1st law: conservation of energy energy is conserved. hence we can make an energy balance for a...
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1st Law: Conservation of Energy
energy flow
boundary
∑ = Δ ( Esystem
)
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
NOTE: sign of flow is positive when into the system
UNITS: Joules, calories, electron volts …
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Types of Work
Electrical
W = – dq
Magnetic
δW= μoV HdM
Surface
W = – dA
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Summary so far
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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
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Example of Equations of State
pV = nRT: for ideal gas
= E for uniaxial elastic deformation
M = H: paramagnetic material
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Properties and State Functions
U = U(variables) e.g. U(T,p) for simple system
Heat capacity
cV =1n
δQdT
⎛ ⎝
⎞ ⎠ V
cp =1n
δQdT
⎛ ⎝
⎞ ⎠ p
Volumetric thermal expansion
Compressibility
αV =1V
dVdT
⎛ ⎝
⎞ ⎠
p
βV = −1V
dVdp
⎛ ⎝ ⎜
⎞ ⎠ ⎟
T
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Some Properties Specific to ideal gasses
∂U∂V
⎛ ⎝
⎞ ⎠ T
=0
1) PV = nRT
2) cp - cv = R
3)
ONLY for IDEAL GASSES
Proof that for ideal gas, internal energy only depends on temperature dU =ncvdT
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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
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Example: Raising temperature under constant pressure
dHp =δQp =ncp dT
cp =∂HdT
⎛ ⎝
⎞ ⎠ p
H
TT
m
Δ H
m
C
,p l
C
,p s
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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.
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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
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Example: Raising temperature under constant pressure
dHp =δQp =ncp dT
cp =∂HdT
⎛ ⎝
⎞ ⎠ p
H
TT
m
Δ H
m
C
,p l
C
,p s
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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.
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Entropy and the Second Law
dS ≥δQT
There exists a Property of systems, called the Entropy (S), for which holds:
How does this solve our problem ?
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The Second Law leads to Evolution Laws
System
S
time
equilibrium
Isolated system
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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
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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.
System
G
time
equilibrium
T,P
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Phase Diagrams: One component
Describes stable phase (the one with lowest Gibbs free energy) as function of temperature and pressure.
WaterCarbon
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Temperature Dependence of the Entropy
dS=δQT
dS( )p =δQ( )p
T=
Cp dT( )p
T∂S∂T
⎛ ⎝
⎞ ⎠ p
=Cp
T
S
TTm
Tb
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Temperature Dependence of the Entropy
dS=δQT
dS( )p =δQ( )p
T=
Cp dT( )p
T∂S∂T
⎛ ⎝
⎞ ⎠ p
=Cp
T
S
TTm
Tb
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What is a solution ?
SYSTEM with multiple chemical components that is mixed homogeneously at the atomic scale
•Liquid solutions
•Vapor solutions
•Solid Solutions
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Composition Variables
MOLE FRACTION:
ATOMIC PERCENT:
CONCENTRATION:
WEIGHT FRACTION:
Xi =ni
ntot
(at%)i =100% Xi
Ci =ni
Vor
Wi
V
wi =Wi
Wtot
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Variables to describe Solutions
G=G(T,p,n1,n2, …, nN)
dG=∂G∂T
⎛ ⎝ ⎜
⎞ ⎠ ⎟
p,ni
dT +∂G∂p
⎛
⎝ ⎜
⎞
⎠ ⎟
T ,ni
dp+∂G∂ni
⎛
⎝ ⎜
⎞
⎠ ⎟
p,T
dni
dG=−SdT+V dp+μi dni
Partial Molar Quantity
V i =∂V∂ni
⎛
⎝ ⎜
⎞
⎠ ⎟
T ,p,nj
H i =∂H∂ni
⎛
⎝ ⎜
⎞
⎠ ⎟
T ,p,n j
S i =∂S∂ni
⎛
⎝ ⎜
⎞
⎠ ⎟
T ,p,nj
Partial Molar Quantity
μi ≡∂G∂ni
⎛
⎝ ⎜
⎞
⎠ ⎟
p,T
chemical potential
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Partial molar quantities give the contribution of a component to a property of the solution
G =i
∑ μini
V =i
∑ V ini
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Properties of Mixing
A,BA
B
Change in reaction:
XA A + XB B -> (XA, XB )
ΔHmix =Hmix −XA H A −XB H B
=Hmix − H A +XB H B −H A( )[ ]
H
HB
HA
XB
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Cu-Pd
Ni-Pt
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Intercept rule with quantity of mixing
ΔVmix
VB - VB
XBX*BVA - VA
ΔGmix
μB - GB
XBX*B
μA - GA
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General Equilibrium Condition in Solutions
μiα =μi
β =μiγ =...
