• topology : knowing one pt on the coexistence line in the potential phase diagram,

the direction of the projected line in the T,P phase diagram

(phase field) is obtained by eliminating from G-D relations

dPVdTSd

dPVdTSd

mmA

mmA

mm

mm

VVSS

dTdP

/ : slope and direction

[Figure 8.6] The T,P phase diagram for carbon, according to a thermodynamic assessment.

1) why the equil line is almost a line except for low temperatures in Fig. 8.6?

2) why the line becomes parallel to the T axis at temperatures?

using the alternative form,

G = ∑iNi = ANA (letting G/NA=Gm)

Gm = Hm - TSm = A

∴ ∆Sm→∆Hm

→

mm

mm

VVHH

TdTdP

1

/ : Clapeyron eq.

S is difficult to determine from experiments, but H is

not !

phase field: geometrical elements of a potential phase diagram

• supposing there is a third phase (and),

points : where three phases are in equil, equil, zero-dim phase field

f=v=c+2-p=1+2-3=0

lines : where two phases are in equil, equil, 1-dim phase field (one var can be changed independently)

f=v=1+2-2=1

surfaces : where a single phase exists, equil, 2-dim phase field (we can change two indep var without leaving this kind of phase field)

f=v=1+2-1=2

(X)

• why do three phases meet at one pt? because f=0, for the plane f=2, and this must be wrong !

• 180° rule : all the angles between three intersecting lines in the phase diagram are than 180°

inconsistency if above 180° !

• binary or multinary potential phase diagramsvar: T, P, A, B

fundamental property diagram : 4-dim can not describe the surface

4-dim 3-dim: removal of one potential (A) and then projection

points: 4 phases equil, lines: 3 phases equilsurfaces: 2 phases equil, volumes: a single phase exists for higher-order systems, the principles will be the same

[Figure 8.11] The T,P,μB phase diagram for a binary system with four phases.

obtained from G-D eq.

• sections of potential phase diagram at a constant value of some potentials ( section) for example sectioning at T=T1 of Fig. 8.11 Fig. 8.12

f=v=c+2-p-nS (nS : the number of sectioning) why –ns ? sectioning means constant (fixed, not variables) values of

potentials - complete potential phase diagram (between indep var) has c+2-1

(from G-D)=c+1 axes after sectioning, c+1-nS axes

[Figure 8.11]

• examples of a sectioned phase diagram

- binary systems equil between the bcc (W) and WC phases from G-D at const P (sectioning), taking W as comp

1 d1=-Sm1

dT+ Vm1dP-∑zidi

where stands for bcc (W) or WC, and i is carbon

∴ dw = – Sm1bccdT – zcdc

= – Hm1bcc/T(dT) – zc

bccdc

= – Hm1WC/T(dT) – zc

WCdc

c

2

• by eliminating dw

bccm

WCmbcc

cWC

c

bccm

WCmc SS

zzT

HH

dTd

1111

)(

)(

0 ,1 bcc

w

cbccc

WC

w

cWCc N

Nz

NN

zusing

by introducing gra as the state of reference for carbon μc = Gc

graphite + RTln aC

grac

bccm

WCm

gracc SSS

dTGd

11

)(

letting x axis: μc-Gcgra /T

y axis: 1/T then the slope is WC

fc H

Ta

R

)/1(ln

c-Gcgra

Tbcc (W)

WC

in the case of reactions involving oxygen, it is natural to use an O2 gas of 1 bar as reference, or express the oxygen potential by the ratio of partial pressures of CO2 and CO of Fig. 8.15

