Thermodynamics of Materials

Thermodynamics of Materials

12th Lecture 2008. 4. 21 (Monday)

o → o

-10 C water (l) -10 C ice (s)

2 7 3K ( l ) 2 7 3K ( s )

2 6 3K ( l ) 2 6 3K ( s )

ⅰ

ⅱ

ⅳ

ⅲ

Freezing of Supercooled Water

Δ = Δ → =

=

=

=

=

2 2

2

m (s l)

H2O(l),298K H2O(s),298K p,H O(l) p,H O(s)

For H O

at 273 K, H H 6008 J / mol S 70. 08 J / K

S 44. 77 J / K C 75. 44 J / K C 38 J / K

Δ = =

Δ = −Δ = −

Δ = = −

∫

∫

ⅰ

ⅱ

ⅲ

273 263

m 263 273

H 75. 44dT 754. 4

H H 6008 J

H 38dT 380

→ → →

ΔG_{(l s)} = ΔH_{(l s)} − ΔT S_{(l s)}

∴ ΔH_{(l s)}→ = ΔH_i + ΔH_ii + ΔH_iii = −5633. 6 J

Δ = −

= +

− + =

Δ = − = −

Δ = − =

∫

∫

i

ii

iii

273(l) 263(l) 273 298(l) 298 p(l)

263 298(l) 298 p(l)

263(s) 273(s)

S S S

(S C d ln T)

(S C d ln T) 75. 44 ln 273 6008 263

S 22. 007

273 263

S S S 38 ln

273

Δ → = Δ + Δ +

∴ S_{(l s)} S_i S_ii ΔS_iii = −20. 61 J K

=

Δ −

∴ G = −5633. 6 263 2+ × 0. 61 21 3 J /mole

→ → →

ΔG_{(l s)} = ΔH_{(l s)} − ΔT S_{(l s)}

∴ ΔH_{(l s)}→ = ΔH_i + ΔH_ii + ΔH_iii = −5633. 6 J Δ → = Δ + Δ +

∴ S_{(l s)} S_i S_ii ΔS_iii = −20. 61 J K

→

= − +

at constant P dG = − dG SdT VdP

SdT

Δ^{G273k 263k}→ = −

∫

→

⎡ ⎛ ⎞⎤

Δ ^{273 263Kwater} = −

∫

G 70. 08 75. 44 ln T dT

298

= + − −

− −

=

700. 8 75. 44[(273 ln 273 273) (263 ln 263 263)] 754. 4 ln 298

620. 7

= ^298k +

∫

C T

S S dT 70. 08 75. 44 ln

T 298

→

⎡ ⎛ ⎞⎤

Δ ^{273 263Kice} = −

∫

G 44. 77 38 ln T dT

298

= −

− − − −

=

447. 7 380 ln 298

38[(263 ln 263 263) 273 ln 273 273)]

407. 4

ΔG_{273k 263k}^water→ = 620. 7 J / mole

ΔG_{273 263K}^ice → = 407. 4 J / mole

Δ

2 9 8

H

S S .

ⅰ ),ⅱ ) 결 과 동 일

ⅰ )의 경 우 상 전 이 시 를 알 아 야 하 고

ⅱ )의 경 우 또 는 특 정 온 도 에 서 를 알 아 야 한 다

→

→ →

∴ Δ = Δ − Δ

= − = −

water ice ice water

273 263K

263k 273k 263k

G G G

407. 4 620. 7 21 3. 3 J / mole

∴ Δ = − G 21 3 J / mole

What would be the driving force for precipitation ( ΔG_290Kvapor→water) when the water vapor saturated at 300K is supercooled to 290K?

What would be the driving force for precipitation (ΔG_290Kvapor→water) when the water vapor saturated at 300K is supercooled to 290K?

