Advanced Physical Metallurgy Advanced Physical Metallurgy
“ “ Phase Phase Equilibria Equilibria in Materials in Materials ” ”
EunEun SooSoo ParkPark
Office: 33-316
Telephone: 880-7221 Email: espark@snu.ac.kr
2009 fall
09.10.2009
2
- Equilibrium
CHAPTER 1
Introductory Thermondynamics - Thermodynamic system
Contents for previous class
: Isolated/closed/open system
- Gibbs and Helmholtz free energies
= 0 dG
The properties of the system-pressure, temperature, volume and concentrations-do not change with time.
No desire to change ad infinitum
Lowest possible value of Gibb’s Free Energy
Helmholtz free energy : F=E−TS
Useful when V is constrained during thermodynamic process.
Gibbs free energy : G=E +PV-TS=H-TS
Useful when P is constrained during thermodynamic process.
CHAPTER 1
Introductory Thermondynamics - Four laws in thermodynamics
Contents for previous class
- Zeroth law : 열역학적 평형
- 열역학적 평형은 열적 평형(열교환과 온도와 관계)과 역학적 평형(일교환과 압력 같은 일 반화된 힘과 관계)과 화학적 평형(물질교환과 화학퍼텐셜과 관계)을 포함한다
- First law: : 에너지 보존 법칙 - Second law : 엔트로피
- Third law of thermodynamics : 절대 0 도
The total entropy of any isolated thermodynamic system tends to increase over time, approaching a maximum value.
q S = T
The entropy of all system and of all states of a system is zero at absolute zero.
( i i) dE =δQ−δw d+ ∑μ N
= th + config S S S
w k S = ln
- Single component system/Binary system
4
* Binary System (two component) Æ A, B
- Mixture ; A – A, B – B ; Æ 각각의 성질 유지, boundary는 존재, 섞이지 않고 기계적 혼합
A B
* Single component system One element (Al, Fe), One type of molecule (H2O)
: 평형 상태 압력과 온도에 의해 결정됨
: 평형 상태 압력과 온도 이외에도 조성의 변화를 고려
- Solution ; A – A – A ; Æ atomic scale로 섞여 있다. Random distribution A – B – A Solid solution : substitutional or interstitial
- Compound ; A – B – A – B ; Æ A, B 의 위치가 정해짐, Ordered state
B – A – B – A
ruthenium 루테늄 《백금류의 금속 원소;기호 Ru, 번호 44》 6
Solid Solution vs. Intermetallic Compounds
Crystal structure
Surface
Pt0.5Ru0.5 – Pt structure (fcc) PtPb – NiAS structure
partial or complete solid solution Alloying: atoms mixed on a lattice
Solid Solution vs. Intermetallic Compounds
Pt0.5Ru0.5 – Pt structure (fcc) PbPt – NiAS structure
8
• Solid solution:
– Crystalline solid
– Multicomponent yet homogeneous
– Impurities are randomly distributed throughout the lattice
• Factors favoring solubility of B in A (Hume-Rothery Rules) – Similar atomic size: ∆r/r ≤ 15%
– Same crystal structure for A and B
– Similar electronegativities: |χA – χB| ≤ 0.6 (preferably ≤ 0.4) – Similar valence
• If all four criteria are met: complete solid solution
• If any criterion is not met: limited solid solution
complete solid solution limited solid solution
10
CHAPTER 1 & 2
Contents for today’s class
- Chemical potential and Activity
- Gibbs Free Energy in Binary System
- Binary System
Ideal solution and Regular solution
- Classification of phase transition
- Gibbs Free Energy as a Function of Temp.
- Pressure Effects
- Driving force for solidification
- Single component system
TS H
G = −
PV E
H = +
System Condensed
E for
H ≅
H : Enthalpy ; Measure of the heat content of the system
E : Internal Energy, Kinetic + Potential Energy of a atom within the system Kinetic Energy :
Atomic Vibration (Solid, Liquid)
Translational and Rotational Energy in liquid and gas.
Potential Energy : Interactions or Bonds between the atoms within the system T : The Absolute Temperature
Relative Stability of a System Gibbs Free Energy
12
Single component system
One element (Al, Fe)
One type of molecule (H2O)
Fe C
(1) Gibbs Free Energy
as a Function of Temp.
TS H
G = −
Fe: α γ δ Liq.
Draw the plots of (a) Cp vs. T, (b) H vs. T and (c) S vs. T.
⎛∂ ⎞
= ⎜⎝ ∂ ⎟⎠
P
P
C H
T =
∫
298T
H C dTP
=
∫
0T CP
S dT
T
How is Cp related with H and S?
H = 0 at 298K for a pure element in its most stable state.
H = ?
