# Thermodynamics l l a system: Some portion of the universe that you wish to study l The surroundings: The adjacent part of the universe outside the system.

## Presentation on theme: "Thermodynamics l l a system: Some portion of the universe that you wish to study l The surroundings: The adjacent part of the universe outside the system."— Presentation transcript:

Thermodynamics l l a system: Some portion of the universe that you wish to study l 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 l l Unstable: falling or rolling l Stable: at rest in lowest energy state l 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 a Phase: a mechanically separable portion of a system F F Mineral F F Liquid F F 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  G =  (n G) products -  (n G) reactants = G C + 4G D - 2G A - 3G B

Thermodynamics 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

Thermodynamics In our calorimeter we can then determine  H for the reaction: Si (metal) + O 2 (gas) = SiO 2  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

Thermodynamics For 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 zz GGVdPSdT TPTP T T P P 211 1 2 1 2 2 

Thermodynamics If V and S are constants, our equation reduces to: G T 2 P 2 - G T 1 P 1 = V(P 2 - P 1 ) - S (T 2 - T 1 ) which ain’t bad!

Thermodynamics In Worked Example 1 we used G T 2 P 2 - G T 1 P 1 = V(P 2 - P 1 ) - S (T 2 - T 1 ) and G 298, 0.1 = -856,288 J/mol to calculate G for quartz at several temperatures and pressures Low quartzEq. 1SUPCRT P (MPa)T (C)G (J) eq. 1G(J)V (cm3)S (J/K) 0.125-856,288-856,64822.6941.36 50025-844,946-845,36222.4440.73 0.1500-875,982-890,60123.2696.99 500 -864,640-879,01423.0796.36 Agreement is quite good (< 2% for change of 500 o and 500 MPa or 17 km)

Thermodynamics Summary thus far: u G is a measure of relative chemical stability for a phase u We can determine G for any phase by measuring H and S for the reaction creating the phase from the elements u We can then determine G at any T and P mathematically F 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 F Why is melt more stable than solids at high T? F Is diamond or graphite stable at 150 km depth? F 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. 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. T eq is the equilibrium temperature. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Free Energy vs. Temperature Slope of G Liq > G sol since S solid < S liquid A: Solid more stable than liquid (low T) B: Liquid more stable than solid (high T)   Slope  P/  T = -S u u Slope S < Slope L Equilibrium at T eq u u G Liq = G Sol Figure 5.3. Relationship between Gibbs free energy and temperature for the solid and liquid forms of a substance at constant pressure. T eq 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: d  G =  VdP -  SdT (again ignoring  X) For a reaction of melting (like ice  water)   V is the volume change involved in the reaction (V water - V ice )  similarly  S and  G are the entropy and free energy changes d  G is then the change in  G as T and P are varied   G is (+) for S  L at point A (G S < G L )   G is (-) for S  L at point B (G S > G L )   G = 0 for S  L at point x (G S = G L )  G for any reaction = 0 at equilibrium

Pick any two points on the equilibrium curve  G = ? at each Therefore d  G from point X to point Y = 0 - 0 = 0 d  G = 0 =  VdP -  SdT X Y dP dT SS  VV

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. P eq 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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