Δ = Δ − ΔG H T S

= − → Δ = −

∫

dG VdP SdT G SdT

→

⎛ ⎞ ⎛ ⎞

Δ = − ⎜ ⎟ = − ⎜ ⎟

⎝ ⎠ ⎝ ⎠

300eq

300 290K eq eq

290 290

P P

G RTln RTln

P P

⎛∂ ⎞ =

⎜ ∂ ⎟

⎝ ⎠ Δ

P,T P,T

Why is the critical nucleus defined by the condition of G 0 instead of

r

the condition of G =0?

Open System

dU = TdS – PdV + dU_matter dG= VdP - SdT + dG_matter

dU_matter or dG_matter ∝ ? dG_matter ∝ dn → dG_matter = μdn

dG= VdP - SdT + Σμ_idn_i

= − + 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

⎛∂ ⎞ ⎛∂ ⎞ ⎛∂ ⎞ ⎛ ∂ ⎞

= μ = = =

⎜∂ ⎟ ⎜∂ ⎟ ⎜∂ ⎟ ⎜∂ ⎟

⎝ i ⎠T,P,n_j ⁱ ⎝ i ⎠S,V,n_j ⎝ i ⎠S,P,n_j ⎝ i ⎠T,V,n_j

G U H F

n n n n

Chemical Potential

α β

surrounding

constant T, P

Close Open

α α β β α α β β

= + = μ + μ

G G n G n n n

Equilibrium

between α and β for one component system

What is the G of α and β system ?

α β → α β

G , G molar free energy of and

( )

, , ,

T P T P T P

G nG n

G G

n n n

μ =^⎛⎜⎝^∂∂ ^′⎞⎟⎠ =^⎡⎢⎣^∂ ∂ ^⎤⎥⎦ = ^⎛⎜⎝^∂∂ ^⎞⎟⎠ =

For one component system

α β β α

= + = → = −

n n n const dn dn

α α β β α α β β

= + = μ + μ

dG G dn G dn dn dn

(

)

= G − G dn < 0

( )

α > β α < μ > μα β α <

if G G , dn 0 if , dn 0

(

)

= μ − μ dn < 0

( )

α < β α > μ < μα β α >

if G G , dn 0 if , dn 0

α α

μ =G → for one component system

α→β β α

Δμ = μ − μ <

→ α β

A A A

Irreversible transfer of A from to 0

α α ←

μ

α

=

α

A A multicomponent system chemical potential of A in

= partial molar Gibbs free energy of A in G

α β

μ = μ

→

α β

A A

Equilibrium condition for the transfer of A between and

Equilibrium between α and β for multi-component system

Pressure-temperature phase diagrams for compounds that may be treated as unary systems: (a) water; (b) SiO₂

Find the expression describing

the boundaries.

Equilibrium in one-component system

Equilibrium between α and β for one component system

α

=

→

=

G G dG dG

Chemical potential or molar Gibbs free energy should be the same.

