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Thermodynamics Free E and Phase D J.D. Price
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Force - the acceleration of matter (N, kg m/s 2 )Force - the acceleration of matter (N, kg m/s 2 ) Pressure (P) - a force applied over an area (N/m 2 )Pressure (P) - a force applied over an area (N/m 2 ) Work (W) - force multiplied by distance (kg m 2 /s 2, Joule)Work (W) - force multiplied by distance (kg m 2 /s 2, Joule) Energy - enables work (J)Energy - enables work (J) Temperature (T) - a measurement relating to the kinetic (movement) energy of the system (units ºC or K)Temperature (T) - a measurement relating to the kinetic (movement) energy of the system (units ºC or K) Heat (Q) - an energy form relatable to temperature (J, but also calories: 1 g water 1 K,)Heat (Q) - an energy form relatable to temperature (J, but also calories: 1 g water 1 K,)
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E.B. Watson
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step 1: heat ice (-20 o C - 0 o C) Q 1 = (100g)(0.50 cal/g o C)(20 o C) = 1.0 kcal step 2: melt ice at 0 o C Q 2 = (100g)(80 cal/g) = 8.0 kcal step 3: heat water (0 o C - 100 o C) Q 3 = (100g)(1.0 cal/g o C)(100 o C) = 10.0 kcal step 4: boil water at 100 o C Q 4 = (100g)(540 cal/g) = 54.0 kcal step 5: heat vapor to 120 o C Q 5 = (100g)(0.48 cal/goC)(20oC) = 0.96 kcal Example calculation: How much energy is required to heat 100 g of ice at -20 o C to water vapor at 120 o C.
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The difference between Q and W is always the same. It is the difference in internal energy (U) between the 2 states. So U 2 - U 1 = Q - W or U = Q - W E.B. Watson
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Ideally, in a heat engine, if heat is put into the system to move from state 1 to state 2 and the engine then returns to state 1, the change in internal energy of the system is zero, so Q in = W out E.B. Watson
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The First Law Energy may be converted from one form to another, but the total amount of energy is the same. U = - U therm - U mech Isolated system Q - heat gained by the system W - work done on the system U = Q - W Thermodynamics – relating heat, work, and energy
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An expression of work can be made using P and V (the steam engine). Thermal energy (H) H = U + PV
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Q - system heat transfer Q = Cp (T 2 - T 1 ) Where Cp is heat capacity W - work done on the system W = P (V 2 - V 1 ) H - Enthalpy, a variable that covers internal energy and the work term U + PV U = U 2 - U 1 = Q - W U 2 - U 1 = Q - P (V 2 - V 1 ) U 2 - U 1 + P (V 2 - V 1 ) = Q U 2 - U 1 + PV 2 - PV 1 = Q (U 2 + PV 2 ) - ( U 1 + PV 1 ) = Q (H 2 - H 1 ) = Cp (T 2 - T 1 )
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dH = dU + PdV +VdP if P is constant dH = dU + PdV
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Reactions A change in phase(s) Phases A and B react to make phase C A + B = C Reversibility - a slight change causes the reaction to proceed, and the opposite change reverses it.
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P is constant Reaction: A + B = C + D H A = U A + PV A H B = U B + PV B H Pr = U Pr +PV Pr H Re = U Re +PV Re
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Products - Reactants H Pr - H Re = H = U +P V H is the latent heat Positive is exothermic Negative is endothermic What of H 2 O solid = H 2 O liquid ?
