Thermodynamics

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# Thermodynamics - PowerPoint PPT Presentation

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.

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## PowerPoint Slideshow about 'Thermodynamics' - reuben-reilly

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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

Changesin 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

aPhase: 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:

2 A + 3 B = C + 4 D

DG = S (n G)products - S(n G)reactants

= GC + 4GD - 2GA - 3GB

Thermodynamics

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 DH 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

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

We can use this equation to calculate G for any phase at any T and P by integrating

z

z

P

T

2

2

-

=

-

G

G

VdP

SdT

T

P

T

P

2

2

1

1

P

T

1

1

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)

Thermodynamics

If V and S are constants, our equation reduces to:

GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)

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

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

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?

Does liquid or solid have a higher entropy?

Figure 5.2.Schematic P-T phase diagram of a melting reaction. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

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.

Does the liquid or solid have the lowest G at point A?

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 increases

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.

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

Equilibriumat 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

dP

DS

=

DV

dT

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

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.