S = ? P = ⎜⎛⎝ ∂∂ ⎞⎟⎠
P
C S
T T
Enropy : q S = T
−2
+ +
= a bT CT CP
(경험식 above room temp)
14
P dP dT G
T dG G
P T G G
T P
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
= ∂
= ( , )
SdT VdP
SdT TdS
VdP PdV
PdV TdS
dG
TS d PV
d dE TS
d dH dG
TS H
G
−
=
−
− +
+
−
=
− +
=
−
=
−
=
) ( )
( )
(
Compare the plots of H vs.T and G vs. T.
H T
T G P V
S G T
G
P T
P
=
∂
= ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
− ∂
⎟ =
⎠
⎜ ⎞
⎝
⎛
∂
∂ ]
1) (
) ( [ ,
,
∫
−∫
+
=
−
=
1 0
1
0 0
0 0
0 1
1, ) ( , ) ( , ) ( , )
( P
P
T
T S P T dT dP
P T V T
P G T
P G
SdT VdP
dG
• Which is larger, HL or HS?
• HL > HS at all temp.
• Which is larger, SL or SS?
• SL > SS at all temp.
→ Gibbs free energy of the liquid
decreases more rapidly with increasing temperature than that of the solid.
• Which is larger, GL or GS at low T?
L S S L
Gibbs Free Energy as a Function of Temp.
16
Considering P, T
∫
−∫
+
=
−
=
=
1 0
1 0 0
0 0 1
1, ) ( , ) ( , ) ( , )
(
)
, (
P P
T
T S P T dT dP
P T V T
P G T
P G
SdT VdP
dG
P T G G
G
T (K)
800 1300
graphite diamond
2.9 kJ
300
P = 1 bar
S(graphite) = 5.74 J/K, S(diamond) = 2.38 J/K,
G
T(K) gas
liquid
0
phase transformation
T S G
P
−
⎟ =
⎠
⎜ ⎞
⎝
⎛
∂
∂
S(water) = 70 J/K S(vapor) = 189 J/K
Different molar volume 을 가진 두 상이 평형을 이룰 때 만일 압력이 변한다면 평형온도 T 또한 압력에 따라 변해야 한다.
α,β 상이 평형이라면
dT S dP V dG
dT S dP V dG
β β
β
α α
α
−
=
−
=
β
α dG
dG =
At equilibrium,
p c
m n
f = − = +2−
α→γ 의 경우 ; ∆V (-), ∆H(+)
⎞ Δ
⎛
γ →liquid 의 경우; ∆V (+), ∆H(+)
>0 Δ
= Δ
⎟⎠
⎜ ⎞
⎝
⎛
V T
H dT
dP
eq
- Clausius-Clapeyron Relation :
(applies to all coexistence curves)
Fcc (0.74)
Bcc (0.68)
hcp (0.74)
(2) Pressure Effects
이므로 여기서
eq eq
T S H
V S V
V S S dT
dP
= Δ Δ
Δ
= Δ
−
= −
⎟⎠
⎜ ⎞
⎝
⎛
α β
α β
V T
H dT
dP
eq
eq Δ
= Δ
⎟⎠
⎜ ⎞
⎝
⎛
18
T
mT G L Δ
= Δ
L : ΔH = HL – HS (Latent heat) T = Tm - ΔT GL = HL – TSL
GS = HS – TSS
ΔG = Δ H -T ΔS ΔG =0= Δ H-TmΔS ΔS=Δ H/Tm=L/Tm
ΔG =L-T(L/Tm)≈(LΔT)/Tm
(eq. 1.17)
Driving force for solidification
20
4 Fold Anisotropic Surface Energy/2 Fold Kinetics, Many Seeds
4. Solidification
: Liquid Solid4.1.1. Homogeneous Nucleation
L V L
S V G
V
G1 = ( + ) G2 = VSGVS +VLGVL + ASLγSL
L V S
V G
G ,
SL SL
S V L
V
S
G G A
V G
G
G = − = − − + γ
Δ
2 1( )
r G
r
G = − 4 π
3Δ + 4 π
2γ
Δ
: free energies per unit volume
for spherical nuclei (isotropic) of radius : r
22
4.1.1. Homogeneous Nucleation
Fig. 4.2 The free energy change associated with homogeneous nucleation of a sphere of radius r.
r < r* : unstable (lower free E by reduce size) r > r* : stable (lower free E by increase size) r* : critical nucleus size
Why r* is not defined by ΔGr= 0?
r* dG=0
Unstable equilibrium
Driving force for solidification
T
mT
G = L Δ
Δ
24
Although nucleation during solidification usually requires
some undercooling, melting invariably occurs at the equilibrium melting temperature even at relatively high rates of heating.
Why?
SV LV
SL
γ γ
γ + <
In general, wetting angle = 0 No superheating required!
4.1.4. Nucleation of melting
(commonly)