Chemical potential for a component has to be the same in all phases
For all components i
OPEN SYSTEM
Components have to have the same chemical potential in system as in environment
e.g. vapor pressure
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Example: Solubility of Solid in Liquid
pure A liquid
B solid
Gliquid
GlBGlA
XB
GsB
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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
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Standard State: Formalism for Chemical Potentials in Solutions
μi =μi0 +RT ln(ai )
chemical potential of i in solution
chemical potential of i in a standard state
effect of concentration
Choice of standard state is arbitrary, but often it is taken as pure state in same phase.
Choice affects value of ai
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Models for Solution: Ideal Solution
An ideal solution is one in which all components behave Raoultian
ai =xi forall i
A,BA
B
ΔGmix =Gmixture−Gcomponents
xAμA +xBμB xAGA +xBGB
ΔGmix =RT xA lnxA +xB lnxB[ ]
ΔGmix =xA μA0 +RT lnaA( )+xB μB
0 +RT lnaB( ) − xAGA −xBGB
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Summary Ideal Solutions
ΔGmix =RT xi lnxii
∑
ΔHmix =0
ΔSmix =−R xi lnxii∑
ΔVmix =0
XB
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Solutions: Homogeneous at the atomic level
Random solutionsOrdered solutions
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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
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Obtaining activity information: Experimental
e.g. vapor pressure measurement
Pure substance i Mixture with component i in it
vapor pressure pi* vapor pressure pi
Δμi =RT lnpi
pi*
⎛
⎝ ⎜
⎞
⎠ ⎟
Δμi =RT ln ai( )
pi
pi* = ai
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Obtaining activity information: Simple Models
Raoultian behavior
Henry’s behavior
ai =xi
ai =kxi
Usually Raoultian holds for solvent, Henry’s for solute at small enough concentrations.
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Intercept rule
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General Equilibrium Condition in Solutions
μiα =μi
β =μiγ =...
Chemical potential for a component has to be the same in all phases
For all components i
OPEN SYSTEM
Components have to have the same chemical potential in system as in environment
e.g. vapor pressure
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Raoultian case and Real case
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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.
G AG B
X B0 1
G
X B0 1
Δ Gmix
μA
- GA
μB
- GB
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Review (continued)
Some simple models for solutions can be me made: Ideal solution, regular solution.Many real solution are much more complex than these !
ΔGmix= ΔHmix – TΔSmix
= ω xB(1−xB) + RT xA lnxA +xB lnxB[ ]
Regular Solution
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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
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The Chord Rule graphically
A B
With pure components Components are solutions
overall composition is XB*
XB
overall composition is XB*
G
0 1XB
XB*
G
0 1XB
XB*
XB
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Regular Solution Model
ΔGmix= ωxB(1−xB) + RT (1−xB)ln(1−xB) +xB lnxB[ ]
0 XB 1< 0
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Regular Solution Model
ΔGmix= ωxB(1−xB) + RT (1−xB)ln(1−xB) +xB lnxB[ ]
0 XB 1
> 0
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Effect of concave portions of G
0 XB* 1
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Single-Phase and Two-phase regions
0 XB* 1XB
XB
Single-phase
Single-phaseTwo-phase
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Two-phase coexistence
When ΔGmix 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.
