1st Law: Conservation of Energy
Energy is conserved. Hence we can make an energy balance for
a system: Sum of the inflows – sum of the outflows is equal to
the energy change of the system
energy flow  ( Esystem )

boundary
NOTE: sign of flow is positive when into the system
UNITS: Joules, calories, electron volts …
Types of Work
Electrical
W = – dq
Magnetic
W  o V HdM
Surface
W = – dA
Summary so far
Variables and Parameters
System is only described by its macroscopic variables
Variables: Temperature + one variable for every work term that
exchanges energy with the system + one variable for every
independent component that can leave or enter the system.
Parameters: Quantities that are necessary to describe the system
but which do not change as the system undergoes changes.
e.g. for a closed system: ni are parameters
for a system at constant volume: V is parameter
Equations of State (Constitutive Relations): equation between the
variables of the system. One for each work term or matter flow
Example of Equations of State
pV = nRT: for ideal gas
= E efor uniaxial elastic deformation
M =  H: paramagnetic material
Properties and State Functions
U = U(variables) e.g. U(T,p) for simple system
Heat capacity
1 Q
cp 
n dT p
cV 
Volumetric
thermal expansion
Compressibility
1 Q 
n dT V
1 dV 
V   
V dT p
1 dV 
V    
V dp T
Some Properties Specific to ideal gasses
1) PV = nRT
ONLY for IDEAL GASSES
2) cp - cv = R
U
3)    0
V T
Proof that for ideal gas, internal energy only depends on
temperature dU  nc vdT
The Enthalpy (H)
H = U + PV
Gives the heat flow for any change of a simple
system that occurs under constant pressure
Example: chemical reaction
Example: Raising temperature under constant
pressure
dHp  Qp  n c p dT
 H 
cp 
 dT p
Enthalpy of Materials
is always relative
Elements: set to zero in their stable state at 298K and 1 atm pressure
Compounds: tabulated. Are obtained experimentally by measuring
the heat of formation of the compounds from the elements under
constant pressure.
The Enthalpy (H)
H = U + PV
Gives the heat flow for any change of a simple
system that occurs under constant pressure
Example: chemical reaction
Example: Raising temperature under constant
pressure
dHp  Qp  n c p dT
 H 
cp 
 dT p
Enthalpy of Materials
is always relative
Elements: set to zero in their stable state at 298K and 1 atm pressure
Compounds: tabulated. Are obtained experimentally by measuring
the heat of formation of the compounds from the elements under
constant pressure.
Entropy and the Second Law
There exists a Property of systems, called the Entropy (S), for
which holds:
dS 
Q
T
How does this solve our problem ?
The Second Law leads to Evolution Laws
Isolated system
Evolution Law for constant Temperature and
Pressure
TdS dU  pdV
 d(TS)  dU  d(pV)
d(U  pV  TS)  0
dG ≤ 0
G (Gibbs free energy is the relevant potential to determine stability
of a material under constant pressure and temperature
Interpretation
For purely mechanical systems: Evolution towards minimal
energy.
Why is this not the case for materials ?
Materials at constant pressure and temperature can
exchange energy with the environment.
G is the most important quantity in Materials Science
determines structure, phase
transformation between them,
morphology, mixing, etc.
Phase Diagrams: One component
Describes stable phase (the one with lowest Gibbs free
energy) as function of temperature and pressure.
Water
Carbon
Temperature Dependence of the Entropy
dS 
Q
T
dSp 
Qp
T
S   C p
T p T

Cp dTp
T
Temperature Dependence of the Entropy
dS 
Q
T
dSp 
Qp
T
S   C p
T p T

Cp dTp
T
What is a solution ?
SYSTEM with multiple chemical components that is mixed
homogeneously at the atomic scale
•Liquid solutions
•Vapor solutions
•Solid Solutions
Composition Variables
MOLE FRACTION:
ATOMIC PERCENT:
CONCENTRATION:
WEIGHT FRACTION:
ni
Xi 
n tot
(at%)i  100% Xi
Ci 
wi
ni
V
or
Wi

