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Teknik Metalurgi
MG-2211 Metallurgical ThermodynamicDr.-Ing. Zulfiadi Zulhan 2010
15. Phase Diagram
Dr.-Ing. Zulfiadi Zulhan
Teknik Metalurgi
Fakultas Teknik Pertambangan dan Perminyakan
Institut Teknologi Bandung
INDONESIA
Termodinamika Metalurgi (MG-2112)
Semester 3 – 2017/2018
Teknik Metalurgi
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Course Content
1. Introduction and Definition of Terms
2. The First Law of Thermodynamics
3. The Second Law of Thermodynamics
4. Property Relationships
5. Equilibrium
6. Chemical Equilibrium and Ellingham Diagrams
7. Electrochemistry and Pourbaix Diagrams
8. Mid Exam
9. Ion Activity
10. Solutions
11. Gibbs-Duhem Equation and Application
12. Application of Electrochemical Methods to Calculate of Thermodynamics
Properties
13. Alternative Standard States
14. Activity Coefficient in Dilute Solutions Multi Components
15. Phase Diagram
16. Final Exam
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Phases
A phase is a portion of matter that is uniform throughout, not only in
chemical but also in physical state.
Example 1: a mixture of ice and pure water → two phases mixture: solid
(ice) and liquid (water).
Example 2: ice – brine mixture exists in equilbrium with air (containing
water vapour) → system have three phases: ice, brine, and gas phase
containing water vapour and air (oxygen, nitrogen and other gases in
small quantities).
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Phase Diagram Water
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Components (Binary System)
One of the factors that determines the number of phases which may exist
in equilibrium is the number of components in a system.
The number of components (C) is defined as the number of chemical
species (N) less the number of indenpendent relationship among them
(R).
C = N - R
In binary solution, solution has two components.
Solution is made of two chemical entities/species (N=2). There is no
relations between two chemical species (R=0):
C = N - R = 2 – 0 = 2
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Components (Binary System)
If now, two chemical entities A and B react to form a compound AB. In this
case, there are three chemical entities: A, B, AB (N = 3).
There is an equilibrium constant for the reaction A + B → AB, which relates
the thermodynamic activities of the three chemical entities at a given
temperature (R=1).
Number of component
C = N – R = 3 – 1 = 2
If there are multiple compound in A-B system, for each additional
compound, N increases → equilibrium contant increases → R increases.
C = N – R = 2
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Specifying A System
Suppose that a system of interest has C components.
To determine the composition of a phase in the system, C-1 pieces of
information about that phase must be specified.
Example: for three-component system, we can specify composition of
phase A by establishing the mole fractions of components 1 and 2 in the
phase. Mole fraction of component 3 will be determined, because the sum
of the mole fractions must be one.
The number of variables to be specified (VAR) is the number of phases (P)
multiplied by (C-1) plus two overall system variables such as temperature
and total pressure of the system:
VAR = P (C-1) + 2
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Equilibrium Conditions
For a system to be in equilibrium, the chemical potential (partial molar
Gibbs free energy) of a component must be the same throughout the
system.
If the system contains P phases, it yields P-1 independent equations:
P phase in i component of potential chemical is where
...
p
i
p
i
4
i
3
i
2
i
1
i
=====
p
i
1p
i
4
i
3
i
3
i
2
i
2
i
1
i ..., , , , ==== −
If the system contains C components, total number of relationship:
REL = C (P-1)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Gibbs Phase Rule
The degree of freedom available in system (F) is the difference between
the number of variables required to specify the system (VAR) and the
number of relationships required by the equilibrium condition (REL)
F = VAR – REL
F = P(C-1) + 2 – C(P-1)
P + F = C + 2
F = C – P + 2
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
One Component System
Pressure-Temperature Component
for a one component system (C=1) .
Point A is in gas region, (P =1),
degree of freedom:
F = C – P + 2
F = 1 - 1 + 2 = 2
In gas region, both temperature and
pressure may be arbitrarily fixed.
