Thermodynamics of Materials

12th Lecture2008. 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 Oat 273 K, H H 6008 J / molS 70. 08 J / KS 44. 77 J / KC 75. 44 J / KC 38 J / K

Δ = =

Δ = −Δ = −

Δ = = −

∫

∫

ⅰ

ⅱ

ⅲ

273

263

m263

273

H 75. 44dT 754. 4

H H 6008 J

H 38dT 380

→ → →Δ = Δ − Δ(l s) (l s) (l s)G H T S

→∴ Δ = Δ + Δ + Δ = −i ii iii(l s)H H H H 5633. 6 J

Δ = −

= +

− + =

−Δ = = −

Δ = − =

∫

∫

i

ii

iii

273(l) 263(l)

273

298(l) p(l)298

263

298(l) p(l)298

263(s) 273(s)

S S S

(S C d ln T)

273(S C d ln T) 75. 44 ln263

6008S 22. 007273

263S S S 38 ln273

→Δ = Δ + Δ +∴ Δ = −i ii iii(l s)S S S S 20. 61 J K

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

→ → →Δ = Δ − Δ(l s) (l s) (l s)G H T S

→∴ Δ = Δ + Δ + Δ = −i ii iii(l s)H H H H 5633. 6 J

→Δ = Δ + Δ +∴ Δ = −i ii iii(l s)S S S S 20. 61 J K

→

= − +

= −at constant P dGdG SdT VdP

SdT

→Δ = −∫273k 263kG SdT

→

⎡ ⎤⎛ ⎞Δ = − + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫263water

273 263K 273

TG 70. 08 75. 44 ln dT298

= + − −

− −

=

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

620. 7

= + = +∫T p

298k 298

C TS S dT 70. 08 75. 44 lnT 298

→

⎡ ⎤⎛ ⎞Δ = − + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫263ice

273 263K 273

TG 44. 77 38 ln dT298

= −

− − − −

=

447. 7 380 ln 29838[(263 ln 263 263) 273 ln 273 273)]407. 4

→Δ =water273k 263kG 620. 7 J / mole

→Δ =ice273 263KG 407. 4 J / mole

Δ

298

H S S .

ⅰ ),ⅱ ) 결 과 동 일ⅰ )의 경 우 상 전 이 시 를 알 아 야 하 고ⅱ )의 경 우 또 는 특 정 온 도 에 서 를 알 아 야 한 다

→→ →∴ Δ = Δ − Δ

= − = −

water ice ice water273 263K263k 273k 263kG 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 ( ΔG290K

vapor→water) when the water vapor saturatedat 300K is supercooled to 290K?

What would be the driving force for precipitation (ΔG290K

vapor→water) when the water vapor saturatedat 300K is supercooled to 290K?

Δ = Δ − ΔG H T S

= − → Δ = −∫dG VdP SdT G SdT

→

⎛ ⎞ ⎛ ⎞Δ = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

eq300

300 290K eq eq290 290

PPG RTln RTlnP P

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠Δ

P,T

P,T

Why is the critical nucleus defined byGthe condition of 0 instead of r

the condition of G =0?

Open System

dU = TdS – PdV + dUmatter

dG= VdP - SdT + dGmatter

dUmatter or dGmatter ∝ ?

dGmatter ∝ dn → dGmatter = μdn

dG= VdP - SdT + Σμidni

= − + μ

= + + μ

= − − + μ

= − + + μ

∑∑∑∑

i i

i i

i i

i i

dU TdS PdV dndH TdS VdP dndF SdT PdV dndG SdT VdP dn

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂= μ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠j j j j

ii i i iT,P,n S,V,n S,P,n T,V,n

G U H Fn n n n

Chemical Potential

α β

surrounding

constant T, P

Close

Open

α α β β α α β β= + = μ + μG G n G n n n

Equilibrium between α and βfor one componentsystem

What is the G of α and β system ?

