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Phase Diagrams- Introduction. The effect of temperature and pressure on a substance in a closed container. Every point represents a possible combination of temperature and pressure for the system. Three areas represent the solid, liquid, and gaseous states of the substance.
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Phase Diagrams- Introduction • The effect of temperature and pressure on a substance in a closed container. • Every point represents a possible combination of temperature and pressure for the system. • Three areas represent the solid, liquid, and gaseous states of the substance.
Along AB line: rate at which solid sublimes to form a gas=rate at which gas condenses to form a solid • Along BC line: rate at which liquid boils to form a gas=rate at which gas condenses to form a liquid • Along BD line: rate at which solid melts to form a liquid=rate at which liquid freezes to form a solid
The Clausius-Clapeyron Equation The relationship between the temperature of a liquid and its vapor pressure is not a straight line. The vapor pressure of water, for example, increases significantly more rapidly than the temperature of the system. This behavior can be explained with the Clausius-Clapeyron equation. If we assume that Hvap does not depend on the temperature of the system, the Clausius-Clapeyron equation can be written in the following integrated form where C is a constant.
Densities of Solid, Liquid, and Gaseous Forms of Three Elements
PHASE DIAGRAMS • STUDY OF PHASE RELATIONSHIPS IMPORTANT IN KNOWING PROPERTIES OF MATERIALS • MAP OF TEMPERATURE, PRESSURE AND COMPOSITION BETWEEN PHASES IN EQUILIBRIUM IN A SYSTEM GIBBS PHASE RULE P + F = C + 2 Eg: states of matter- gas, liquid and solid – single phase Liquid mixture- oil and water- 2 phases In solid , several phases depending on crystal structure • STUDY IMPORTANT IN ALLOYS • ALLOY- SUBSTANCE COMPOSED OF 2 OR MORE CHEMICAL ELEMENTS • MAIN CONSTITUENT- BASE METAL AND OTHERS ALLOYING ELEMENTS
CLASSIFICATION • SINGLE COMPONENT- UNARY • TWO COMPONENT- BINARY • THREE COMPONENT- TERNARY, QUARTERNARY ETC. • EQUILIBRIUM APPROACHED BY VERY SLOW HEATING/COOLING
COOLING CURVES For pure metal or compound TEMPERATURE Cooling of Liquid Latent heat of solidification given off during freezing- At constant temperature Freezing begins Freezing ends Liquid + Solid Cooling of solid Liquid Solid TIME, log scale
COOLING CURVES For Binary solid solutions TEMPERATURE Freezing with drop in temperature TIME, log scale
For Binary solid solutions- composition 2 composition1 TEMPERATURE Ts1 Ts2 Te1 Te2 TIME, log scale
FUSION LINE ALMOST VERTICAL- VARIATION IN PRESSURE –NO EFFECT ON M.P. OF ICE Phase Rule FUSION WATER ICE Pressure 76cm B VAPORISATION 30 cm WATER VAPOUR SUBLIMATION T A 0 100 50 Temperature o C
At ‘A’ , water vapour - 1 phase • At ‘B’ , water and water vapour co exist -2 phases • At ‘T’ , ice, water and water vapour exist – 3 phases • At ‘A’ • 1 + F = chemical compound H2O + 2 • F = 2 …. BIVARIANT • At ‘B’ • 2+ F = 1 +2,F = 1… UNIVARIANT • At ‘T’ • All three phasesP = 3, 3 + F = 1 + 2; F = 0 INVARIANT
Equilibrium Diagram Case 1: Binary Alloy with COMPLETE SOLUBILITY IN BOTH LIQUID AND SOLID PHASES in all compositions Eg: Ag-Au Cu-Ni Ge-Si Al2O3-Cr2O3 Sb-Bi Silver-Palladium Co-Ni Cu-Pt Fe-Pt Ni-Pt Ta-Ti HUME ROTHERY’S RULE- FICK’S LAWS OF DIFFUSION FIRST LAW SECOND LAW
Elements A and B in a Binary Alloy Cooling Curves & Phase Diagram
Phase (Equilibrium) Diagram Liquidus curve L + α Solidus curve Composition, C (% wt of B)
