재 료 상 변 태
Phase Transformation of Materials
2008.09.09.
박 은 수
서울대학교 재료공학
Contents for previous class
•
상,
상태도,
상변태에 대한 이해•
열역학 계와 열역학 법칙에 대한 이해• Gibbs
와Helmholtz Free Energy
의 유도•
평형의 개념•
단일성분계의 이해와 변태거동• 1
차 변태, 2
차 변태와 변태시 압력효과에 관한 이해•
응고 구동력과 과냉의 역할에 대한 이해Contents for today’s class
• Ideal solution 과 regular solution 의 이해
• 이상 용액과 규칙 용액에서 Gibbs Free Energy
• Chemical potential 과 Activity 의 이해
• Real solution 의 이해
*
Binary System (two component)Æ A, B
- Mixture ; A – A, B – B ; Æ 각각의 성질 유지, boundary는 존재, 섞이지 않고 기계적 혼합
A B
- 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, B의 위치가 정해짐, Ordered state B – A – B – A
*
Single component system One element (Al, Fe), One type of molecule (H2O):
평형 상태 압력과 온도에 의해 결정됨:
평형 상태 온도와 압력 이외에도 조성의 변화를 고려- Mixture ; A – A, B – B ; Æ 각각의 성질 유지, boundary는 존재, 섞이지 않고 기계적 혼합
A B
- 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, B의 위치가 정해짐, Ordered state B – A – B – A
MgCNi3
ruthenium 루테늄 《백금류의 금속 원소;기호Ru, 번호44》
• Solid solution:
– Crystalline solid
– Multicomponent yet homogeneous
– Impurities are randomly distributed throughout the lattice
• Factors favoring solubility of B in A
– Similar atomic size: ∆r/r ≤ 15%
– Same crystal structure for A and B
– Similar electronegativities: |χA – χB| ≤ 0.6 (preferably ≤ 0.4) – Similar valency
• If all four criteria are met: complete solid solution
• If any criterion is not met: limited solid solution
: 합금에 대한 열역학 기본 개념 도입 위해 2원 고용체, 1기압의 고정된 압력 조건 고려
* Composition in mole fraction XA, XB XA + XB = 1
1. bring together XA mole of pure A and XB mole of pure B
2. allow the A and B atoms to mix together to make a homogeneous solid solution.
Gibbs Free Energy of Binary Solutions
Gibbs Free Energy of The System
In Step 1
- The molar free energies of pure A and pure B pure A;
pure B;
;X
A, X
B(mole fraction
) ), (T P GA
) , (T P GB
G
1= X
AG
A+ X
BG
BJ/mol
Free energy of mixture
Gibbs Free Energy of The System
In Step 2
G
2= G
1+ Δ G
mixJ/mol
Since G
1= H
1– TS
1and G
2= H
2– TS
2And putting ∆ H
mix= H
2- H
1∆ S
mix= S
2- S
1∆ G
mix= ∆ H
mix- T ∆ S
mix∆
H
mix: Heat of Solution i.e. heat absorbed or evolved during step 2
∆
S
mix: difference in entropy between the mixed and unmixed state
.How can you estimate Δ H
mixand Δ S
mix?
Mixing free energy Δ G
mix가정
1 ; ∆ H
mix=0 :
; A
와B
가complete solid solution ( A,B ; same crystal structure)
; no volume change
- Ideal solution
w k
S = ln
w : degree of randomness, k: Boltzman constant Æ thermal; vibration ( no volume change )Æ Configuration; atom 의 배열 방법 수 ( distinguishable )
Δ G
mix= -T Δ S
mixJ/mol
∆ G
mix =∆ H
mix- T ∆ S
mixEntropy can be computed from randomness by Boltzmann equation, i.e.,
=
th+
configS S S
)]
ln (
) ln
( ) ln
[(
,
,
1
! ln
!
)!
ln (
0 0
0
o B o
B o
B o
A o
A o
A o
o o
B A
B B
A A
B A
B before A
after mix
N X N
X N
X N
X N
X N
X N
N N
k
N N
N N
X N
N X N
N k N
N k N
S S
S
−
−
−
−
−
=
= +
=
=
→
+ −
=
−
= Δ
이므로
N
N N
N ! ≈ ln − ln
Excess mixing Entropy
) , _(
_
!
!
)!
(
) _
_(
_
1
B A B
A B A
config config
N N solution after
N N
N w N
pureB pureA
solution before
w
+ →
=
→
=
If there is no volume change or heat change,
Number of distinguishable way of atomic arrangement
kN
0R =
using Stirling’s approximation and
Excess mixing Entropy
) ln
ln
(
A A B Bmix
R X X X X
S = − +
Δ
) ln
ln
(
A A B Bmix
RT X X X X
G = +
Δ
Δ G
mix= -T Δ S
mixSince Δ H = 0 for ideal solution,
Compare G
solutionbetween high and low Temp.
