Phase diagram

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Phase diagram. Need to represent how mineral reactions at equilibrium vary with P and T. P-X stability and mixing. Gibbs Phase Rule. The number of variables which are required to describe the state of a system: p+f=c+2 f=c-p+2 Where p=# of phases, c= # of components, - PowerPoint PPT Presentation

Transcript of Phase diagram

Phase diagram• Need to represent how mineral reactions

at equilibrium vary with P and T

R

R

G V

S

T

P

0P

R

R

R

T

V

V

S

T

C

T

S P

P

R

P-X stability and mixing

Gibbs Phase Rule

• The number of variables which are required to describe the state of a system:

• p+f=c+2 f=c-p+2– Where p=# of phases, c= # of components,

f= degrees of freedom– The degrees of freedom correspond to the

number of intensive variables that can be changed without changing the number of phases in the system

Variance and f• f=c-p+2

• Consider a one component (unary) diagram

• If considering presence of 1 phase (the liquid, solid, OR gas) it is divariant

• 2 phases = univariant

• 3 phases = invariant

Melts• Liquid composed of predominantly silica and

oxygen. Like water, other ions impart greater conductivity to the solution

• Si and O is polymerized in the liquid to differing degrees – how ‘rigid’ this network may be is uncertain…

• Viscosity of the liquid increases with increased silica content, i.e. it has less resistance to flow with more SiO2… related to polymerization??

• There is H2O is magma 2-6% typically – H2O decreases the overall melting T of a magma, what does that mean for mineral crystallization?

Thermodynamic definitions

• Gi(solid) = Gi(melt)

• Ultimately the relationships between these is related to the entropy of fusion (S0

fus), which is the entropy change associated with the change in state from liquid to crystal

• These entropies are the basis for the order associated with Bowen’s reaction series greater bonding changes in networks, greater entropy change lower T equilibrium

0fus

fus

i S

RT

dX

dT

Melt-crystal equilibrium• Precipitated crystals

react with cooling liquid, eventually will re-equilibrate back, totally cooled magma xstals show same composition

• UNLESS it cools so quickly the xstal becomes zoned or the early precipitates are segregated and removed from contact with the bulk of the melt

Why aren’t all feldspars zoned?

• Kinetics, segregation

• IF there is sufficient time, the crystals will re-equilibrate with the magma they are in – and reflect the total Na-Ca content of the magma

• IF not, then different minerals of different composition will be present in zoned plagioclase or segregated from each other physically

ExsolutionP-X stability and mixing

• More than 1 crystal can precipitate from a melt – different crystals, different stabilities…– 2+ minerals that do not share equilibrium in a melt are

immiscible (opposite of a solid solution)– Liquidus Line describing equilibrium between melt and

one mineral at equilibrium– Solidus Line describing equilibrium with melt and solid– Eutectic point of composition where melt and solid can

coexist at equilibrium

Diopside is a pyroxeneAnorthite is a feldspar

Solidus

Liquidus

Eutectic

• Melt at composition X cools to point Y where anorthite (NOT diopside at all) crystallizes, the melt becomes more diopside rich to point C, precipitating more anorthite with the melt becoming more diopside-rich

• This continues and the melt continues to cool and shift composition until it reaches the eutectic when diopside can start forming

A

B S1

Z

C S2

At eutectic, diopside AND anorhtite crystals precipitateLever Rule diopside/anorthite (42%/58%) crystallize until last of melt precipitates and the rock composition is Z

• Melting when heated to eutectic, the rock would melt such that all the heat goes towards heat of fusion of diopside and anorthite, melts so that 42% diopside / 58% anorthite…

• When diopside gone, temperature can increase and rest of anorthite can melt (along liquidus)

• How does free energy change with T and P?• From G=H-TS:

• T and P changes affect free energy and can drive reactions!!

2

1

2

1

2

)1(

2

1

)1(11222)( 121,1,,

T

T

P

P

T

PT

T

PTPTPTP dPVdTT

CTdTCTTSGG P

P

Volume Changes (Equation of State)

VdP

dG

T

PT

V

V

1

TP

V

V

1

Volume is related to energy changes:

Mineral volume changes as a function of T: , coefficient of thermal expansion

Mineral volume changes as a function of P: , coefficient of isothermal expansion

For Minerals:

Volume Changes (Equation of State)

• Gases and liquids undergo significant volume changes with T and P changes

• Number of empirically based EOS solns..• For metamorphic environments:

– Redlich and Kwong equation:

• V-bar denotes a molar quatity, aRw and bRK are constants

)(2/1RK

Rw

RK bVVT

a

bV

RTP

Phase Relations• Rule: At equilibrium, reactants and products have

the same Gibbs Energy– For 2+ things at equilibrium, can investigate the P-T

relationships different minerals change with T-P differently…

• For GR = SRdT + VRdP, at equilibrium, Grearranging:

R

R

G V

S

T

P

0

Clausius-Clapeyron equation

Remember that a line on a phase diagram describes equilibrium, GR=0!!

V for solids stays nearly constant as P, T change, V for liquids and gases DOES NOT

• Solid-solid reactions linear S and V nearly constant, S/V constant + slope in diagram

• For metamorphic reactions involving liquids or gases, volume changes are significant, V terms large and a function of T and P (and often complex functions) – slope is not linear and can change sign (change slope + to –)

R

R

G V

S

T

P

0

P

R

TR

R

T

V

V

S

T

C

T

S P

P

R

SR change with T or P?

V = Vº(1-P)

21

22

00

00

(2

1

2

1

2

PPPVS

VdPSdPP

SSS

P

PT

P

P

P

R

R

G V

S

T

P

0

Example – Diamond-graphite• To get C from

graphite to diamond at 25ºC requires 1600 MPa of pressure, let’s calculate what P it requires at 1000ºC:

graphite diamond

(K-1)

1.05E-05 7.50E-06

(MPa-1)

3.08E-05 2.27E-06

(J/mol K)

5.74 2.38

(cm3/mol)

5.2982 3.417

Clausius-Clapyron Example