Lecture 3: Solutions: Activities and Phase...

Lecture 3: Solutions: Activities and Phase Diagrams

21-09-2010

• Lecture plan:– Gibbs phase rule– vapour composition– two-component phase diagrams– phase diagrams in material science:

• microstructures in isomorphous binary systems• microstructures in eutectic alloys• liquid crystals

– problems

Phase diagrams

iron-carbon diagram

what is the composition (number of phases and their amount and composition) at equilibrium at a given temperature;

what happens to the system when is cools down/heats up

we can predict the structure and the properties of the system at low temperature.

we can understand development and preservation of non-equilibrium structures

design materials of required properties

Phase diagrams

iron-carbon diagram

That’s the base of all modern engineering from swiss knife to food and cosmetics!

water-surfactant-oil

Phase diagrams• Constituent – a chemical species that is present• Component – a chemically independent constituent of the system (i.e.

not connected by a chemical reaction)

variance

number of components

number of phases

Indeed: number of variables would be: P*(C-1)+2number of equations: C*(P-1)

• Variance – the number of intensive variables that can be changed independently without disturbing the number of phases at equilibrium.

• Phase rule (J.W. Gibbs):F=C-P+2

3 2( ) ( ) ( )CaCO s CaO s CO g⎯⎯→ +←⎯⎯Phase1 Phase2 Phase3

2C =

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

One component diagramsDetection of phase transitions and building a phase diagram is based on calorimetry measurements

Two-components diagramsC=2 therefore F=4-P. We have to reduce degree of freedom e.g. by fixing T=const

• Vapour pressure diagrams

* *A A A B B Bp x p p x p= =

Raoult’s Law

* * *( )A B B A A Bp p p p x p p= + = + −

Two-components diagrams

• The composition of vapour

From Dalton’s law: ;

*; *

A BA B

A A A B B B

p py yp p

p x p p x p

= =

= =From Raoult’s law:

* ; 1* ( * *)

AA B A

B A B A

py y yp p p x

= = −+ −

*

*A

B

pp

Two components diagrams


The lever rule

n l n lα α β β=

Relative amount and the composition of phases in equilibrium can be found on the phase diagram

Two-components diagramsTemperature-composition diagrams

• Distillation of mixtures

number of theoretical plates

Fractional distillation

Two-components diagramsTemperature-composition diagrams

•Azeotropes

A-B interacation stabilize the mixture

Azeotrope, evaporation w/o change in composition

n lα α

A-B interacation destabilize the mixture


• Immiscible liquids

Will boil at lower temperature!

1A Bp p p atm= + =boiling condition

• can be used for steam distillation of heat sensitive components

Two components diagrams• Liquid-liquid phase diagrams: partially miscible

liquids

hexane nitrobenzene


Upper critical solution T

( ln ln )mix A A B B A BG nRT κ κ κ κ βκ κΔ = + +

0 ln (1 2 ) 01mixG

κ β κκ κ∂

Δ = + − =∂ −


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


Upper critical temperature is less than the boiling point

Boiling occur before liquids are fully miscible

Liquid-solid phase diagrams

Eutectic composition

Liquid-solid phase diagrams

Eutectic halt

Liquid-solid phase diagrams• Reacting systems

Incongruent melting: compounds melts into components

peritectic line: 3 phases are in equilibrium

Phase diagrams and Microstructure

Binary phase diagrams• Phase diagram with total solubility in both liquid and solid

state: isomorphous system

• 2 phases: L (liquid) α (FCC solid solution)

• 3 phase fields: L L + α α

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

α (FCC solid solution)

L + αliquidus

solidus

Cu-Ni phase diagram

homogeneous liquid solution of Cu and Ni.

homogeneous solid solution of Cu and Ni.

Cu-Ni phase diagram

wt% Ni

20

1200

1300

T(°C)

L (liquid)

α (solid)

L + α

liquidus

solidus

30 40 50

TAA

DTD

TBB

tie line

L + α

433532CoCL Cα

R S

Information we can extract from the diagram:the phases present;composition of the phasespercentage of fraction of the phases

Composition at TB:• Liquid phase (L) of 32% Ni• Solid phase (α) of 43% Ni• Weight ratio:

(43 35); 73%(43 32)

LW S RWW R S Rα

α

−= = = =

+ −

C0=35 wt% Ni

at TA: Only liquid, composition of liquid is given by the overall composition (C0=35 wt% Ni)at TD: Only liquid, composition of liquid is given by the overall composition (C0=35 wt% Ni)at TB: Both L and α are present

Development of microstructure in a Cu-Ni alloy

Equilibrium case(very slow cooling)

Development of microstructure in a Cu-Ni alloy

Non-Equilibrium case(real)

How we can prevent coring and get equilibrium structure?