α β → α βG , G molar free energy of and

( ), ,,T P T PT P

nGG nG Gn n n

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

For one component system

α β β α= + = → = −n n n const dn dnα α β β α α β β= + = μ + μdG G dn G dn dn dn

( )α β α= − <G G dn 0 for an irreversible process

( )α β α α β α> < μ > μ <if G G , dn 0 if , dn 0

( )α β α= μ − μ <dn 0

( )α β α α β α< > μ < μ >if G G , dn 0 if , dn 0

α αμ = →G for one component system

α→β β αΔμ = μ − μ <

→ α βA A A

Irreversible transfer of A from to 0

α α ←μ

α

=

αA A multicomponent system

chemical potential of A in = partial molar Gibbs free energy of A in

G

α βμ = 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 expressiondescribing the boundaries.

Equilibrium in one-componentsystem

Equilibrium between α and βfor one component system

α β α β= → =G G dG dG

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

α βμ = 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 dPd S dT V dP

α→β α→βΔ = ΔS dT V 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,TClausius-Clapeyron equation

α→β α→β

α→β α→β

Δ Δ= → =

Δ ΔdP S dP HdT V dT T V

α→β

α→β

Δ=

ΔdP HdT T V

→

→

Δ=

Δ

l s

l sdP HdT T V

→

→

Δ=

Δ

l s

l sdP HdT T V

equilibriumvapor pressure

Equilibrium Vapor Pressure

α→ αΔ = − ≅ =G G G RTV V V VP

Δ Δ Δ= = =

⎛ ⎞⎜ ⎟⎝ ⎠

2dP H H P H

RTdT TV RTTP

⎛ ⎞ ⎛ ⎞Δ Δ= → = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠2

0 0

dP H P H 1 1dT lnP RT P R T T

→

→

Δ=

Δ

l s

l sdP HdT T V

⎛ ⎞Δ Δ Δ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

00

H H HP P exp exp AexpRT RT RT

Equilibrium Vapor Pressure → Boltzman distribution

⎛ ⎞ ⎛ ⎞Δ= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠0 0

P H 1 1lnP R T T

Δ⎛ ⎞= −⎜ ⎟⎝ ⎠

HP AexpRT

⎛ ⎞ ⎛ ⎞Δ= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠0 0

P H 1 1lnP R T T

⎛ ⎞ ⎛ ⎞Δ= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠0 0

P H 1 1lnP R T T

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

Intensive parameters of P, T, and μare the same at equilibrium. → Equilibrium along the phase boundary

P-T diagram

P : intensive parameterT : intensive parameter

P-V diagram V : extensive parameter

Alternative Representations of Unary Phase Diagrams

T

PS L

V

C

O

LPP =α LTT =α

In two phase equilibrium

No tie-line

LPP =α LVV ≠α

In two phase equilibrium

P

V

S LV

C

Tie Tie LineLine

Horizontal tie-line

LSS ≠α LVV ≠α

In two phase equilibrium

S

V

SL 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-Xdiagramsat 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

Dickersonideal gas, ideal solution

Dickerson, , , ,A B

B AB AP T n P T n

V VV Vn n

⎛ ⎞ ⎛ ⎞∂ ∂= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Dickerson

partial pressure (ideal gas) pi = XiP

partial volume (ideal solution) Vi = XiV

partial Gibbs free energy Gi ≠ XiG

Gas mixture or Solution (solid or liquid)

≠ +o oA Amixture B BV V X V X

≠ +o oA Amixture B BG G X G X

partial volume (real solution) Vi ≠ XiV

Rule of mixture does not hold!

How should we handle the mixture?

mixVΔ

6000

00 12X

oX2

P B

A

mG

BX0 1

mixVΔ

6000

00 1

2XoX 2

P B

C

A

2VΔ

1VΔ

Express ΔVmix at X2o in terms of and . 2VΔ1VΔ

partial molar volume j

ii P ,T ,n

VVn

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

partial molar Gibbs free energy

j

ii P ,T ,n

GGn

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

partial molar quantity

partial molar operatorj

i P ,T ,nn⎛ ⎞∂⎜ ⎟∂⎝ ⎠