ThermodynamicsFree E and Phase D

J.D. Price

•Force - the acceleration of matter (N, kg m/sForce - the acceleration of matter (N, kg m/s22))

•Pressure (Pressure (PP) - a force applied over an area ) - a force applied over an area (N/m(N/m22))

•Work (W) - force multiplied by distance (kg Work (W) - force multiplied by distance (kg mm22/s/s22, Joule), Joule)

•Energy - enables work (J)Energy - enables work (J)

•Temperature (Temperature (TT) - 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 0oCQ2 = (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 100oCQ4 = (100g)(540 cal/g) = 54.0 kcal

step 5: heat vapor to 120oCQ5 = (100g)(0.48 cal/goC)(20oC) = 0.96 kcal

Example calculation: How much energy is required to heat 100 g of 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 - Wor

U = Q - WE.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 LawThe 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 systemQ - heat gained by the systemW - 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 - WU2 - U1 = Q - P (V2 - V1)U2 - U1 + P (V2 - V1) = QU2 - 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 opposite 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 LawThe 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 energymechanical 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 = Hfo - TSo + PVo

Isothermal and Isobaric

dG = dU +PdV - TdS

dG = 0 if reversible

dU = TdS - PdV

dG < 0 if irreversible

dU < TdS - PdV

CaCO3Al2SiO5

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

ddG = G = VVddP P - - SSddTT

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!!!

BombReaction VesselCalorimeter

BombReaction VesselCalorimeter

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.

Erxn = -Cp x T Univ. of Maine

Graphite - Diamond298 K Ho

( /Kcal )mole

So ( /cal

)mole K

Go ( /Kcal

)mole

Vo (cm3/

)mole Graphite 0 1.372 0 5.2982

Diamond 0.453 0.568 0.693 3.4166

Reaction is Cgraphite= Cdiamond

Hfo

di Hfo

gr = 453 - 0 = 453 (cal/mole)

S = Sodi - So

gr = 0.568 - 1.372 = -0.804 (cal/mole K)

V = Vodi - Vo

gr = 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 = Hfo - TSo + PVo

G = Ho - TSo + PVo

dP/dT = S/ V