First α to solidfy: 46wt%Ni

Uniform Cα:

35wt%Ni

Last α to solidfy: < 35wt%Ni

• Fast cooling:Cored structure

• Slow cooling:Equilibrium structure

Binary Eutectic Systems: Sn-Pb

L + α L+β

α + β

200

T(°C)

18.3

Co, wt% Sn 20 40 60 80 100 0

Co

300

100

L (liquid)

α 183°C 61.9 97.8

150

11 99

R S

β

• For a 40wt%Sn-60wt%Pb alloy at 150C, find...--the compositions of

the phases:Ca = 11wt%SnCb = 99wt%Sn

W α = 59 88

= 67 wt %

W β = 29 88

= 33 wt %

Sn-Pb system:limited solubility in solid state3 single phase regions (L, a, b);TE=183 0C, no liquid below TE.Eutectic composition 61.9%

Pb Sn

At the eutectic temperature:

( ) ( ) ( )E E EL C C Cα βα β+

Microstructures in binary systems

L + α200

T(°C)

Co, wt% Sn10

2

200Co

300

100

L

α

30

L: Cowt%Sn

αL

α: Cowt%Sn

α + β

400

(room T solubility limit)

TE(Pb-Sn System)

• Co < 2wt%Sn• Result:

--polycrystal of α grains.

Microstructures in binary systems

• 2wt%Sn < Co < 18.3wt%Sn• Result:

--α polycrystal with fineβ crystals. α: Cowt%Sn

L + α

200

T(°C)

Co, wt% Sn10

18.3

200Co

300

100

L

α

30

L: Cowt%Sn

α + β

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

Lα

αβ

Microstructures in binary systems• Eutectic composition

Microstructures in binary systems: eutectic and around

T(°C)

(Pb-Sn System)

L + α200

Co, wt% Sn20 400

300

100

L

α

60

α + β

TE β

080 100

L + β

18.361.9

97.8

Cohypoeutectic

Cohypereutectic

eutectic

hypereutectic: (illustration only)

160μm

eutectic: Co=61.9wt%Sn

175μm

β

ββ

ββ

β

α

α

α

αα

α

hypoeutectic: Co=50wt%Sn

eutectic micro-constituentFrom: W.D. Callister, “Materials Science and Engineering: An Introduction”, 6e.

21

Adapted from Fig. 9.21,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed.,Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

(Adapted from Fig. 9.24, Callister 6e. (Fig. 9.24 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)

Result: Pearlite = alternating layers of α and Fe3C phases.

120μm

• 2 important points

-Eutectic (A):

-Eutectoid (B): L ⇒ γ + Fe3C

γ ⇒ α +Fe3C

Fe

3C

(c

em

en

tite

)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ+Fe3C

α+Fe3C

α+γ

L+Fe3C

δ

(Fe) Co, wt% C0.77 4.30

727°C = Teutectoid

1148°C

T(°C)

A

B

SR

R S

γ γγγ

Fe3C (cementite-hard)α (ferrite-soft)

α

Ce

ute

cto

id

IRON-CARBON (Fe-C) PHASE DIAGRAM

22

Adapted from Figs. 9.21 and 9.26,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

Adapted fromFig. 9.27,Callister6e. (Fig. 9.27 courtesy Republic Steel Corporation.)

HYPOEUTECTOID STEEL

(Fe-C System)

Co

Fe

3C

(c

em

en

tite

)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ+Fe3C

α+Fe3C

L+Fe3C

δ

Co, wt% C0.7

7

727°C

1148°C

T(°C)

R S

γ γγγ

α

γγγ γ

γγ γ

γ r s

wα =s/(r+s)wγ =(1-wα)

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγ

α

αα

αα

α pearlite

100μm Hypoeutectoid steel

(Fe-C System)

Co

Fe

3C

(c

em

en

tite

)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ+Fe3C

α+Fe3C

L+Fe3C

δ

Co, wt% C0.7

7

1148°C

T(°C)

R S

γ γγγ

αs

wFe3C =r/(r+s)wγ =(1-wFe3C)

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγpearlite

60μm Hypereutectoid steel

rγγ

γ γ

γγγ γ

Fe3C

23

Adapted from Figs. 9.21 and 9.29,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

Adapted fromFig. 9.30,Callister6e. (Fig. 9.30copyright 1971 by United States Steel Corporation.)

HYPEREUTECTOID STEEL

Liquid crystals• Mesophase – an intermedediate phase between

solid and liquid. Example: liquid crystal• Liquid crystal – substance having a liquid-like

imperfect order in at least one direction and long-range positional or orientational order in at least one another direction

NematicNematic

SmecticSmectic

CholestericCholesteric

calamitic(rod-like)

discotic

Nematic crystals in LCD

Problems (to solve in the class)• 6.1a: At 90ºC the vapour pressure of

methylbenzene is 53.3kPa and that of 1.2-dimethylbenzene is 20kPa. What is the composition of a liquid mixture that boils at 90ºC when the pressure is 0.5 atm. What is the composition of the vapour produced. down

• 6.9b: sketch the phase diagram of the system NH3/N2H4 given that the two substances do not form a compound and NH3 freezes at -78C, N2H4 freezes at +2C, eutectic formed with mole fraction of N2H40.07 and melts at -80C.

• 6.10b Describe the diagram and what is observed when a and b are cooled down

Lecture 3: Solutions: Activities and Phase...

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