Thermodynamics Free E and Phase D J.D. Price Force - the - PowerPoint PPT Presentation

Thermodynamics Free E and Phase D J.D. Price

• Force - the acceleration of Force - the acceleration of matter (N, kg m/s matter (N, kg m/s 2 ) • 2 ) • Pressure ( Pressure (P P) - a force applied over an area ) - a force applied over an area • (N/m 2 2 ) ) (N/m • Work (W) - Work (W) - force multiplied by distance ( force multiplied by distance (kg kg • m 2 2 /s /s 2 2 , Joule) , Joule) m • Energy - Energy - enables work (J) enables work (J) • • Temperature ( Temperature (T T) - a measurement relating to the ) - a measurement relating to the • kinetic (movement) energy of the system (units kinetic (movement) energy of the system (units ºC or K) ºC or K) • Heat (Q) - an energy form relatable to Heat (Q) - an energy form relatable to • temperature (J, but also calories: 1 g water 1 K,) temperature (J, but also calories: 1 g water 1 K,)

E.B. Watson

Example calculation: How much energy is required to heat 100 g of ice at -20 o C to water vapor at 120 o C. 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

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 E.B. Watson � U = Q - W

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 E.B. Watson state 1, the change in internal energy of the system is zero, so Q in = W out E.B. Watson

Thermodynamics – relating heat, work, and energy The First Law The First Law Energy may be converted from one form to another, but Energy may be converted from one form to another, but the total amount of energy is the same. 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

An expression of work can be made using P and V (the steam engine). Thermal energy (H) H = U + PV

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 )

d H = d U + P d V +V d P if P is constant d H = d U + P d V

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.

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

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 ?

E.B. Watson � U = 0 = Q – W Okay, but could you go backwards?

Thermodynamics – relating heat, work, and energy The Second Law The Second Law Heat flows from warmer to cooler bodies. To go Heat flows from warmer to cooler bodies. To go backwards requires energy or work. backwards requires energy or work. The second law is also stated: Mechanical Mechanical energy can be converted 100% into heat, but energy can be converted 100% into heat, but heat cannot be converted 100% into mechanical heat cannot be converted 100% into mechanical energy. energy. "You can't break even." "You can't break even."

Thermodynamics – relating heat, work, and energy Heat cannot be converted 100% into Heat cannot be converted 100% into mechanical energy mechanical energy Some of the heat is lost, because it creates disorder in the system. Entropy � S U = d Q / T (rev) = 0

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 = d q/T (rev) � U = T � S - P � V or d U = T d S - P d V

E.B. Watson 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.

Thermodynamics – relating heat, work, and energy 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.

Total Energy = bound energy + free energy Gibbs Free Energy (G) G = H - TS -or- G = U + PV -TS d G = d U +P d V + V d P - T d S - S d T

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

Isothermal and Isobaric d G = d U +P d V - T d S d G = 0 if reversible d U = T d S - P d V d G < 0 if irreversible d U < T d S - P d V

Al 2 SiO 5 CaCO 3

Some substituting Reversible d G = d U+P d V + V d P - T d S - S d T d U = T d S - P d V d G = G = V V d d P P - - S S d d T T d d G Re = V Re d P - S Re d T and d G Pr = V Pr d P - S Pr d T d G Re - d G Pr = V Re d P - S Re d T - V Pr d P + S Pr d T � d G = � V d P - � S d T

� d G = � V d P - � S d T At equilibrium, � d G = 0 d P/ d T = � S/ � V = � H/ (T � V) Clausius-Clapeyron equation - the slope of a reaction boundary in P-T space!!!

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

Graphite - Diamond S o V o � H o � G o 298 K (cm 3 / (cal/ (Kcal/ (Kcal/ mole K) mole) mole) mole) Graphite 0 1.372 0 5.2982 Diamond 0.453 0.568 0.693 3.4166 41.8 bar cm 3 = 1 calorie Reaction is C graphite = C diamond �� = � H fodi � � H fogr = 453 - 0 = 453 (cal/mole) � S = S odi - S ogr = 0.568 - 1.372 = -0.804 (cal/mole K) � V = � V odi - � V ogr = 3.4166 - 5.2982 = -1.881 (cm 3 /mole) -1.881 / 41.8 = -0.0450 (cal/mole)

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

d P/ d T = � S/ � V -0.804 cal/mole K X 41.8 bar cm 3 /cal -1.881 cm 3 /mole = 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

Which phase is stable at 1 bar and 25 o C? G = � H fo -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)

Which phase is stable at 20 kbar and 25 o C? G = � H fo -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)

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

Why the discrepancy between the three curves?

Phase stability G = � H f o - TS o + PV o � G = � H o - T � S o + P � V o d P/ d T = � S/ � V