Ch 05 Thermodynamics

8/2/2019 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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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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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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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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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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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.

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Ch 05 Thermodynamics