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thermodynamics&amp;metamorphism

# thermodynamics&amp;metamorphism - Thermodynamics and...

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This document last updated on 18-Mar-2010 EENS 212 Petrology Prof. Stephen A. Nelson Tulane University Thermodynamics and Metamorphism Equilibrium and Thermodynamics Although the stability relationships between various phases can be worked out using the experimental method, thermodynamics gives us a qualitative means of calculating the stabilities of various compounds or combinations of compounds (mineral assemblages). We here give an introductory lesson in thermodynamics to help us better understand the relationships depicted on phase diagrams. The First Law of Thermodynamics states that "the internal energy, E, of an isolated system is constant". In a closed system, there cannot be a loss or gain of mass, but there can be a change in energy, dE. This change in energy will be the difference between the heat, Q, gained or lost, and the work , W done by the system. So, dE = dQ - dW (1) Work, W, is defined as force x distance. Since Pressure, P, is defined as Force/surface area, Force = P x surface area, and thus W = P x surface area x distance = P x V, where V is volume. If the work is done at constant pressure, then W = PdV. Substitution of this relationship into (1) yields: dE = dQ - PdV (2) This is a restatement of the first law of thermodynamics. The Second Law of Thermodynamics states that the change in heat energy of the system is related to the amount of disorder in the system. Entropy is a measure of disorder, and so at constant Temperature and Pressure: dQ = TdS Thus, substituting into (2) we get: dE = TdS - PdV (3) The Gibbs Free Energy, G , is defined as the energy in excess of the internal energy as follows: G = E + PV - TS (4) Differentiating this we get: dG = dE +VdP + PdV - TdS - SdT Thermodynamics and Metamorphism 3/18/2010 Page 1 of 14

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Substituting (3) into this equation then gives: dG = TdS - PdV + VdP + PdV - SdT - TdS or dG = VdP - SdT (5) For a system in equilibrium at constant P and T , dG = 0. If we differentiate equation (5) with respect to P at constant T, the result is: (6) and if we differentiate equation (5) with respect to T at constant P we get: (7) Equation (6) tells us that phases with small volume are favored at higher pressure, and equation (7) tells us that phases with high entropy (high disorder) are favored at higher temperature. Equation (5) tells us that the Gibbs Free Energy is a function of P and T. We can see this with reference to the diagram below, which shows diagrammatically how G, T, and P are related in a system that contains two possible phases, A and B. In the diagram, phase A has a steeply sloping free energy surface. Phase B has a more gently sloping surface. Where the two surfaces intersect, Phase A is in equilibrium with phase B, and G A = G B . Next we look at 3 cross-sections through this figure.
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thermodynamics&amp;metamorphism - Thermodynamics and...

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