- ternary systems for the case of Ti-O-Cl letting axes O/RT & Cl/RT

ClClOOTi

ClClCl

OOO

dzdzd

GPRTG

GPRTG

PT, constant at D-G

atm) 1 standard, :(ln21

21

atm) 1 standard, :(ln21

21

00

00

22

22

, arbitrary phase in Ti-O-Cl, but difficult to express μo/RT

Therefore,

ClCl

oo

O

Cl

O

Cl

ClClooTi

ClClooTi

zzzz

T

T

TdzTdzTd

TdzTdzTd

)/(

)/(

)/()/()/(

)/()/()/(

deleting d(Ti/T)

of course, zia = Ni

a/ NTia and the slope can be

calculated directly from the comp involved

this kind of sectioned potential diagram is called (M-Gas1-Gas2) diagram

applicable to hot corrosion, high temp oxidation of

Cu-O-S, Ce-O-S

• direction of phase fields in potential phase diagrams

G-D: -SmdT+Vm

dP-∑xidi=0

- if all the phases in equil, then each of di,dT,dP must have the same value for all phases

- with p phases we have p G-D eq and thus eliminate p-1 of var, choosing to eliminate i (i from 1 to p-1)

- multiply the G-D eq for the phase by a factor represented by the determinant (Cramer’s eq)

xi, i from 1 to p-1

p phases∴p-1 phases

121

121

121

,,,

,,

,,,

p

p

p

XXX

XXX

XXX

as a short-handed notation, let the determinant

121 ,,, pXXX

by adding the eq for all the phases, we obtain

using an alternative form

0)/(

/)/1(

121

121121

Tdxxxx

TdPxxxVTdxxxH

ipi

pmpm

0121121121 ipipmpm dxxxxdPxxxVdTxxxS

= 0

for i from 1 to p-1 ∵ xi is one of x1 to xp-1

① if p=c+1, f=v=c+2-p=c+2-(c+1)=1 univariant equil, (linear phase field in the potential diagram) all the di or d(i/T) can be eliminated math

because c=p-1 (other potentials, too), the remaining term is dT/dP

0/1)/1(1 22 TdPxxVTdxxH cmcm

for a binary sys with three phases

2

2

2

22

1

1

//

)/1(/

0/1)/1(1

xV

xH

dTTdP

TdTTdP

TdTdP

TdPxVTdxH

m

m

mm

)()()(

1111

222222

222

2

xxHxxHxxH

xxx

HHH

xH

mmm

mmm

m

likewise ㅣ Vm1x2

ㅣ V used instead of H

1

TdTdP

why x1 becomes 1 ?

② if p=c v=f=2 divariant equil phase field is a surface, obtain a relation between one of di (e.g., dmc), dT & dP

③ if p=c-1 v=f=3, two i & dT, dP survive, under isobarothermal cond we obtain a relation between dc and dc-1

• extremum in T & P : Konovalov’s rule

when p=c, for example, in a binary or ternarycc

mcmc

dxx

dPVxxxdTSxxx

1

121121

supposing ㅣ x1 x2

… xcㅣ = 0

under isobaric cond, dT/dc = 0 for a linear phase field

the phase field must go through a temp extremum in the T-c phase diagram

the requirement for this to occur is

ㅣ…ㅣ = 0 → ㅣ 1 x2… xc

ㅣ = 0

this shows that the phases fall on the same point (i.e. have the same composition) for c=p=2

for c=p=2, phases fall on the same pt, congruent melting

for c=p=3, they fall on a straight line for c=p=4, they fall on a plane surface

Konovalov’s rule, 1881

theorem, 1893

c

c

c

c

ci

ci

ci

c

c

c

c

c

c

c

c

c

c

c

c

c

xx

xx

xx

xx

xxx

xxx

xxx

xxxx

xxxx

xxxx

xxx

xxx

xxx

xxx

xxx

xxx

xxx

xxx

xxx

xxx

2

2

2

2

2

2

2

221

221

221

22

22

22

21

21

21

21

21

21

21

1

1

1

1

0

0

by adding more columns

dT/dc= 0

for p=c=2

1122

222

22

,

1

110

xxxx

xxx

xx

for p=c=3

0

1

1

1

1

32

32

32

32

xx

xx

xx

xx

a linear relation exists between three equil comp of three phases

they fall on a line, 직선식

its meaning ?

ex) congruent melting !

)(

)(

)(

321

321

321

XXX

XXX

XXX