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
One Component System
In two-phase regions, such as point B
on liquid vapor line:
F = C – P + 2
F = 1 - 2 + 2 = 1
On liquid vapor line, or on any other
of the two-phase lines, we may
arbitrarily fix either pressure or
temperature.
There is only one degree of freedom.
If temperature is fixed then pressure
is also fixed.
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
One Component System
At triple point, where three phases:
solid, liquid and gas exist in
equilibrium, the degree of freedom:
F = C – P + 2
F =1 - 3 + 2 = 0
There is no degree of freedom, there
is only one combination of
temperature and pressure at which
all three phases may coexist in a
single-component system.
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Example 1:
Consider the decomposition of calcium carbonate into calcium oxide and
carbon dioxide:
CaCO3 (s) → CaO (s) + CO2 (g)
C = N – R
Chemical species:
Relationship :
Number of “Component” = N – R =
Degree of freedom, F = C – P + 2 =
CaCO3(s), CaO(s), and CO2(g) = three chemical
species
a
a p K
3
2
CaCO
CaOCO= = one relationship
2
2 – 3 + 2 = 1
If T is fixed, then
pressure of carbon
dioxide is also fixed
(decomposition pressure
of calcium carbonate)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Example 2:
Suppose that an inert material, such as nitrogen, is added to the system
considered in example 1. Number of species will be:
Degree of freedom, F = C – P + 2 = 3 – 3 + 2 = 2
Temperature and total pressure can each be varied independently.
Pressure of carbon dioxide is fixed at a specific temperature, but the total
pressure of the system can be varied by varying the nitrogen pressure.
Four (CaO, CaCO3,
CO2 and N2)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Example 3:
Consider the equilibrium between solid carbon, carbon monoxide and
carbon dioxide:
CO2 (g) + C(s)→ 2CO (g)
There are two phases present: solid carbon and the gas.
Chemical species:
Relationship :
Number of “Component” = N – R =
Degree of freedom, F = C – P + 2 =
CO2(g), C (s), and CO(g) = three chemical species
( )
a p
p K
CCO
2
CO
2
= = one relationship
2
2 – 2 + 2 = 2 If T and P are fixed,
then the system is
specified.
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Phase Rule for Condensed System
For condensed system, such as solids, it is common to assume that the
pressure of the system is one atmosphere.
The pressure variable might be ignored for small pressure changes, the
Phase Rule for condensed system:
F + P = C + 1
F = C – P + 1
F = C – P + 2 (temperature and pressure) (For gas)
F = C – P + 1 (temperature) (for condensed system, solid, liquid)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
LIQUID
SOLID
Freezing Point Depression
At temperature Tm, the two phases, solid
and liquid, are in equilibrium.
Temperature interest (for freezing point
depression) is in the region below Tm,
pure solid A is taken as standard state.
At temperature lower than Tm:
T
Tm (melting temperature)
Phase diagram of single-
component system
ls AA =
==
pures,
purel,
s
l
a
a ln RT
f
f ln RTG
1 a if a ln RTG pures,purel, ==
meltingmeltingmeltingmelting S T - L S T - HG ==
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
LIQUID
SOLID
Freezing Point Depression
At temperature lower than Tm:T
Tm (melting temperature)
Phase diagram of single-
component system
melting
meltingmeltingmelting
S T - L
S T - H G
=
=
L is latent heat of fusion. For
simplicity, assume that there is no
difference in heat capacity between
liquid and solid.
0 G e,temperatur melting At melting =
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
LIQUID
SOLID
Freezing Point Depression
T
Tm (melting temperature)
Phase diagram of single-
component system
0 G e,temperatur melting At melting =
meltingm S T L =
( )
m
m
m
meltingT
TT L
T
T-1 L G
−=
=
( )
m
m
pures,
purel,
meltingT
TT L
a
a ln RT G
−=
=
meltingmeltingmelting S T - HG =
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Addition of Material B to A
At temperature T’ (below Tm), activity of
A in an ideal A-B solution is a function of
composition.