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Common tangent does not need to be horizontal
0 XB* 1
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Temperature Dependence of Two-phase region
0XB
1
ΔHmix
0XB
1
ΔHmix
0XB
1
ΔHmix
Low temperature Intermediate temperature High temperature
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Phase Diagram of Regular Solution Model with > 0
0 XB 1
T
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Example: Cr-W
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Miscibility gap does not have to be symmetric
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Lens-Type Diagram
Liquid
Solid
L+S
XB
0 1
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Free energy curves for liquid and solid
0
XB
1
G
0
XB
1
G
0
XB
1
G
T > TBM > TA
M TBM > T > TA
M
TBM > TA
M > T
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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
XB* = fαXB
α +fβ XBβ
1= fα +fβ-> solve for f and f
fα =XB
β −XB*
XBβ −XB
α fβ =XB
* −XBα
XBβ −XB
α
The result is known as the Lever Rule
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Lever Rule
x x
x
x x
x
f
f
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In a two-phase region chemical potential is constant
0 XB 1
0 XB 1
G
μ
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Comparison between Eutectic and Peritectic
L
L
Eutectic Peritectic
Cooling: L -> Heating: -> L +
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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
nr : average number of spherical clusters with radius r.
n0: total number of atoms in the system.
ΔGr: excess free energy associated with the cluster.
n nG
kTrr= −⎛
⎝⎞⎠0 xp
Δ
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(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 ΔGV = GL- GS = HL-HS-T(SL-SS) =ΔH-TΔS,
ΔGV: volume free energy.
where ΔH and ΔS are approximately independent of temperature.
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For a spherical solid of radius r,
Δ ΔG r G rV SL= − +4
343 2π π γ
For a given ΔT, the solid reach a critical radio r*, whend G
dr
( )Δ=0
rG
SL
V
* =2γΔ
ΔΔ
GG
SL
V
∗ =16
3
3
2
πγ
( )
ΔΔ
G LT
TV mm
=
rT
L TSL m
m
* =2 1γ
Δ( )22m
3SL
T
1
L3
Tm16G
Δ
πγ=Δ ∗
)(
We get
If
and
then,
and
When r < r, the solid is not stable, and when r > r, the solid is stable.
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The effect of temperature on the size of critical nucleation and the actual shape of a nucleus.
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(4) Heterogeneous nucleation If a solid cap is formed on a mould, the interfacial tensions balance in the plane of the mould wall, or
γSL, γSM and γML: the surface tensions of the solid/liquid, solid/mould and mould/liquid interfaces
respectively.
γ γ γ θML SM SL= + coscosθ γ γγ=−ML SM
SL
The excess free energy for formation of a solid spherical cap on a mould is ΔGhet = -VSΔGV + ASLγSL + ASMγSM - ASMγML
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, where
Δ ΔG r G r fhet V SL= − + ×( ) ( )43 43 2π π γ θ
f ( )( cos cos )θ θ θ
=− +2 3
43
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(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.
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600620640660680700720740760780800820840860
100 1000 10000
Actual examples of TTT diagram
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Nucleation in Solid
Interface structure
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Coherency loss
24
3
μ
γ= st
critr
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Homogeneous Nucleation
(a) Homogeneous nucleation
SV GVAGVG ΔΔΔ +γ+−=
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 VΔGv.
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 Aγ.
