Chapter 1 Introduction

bronzekerplunkMechanics

Oct 27, 2013 (3 years and 10 months ago)

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Thermodynamics


a
system:

Some portion of the universe that you wish to study


The
surroundings:

The adjacent part of the universe outside the
system

Changes

in a system are associated with the
transfer of
energy

Natural systems tend toward states of minimum energy

Energy States


Unstable:

falling or rolling


Stable:

at rest in lowest
energy state


Metastable:

in low
-
energy
perch

Figure 5
-
1.
Stability states. Winter (2001) An Introduction to
Igneous and Metamorphic Petrology. Prentice Hall.

Gibbs Free Energy

Gibbs free energy is a measure of
chemical

energy

All chemical systems tend naturally toward states
of minimum Gibbs free energy

G = H
-

TS

Where:

G = Gibbs Free Energy

H = Enthalpy (heat content)

T = Temperature in Kelvins

S = Entropy (can think of as randomness)

Thermodynamics

a
Phase:

a mechanically separable portion of a system


Mineral


Liquid


Vapor

a
Reaction
:
some change in the nature or types of phases
in a system

reactions are written in the form:




reactants = products

Thermodynamics

The change in some property, such as G for a
reaction of the type:




2 A + 3 B = C + 4 D


D
G =
S

(n G)
products

-

S
(n G)
reactants



= G
C

+ 4G
D

-

2G
A

-

3G
B

The side of the reaction with lower G will be more stable

Thermodynamics

For a
phase

we can determine V, T, P, etc., but not G or H

We can only determine
changes

in G or H as we change
some other parameters of the system

Example: measure
D
H for a reaction by calorimetry
-

the heat
given off or absorbed as a reaction proceeds

Arbitrary

reference state

and assign an equally arbitrary
value of H to it:



Choose
298.15 K and 0.1 MPa

(lab conditions)


...and assign
H = 0 for pure elements

(in their natural state
-

gas, liquid, solid) at that reference

Thermodynamics

In our calorimeter we can then determine
D
H for the reaction:


Si (metal) + O
2

(gas) = SiO
2

D
H =
-
910,648 J/mol

=
molar

enthalpy of formation

of quartz
(at 298, 0.1)

It serves quite well for a standard value of H for the phase

Entropy

has a more universal reference state: entropy of every
substance = 0 at 0K, so we use that (and adjust for temperature)

Then we can use
G = H
-

TS

to determine G of quartz



=
-
856,288 J/mol

Thermodynamics

For other temperatures and pressures we can use the equation:




dG = VdP
-

SdT


(ignoring
D
X for now)


where V = volume and S = entropy (both molar)

We can use this equation to calculate G for any phase at any T and P
by integrating

G

G

VdP

SdT

T

P

T

P

T

T

P

P

2

1

1

1

2

1

2

2

-

=

-

If V and S are constants, our equation reduces to:



G
T2 P2

-

G
T1 P1

= V(P
2

-

P
1
)
-

S (T
2

-

T
1
)



Thermodynamics

If V and S are constants, our equation reduces to:



G
T2 P2

-

G
T1 P1

= V(P
2

-

P
1
)
-

S (T
2

-

T
1
)




which ain’t bad!

Thermodynamics

In Example 1 we use



G
T2 P2

-

G
T1 P1

= V(P
2

-

P
1
)
-

S (T
2

-

T
1
)



and G
298, 0.1

=
-
856,288 J/mol to calculate G for quartz at several
temperatures and pressures

Low quartz

Eq. 1

SUPCRT

P (MPa)

T (C)

G (J) eq. 1

G(J)

V (cm3)

S (J/K)

0.1

25

-
856,288

-
856,648

22.69

41.36

500

25

-
844,946

-
845,362

22.44

40.73

0.1

500

-
875,982

-
890,601

23.26

96.99

500

500

-
864,640

-
879,014

23.07

96.36

Agreement is quite good

(< 2% for change of 500
o

and 500 MPa or 17 km)

Thermodynamic

Summary thus far:


G is a measure of relative chemical stability for a phase


We can determine G for any phase by measuring H and S for
the reaction creating the phase from the elements


We can then determine G at any T and P mathematically


Most accurate if know how V and S vary with P and T


dV/dP is the coefficient of isothermal compressibility


dS/dT is the heat capacity (Cp)

Use?

If we know G for various phases, we can determine which is
most stable


Why is melt more stable than solids at high T?


Is diamond or graphite stable at 150 km depth?


What will be the effect of increased P on melting?

Free Energy vs. Temperature

dG = VdP
-

SdT at constant pressure:
dG/dT =
-
S

Because S must be (+)
G for a phase decreases as T
increases

Would the slope for the
liquid be steeper or
shallower than that for
the solid?

Figure 5
-
3.
Relationship between Gibbs free energy and
temperature for a solid at constant pressure. T
eq

is the equilibrium
temperature.
Winter (2001) An Introduction to Igneous and
Metamorphic Petrology. Prentice Hall.