G2 = G1 + ΔGmix
Chemical potential
= μ
A AdG ' dn
⎛ ∂ ⎞
μ = ⎜
A⎝ ∂
A⎟ ⎠
T, P, nBG' n
The increase of the total free energy of the system by the increase of very small quantity of A, dn
A, will be proportional to dn
A.
(T, P, n
B: constant )
μ
A: partial molar free energy of A or chemical potential of A
⎛ ∂ ⎞
μ = ⎜
B⎝ ∂
B⎟ ⎠
T, P, nAG'
n
Chemical potential
= μ
A A+ μ
B BdG ' dn dn
= − + + μ
A A+ μ
B BdG ' SdT VdP dn dn
for A-B binary solution, dG’
for variable T and P
dG = -SdT + VdPChemical potential
⎛ ∂ ⎞ = μ = ⎛ ∂ ⎞ = ⎛ ∂ ⎞ = ⎛ ∂ ⎞
⎜ ∂ ⎟ ⎜ ∂ ⎟ ⎜ ∂ ⎟ ⎜ ∂ ⎟
⎝
i⎠
T,P,nj i⎝
i⎠
S,V,nj⎝
i⎠
S,P,nj⎝
i⎠
T,V,njG U H F
n n n n
• The chemical potential is the change in a characteristic thermodynamic state function (depending on the
experimental conditions), the characteristic thermodynamic state function is either: internal energy (U), enthalpy (H), Gibbs free energy (G), or Helmholtz free energy (F)) per change in the number of molecules.
From Wikipedia, the free encyclopedia
= − + μ
= + + μ
= − − + μ
= − + + μ
∑ ∑
∑ ∑
i i
i i
i i
i i
dU TdS PdV dn
dH TdS VdP dn
dF SdT PdV dn
dG SdT VdP dn
−1
= μ
A A+ μ
B BG X X Jmol
For 1 mole of the solution dG' = μ
Adn
A+ μ
Bdn
Bμ = + μ = +
A A A
B B B
G RT ln X G RT ln X
μ = + μ = +
A A A
B B B
G RT ln X G RT ln X
μ
A’
μ
B’
X
B’
Ideal solution : ΔH
mix= 0
Quasi-chemical model assumes that heat of mixing, ΔH
mix, is only due to the bond energies between adjacent atoms.
Structure model of a binary solution
Regular Solutions
∆ G
mix =∆ H
mix- T ∆ S
mixBond energy Number of bond A-A ε
AAP
AAB-B ε
BBP
BBA-B ε
ABP
AB=
AA AAε +
BB BBε +
AB ABε
E P P P
Δ
mix=
ABε ε = ε −
AB1 ε + ε
AA BBH P where ( )
2
Internal energy of the solution
Regular Solutions
Δ H
mix= 0
Completely random arrangement
ideal solution
AB
=
a A Ba
P N zX X bonds per mole N : Avogadro ' s number
z : number of bonds per atom
∆ H
mix= Ω X
AX
Bwhere Ω = N
az ε
Δ H
mix= P
ABε 0
Regular Solutions
ε = ( )
2 1
BB AA
AB
ε ε
ε = +
↑
→
< 0 P
ABε ε > 0 → P
AB↓
≈ 0 ε
Ω >0 인 경우
Regular Solutions
Reference state
Pure metal
G
0A= G
B0= 0
) ln
ln
(
A A B BB A B
B A
A
G X G X X RT X X X X
X
G = + + Ω + +
G
2= G
1+ Δ G
mix∆ G
mix =∆ H
mix- T ∆ S
mixPhase separation in metallic glasses
Regular Solutions
) ln
ln
(
A A B BB A B
B A
A
G X G X X RT X X X X
X
G = + + Ω + +
B B
B B
A A
A A
X RT
X G
X RT
X G
ln )
1 (
ln )
1 (
2 2
+
− Ω
+
=
+
− Ω
+
= μ μ
Free Energy Change on Mixing
A B B
A B
AX X X X X
X = 2 + 2
−1
= μ
A A+ μ
B BG X X Jmol
Activity, a
ideal solution regular solution
• μ
A= G
A+ RTlna
Aμ
B= G
B+ RTlna
B⎛ ⎞ = Ω −
⎜ ⎟
⎝ ⎠
B
B B
ln a (1 X )
X RT
γ =B B
B
a X
μ = + μ = +
A A A
B B B
G RT ln X G RT ln X
• For a dilute solution of B in A (X
B→0)
γ = ≅ γ = ≅
B B
B A A
A
a cons tan t (Henry ' s Law) X
a 1 (Rault ' s Law)
X