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 ΔGv, per unit volume of . (It was shown in Chapter 3 that, for both coherent and incoherent inclusions, ΔGv, is proportional to the volume of the inclusion.) Summing all of these gives the total free energy change as
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If we assume the nucleus is spherical with a radius r, we have
γp+−p−= 23 43
4rGGrG SV )( ΔΔΔ
)(*
SV GGr
ΔΔ −γ
=2
2
3
3
16
)( SV GGG
ΔΔΔ
−
pγ=∗
Similarly we have
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Nucleation at Grain Boundary
The excess free energy associated with the embryo at a grain boundary will be given by
ΔG = -VΔGv + Aγ - Aγ
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.
γγ
=ϑ2
cos
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Fick I
It would be reasonable to take the flux across a given plane o be proportional to the concentration gradient across the plane:
⎥⎦
⎤⎢⎣
⎡∂∂
−=x
cDJ
J: the flux, [quantity/m2s]
D: diffusion coefficient, [m2/s]
⎥⎦
⎤⎢⎣
⎡∂∂x
cConcentration gradient, [quantity /m-4]
Here, quantity can be atoms, moles, kg etc.
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Fick II:
2
2
x
CD
t
C
∂
∂=
∂
∂
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1. Thin Film (Point Source) Solution
M( )x
X
1D
X
2Dy
M( ) x per unit y
X
3D
M( ) x per unit area
c(x,t) =M
4πDtexp−
x2
4Dt
⎛
⎝ ⎜
⎞
⎠ ⎟
assume boundaries at infinity c(x,0) =Mδ(x)
c(∞,t)=c(−∞,t)=0
Solution to Fick II
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c(x,t) =M
4πDtexp−
x2
4Dt
⎛
⎝ ⎜
⎞
⎠ ⎟
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Measurement of diffusion coeffiecient
c(x,t) =M
4πDtexp−
x2
4Dt
⎛
⎝ ⎜
⎞
⎠ ⎟
Dt4
xtconsc
2−= tanln
If c versus x is known experimentally, a plot of ln(c) versus x2 can then be used to determine D.
x
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Error functions
2π
exp(−x2)
2π
exp(−u2)0
x
∫ du =erf(x)
)()(
)()(
)(
)(
zerfzerf
xerf1xerfc
1erf
00erf
−=−−=
=∞=
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The final solution should be:
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛+=
Dt2
xerf1
2
ctxc
'),(
( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −−
p= ∫∞ d
Dt4
x
Dt2
ctxc 0
2exp
'),(
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⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+=
Dt2
xerf1Ctxc s),(
When x=0, the composition is always kept at c=c’/2
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛+=
Dt2
xerf1
2
ctxc
'),(
erf(-z)=-erf(z)
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛−=
Dt2
xerf1Ctxc s),(
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Diffusion Mechanis
Interstitial mechanism Self Diffusion with Vacancy
Ring Mechanism Interstitialcy
These are the two most common mechanisms
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A Random Jump Process
1B1 n6
1J Γ=
2n
6
1J B2 Γ=
)( 21B21B nn6
1JJJ −Γ=−=
⎟⎠
⎞⎜⎝
⎛∂∂
−=−x
C2C1C B
BB )()(
x
C
6
1J B2
BB ∂∂
⎟⎠
⎞⎜⎝
⎛ Γ−=
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Diffusion by Vacancy mechanism
⎟⎠
⎞⎜⎝
⎛ Δ−υ=Γ
RT
GXz m
vB exp
⎟⎠
⎞⎜⎝
⎛−=RT
GX v
v exp
2BA 6
1D Γ=
⎟⎠
⎞⎜⎝
⎛ Δ+Δ−⎟
⎠
⎞⎜⎝
⎛ Δ+Δυ=
RT
HH
R
SSz
6
1D vmvm2
A expexp
⎟⎠
⎞⎜⎝
⎛−=RT
QDD 0A exp
⎟⎠
⎞⎜⎝
⎛ Δ+Δυ=
R
SSz
6
1D vm2
0 exp
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Effeect of temperature on diffusion
⎟⎠
⎞⎜⎝
⎛−=RT
QDD 0A exp
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Effect of defect on diffusion