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MSE 3050, Phase Diagrams and Kinetics, Leonid Zhigilei

Phase Transitions and Phase Diagrams

One-component systems

Enthalpy and entropy dependence on P and TGibbs free energy dependence on P and TClapeyron equationUnderstanding phase diagrams for one-component systemsPolymorphic phase transitions

Driving force for a phase transitionFirst order and second-order phase transitions

Reading: 1.2 of Porter and EasterlingChapter 7.1 – 7.4 of Gaskell

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PVT Surface of a Pure Substance

http://www.eng.usf.edu/~campbell/ThermoI/ThermoI_mod.html

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A pure substance is heated at constant pressure

T

T b

V

P

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H and S as function of T at constant P

In a closed one-component system equilibrium, at temperature T and pressure P, corresponds to the state with minimum Gibbs free energy G.Therefore, in order to predict what phases are stable under different

conditions we have to examine the dependence of G on T and P.Let’s use thermodynamic relations to predict the temperaturedependence of H, S, and G at constant P.

For H(T) we have ( ) ∫+=T

298P298 dTCHTHP

P

CTH =⎟

⎠ ⎞

⎜⎝ ⎛

∂∂

For S(T) we have ( ) ∫=T

0

P dTT

CTSTCTSP

P=⎟ ⎠ ⎞⎜⎝ ⎛ ∂∂

0

0

0

CP

H

S

T, K

T, K

T, K 298

Slope = C P

Slope = C P/T

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G as function of T at constant P

For G = H – TS we have dG = -SdT +VdP and for P = const

ST

G

P

−=⎟ ⎠

⎞⎜

⎝

⎛

∂

∂for the slope

Tc

TS

TG P

PP2

2

−=⎟ ⎠ ⎞

⎜⎝ ⎛ ∂∂−=⎟⎟ ⎠

⎞⎜⎜

⎝ ⎛

∂∂ for the curvature

0

H

T, K TS

Slope = C P

Slope = -SG

G(T) for a single phase at P = const

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G as function of T at constant P for liquid and solid phases

At all temperatures the liquid has a higher internal energy U andenthalpy H as compared to the solid. Therefore G l > G s at low T.

The liquid phase, however, has a higher entropy S than the solid phase atall T. Therefore G l decreases more rapidly with T as compared to G s.

At T m G l(T) crosses G s(T) and both liquid and solid phases can co-existin equilibrium (G l = G s)

0

H l

T, K Tm

G l

At T m the heat supplied to the system will not rise its temperature butwill be used to supply the latent heat of melting ΔHm that is required toconvert solid into liquid. At T m the heat capacity C p = ( H/ T) P isinfinite – addition of heat does not increase T.

Hs

Gs

ΔHm

H l > H s

Sl > S s

at all T

ΔHm = T mΔSm

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A typical P-T phase diagram for a pure material

The red lines on the phase diagram show the conditions where different phases coexist in equilibrium: G phase1 = G phase2

liquid

solid

gas

T

P

triple point

1 atm

normalfreezing

point

critical

point

normal boiling point

G

P = 1 atm vapor

T

liquid

solid

liquid isstable

vapor isstable

solid isstable

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G as function of P at constant T for liquid and solid phases

As we can see from the fundamental equation, dG = VdP – SdT, the freeenergy of a phase increases with pressure:

If the two phases have different molar volumes, their free energies willincrease by different amounts when pressure changes at a fixed T.

0VP

G

T

>=⎟ ⎠

⎞⎜

⎝

⎛

∂

∂

G

T = 0ºC

1 atm

l i q u i d H 2 O

i c e

P

V l < V s for water

V l > V s for most materials

How the unusual change of V upon melting of water could be related toice-skating?

What is the curvature of the G(P) at constant T?

TT P

VV1

k ⎟ ⎠ ⎞

⎜⎝ ⎛

∂∂−=

TVP

VB ⎟ ⎠ ⎞

⎜⎝ ⎛ ∂∂−=- isothermal

compressibility- bulkmodulus

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Equilibrium between two phases: Clapeyron equation

If two phases in equilibrium have different molar volumes, their freeenergies will increase by different amounts when pressure changes at afixed T. The equilibrium, therefore will be disturbed by the change in

pressure. The only way to maintain equilibrium at different pressures isto change temperature as well.

For two phases in equilibrium G l = G s and dG l = dG s for infinitesimalchange in T and P (so that the system remains in equilibrium)

dTS-dPVdG lll =

At equilibrium

Δ V

Δ SVVSS

dTdP

ls

ls

eq.

=−−=⎟

⎠ ⎞

⎜⎝ ⎛

dTS-dPVdG sss =dTS-dPVdTS-dPV

ssll =

0STΔΔ HΔ G =−= STΔΔ H =and

ThereforeVTΔ

Δ HdTdP

eq.

=⎟ ⎠ ⎞

⎜⎝ ⎛

- the Clapeyron equation

The Clapeyron equation gives the relationship between the variations of

pressure and temperature required for maintaining equilibrium betweenthe two phases.

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G as function of P and T for liquid and solid phases

Schematic representation of the equilibrium surfaces of the solid andliquid phases of water in G-T-P space.

The planes show the free energies of liquid and solid phases, theintersections of the planes correspond to the (P, T) conditions needed formaintaining equilibrium between the phases, G l = G s.

G

T

P

liquid

solid

0ºC

1 atm

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For liquid to gas transition: ΔV =Vg - V l >> 0

ΔH = H g - H l > 0 – we have to addheat to convert liquid to gas.

Therefore 0VTΔ

Δ HdTdP

eq.

>=⎟ ⎠ ⎞

⎜⎝ ⎛

Clapeyron equation: examples

A typical diagram for a pure material:liquidsolid

gas

T

P

For liquid to solid transition: ΔV = V s – V l < 0 for most materials

ΔH = H s - H l < 0 – heat is released upon crystallization.

