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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. - PowerPoint PPT Presentation
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ThermodynamicsThermodynamics 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 StatesEnergy 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 EnergyGibbs 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 - TSWhere:
G = Gibbs Free Energy
H = Enthalpy (heat content)
T = Temperature in Kelvins
S = Entropy (can think of as randomness)
ThermodynamicsThermodynamicsa 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
ThermodynamicsThermodynamics
The change in some property, such as G for a reaction of the type:
2 A + 3 B = C + 4 D
G = (n G)products - (n G)reactants
= GC + 4GD - 2GA - 3GB
ThermodynamicsThermodynamics
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 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
ThermodynamicsThermodynamicsIn our calorimeter we can then determine H for the reaction:
Si (metal) + O2 (gas) = SiO2 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
ThermodynamicsThermodynamicsFor other temperatures and pressures we can use the
equation:
dG = VdP – SdT (ignoring 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
zzG G VdP SdTT P T PT
T
P
P
2 1 11
2
1
2
2
ThermodynamicsThermodynamics
If V and S are constants, our equation reduces to:
GT2 P2 - GT1 P1
= V(P2 - P1) - S (T2 - T1)
which ain’t bad!
ThermodynamicsThermodynamics
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)
ThermodynamicsThermodynamicsSummary 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?
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.
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.
Free Energy vs. TemperatureFree 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.
Free Energy vs. TemperatureFree 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 P/T = -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:
dG = VdP - SdT (again ignoring X)
For a reaction of melting (like ice water) V is the volume change involved in the reaction (Vwater - Vice) similarly S and G are the entropy and free energy changes
dG is then the change in G as T and P are varied G is (+) for S L at point A (GS < GL)
G is (-) for S L at point B (GS > GL)
G = 0 for S L at point x (GS = GL)
G for any reaction = 0 at equilibrium
Pick any two points on the equilibrium curve G = ? at each Therefore dG from point X to point Y = 0 - 0 = 0
dG = 0 = VdP - SdT
X
Y
dPdT
SV
Figures I don’t use in classFigures 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 classFigures 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.