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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.

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## Thermodynamics

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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) which ain’t bad!**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?**What about at pointB? 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.

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