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U = 0 = Q – W Okay, but could you go backwards? E.B. Watson
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The Second Law Heat flows from warmer to cooler bodies. To go backwards requires energy or work. Mechanical energy can be converted 100% into heat, but heat cannot be converted 100% into mechanical energy. The second law is also stated: Mechanical energy can be converted 100% into heat, but heat cannot be converted 100% into mechanical energy. "You can't break even." Thermodynamics – relating heat, work, and energy
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Heat cannot be converted 100% into mechanical energy Some of the heat is lost, because it creates disorder in the system. Entropy Thermodynamics – relating heat, work, and energy S U = dQ / T (rev) = 0
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Entropy - the possible ways to combine the properties of individual particles to produce the observable properties of the whole system. Solids - low S, Liquids - higher S S = dq/T (rev) U = T S - P V or dU = TdS - PdV
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Plausible conclusion: the total entropy of the entire universe is continually increasing. The "heat death of the universe." At some point, the universe may run out of heat. E.B. Watson
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The Third Law The entropy of a perfect crystal at 0 K is zero. The "Zeroth" Law Two systems in thermal equilibrium with the same third system are in thermal equilibrium with each another. Thermodynamics – relating heat, work, and energy
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Total Energy = bound energy + free energy Gibbs Free Energy (G) G = H - TS -or- G = U + PV -TS dG = dU +PdV + VdP - TdS - SdT
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G = H - TS Provided all terms are at the same conditions! Since the earth includes a wide range of T and P (easily measured), and H, S, and V are often difficult to measure, we would like to calculate G at different P and T using steps of H, S, and V. G = H f o - TS o + PV o
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Isothermal and Isobaric dG = dU +PdV - TdS dG = 0 if reversible dU = TdS - PdV dG < 0 if irreversible dU < TdS - PdV
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CaCO 3 Al 2 SiO 5
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Some substituting Reversible dG = dU+PdV + VdP - TdS - SdT dU = TdS - PdV dG Re = V Re dP - S Re dT and dG Pr = V Pr dP - S Pr dT dG Re - dG Pr = V Re dP - S Re dT - V Pr dP + S Pr dT dG = VdP - SdT dG = VdP - SdT
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dG = VdP - SdT At equilibrium, dG = 0 dP/dT = S/ V = H/ (T V) Clausius-Clapeyron equation - the slope of a reaction boundary in P-T space!!!
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Bomb Reaction Vessel Calorimeter Bomb Reaction Vessel Calorimeter The vessel is strong such that there is constant P With a known heat capacity (Cp) for all of the calorimeter parts, we can determine the energy of reaction. E rxn = -Cp x T Univ. of Maine
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Graphite - Diamond Reaction is C graphite = C diamond H f o di H f o gr = 453 - 0 = 453 (cal/mole) S = S o di - S o gr = 0.568 - 1.372 = -0.804 (cal/mole K) V = V o di - V o gr = 3.4166 - 5.2982 = -1.881 (cm 3 /mole) -1.881 / 41.8 = -0.0450 (cal/mole) 41.8 bar cm 3 = 1 calorie
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Graphite becomes diamond, G = 0 G = 0 = H o -T S o + P V o G = 0 = 453 (cal/mole) - -0.804 (cal/mole K) T - - 0.045 (cal/mole bar) P P = ( H o -T S o ) / - V o P = (453 (cal/mole) + -0.804 (cal/mole K) T ) / 0.045 (cal/mole bar) T = 298.15 K {note: this is 25 o C} P = (453 - -0.804(298))/ 0.045 = 15,389 bars
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dP/dT = S/ V -0.804 cal/mole K -1.881 cm 3 /mole X 41.8 bar cm 3 /cal = 17.9 bar/K {Line: y = mx + b} P = 17.9 T + b 15,389 bars - 17.9 (298.15 K) = b b = 1.006 kb
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Which phase is stable at 1 bar and 25 o C? G = H f o -TS o + PV o G gr = 0 - 298.15 K x (1.372 (cal/mole K) ) + 1 bar (5.2982 / 41.8) (cal/mole bar) = -408.9 (cal/mole) G di = 453 (cal/mole) - 298.15 K x (0.568 (cal/mole) ) + 1 bar (3.4166 / 41.8) (cal/mole bar) = 283.7 (cal/mole)
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Which phase is stable at 20 kbar and 25 o C? G = H f o -TS o + PV o G gr = 0 - 298.15 K x (1.372 (cal/mole K) ) + 20000 bar (5.2982 / 41.8) (cal/mole bar) = 2126.0 (cal/mole) G di = 453 (cal/mole) - 298.15 K x (0.568 (cal/mole) ) + 20000 bar (3.4166 / 41.8) (cal/mole bar) = 1918.4 (cal/mole)
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Phase Diagram Recall that as you go into the Earth, both P and T increase These two variables control phase stability of compositions in the earth. On the left is a map for phases of carbon
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Why the discrepancy between the three curves?
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Phase stability G = H f o - TS o + PV o G = H o - T S o + P V o dP/dT = S/ V
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