into ocl and d2 determined by the common tangent construction There is large difference in composition between ocl and d2 and large composition uctuations are required in order to decrease the free energy A process of formation of a large composition uctuation is called nucleation The phase separation is occurring in this case by nucleation and growth will be discussed later Spinodal decomposition Nucleation and growth XEZ Batoms XEZ X0 X0 Batoms B B XE XE XEZ Batoms X2 X X X0 X0 X5 X X X spatial coordinate 7 spatial coordinate V MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Region of spinodal decomposition on a phase diagram With a miscibility gap G O T T1 0 1 0 2 0 XB 1 2 2 a j gt 0 a j gt 0 6X3 32G aXB 2 lt 0 T aXB Nucleation and growth Spinodal a2 decomposition T1 XB 1 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Driving force for diffusion In practice the case when atoms diffuse from regions of high concentration towards the regions of low concentration is much more common Because of this and also since concentration is much easier to measure than the chemical potential it the discussion below we will relate diffusion to concentration gradient eg consider ideal solution Atoms here lump But there are not randomly bOth ght many atoms here to and left jump to the left As a result there is a net ux of atoms from left to right MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Numerical integration of the diffusion equation 1 Finite difference method Spatial Discretization Internal nodes For 1D thermal conduction let s discretize the 1D spatial domain into N small nite spans i 1 N Ji1lt gti Jilt gti1 11 1 11 39 quot7 X I g I 1 I I AL Vi AR r 1 I I l I I Balance of particles for an internal i 2N1 volume Vi change in concentration during At due to the particle exchange with adjacent cells CiAt 1 At 1 Vi Ji 1 gtiAL Ji gti1AR CF C D 1 1 1 F1rst F1ck slaw l Im AX J DM 1 gt11 AX tAt Ci C1 Vi D C1 C1 1ALDC11 C1 AR At AX AX ctAtCt ct 2ctCt i i i1 i i 1 Vi AXAyAZ ALARAyAZ At D AX 2 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Numerical integration of the diffusion equation 11 Finite difference method Spatial Discretization Internal nodes C CI CI1 2C Ci D 2 At AX CtAt Ci1 Cit l AX2 Using this equation we can calculate unknown concentration at time tAt if we know concentrations at previous timestep t The nite difference scheme that we discussed is called the explicit Forward Time Centered Space FTCS method since the unknown temperature at node i is given explicitly in terms of only known temperatures computed at the previous timestep new point C torn known points xorl39 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Numerical integration of the diffusion equation III Finite difference method Spatial Discretization End nodes Let s consider the first node i 1 and let s place the node at the surface Node 1 Node 2 Node 3 l J1lt gt2 l J2lt gt3 l quotquotquotquotquotquot quot3Tquotquotquotquot39quotquotquot7I quotfa quota A v i a a i I I x l 39 I AX Balance of particles for volume V1 that is represented by the end node tAt t MVI J1 gt2AR At ct t First Fick s law JHZ DAR AX tAt t t t C1 C1 VIZDCZ C1 AR At AX V1 VN Vi 2N12 AXAR AXAyAz Vi 2N1 2V1 crm CIZZch c At AX 2 Numerical integration of the diffusion equation IV Finite difference method Spatial Discretization End nodes Cr c 2 2D c c At AX2 Ct Ct Cf CAt2D 2 1 AX Similarly for another end node N one can derive Ct Ct Cg Cg At2D N X21 If an initial concentration profile at time t0 is given we can use the derived equations for all nodes from 1 to N to calculate concentration profile at time At 2At 3At etc up to the time that we are interested in MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Stability criterion Assuming that the coefficients of the differential equation are constant or so slowly varyng as to be considered constant the stability criterion for Ct 2C1t C111 11 2 X cf C AtD 2DAt lt 1 This condition is referred to as the Von AXZ Neumann stability condition Physical interpretation the maximum timestep is up to a numerical factor the diffusion time across a cell of Width AX remember Einstein relation D szZdt Position C T dependent DXt For a positiondependent diffusion coefficient it should be evaluated at the interfaces between