Thermodynamics: Study of interactions among work, energy, and heat. Calorimetry

1. 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

2. 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.

3. 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)

4. 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

5. Thermodynamics
The change in some property, such as G for a
reaction of the type:
2 A + 3 B = C + 4 D
DG = S (n G)products - S(n G)reactants
= GC + 4GD - 2GA - 3GB

6. 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

7. 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

8. 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
z
z
G G VdP SdT
T P T P
T
T
P
P
2 1 1
1
2
1
2
2
- = -

9. 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!

10. 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)

11. 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?

12. Does the liquid or
solid have the larger
volume?
High pressure favors
low volume, so which
phase should be stable
at high P?
Does liquid or solid have a
higher entropy?
High temperature favors
randomness, so which
phase should be stable at
higher T?
We can thus predict that the slope
of solid-liquid equilibrium should
be positive and that increased
pressure raises the melting point.
Figure 5.2. Schematic P-T phase diagram of a melting reaction.
Winter (2001) An Introduction to Igneous and Metamorphic
Petrology. Prentice Hall.

13. Does the liquid or solid
have the lowest G at
point A?
What about at point B?
The phase assemblage with the lowest G under a specific set of
conditions is the most stable
Figure 5-2. Schematic P-T phase diagram of a melting reaction.
Winter (2001) An Introduction to Igneous and Metamorphic
Petrology. Prentice Hall.

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

15. 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.

16. 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

17. 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
X
Y
dP
dT
DS
=
DV

18. 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.

19. 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.