Consider that A and B are immiscible in
the solid state, but form ideal solution in
liquid state.
Liquid of composition XA,l (where activity aA,l =1) is in equilibrium with pure
solid A at Temperature T’.
( )
m
m
pures,
purel,
meltingT
TT L
a
a ln RT G
−=
=
Activity of pure liquid A is greater than
one at temperatures below Tm (Solid A
was considered as standard state).
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Addition of Material B to A
Consider: dissolving of pure liquid
A in the liquid solution:
Dissolution of pure solid A in
liquid solution is the sum of:
melting of pure A and dissolution
of pure liquid A in liquid solution
solutionl,pure,l A A =
=
purel,
solutionl,
a
a ln RT G
solutionl,pure,s A A =
If liquid solution is in equilibrium
with pure solid, then G=0
( )solutionl,A,
m
m X ln -RTT
TT L =
−
If solution is ideal
If T is close to Tm
( )( )2m
msolutionl,A,
T R
TTL-X ln
−=
( )0
a
a ln RT
T
TT L G
purel,
solutionl,
m
m =
+
−=
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Addition of Material B to A
Note: XA = 1 - XB
( )( )2m
mB
T R
TTL-)X-(1 ln
−=
If variable z is small: ln (1-z) = -z
( )( )2m
mB
T R
TTLX
−=
T = Tm – T (melting point
depression)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Melting Point Depression
( )( )2m
mB
T R
TTLX
−=
T = Tm – T (melting point
depression)
This expression can be plotted
on a phase in which temperature
is ordinate and composition in
absisca.
In two phase region labelled
“L+S” (liquid plus solid), pure
solid A is in equilibrium with a
liquid solution.
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Simple Eutectic Phase Diagram
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Example
Calculate the lowering of melting point of silver caused by addition of one
mole percent of lead. Although there is small solubility of lead in solid
silver, it was assummed that silver lead system follows the following
equation:
For silver: Tm = 1234 K, and L = 11 300 J/mol
( )( )2m
mB
T R
TTLX
−=
T = 11.2 K
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
http://www.ques10.com
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Lever Rule
At T1, the phase diagram tells us that the
equilibrium liquid composition is XB,l. The
equilibrium solid composition is XB,S.
Relative quantities or fractions can be
calculated using mass balance
If Fl = fraction liquid and FS =
fraction solid, then Fl + FS = 1
Based on a mass balance for B:
sB,slB,lB X F X FX +=
sB,slB,lBSl X F X FX )FF( +=+
)X (X F )XX(F sB,BSBlB,l −=−
Liquid
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Lever Rule
Lever rule can be expressed as a ratio of
fraction liquid to the fraction of solid:
For fraction of liquid:
XX
XX
F
F
BlB,
sB,B
s
l
−
−=
XX
XXF
sB,lB,
sB,B
l−
−=
Note: Fl + FS = 1
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Lever Rule: Example
Calculate the fraction of liquid at 500K in a lead-tin binary alloy containing
50 mol% tin. At 500 K, the equilibrium composition of the liquid is 59
mol% tin, and the composition of the solid is 24 mol% tin.
74.0Fl =
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Simple Eutectic Diagram
Liquid = solid A + solid BConsider: material A
and B are immiscible
in solid state,
completely miscible in
liquid state.
Addition of B to A
lowers the melting
point of A. addition of
A to B lowers the
melting point of B
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Cooling Curve (pure material)
If a pure material (pure A), is
cooled from a temperature Tm
(its melting temperature) to
below Tm by removing thermal
energy at a constant rate,
temperature of material as a
function of time follows a pattern
illustrated in the figure 9.6.
During solidification, liquid A and
solid A are in equilibrium,
temperature of the system does
not change (thermal arrest).
Once all material A has solidified, temperature decrease continue.
