Download - Ch 05 Thermodynamics
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ThermodynamicsThermodynamicsq a system:
Some portion of the universe that you wish to
studyq The surroundings:
The adjacent part of the universe outside thesystem
Changesin a system are associated with the transfer of
energy
Natural systems tend toward states of minimum energy
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Energy StatesEnergy States
q Unstable: falling or rolling
q Stable: at rest in lowestenergy state
q
Metastable: in low-energyperch
Figure 5.1. Stability states. Winter (2001) An Introduction to Igneous
and Metamorphic Petrology. Prentice Hall.
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ThermodynamicsThermodynamics
aPhase: a mechanically separable portion of a system3 Mineral3 Liquid3 Vapor
a Reaction: some change in the nature or types of phasesin a system
reactions are written in the form:
reactants = products
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ThermodynamicsThermodynamics
The change in some property, such as G for areaction of the type:
2 A + 3 B = C + 4 D
G = (n G)products - (n G)reactants
= GC + 4GD - 2GA - 3GB
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ThermodynamicsThermodynamics
For a phasewe 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
Arbitraryreference stateand assign an equally arbitrary
value of H to it: Choose298.15 K and 0.1 MPa (lab conditions)
...and assignH = 0 for pure elements(in their natural
state - gas, liquid, solid) at that reference
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ThermodynamicsThermodynamics
In 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
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ThermodynamicsThermodynamics
If V and S are constants, our equation reduces to:
GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)
which aint bad!
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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)
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ThermodynamicsThermodynamicsSummary thus far:
x G is a measure of relative chemical stability for a phasex We can determine G for any phase by measuring H and S for
the reaction creating the phase from the elements
x We can then determine G at any T and P mathematically
3 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 ismost stable
3 Why is melt more stable than solids at high T?
3 Is diamond or graphite stable at 150 km depth?
3 What will be the effect of increased P on melting?
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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.
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Does the liquid or solid
have the lowest G at
point A?
What about at pointB?
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 MetamorphicPetrology. Prentice Hall.
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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 theliquid 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.
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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)
x Slope P/T = -Sx Slope S < Slope L
Equilibriumat Teq
x 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.
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Now consider a reaction, we can then use the equation:
dG = VdP - SdT (again ignoring X)
For a reaction of melting (like ice water)
x V is the volume change involved in the reaction (Vwater- Vice)
x
similarly
S and
G are the entropy and free energy changes
dG is then the change in G as T and P are variedx G is (+) for S L at point A (GS < GL)
x G is (-) for S
L at point B (GS > GL)x G = 0 for S L at point x (GS = GL)
G for any reaction = 0 at equilibrium
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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
dP
dT
S=
V
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Figures I dont use in classFigures I dont 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 toIgneous and Metamorphic Petrology. Prentice Hall.
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Figures I dont use in classFigures I dont 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.