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:
2A + 3B =C +4D
DG = S (n G)products - S(n G)reactants
= GC + 4GD - 2GA - 3GB
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 DH 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 DH for the reaction:
Si (metal) + O2 (gas) = SiO2
DH = -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 DX 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
GT
2
P2
- GT
1
P1
=
z
P2
P1
VdP -
z
T2
T1
SdT
Thermodynamics
If V and S are constants, our equation reduces to:
GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)
which ain’t bad!
Thermodynamics
In Worked Example 1 we used
GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)
and G298, 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 500o and 500 MPa or 17 km)
Thermodynamics
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?
Does the liquid or
solid have the larger
volume?
High pressure favors
low volume, so which
phase should be stable
at high P?
Figure 5.2. Schematic P-T phase diagram of a melting reaction.
Winter (2001) An Introduction to Igneous and Metamorphic
Petrology. Prentice Hall.
We can thus predict that the slope
of solid-liquid equilibrium should
be positive and that increased
pressure raises the melting point.
Does liquid or solid have a
higher entropy?
High temperature favors
randomness, so which
phase should be stable at
higher T?
Does the liquid or solid
have the lowest G at
point A?
What about at point B?
Figure 5-2. Schematic P-T phase diagram of a melting reaction.
Winter (2001) An Introduction to Igneous and Metamorphic
Petrology. Prentice Hall.
The phase assemblage with the lowest G under a specific set of
conditions is the most stable
Free Energy vs. Temperature
dG = VdP - SdT
at constant pressure: dG/dT = -S
Because S must be (+) G for a phase decreases as T
Figure 5.3. Relationship between Gibbs free energy and temperature
increases
for a solid at constant pressure. T is the equilibrium temperature.
eq
Winter (2001) An Introduction to Igneous and Metamorphic
Petrology. Prentice Hall.
Would the slope for the
liquid be steeper or
shallower than that for
the solid?
Free Energy vs. Temperature
Slope of GLiq > Gsol since
Ssolid < Sliquid
A: Solid more stable than
liquid (low T)
B: Liquid more stable than
solid (high T)
 Slope dP/dT = -S
 Slope S < Slope L
Equilibrium at Teq
 GLiq = GSol
Figure 5.3. Relationship between Gibbs free energy and temperature
for the solid and liquid forms of a substance at constant pressure. Teq
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:
dDG = DVdP - DSdT
(again ignoring DX)
For a reaction of melting (like ice  water)
 DV is the volume change involved in the reaction (Vwater - Vice)
 similarly DS and DG are the entropy and free energy changes
dDG is then the change in DG as T and P are varied



DG is (+) for S  L at point A (GS < GL)
DG is (-) for S  L at point B (GS > GL)
DG = 0 for S  L at point x (GS = GL)
DG for any reaction = 0 at equilibrium
Y
X
Pick any two points on the equilibrium curve
DG = ? at each
Therefore dDG from point X to point Y = 0 - 0 = 0
dDG = 0 = DVdP - DSdT
dP DS
=
dT DV
Figures I don’t use in class
Figure 5.4. Relationship between Gibbs free energy and pressure for
the solid and liquid forms of a substance at constant temperature.
Peq is the equilibrium pressure. Winter (2001) An Introduction to
Igneous and Metamorphic Petrology. Prentice Hall.
Figures I don’t use in class
Figure 5.5. Piston-and-cylinder apparatus to compress a gas. Winter
(2001) An Introduction to Igneous and Metamorphic Petrology. Prentice
Hall.