The same type of cooling curve is observed if one cools a liquid of
eutectoic composition
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Cooling Curve (pure material)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Cooling Curve (other than eutectic composition)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility
Consider two component system A-B, where A and B are completely
miscible in both solid and liquid state, and form ideal solution in both.
( ) p
s
p
l
m
mmelting -
T
TT L G =
−=
( )
m
mp
l
p
sT
TT L -
−=
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Phase Diagram
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Fe-Ti Diagram
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Phase Diagram
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
A
o
AA
rel
A a ln RT G G G =−=
i
nnS,V,innS,P,innV,T,innP,T,iijijijij
n
U
n
H
n
F
n
G=
=
=
=
i = chemical potential of component i
BBAA X V X V V +=
Bn,P,TA
A
n
VV
=
l,A
p
lA,l,A a ln RT += solution) (ideal Xln RT l,A
p
lA,l,A +=
l,BBl,AAl X XG +=
) Xln RT ( X ) Xln RT (XG l,B
p
lB,Bl,A
p
lA,Al +++=
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Graphical Representation: Ideal Solution
AABA
PT,B
A VXVX dX
VdX :1 −=
ABBB
PT,B
B VXVX dX
VdX :2 −=
BBAA X V X V V :3 +=
B
PT,B
A V dX
VdXV :31 =
++
A
PT,B
B V dX
VdXV :23 =
−−
10. Solutions
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
s,BBs,AAs X XG +=
( )
m
mp
l
p
sT
TT L -
−=
) Xln X Xln X(RTG l,BBl,AAl +=
) Xln RT ( X ) Xln RT (XG s,B
p
sB,Bs,A
p
sA,As +++=
−−
−−
+=
)TT(T
LX)TT(
T
LX
) Xln X Xln X(RTG
B,m
B,m
Bs,BA,m
A,m
As,A
s,Bs,Bs,As,As
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
T > Tm,A and T > Tm,B
solid
liquid
Tm,A = 900 K
Tm,B = 1300 K
L/Tm = 9 J/K (for A
and B)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
T < Tm,A and T < Tm,B
solid
liquid
Tm,A = 900 K
Tm,B = 1300 K
L/Tm = 9 J/K (for A
and B)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
T = Tm,A and T < Tm,B
solid
liquid
Tm,A = 900 K
Tm,B = 1300 K
L/Tm = 9 J/K (for A
and B)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
solid liquid
Tm,B > T > Tm,A
liquid solidliquid
+ solid
xB1 xB2
Tm,A = 900 K
Tm,B = 1300 K
L/Tm = 9 J/K (for A
and B)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
Phase diagram: complete range
of liquid and solid miscibiliy, ideal
solution
solid
liquid
Tm,A
Tm,B
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Ideal Solution
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Example (cont.): Complete Miscibility, Ideal
Solution
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Complete Miscibility, Non Ideal Behavior
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
p
i
4
i
3
i
2
i
1
i ... =====
at certain T
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Specific Gibbs
free energy of
pure A and B at
different levels
IDEAL SOLUTIONS:
10.
So
luti
on
s
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Ph
ase
mis
cib
ility
ga
p in
so
lid s
tate
: tw
o
so
lid s
olu
tio
ns (
an
d
)
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Zone Refining
Liquid Co is in equilibrium
with solid kCo which has a
lower solute content.
On soldification of an
impure metal with initial
concentration Co, the first
solid formed has
composition kCo.
Enrichment of liquid stops
after the composition of
solute in molten zone Co/k.
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Zone Refining
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Zone Refining
Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi
Teknik MetalurgiThank you for your attention!
Dr.-Ing. Zulfiadi Zulhan
Department of Metallurgical Engineering
Institute of Technology Bandung
Jl. Ganesha No. 10
Bandung 40132
INDONESIA
Telefon : +62 (0) 22 250 2239
Fax : +62 (0) 22 250 4209
Mobile : +62 (0) 813 22 93 94 70
E-Mail: [email protected]