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재 료 상 변 태

Phase Transformation of Materials

2008.09.09.

박 은 수

서울대학교 재료공학

(2)

Contents for previous class

,

상태도

,

상변태에 대한 이해

열역학 계와 열역학 법칙에 대한 이해

• Gibbs

Helmholtz Free Energy

의 유도

평형의 개념

단일성분계의 이해와 변태거동

• 1

차 변태

, 2

차 변태와 변태시 압력효과에 관한 이해

응고 구동력과 과냉의 역할에 대한 이해
(3)

Contents 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

(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

- 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
(5)

- Mixture ; A – A, B – B ; Æ 각각의 성질 유지, boundary는 존재, 섞이지 않고 기계적 혼합

A B

(6)

- Solution ; A – A – A ; Æ atomic scale로 섞여 있다. Random distribution A – B – A Solid solution : substitutional or interstitial

(7)

- compound ; A – B – A – B ; Æ A, B의 위치가 정해짐, Ordered state B – A – B – A

MgCNi3

(8)

ruthenium 루테늄 《백금류의 금속 원소;기호Ru, 번호44

(9)

• 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

(10)

* 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

(11)

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

A

G

A

+ X

B

G

B

J/mol

Free energy of mixture

(12)

Gibbs Free Energy of The System

In Step 2

G

2

= G

1

+ Δ G

mix

J/mol

Since G

1

= H

1

– TS

1

and G

2

= H

2

– TS

2

And 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

mix

and Δ S

mix

?

(13)

Mixing free energy Δ G

mix

가정

1 ; ∆ H

mix

=0 :

; A

B

complete 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

mix

J/mol

∆ G

mix =

∆ H

mix

- T ∆ S

mix

Entropy can be computed from randomness by Boltzmann equation, i.e.,

=

th

+

config

S S S

(14)

)]

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

0

R =

using Stirling’s approximation and

(15)

Excess mixing Entropy

) ln

ln

(

A A B B

mix

R X X X X

S = − +

Δ

) ln

ln

(

A A B B

mix

RT X X X X

G = +

Δ

Δ G

mix

= -T Δ S

mix
(16)

Since Δ H = 0 for ideal solution,

Compare G

solution

between high and low Temp.

G2 = G1 + ΔGmix

(17)

Chemical potential

= μ

A A

dG ' dn

⎛ ∂ ⎞

μ = ⎜

A

⎝ ∂

A

⎟ ⎠

T, P, nB

G' 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, nA

G'

n

(18)

Chemical potential

= μ

A A

+ μ

B B

dG ' dn dn

= − + + μ

A A

+ μ

B B

dG ' SdT VdP dn dn

for A-B binary solution, dG’

for variable T and P

dG = -SdT + VdP
(19)

Chemical potential

⎛ ∂ ⎞ = μ = ⎛ ∂ ⎞ = ⎛ ∂ ⎞ = ⎛ ∂ ⎞

⎜ ∂ ⎟ ⎜ ∂ ⎟ ⎜ ∂ ⎟ ⎜ ∂ ⎟

i

T,P,nj i

i

S,V,nj

i

S,P,nj

i

T,V,nj

G 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

(20)

1

= μ

A A

+ μ

B B

G X X Jmol

For 1 mole of the solution dG' = μ

A

dn

A

+ μ

B

dn

B
(21)

μ = + 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

(22)

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

mix
(23)

Bond energy Number of bond A-A ε

AA

P

AA

B-B ε

BB

P

BB

A-B ε

AB

P

AB

=

AA AA

ε +

BB BB

ε +

AB AB

ε

E P P P

Δ

mix

=

AB

ε ε = ε −

AB

1 ε + ε

AA BB

H P where ( )

2

Internal energy of the solution

Regular Solutions

(24)

Δ H

mix

= 0

Completely random arrangement

ideal solution

AB

=

a A B

a

P N zX X bonds per mole N : Avogadro ' s number

z : number of bonds per atom

∆ H

mix

= Ω X

A

X

B

where Ω = N

a

z ε

Δ H

mix

= P

AB

ε 0

Regular Solutions

ε = ( )

2 1

BB AA

AB

ε ε

ε = +

< 0 P

AB

ε ε > 0 → P

AB

≈ 0 ε

Ω >0 인 경우

(25)

Regular Solutions

Reference state

Pure metal

G

0A

= G

B0

= 0

) ln

ln

(

A A B B

B 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

mix
(26)

Phase separation in metallic glasses

(27)

Regular Solutions

) ln

ln

(

A A B B

B 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 B

G X X Jmol

(28)

Activity, a

ideal solution regular solution

• μ

A

= G

A

+ RTlna

A

μ

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

(29)

• 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

Referensi

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