the volumes V1 in order to preserve the conservation of the total number of particles CitAt Cit Di12Cit1 Ci Di 12 Cit 1 At AX2 eg for sample with nonuniform T distribution Di12 DT39t MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Simple boundary conditions No mass transfer through the boundary 6C 6X XZL 0 Constant concentration at the boundary CL Cf and we can use the following boundary condition C LCf X MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Analytical Solutions of the Diffusion Equation While the numerical method described above can be applied to solve the diffusion equation for any initial and boundary conditions in many special cases it is possible to derive analytic solutions as well The advantage of analytical solution does not involve long computations gives clear picture of the dependence of the solution on different parameters in the Whole range of X and t acx t D 62C x t 8t 8 X 2 Let s consider the simple case of solute atoms initially inserted into the middle of an infinite onedimensional system plane source gt I 5 I All solute is in one plane initially With time the solute atoms will diffuse from the plane The redistribution of the atoms can be described by the following solution of the diffusion equation Cx t exp MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Analytical solution for plane source We can show by differentiation that for the plane source in an infinite system A X 2 CXt exp J 4Dt is the solution of the ac x t 62C x t D diffu510n equation at 5X 2 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Analytical solution for plane source Coef cient A we can nd from the condition that the total quantity of solute per unit crosssection area S is 03 00 A 2 M SJCX tix Sijfexp Using substitution of variable y Jf dx I 4Dt dy M 1stth Texp7y2gt1y ASWXJ Lap i S 21in Dt 4Dt The solution is a Gaussian concentration pro le Whose 1 gt t Width increases With time 1 MS 3051711332 Dizgxams and Kinetics Lennirl Zhigjlei Thin lm deposited on a semiin nite piece of material 60 0 s ax Using the same solution we have a different condition for coef cientA T A E X2 M S eXp dx 0 J 4Dt Using substitution of variable y dx Wdy M gifexp yzly 2 AS 4D X J M M X2 A c t lertD X anpteXpi 4m The solution is a half of a Gaussian concentration pro le At the surface the boundary condition gt t t 1 ac 0 is satis ed 0 X 6X x0 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei Analytical solutions Pair of Semiinfinite Solids I gtquotB AS Mgt t1 0 L r X0 X Think of the block of B as being made up of many thin layers of thickness AX and crosssectional area S Each layer initially contains CIAX of B atoms We know that if the areas surrounding a layer are initially Bfree the distribution of solute atoms that originate from a thin layer can be described by the Gaussian pro le Assuming that the presence of other solutes atoms of type B do not affect the diffusion of atoms that originate from the layer the overall solution can be found as superposition of the Gaussian pro les om the layers Analytical solutions Pair of Semiinfinite Solids 11 n 2 A CXtZ C1AXS exp XXi i1 S 239TC I I I In the limit of 11 going to in nity and AX going to zero we can go from the sum to an integral CXt LTeXp mdw 4Dt XW Substitutin g 2 Dt u we can rewrite the integral J CX t 37 Jexp u2du Using definition of the error function erf z j exp y2dy and taking into account th at erf oo 1 and erf z erf z Cxt C12 1erf Analytical solutions Constant surface composition 1 X0CS A similar solution of the diffusion equation can be derived OOZCO X for a constant surface concentration If the initial concentration of solute atoms in the specimen is C0 and concentration at the surface is CS the analytical solution of the Fick s second law is C CsCs C0erf Where erf is the error function that is an indefinite integral 2 Z de ned by the equatlon erf Z Z J exp y 2 dy V n 0 The error function is tabulated in books of standard mathematical functions and is defined as an internal function in some of programming languages erf z erf z 1 for z 00 1 39393939 quotquotquotquotquotquotquotquotquot quot erf z O for z O 0395 i erf z 05 for Z 05 05 1 MSE 305 Phase Diagrams and Kinetics Leonid Zhigilei

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.