phase transitions and phase diagrams

MSE 3050, Phase Diagrams and Kinetics, Leonid Zhigilei

Phase Transitions and Phase Diagrams

One-component systems

Enthalpy and entropy dependence on P and T

Gibbs free energy dependence on P and T

Clapeyron equation

Understanding phase diagrams for one-component systems

Polymorphic phase transitions

Driving force for a phase transition

First order and second-order phase transitions

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


PVT Surface of a Pure Substance

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


A pure substance is heated at constant pressure

T

Tb

V

P


H and S as function of T at constant P

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

298

P298 dTCHTHPP

CT

H

For S(T) we have T

0

P dTT

CTS

T

C

T

S P

P

0

0

0

CP

H

S

T, K

T, K

T, K298

Slope = CP

Slope = CP/T


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

T

c

T

S

T

G P

PP

2

2

for the curvature

0

H

T, KTS

Slope = CP

Slope = -S

G

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


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 Gl > Gs at low T.

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

At Tm Gl(T) crosses Gs(T) and both liquid and solid phases can co-existin equilibrium (Gl = Gs)

0

Hl

T, KTm

Gl

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

Hs

Gs

Hm

Hl > Hs

Sl > Ss

at all T

Hm = TmSm


A typical P-T phase diagram for a pure material

The red lines on the phase diagram show the conditions where differentphases coexist in equilibrium: Gphase1 = Gphase2

liquid

solid

gas

T

P

triple point

1 atm

normal freezing point

criticalpoint

normal boilingpoint

G

P = 1 atmvapor

T

liquid

solid

liquid is stable

vapor is stable

solid is stable


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 P

Vl < Vs for water

Vl > Vs 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

V

V

1k

TV

PVB

- isothermal compressibility

- bulk modulus


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

For two phases in equilibrium Gl = Gs and dGl = dGs for infinitesimalchange in T and P (so that the system remains in equilibrium)

dTS-dPVdG lll

At equilibrium

ΔV

ΔS

VV

SS

dT

dP

ls

ls

eq.

dTS-dPVdG sss dTS-dPVdTS-dPV ssll

0S TΔΔHΔG S TΔΔH and

Therefore V TΔ

ΔH

dT

dP

eq.

- the Clapeyron equation

The Clapeyron equation gives the relationship between the variations ofpressure and temperature required for maintaining equilibrium betweenthe two phases.


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, Gl = Gs.

G

T

P

liquid

solid

0ºC

1 atm


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

H = Hg - Hl > 0 – we have to addheat to convert liquid to gas.

Therefore 0V TΔ

ΔH

dT

dP

eq.

Clapeyron equation: examples

A typical diagram for a pure material:liquidsolid

gas

T

P

For liquid to solid transition: V = Vs – Vl < 0 for most materials

H = Hs - Hl < 0 – heat is released upon crystallization.

Therefore 0V TΔ

ΔH

dT

dP

eq.

For some materials, however, V = Vs – Vl > 0 and 0dT

dP

eq.

Si, Ge, Ga, Bi, GaAs, diamond …

In general, it takes heat (H > 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

TFe, Ni, Au, CuZn, Ar, …

0dT

dP

m

P

0dT

dP

m


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 -Fe has asmaller molar volume than BCC-Fe: V = V - V < 0

At the same timeH = H - H > 0

Therefore 0V TΔ

ΔH

dT

dP

eq.

The diagram shows the stable phases for pure iron at varyingtemperature and pressure.

-Fe

-Fe

The effect of increasing pressure is to increase the area of the phasediagram over which the phases of the smaller molar volume (higherdensity) is stable.

-Fe

G

T

BCCFCC

P = constliquid

vapor


-Fe

-Fe

-Fe

V TΔ

ΔH

dT

dP

eq.


V TΔ

ΔH

dT

dP

eq.


V TΔ

ΔH

dT

dP

eq.

Can the volume expansion of water upon melting explain ice-skating?

kg 100m

lengthcontact cm 1 wide,mm 2 :skateJ/mol 5636

/molcm 0.18/molcm 63.193

3

m

liquid

ice

HVV


Squeezing diamond from graphite: What pressure should we apply to transform graphite to diamond at 298 K?

In reference books we can find that at 298 K and 1 atm:

Hdiamond = 1900 J/molSgraphite = 5.73 J/KSdiamond = 2.43 J/Kgraphite = 2.22 g/cm3

diamond = 3.515 g/cm3


The driving force for the phase transformation

If solid and liquid are in equilibrium, Gs = Gl 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 = nl Gl + ns Gs = const, where nl and ns are the numbers of moles ofliquid and solid phases, and Gl and Gs 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*

G

Gs

At temperature T*

Gl

Tm

T

l*

ll ST-HG

s*

ss ST-HG

ST-HG *

At temperature Tm

0ST-HG m

m

m

T

HS

For small undercooling T we can neglect the difference in Cp ofliquid and solid phases and assume that H and S are independent oftemperature.

m

m*m T

ΔHTΔHΔG

m

m

T

ΔTΔHΔG The driving force for solidification

At temperature T*:


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

Ttrs

trs

trstrs T

HS

0ΔGtrs

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

Ttrs T

S

Ttrs T

H Cp

-ST

G

P

VP

G

T

- discontinuous

Ttrs T Ttrs T 0Strs

0Htrs

PP dT

dHC

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


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

G

Ttrs

0ΔGtrs

Second-order phase transition

T

V

Ttrs T

S

Ttrs T

H Cp

-ST

G

P

VP

G

T

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

Ttrs T Ttrs T

0Strs

0Htrs

PP dT

dHC

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

0Vtrs

P

2

T

V

PT

G

T

2

P

S

TP

G

- discontinuous


Summary

Make sure you understand language and concepts:

Enthalpy and entropy dependence on P and T

Gibbs free energy dependence on P and T

Clapeyron equation

Understanding phase diagrams for one-component systems

Driving 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.)

phase transitions and phase diagrams

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