Wtot
Wi
V
Variables to describe Solutions
G=G(T,p,n1,n2, …, nN)
G 
G 
G 
dG    dT    dp    dni
T p,n
p T , n
ni p,T
i
i
Partial Molar Quantity
chemical
potential
 G 
i   
 ni p,T
dG  S dT  V dp  i dni
Partial Molar Quantity
V 
Vi   
 ni T , p,n
j
 H 
Hi   
ni T , p,n
S 
S i   
n i T ,p, n
j
j
Partial molar quantities give the contribution of a component to a property of the solution
G    i ni
i
V   Vi ni
i
Properties of Mixing
Change in reaction:
XA A + XB B -> (XA, XB )
 H mix  H mix  X A H A  X B H B
 H mix  H A  X B H B  H A 
Cu-Pd
Ni-Pt
Intercept rule with quantity of mixing
General Equilibrium Condition in Solutions
Chemical potential for a component has to be the same in all phases
i  i  i  ...



For all components i
OPEN SYSTEM
Components have to have the
same chemical potential in system
as in environment
e.g. vapor pressure
Example: Solubility of Solid in Liquid
Summary so far
1) Composition variables
2) Partial Molar Quantities
3) Quantities for mixing reaction
4) Relation between 2) and 3): Intercept rule
5) Equilibrium between Solution Phases
Standard State: Formalism for Chemical Potentials in Solutions
i    RT ln( ai )
0
i
chemical potential of i in solution
effect of concentration
chemical potential of i in a standard state
Choice of standard state is arbitrary, but often it is taken as pure
state in same phase.
Choice affects value of ai
Models for Solution: Ideal Solution
An ideal solution is one in which all components behave Raoultian
ai  xi
for all i
G mix  G mixture  G component s
x A A  xB B
x A GA  x B GB
Gmix  x A  A0  RT ln aA   xB  0B  RT ln aB   x A GA  x B GB
Gmix  RTx A ln x A  xB ln x B 
Summary Ideal Solutions
G mix  RT  xi ln xi
i
 H mix  0
 S mix  R  xi ln xi
i
V mix  0
Solutions: Homogeneous at the atomic level
Random solutions
Ordered solutions
Summary so far
1) Composition variables
2) Partial Molar Quantities
3) Quantities for mixing reaction
4) Relation between 2) and 3): Intercept rule
5) Equilibrium between components in Solution Phases
For practical applications it is important to know relation between
chemical potentials and composition
Obtaining activity information: Experimental
e.g. vapor pressure measurement
 pi 
 i  RT ln * 
pi 
*
vapor pressure pi
Pure substance i
i  RT lnai 
pi
*  ai
pi
vapor pressure pi
Mixture with
component i in it
Obtaining activity information: Simple Models
Raoultian behavior
ai  xi
Henry’s behavior
ai  k xi
Usually Raoultian holds for solvent, Henry’s for solute at small enough concentrations.
Intercept rule
General Equilibrium Condition in Solutions
Chemical potential for a component has to be the same in all phases
i  i  i  ...



