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.
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