Introductory Thermondynamics - Thermodynamic system

Advanced Physical Metallurgy Advanced Physical Metallurgy

“ “ Phase Phase Equilibria Equilibria in Materials in Materials ” ”

EunEun SooSoo ParkPark

Office: 33-316

Telephone: 880-7221 Email: [email protected]

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

Pt_0.5Ru_0.5 – Pt structure (fcc) PtPb – NiAS structure

partial or complete solid solution Alloying: atoms mixed on a lattice

Solid Solution vs. Intermetallic Compounds

Pt_0.5Ru_0.5 – 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: |χ_A – χ_B| ≤ 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) C_p vs. T, (b) H vs. T and (c) S vs. T.

⎛∂ ⎞

= ⎜⎝ ∂ ⎟⎠

P

P

C H

T =

∫

T

H C dTP

=

∫

T CP

S dT

T

How is C_p 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 C_P

(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, H^L or H^S?

• H^L > H^S at all temp.

• Which is larger, S^L or S^S?

• S^L > S^S at all temp.

→ Gibbs free energy of the liquid

decreases more rapidly with increasing temperature than that of the solid.

• Which is larger, G^L or G^S 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

T G L Δ

= Δ

L : ΔH = H^L – H^S (Latent heat) T = T_m - ΔT G^L = H^L – TS^L

G^S = H^S – TS^S

ΔG = Δ H -T ΔS ΔG =0= Δ H-T_mΔS ΔS=Δ H/T_m=L/T_m

ΔG =L-T(L/T_m)≈(LΔT)/T_m

(eq. 1.17)

Driving force for solidification

20

4 Fold Anisotropic Surface Energy/2 Fold Kinetics, Many Seeds

4. Solidification

4.1.1. Homogeneous Nucleation

L V L

S V G

V

G₁ = ( + ) G₂ = V_SG_V^S +V_LG_V^L + A_SLγ_SL

L V S

V G

G ,

SL SL

S V L

V

S

G G A

V G

G

G = − = − − + γ

Δ

( )

r G

r

G = − 4 π

Δ + 4 π

γ

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 ΔG_r= 0?

r* dG=0

Unstable equilibrium

Driving force for solidification

T

T

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)