Therefore 0VTΔ

Δ HdTdP

eq.

>=⎟ ⎠ ⎞

⎜⎝ ⎛

For some materials, however, ΔV = V s – V l > 0 and 0dTdP

eq.

0) to proceed from a low-temperature to ahigh-temperature phase (entropy of a high-temperature phase is higherthan the entropy of a low-temperature phase). Therefore, the slope ofthe equilibrium lines in a P-T phase diagram of a pure materialreflects the relative densities of the two phases.

liquidsolid

gas

T

P

liquid

solid gas

T

Fe, Ni, Au, CuZn, Ar, …

0dTdP

m>⎟ ⎠

⎞⎜⎝ ⎛

P

0dTdP

m

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Clapeyron equation: more examples

Some materials may exist in more than one crystal structure, this iscalled polymorphism . If the material is an elemental solid, it is calledallotropy .

Close-packed FCC γ-Fe has asmaller molar volume than BCCα-Fe: ΔV = V γ - Vα < 0

At the same timeΔH = H γ - Hα > 0

Therefore 0VTΔ

Δ HdTdP

eq.

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γ-Fe

α-Fe

δ-Fe

0VTΔ

Δ HdTdP

eq.

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VTΔΔ H

dTdP

eq.

=⎟ ⎠ ⎞

⎜⎝ ⎛

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VTΔΔ H

dTdP

eq.

=⎟ ⎠ ⎞

⎜⎝ ⎛

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VTΔΔ H

dTdP

eq. =⎟ ⎠ ⎞

⎜⎝ ⎛

Can the volume expansion of water upon melting explainice-skating?

kg100=m

lengthcontactcm1wide,mm2:skateJ/mol5636

/molcm0.18

/molcm63.193

3

=Δ =

=

m

liquid

ice

H V

V

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Example: squeezing diamond from graphite

What pressure should we apply to transform graphite to diamond at298 K?

In a reference book we can find that at 298 K and 1 atm:Hdiamond = 1900 J/molSgraphite = 5.73 J/K Sdiamond = 2.43 J/K ρgraphite = 2.22 g/cm 3ρdiamond = 3.515 g/cm 3

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The driving force for the phase transformation

If solid and liquid are in equilibrium, G s = G l and a slow addition of heatleads to the melting of some part of the solid, but do not change the totalG of the system:

G = n l G l + n s Gs = const, where n l and n s are the numbers of moles ofliquid and solid phases, and G l and G s are the molar Gibbs free energies.

If energy is added/removed quickly , the system can be brought out ofequilibrium (overheated or undercooled) – the melting/freezing processis spontaneous/irreversible and G is decreasing.

G

T*

ΔG

Gs

At temperature T *

G l

Tm

ΔT

l*

ll ST-HG =

s*

ss ST-HG =

ST-HG*

ΔΔ=ΔAt temperature T m

0ST-HG m =ΔΔ=Δ

m

m

TH

S Δ=Δ

For small undercooling ΔT we can neglectthe difference in C p of liquid and solid

phases and assume that ΔH and ΔS areindependent of temperature.

m

m*m T

Δ HTΔ HΔ G −≈

m

m

T

Δ TΔ HΔ G ≈The driving force for solidification

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First-order and second-order phase transitions (I)

The classification of phase transitions proposed by Ehrenfest is based onthe behavior of G near the phase transformation.

G

T trs

trs

trstrs T

HS

Δ=Δ

0Δ G trs =

First-order phase transition: first derivatives of G arediscontinuous.

Second-order phase transition: first derivatives of G arecontinuous, but second derivatives of G are discontinuous.

First-order phase transition

T

V

T trs T

S

T trs T

H C p

-STG

P

=⎟ ⎠ ⎞

⎜⎝ ⎛

∂∂

VPG

T

=⎟ ⎠ ⎞

⎜⎝ ⎛

∂∂ - discontinuous

T trs T T trs T0S

trs ≠Δ

0H trs ≠Δ

PP dT

dHC ⎟

⎠ ⎞

⎜⎝ ⎛ =

e.g. melting, boiling, sublimation, some polymorphous phase transitions.

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First-order and second-order phase transitions (II)

G

T trs

0Δ G trs =

Second-order phase transition

T

V

T trs T

S

T trs T

H C p

-STG

P

=⎟ ⎠ ⎞

⎜⎝ ⎛

∂∂

VPG

T

=⎟ ⎠ ⎞

⎜⎝ ⎛

∂∂

- continuous (S and V do not jump at transition)

T trs T T trs T

0S trs =Δ

0H trs =Δ

PP dT

dHC ⎟

⎠ ⎞

⎜⎝ ⎛ =

e.g. conducting-superconducting transition in metals at lowtemperatures.

0Vtrs =Δ

P

2

TV

PTG ⎟

⎠ ⎞⎜

⎝ ⎛

∂∂=⎟⎟ ⎠

⎞⎜⎜⎝ ⎛

∂∂∂

T

2

PS

TPG ⎟

⎠ ⎞⎜

⎝ ⎛ ∂∂−=⎟⎟ ⎠

⎞⎜⎜⎝ ⎛

∂∂∂ - discontinuous

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MSE 3050 Phase Diagrams and Kinetics Leonid Zhigilei

Summary

Make sure you understand language and concepts:

Enthalpy and entropy dependence on P and TGibbs free energy dependence on P and TClapeyron equationUnderstanding phase diagrams for one-component systemsDriving force for a phase transition

First order and second-order phase transitions

Make sure you understand P-T, G-P, G-T 2D phase diagrams fora one-component system (what is shown, what are the linesseparating different regions, how to predict the slopes of thelines, etc.)

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