LEVER RULE • With Fulcrum atP, weightsWAandWBat the end of a lever, for equilibrium, the lever rule states: WA / WB= b/a WB WA P a b
Liquid P1 P Y X Liquid + Solid Solid 47.5 16 37 58 For P: SS/LS = (37-16)/(58-37)= 1/1 For P1: SS/LS = 31.5/ 10.5= 3/1
Liquid P1 P Y X Liquid + Solid Solid 31.5/ 10.5= 3 47.5 16 37 58 = (37-16)/(58-37)
60% Ni, 40 % Cu- Liquid Phase Solid Solution, with L and S phases Nickel Rich Solid Solution
At A: 60/40 composition - SS formed as with BAt X, L + α, α with SS B1, rich in Ni, LS rich in CuAt B2: 60/40 composition- SS formed as (60/40)
Liquidus A Solidus Cu Ni Ni 60 Ni/40 Cu
There areThree variables,one of these can be chosen as independent If fl and fs are the liquid and solid fractions,
At A: 60/40 composition - SS formed as with BAt X, L + α, α with SS B1, rich in Ni, LS rich in CuAt B2: 60/40 composition- SS formed as (60/40)
Case 2: Binary Alloy with COMPLETE SOLUBILITY IN LIQUID STATE in all compositions, but COMPLETELY INSOLUBLE IN THE SOLID STATE • A very doubtful situation in practice, since most solid metals appear to dissolve small quantities of other metals • In Bismuth-Cadmium, mutal solid solubility is negligible. • Bi- heavy, brittle- positioned near to non metals in periodic table- Rhombic type structure-covalent bond • Cadmium- HCP-
When two metals show complete solubility in liquid state, and complete insolubility in the solid state,they do so by crystallising out as alternate layers of the two pure metals.This laminated structure termed as EUTECTIC Te 40Cd/60Bi
When two metals show complete solubility in liquid state, and complete insolubility in the solid state, they do so by crystallising out as alternate layers of the two pure metals.This laminated structure termed as EUTECTIC INVARIANT REACTION Te (EUTECTIC Temperature) 40Cd/60Bi
At E, solid Cadmium (40%) and solid Bismuth(60%)co-exist EUTECTIC A: Molten homogeneous alloy – 1 phase with 2 components, Bi and Cd 1+F = 2 +1 (only temperature is the variable, not pressure) , F=2 • B: 2 + F = 2 + 1, F= 1 • C: 3 + F = 2 + 1, F=0
Eutectic is considered as an intimate mixture of two metals Phase Rule applied, P+F = C+ 1 3 + F = 2 +1, F = 0
For compositions to left /right of Eutectic Temperature o C Time
Gold- Silver, Copper- Nickel, Germanium- Silicon, Antimony- Bismuth, Aluminium Oxide- Chromium Oxide etc. are examples
FICK’S LAWS OF DIFFUSION Mass Flow Process by which atoms (molecules) change their positions relative to their neighbours in a given phase under the influence of thermal energy and gradient :
FIRST LAW dn/dt = no. of moles of B atoms crossing per unit time D= Diffusion coefficient A= Planar area dc/dx= concentration gradient If J = flux flow / unit area per unit time,
SECOND LAW If D is independent of concentration,
Case 3: Two metals completely soluble in all proportions in liquid state, but partially soluble in solid state • Melting Point of Lead:3270C • Melting Point of Tin: 2320C • Eutectic Temperature: 1830C • Eutectic Composition: 62% Sn, 38%Pb • Max. solid solubility tin in lead at 1830C: 19.5% tin • Max. Solid solubility of lead in tin at 1830C: 2.6% lead • Eutectic of two solid solutions α and β (instead of two metals) form
Melting Point of Tin (Pb) : 2320C Melting Point of Lead (Sn) :3270C Eutectic Temperature: 1830C Eutectic Composition: 38%Pb, 62% Sn Max. solid solubility tin in lead at 1830C: 19.5% tin Max. Solid solubility of lead in tin at 1830C: 2.6% lead
Liquid solubility of salt in water & partial solid solubility of one metal in another- ( (similarity schematically represented)