For all components i
OPEN SYSTEM
Components have to have the
same chemical potential in system
as in environment
e.g. vapor pressure
Raoultian case and Real case
Review
• At constant T and P, a closed system strives to minimize its Gibbs free energy:
G = H - TS
• Mixing quantities are defined as the difference between the quantity of the
mixture and that of the constituents. All graphical constructions derived for
the quantity of a mixture can be used for a mixing quantity with appropriate
adjustment of standard (reference) states.
0
GA
XB
0
1
XB
1
GB
A - G A
B - G B
G mix
G
Review (continued)
Some simple models for solutions can be me made: Ideal
solution, regular solution.
Many real solution are much more complex than these !
Regular Solution
G mix   Hmix – TS mix
  xB (1  x B ) 
RTx A ln x A  x B ln x B 
The Chord Rule
What is the free energy of an inhomogeneous systems
( a system that contains multiple distinct phases) ?
CHORD RULE: The molar free energy of a two-phase system
as function of composition of the total system, is given by the
chord connecting the molar free energy points of the two
constituent phases
The Chord Rule graphically
With pure components
Components are solutions
B
A
XB
overall composition is XB*
overall composition is XB*
G
0
XB
G
XB*
XB
1
0
XB*
XB
1
Regular Solution Model
G mix   xB (1 xB ) 
< 0
0
RT(1  xB )ln(1 xB )  xB ln xB 
XB
1
Regular Solution Model
G mix   xB (1 xB ) 
RT(1  xB )ln(1 xB )  xB ln xB 
 > 0
0
XB
1
Effect of concave portions of G
0
XB*
1
Single-Phase and Two-phase regions
0
XB
Singlephase
XB*
Two-phase
1
XB
Singlephase
Two-phase coexistence
When Gmix has concave parts (i.e. when the second
derivative is negative) the coexistence of two solid solutions
with different compositions will have lower energy.
The composition of the coexisting phases is given by the
Common Tangent construction
The chemical potential for a species is identical in both
phases when they coexist.
Common tangent does not need to be horizontal
0
XB*
1
Temperature Dependence of Two-phase region
Hmix
0
XB
Low temperature
Hmix
1
0
XB
Hmix
1
Intermediate temperature
0
XB
High temperature
1
Phase Diagram of Regular Solution Model with 
>0
T
0
XB
1
Example: Cr-W
Miscibility gap does not have to be symmetric
Lens-Type Diagram
Free energy curves for liquid and solid
T > TBM > TAM
G
0
G
1
XB
TBM > T > TAM
0
1
XB
G
TBM > TAM > T
0
1
XB
How much of each phase ? - The Lever Rule
Since the composition of each phase is fixed by the common
tangent, the fraction of each phase can be determined from
requiring that the system has the given overall composition
X *B  f X B  f X B
1  f  f
-> solve for f and f
The result is known as the Lever Rule
X B  X *B
f  
X B  X B
XB*  X B
f  
X B  X B
Lever Rule
f
x
x
x
f
x
x
x
0
In a two-phase region chemical potential is
constant
XB
1
XB
1
G
0

Comparison between Eutectic and Peritectic
Eutectic
Peritectic
L


L


Cooling: L -> 
Heating:  -> L + 
Nucleation
(1) The structure of the liquid
Many small closed packed solid clusters are present in the liquid. These clusters
would form and disperse very quickly. The number of spherical clusters of
radius r is given by
 Gr 
nr  n0 exp
kT 
nr : average number of spherical clusters with radius r.
n0: total number of atoms in the system.
Gr: excess free energy associated with the cluster.
(2) The driving force for nucleation
The free energies of the liquid and solid at a temperature T are given by
GL = HL- TSL and GS = HS- TSS
GL and GS are the free energies of the liquid and solid respectively.
HL and HS are the enthalpy of the liquid and solid respectively.
SL and SS are the entropy of the liquid and solid respectively.
At temperature T, we have
GV = GL- GS = HL-HS-T(SL-SS) =H-TS,
GV: volume free energy.
where H and S are approximately
independent of temperature.
For a spherical solid of radius r,
4
G   r 3GV  4r 2 SL
3
For a given T, the solid reach a critical radio r*, when
3
16 SL
G 
3( GV ) 2

We get
d ( G)
0
dr
and
r* 
If
G  
T
Tm
GV  Lm
3
16 SL
Tm
2
3( Lm )
1
T
2
2 SL
GV
then,
*
and r 
2 SL Tm 1
Lm T
When r < r, the solid is not stable, and
when r > r, the solid is stable.
The effect of temperature on the size of critical nucleation and the actual shape of a nucleus.
(4) Heterogeneous nucleation
If a solid cap is formed on a mould, the interfacial tensions balance in the plane of the mould
wall,
or
  SM
 ML   SM   SL cos
cos  ML
 SL
SL, SM and ML: the surface tensions of the solid/liquid, solid/mould and mould/liquid interfaces
respectively.
The excess free energy for formation of a
solid spherical cap on a mould is
Ghet = -VSGV + ASLSL + ASMSM - ASMML
Where VS: the volume of the spherical cap.
ASL and ASM: the areas of the solid/liquid and
solid/mould interfaces.
It can be easily shown that,
4
Ghet  ( r 3GV  4r 2 SL )  f ( )
3
where
(2  3 cos  cos3  )
f ( ) 
4
(c) Nucleation rate and nucleation time as a function of temperature
The overall nucleation rate, I, is influenced both by the rate
of cluster formation and by the rate of atom transport to the
nucleus, which are both influenced in term by temperature.
TTT diagram gives time required for nucleation, which is
inversely proportional to the nucleation rate.
860
840
820
800
780
760
740
720
700
680
660
640
620
600
100
1000
Actual examples of TTT
diagram
10000
Nucleation in Solid
Interface structure
Coherency loss
rcrit 
3 st
42
Homogeneous Nucleation
(a) Homogeneous nucleation
The free energy change associated with the nucleation process will have the following three
contributions.
1 . At temperatures where the  phase is stable, the creation of a volume V of  will cause a
volume free energy reduction of VGv.
2. Assuming for the moment that the / interfacial energy is isotropic the creation of an
area A of interface will give a free energy increase of A.
3. In general the transformed volume will not fit perfectly into the space originally
occupied by the matrix and this gives rise to a misfit strain energy Gv, per unit volume of
. (It was shown in Chapter 3 that, for both coherent and incoherent inclusions, Gv, is
proportional to the volume of the inclusion.) Summing all of these gives the total free
energy change as
G  VGV  A  VGS
If we assume the nucleus is spherical with a radius r, we
have
4
G  
3
r 3 ( GV  G S )  4r 2 
Similarly we have
r* 

