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/s2

2)

)

• Pressure (

Pressure (P P) - a force applied over an area ) - a force applied over an area (N/m (N/m2

2)

)

• Work (W) -

Work (W) - force multiplied by distance ( force multiplied by distance (kg kg m m2

2/s

/s2

2, Joule)

, Joule)

• 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

step 1: heat ice (-20oC - 0oC) Q1 = (100g)(0.50 cal/goC)(20oC) = 1.0 kcal step 2: melt ice at 0oC Q2 = (100g)(80 cal/g) = 8.0 kcal step 3: heat water (0oC - 100oC) Q3 = (100g)(1.0 cal/goC)(100oC) = 10.0 kcal step 4: boil water at 100oC Q4 = (100g)(540 cal/g) = 54.0 kcal step 5: heat vapor to 120oC Q5 = (100g)(0.48 cal/goC)(20oC) = 0.96 kcal Example calculation: How much energy is required to heat 100 g

• f ice at -20oC to water vapor at 120oC.

The difference between Q and W is always the same. It is the difference in internal energy (U) between the 2 states. So U2 - U1 = Q - W

• r

U = Q - W

E.B. Watson

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 Qin = Wout

E.B. Watson E.B. Watson

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 = -Utherm - Umech Isolated system Q - heat gained by the system W - work done on the system U = Q - W

Thermodynamics – relating heat, work, and energy

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 (T2 - T1) Where Cp is heat capacity W - work done on the system W = P (V2 - V1) H - Enthalpy, a variable that covers internal energy and the work term U + PV

U = U2 - U1 = Q - W U2 - U1 = Q - P (V2 - V1) U2 - U1 + P (V2 - V1) = Q U2 - U1 + PV2 - PV1 = Q (U2 + PV2 ) - ( U1 + PV1 ) = Q (H2 - H1) = Cp (T2 - T1)

dH = dU + PdV +VdP if P is constant dH = dU + PdV

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

• pposite change reverses it.

P is constant Reaction: A + B = C + D HA = UA + PVA HB = UB + PVB HPr = UPr +PVPr HRe = URe +PVRe

Products - Reactants HPr - HRe = H = U +PV H is the latent heat Positive is exothermic Negative is endothermic What of H2Osolid = H2Oliquid?

U = 0 = Q – W Okay, but could you go backwards?

E.B. Watson

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 Thermodynamics – relating heat, work, and energy

SU = dQ / 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 = dq/T (rev) U = TS - PV or dU = TdS - PdV

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

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

Total Energy = bound energy + free energy Gibbs Free Energy (G) G = H - TS -or- G = U + PV -TS dG = dU +PdV + VdP - TdS - SdT

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 = Hf

• - TSo + PVo

Isothermal and Isobaric dG = dU +PdV - TdS dG = 0 if reversible dU = TdS - PdV dG < 0 if irreversible dU < TdS - PdV

CaCO3 Al2SiO5

Some substituting Reversible dG = dU+PdV + VdP - TdS - SdT dU = TdS - PdV dGRe = VRedP - SRedT and dGPr = VPrdP - SPrdT dGRe - dGPr = VRedP - SRedT - VPrdP + SPrdT dG = VdP - SdT

d dG = G = V Vd dP P -

• S

Sd dT T

dG = VdP - SdT At equilibrium, dG = 0 dP/dT = S/ V = H/ (TV) 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

• f the calorimeter

parts, we can determine the energy

• f reaction.

Erxn = -Cp x T

• Univ. of Maine

Graphite - Diamond

298 K Ho (Kcal/ mole) So (cal/ mole K) Go (Kcal/ mole) Vo (cm3/ mole) Graphite 0 1.372 5.2982 Diamond 0.453 0.568 0.693 3.4166

Reaction is Cgraphite= Cdiamond = Hfodi Hfogr = 453 - 0 = 453 (cal/mole) S = Sodi - Sogr = 0.568 - 1.372 = -0.804 (cal/mole K) V = Vodi - Vogr = 3.4166 - 5.2982 = -1.881 (cm3/mole)

• 1.881 / 41.8 = -0.0450 (cal/mole)

41.8 bar cm3 = 1 calorie

Graphite becomes diamond, G = 0 G = 0 = Ho -TSo + PVo G = 0 = 453 (cal/mole) - -0.804 (cal/mole K) T - - 0.045 (cal/mole bar) P P = (Ho -TSo) / -Vo P = (453 (cal/mole) + -0.804 (cal/mole K) T ) / 0.045 (cal/mole bar) T = 298.15 K {note: this is 25oC} P = (453 - -0.804(298))/ 0.045 = 15,389 bars

dP/dT = S/ V

• 0.804 cal/mole K
• 1.881 cm3/mole

X 41.8 bar cm3/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

Which phase is stable at 1 bar and 25 oC?

G = Hfo -TSo + PVo Ggr = 0 - 298.15 K x (1.372 (cal/mole K) ) + 1 bar (5.2982 / 41.8) (cal/mole bar) = -408.9 (cal/mole) Gdi = 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 oC?

G = Hfo -TSo + PVo Ggr = 0 - 298.15 K x (1.372 (cal/mole K) ) + 20000 bar (5.2982 / 41.8) (cal/mole bar) = 2126.0 (cal/mole) Gdi = 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 = Hf

• - TSo + PVo

G = Ho - TSo + PVo dP/dT = S/ V