MSAE 530
Binary Phase Diagram II
2
Overview of this lecture
• How to calculate phase equilibrium?
• Reference states
• Different types of phases diagrams with two
phases involved
* Ideal solution model for both phases
* Liquid phase ideal, solid phase regular
* Both phases strictly regular
3
Introduction
Consider only two phases (liquid and α) at a time:
)()( ),()( :T p,constant at
BBAA
== ll
Select reference states: Take pure A(l) at the p, T of interest as the
reference state for A in both phases
)()()()(
*
AA
*
AA
lll −=−
)A(G)()()()(
*
fus
*
AA
*
AA
−−=− ll
)A(G)()(
*
fusAmixAmix
−= l
4
Introduction
Introduce approximations/models in the general relations:
(i) Materials science approximation: pressure independence of the
thermodynamic properties of condensed phases, replace * with .
(ii) The linear approximation for G as a function of T:
)T(ST)T(H)T(G
fusfusfus
−=
)T(ST)T(H0
fusfusfusfus
−=
S}TT{)T(G
fusfusfus
−
(iii) An appropriate solution model:
)()()(
E
A
id
AmixAmix
lll +=
5
Reference States
The phase compositions in equilibrium are independent of
the choice of reference state for the elements.
0.0 0.2 0.4 0.6 0.8 1.0
-2000
-1000
0
1000
2000
3000
Gibbs Energy of fcc phase at 400K
Zn(liq)
Zn(fcc)
Zn(hcp)
Zn
Al
G
m
/ Jmol
-1
X
Zn
6
Ideal Solution Model for Both Phases
0)()()()( :T p,constant At
E
B
E
B
E
A
E
A
==== ll
)B(G)()(
)A(G)()(
fus
id
Bmix
id
Bmix
fus
id
Amix
id
Amix
−=
−=
l
l
)B(G)(xlnRT)(xlnRT
)A(G)(xlnRT)(xlnRT
fusBB
fusAA
−=
−=
l
l
)B(G)}B(TT{
)T,(x
)T,(x
lnRT
)A(G)}A(TT{
)T,(x
)T,(x
lnRT
fusfus
B
B
fusfus
A
A
−+=
−+=
l
l
7
Example
0.0 0.2 0.4 0.6 0.8 1.0
700
800
900
1000
1100
x
2
x
2
L
Liquid
B
A
T(K)
x
2
0.0 0.2 0.4 0.6 0.8 1.0
-8000
-6000
-4000
-2000
0
2000
x
2
x
2
L
Liquid
T=900K
B
A
G(J/mol)
x
2
8
Liquid Phase Ideal, Solid Phase Regular
0)()(
E
B
E
A
== ll
Liquid phase l is ideal:
)(x)(
)(x)(
2
A0
E
B
2
B0
E
A
=
=
Solid phase α is strictly regular:
)B(G)}B(TT{)(x
)T,(x
)T,(x
lnRT
)A(G)}A(TT{)(x
)T,(x
)T,(x
lnRT
fusfus
2
A0
B
B
fusfus
2
B0
A
A
−+=
−+=
l
l
Equations to be solved:
9
Examples (Δ
fus
S
m
10 JK
-1
mol
-1
)
T[K]
X[B]
1200
1400
1600
1800
2000
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
BA
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1200
1400
1600
1800
2000
X[B]
T[K]
ε
0
(α) = -10 kJmol
-1
Liquid
T[K]
X[B]
0
400
800
1200
1600
2000
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
BA
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
400
800
1200
1600
2000
X[B]
T[K]
ε
0
(α) = +10 kJmol
-1
Liquid
1
+
2
10
Examples (Δ
fus
S
m
15 JK
-1
mol
-1
)
T[K]
X[B]
400
800
1200
1600
2000
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
BA
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
400
800
1200
1600
2000
X[B]
T[K]
ε
0
(α) = -25 kJmol
-1
Liquid
T[K]
X[B]
400
800
1200
1600
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
BA
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
400
800
1200
1600
X[B]
T[K]
ε
0
(α) = -15 kJmol
-1
Liquid
11
Examples (Δ
fus
S
m
15 JK
-1
mol
-1
)
T[K]
X[B]
200
400
600
800
1000
1200
1400
1600
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
BA
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
200
400
600
800
1000
1200
1400
1600
X[B]
T[K]
ε
0
(α) = +8 kJmol
-1
Liquid
1
+
2
T[K]
X[B]
200
400
600
800
1000
1200
1400
1600
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
BA
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
200
400
600
800
1000
1200
1400
1600
X[B]
T[K]
ε
0
(α) = +10 kJmol
-1
Liquid
1
+
2
12
Both Phases Strictly Regular
)(x)(
)(x)(
2
A0
E
B
2
B0
E
A
ll
ll
l
l
=
=
)(x)(
)(x)(
2
A0
E
B
2
B0
E
A
=
=
Both Liquid and solid phases are strictly regular
)B(G)}B(TT{)(x)(x
)T,(x
)T,(x
lnRT
)A(G)}A(TT{)(x)(x
)T,(x
)T,(x
lnRT
fusfus
2
A0
2
A0
B
B
fusfus
2
B0
2
B0
A
A
−+−=
−+−=
l
l
l
l
l
l
Equations to be solved:
13
Examples
14
END !