G 
2
( GV  G S )
16 3
3( GV  G S )2
Nucleation at Grain Boundary
The excess free energy associated with the embryo at a grain
boundary will be given by
G = -VGv + A - A
Where V is the volume of the embryo, A is the area of / interface of energy  created, and
A the area of / grain boundary of energy  destroyed during the process.

cos  
2  
Fick I
It would be reasonable to take the flux across a given
plane o be proportional to the concentration gradient
across the plane:
 c 
J  D 
 x 
J: the flux, [quantity/m2s]
D: diffusion coefficient, [m2/s]
 c 
 x 
 
Concentration gradient, [quantity /m-4]
Here, quantity can be atoms, moles, kg etc.
Fick II:
C
t
D
 2C
x 2
Solution to Fick II
1.
Thin Film (Point Source) Solution
assume boundaries at infinity
c(x,t) 
c(x,0)  M  (x)
c(,t)  c(,t )  0
2

M
x 
exp 

4 Dt
 4Dt 
c(x,t) 
2

M
x 
exp 

4 Dt
 4Dt 
Measurement of diffusion coeffiecient
c(x,t) 
2

M
x 
exp 

4 Dt
 4Dt 
x
x2
ln c  constant 
4Dt
If c versus x is known experimentally, a plot of ln(c) versus x2
can then be used to determine D.
Error functions
2

2
exp(x )
x

0
2

exp(u2 ) du  erf (x)
e rf( 0 )  0
e rf(  )  1
e rfc( x )  1  e rf( x )
e rf(  z )  e rf( z )
c( x , t ) 
c'
2 Dt


 exp  x  
0

The final solution should be:
c' 
 x 
c( x , t )  1  e rf

2
 2 Dt 
2 d

4Dt 
When x=0, the
composition is always
kept at c=c’/2
c( x , t ) 

 x 
 
c( x , t )  C s 1  e rf
2
Dt



erf(-z)=-erf(z)

 x 
c( x , t )  Cs 1  e rf

 2 Dt  

c' 
 x 
1

e
rf



2
 2 Dt 
Diffusion Mechanis
Interstitial mechanism
Self Diffusion with Vacancy
These are the
two most
common
mechanisms
Ring Mechanism
Interstitialcy
A Random Jump Process
J1 
1
Bn1
6
1
J 2  Bn 2
6
J B  J1  J 2 
1
B ( n1  n 2 )
6
 C 
CB ( 1 )  CB ( 2 )   B 
 x 
1
 C
J B   B 2  B
6
 x
Diffusion by Vacancy mechanism
DA 
1
B 2
6
 Gm 
B  zX v exp 

RT


  Gv 
Xv  exp

 RT 
DA 
1 2
 S  S v 
 Hm  H v 
 z exp m
 exp 

6
R
RT




 Q 
DA  D0 exp 

 RT 
D0 
1 2
 S  S v 
 z exp m

6
R


Effeect of temperature on diffusion
 Q 
DA  D0 exp 

RT


Effect of defect on diffusion