α β

μ = μ

= − +

dG SdT VdP

→ G, S, V : molar quantity

α α α α α

β β β β β

μ = − + μ = − +

d S dT V dP d S dT V dP

α→β α→β

ΔS dT = ΔV dP

α α α α β β β β

−S dT +V dP = −S dT +V dP

β − α = β − α

(S S )dT (V V )dP

⇒

note conjugates of P,V and S,T Clausius-Clapeyron equation

α→β α→β

α→β α→β

Δ Δ

= → =

Δ Δ

dP S dP H

dT V dT T V

α→β α→β

= Δ Δ dP H dT T V

→

→

= Δ Δ

l s l s

dP H

dT T V

→

→

= Δ Δ

l s l s

dP H

dT T V

equilibrium vapor pressure

Equilibrium Vapor Pressure

α→ α

ΔV ^G = V^G −V ≅ V^G = RT P

Δ Δ Δ

= = =

⎛ ⎞

⎜ ⎟

⎝ ⎠

2

dP H H P H

dT TV T RT RT

P

⎛ ⎞ ⎛ ⎞

Δ Δ

= → ⎜ ⎟ = ⎜ − ⎟

⎝ ⎠ ⎝ ⎠

2 0 0

dP H dT ln P H 1 1

P RT P R T T

→

→

= Δ Δ

l s l s

dP H

dT T V

⎛ Δ ⎞ ⎛ Δ ⎞ ⎛ Δ ⎞

= ⎜ ⎟ ⎜− ⎟ = ⎜− ⎟

⎝ ⎠ ⎝ ⎠

⎝ ⎠

0 0

H H H

P P exp exp Aexp

RT RT RT

Equilibrium Vapor Pressure

→ Boltzman distribution

⎛ ⎞ = Δ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠0 ⎝ 0 ⎠

P H 1 1

ln P R T T

⎛ Δ ⎞

= ⎜⎝− ⎟⎠ P Aexp H

RT

⎛ ⎞= Δ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠0 ⎝ 0 ⎠

P H 1 1

ln P R T T

⎛ ⎞ = Δ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠0 ⎝ 0 ⎠

P H 1 1

ln P R T T

Why is P-T diagram different from P-V or T-X diagram?

Intensive parameters of P, T, and μ are the same at equilibrium.

→ Equilibrium along the phase boundary

P-T diagram

P : intensive parameter T : intensive parameter

P-V diagram V : extensive parameter

Alternative Representations of Unary Phase Diagrams

T

P S L

V

C

O

PL

P^α = T^α=T^L

In two phase equilibrium

No tie-line

PL

P^α = V^α≠V^L

In two phase equilibrium

P

V

S L V

C Tie

Tie Line Line

Horizontal tie-line

SL

S^α≠ V^α ≠V^L

In two phase equilibrium

S

V

S L V

C Tie Tie LineLine

Inclined tie-line Intensive-Intensive Intensive-Extensive Extensive-Extensive

Compare with T-X and Ternary diagrams.

P₃

Draw G-T diagram at P = 1, 0.006 and P₃.

Draw G-P diagram at temperatures above, at and below the triple point.

T₂

3

T₄

Draw G-X and a-X diagrams at T₁, T₂, T₃ and T₄.

What would be the volume of the mixture if half a mole of each is mixed?

The volume per one mole of pure water : 18 cm³ The volume per one mole of pure ethanol : 58 cm³

9 + 29 = 38 cm³

However, experimentally 37.1 cm³

→ if it is ideal.

How do we handle a non-ideal mixture?

How do we handle gas mixture or solution (solid or liquid)? ex) volume of mixture

Dickerson ideal gas, ideal solution

Dickerson

, , _A , , _B

B A

B P T n A P T n

V V

V V

n n

⎛ ∂ ⎞ ⎛ ∂ ⎞

= ⎜⎝∂ ⎟⎠ = ⎜⎝∂ ⎟⎠

Dickerson

partial pressure (ideal gas) p_i = X_iP partial volume (ideal solution) V_i = X_iV

partial Gibbs free energy G_i ≠ X_iG

Gas mixture or Solution (solid or liquid)

≠

+

mixture B B

V V X V X

≠

+

mixture B B

G G X G X

partial volume (real solution) V_i ≠ X_iV

Rule of mixture does not hold!

How should we handle the mixture?

Vmix

Δ 6000

0

0 1

X2

X₂o

P B

A

Gm

XB

0 1

Vmix

Δ

6000

0 0 1

X2

X₂o

P B

C

A

V2

Δ

V1

Δ

Express ΔV_mixat X₂^o in terms of and . ΔV1 ΔV₂

partial molar volume

j

i

i P ,T ,n

V V

n

⎛∂ ⎞

= ⎜⎝∂ ⎟⎠ partial molar Gibbs free energy

j

i

i P ,T ,n

G G

n

⎛∂ ⎞

= ⎜⎝∂ ⎟⎠

partial molar quantity

partial molar operator

i P ,T ,nj

n

⎛ ∂ ⎞

⎜∂ ⎟

⎝ ⎠