Free Energy vs. Temperature

Slope of G
Liq

> G
sol
since
S
solid

< S
liquid

A:

Solid more stable than
liquid (low T)

B:

Liquid more stable than
solid (high T)


Slope
d
P/
d
T =
-
S


Slope S < Slope L

Equilibrium

at T
eq


G
Liq

= G
Sol

Figure 5
-
3.
Relationship between Gibbs free energy and
temperature for the solid and liquid forms of a substance at constant
pressure. T
eq

is the equilibrium temperature.
Winter (2001) An
Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Now consider a
reaction,

we can then use the equation:




d
D
G =
D
VdP
-

D
SdT

(again ignoring
D
X)

For a reaction of melting (like ice


water)


D
V is the volume change involved in the reaction (V
water

-

V
ice
)


similarly
D
S and
D
G are the entropy and free energy changes

d
D
G is then the change in
D
G as T and P are varied


D
G is (+) for S


L at point A (G
S

< G
L
)


D
G is (
-
) for S


L at point B (G
S

> G
L
)


D
G = 0 for S


L at point x (G
S

= G
L
)

D
G for any reaction = 0 at equilibrium

Worked Problem #2 used:




d
D
G =
D
VdP
-

D
SdT


and G, S, V values for albite, jadeite and quartz to
calculate the conditions for which
D
G of the reaction:



Ab + Jd = Q


is equal to 0


from G values for each phase at 298K and 0.1 MPa calculate
D
G
298, 0.1

for the
reaction, do the same for
D
V and
D
S


D
G at equilibrium = 0
, so we can calculate an isobaric change in T that would
be required to bring
D
G
298, 0.1

to 0



0
-

D
G
298, 0.1

=
-
D
S (
T
eq

-

298)

(at constant P)


Similarly we could calculate an isothermal change



0
-

D
G
298, 0.1

=
-
D
V (
P
eq

-

0.1)

(at constant T)

Method:

NaAlSi
3
O
8

= NaAlSi
2
O
6

+ SiO
2

P
-

T phase diagram of the equilibrium curve

How do you know which side has which phases?

Temperature
-
pressure phase
diagram for the reaction: Albite
= Jadeite + Quartz calculated
using the program TWQ of
Berman (1988, 1990, 1991).

Winter (2001) An Introduction to
Igneous and Metamorphic
Petrology. Prentice Hall.

pick any two points on the equilibrium curve

d
D
G = 0 =
D
VdP
-

D
SdT

Thus

dP

dT

S

V

=

D

D

Temperature
-
pressure phase
diagram for the reaction: Albite
= Jadeite + Quartz calculated
using the program TWQ of
Berman (1988, 1990, 1991).

Winter (2001) An Introduction to
Igneous and Metamorphic
Petrology. Prentice Hall.

Return to dG = VdP
-

SdT, for an isothermal process:

G

G

VdP

P

P

P

P

2

1

1

2

-

=

Gas Phases

For solids it was fine to ignore V as f(P)

For gases this assumption is a TERRIBLE MISTAKE!!!!


You can imagine how a gas compresses as P increases

How can we define the relationship between V and P for a gas?

Ideal Gas


As P increases V decreases


PV=nRT

Ideal Gas Law


P = pressure


V = volume


T = temperature


n = # of moles of gas


R = gas
constant



= 8.3144 J mol
-
1

K
-
1

P x V is a constant at constant T

Gas Pressure
-
Volume Relationships

Figure 5
-
5.
Piston
-
and
-
cylinder apparatus to
compress a gas.
Winter (2001) An Introduction
to Igneous and Metamorphic Petrology. Prentice
Hall.

Gas Pressure
-
Volume Relationships

Since



we can substitute RT/P vor V (for a single mole of gas), thus:



and, since R and T are certainly independent of P:

G

G

VdP

P

P

P

P

2

1

1

2

-

=

G

G

RT

P

dP

P

P

P

P

2

1

1

2

-

=

G

G

RT

P

dP

P

P

P

P

2

1

1

2

-

=

1

Dehydration Reactions


Mu + Q = Kspar + Sillimanite + H
2
O


We can treat the solids and gases separately



G
P, T

-

G
T

=
D
V
solids

(
P

-

0.1) + RT

ln

(
P
/0.1)

(isothermal)


The treatment is then quite similar to solid
-
solid reactions, but
you have to solve for the equilibrium P by iteration

Dehydration Reactions




(qualitative analysis)

dP

dT

S

V

=

D

D

Pressure
-
temperature phase diagram
for the reaction muscovite + quartz =
Al
2
SiO
5

+ K
-
feldspar + H
2
O,
calculated using SUPCRT (Helgeson
et al
., 1978).
Winter (2001) An
Introduction to Igneous and
Metamorphic Petrology. Prentice
Hall.