Thermodynamics

Free E and Phase D

J.D. Price

••Force - the acceleration ofForce - the acceleration of matter (N, kg m/smatter (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) -Work (W) - force multiplied by distance (force multiplied by distance (kgkg

mm22/s/s22, Joule), Joule)

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

••Temperature (Temperature (TT) - a measurement relating to the) - a measurement relating to the

kinetic (movement) energy of the system (unitskinetic (movement) energy of the system (units

ºC or K)ºC or K)

••Heat (Q) - an energy form relatable toHeat (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

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 - W

or!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 LawThe First Law

Energy may be converted from one form to another, butEnergy 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

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 +P!V

!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 goHeat flows from warmer to cooler bodies. To go

backwards requires energy or work.backwards requires energy or work.

The second law is also stated: MechanicalMechanical

energy can be converted 100% into heat, butenergy can be converted 100% into heat, but

heat cannot be converted 100% into mechanicalheat 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% intoHeat 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 = T!S - P!V 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/ (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.

!Erxn = -Cp x !TUniv. of Maine

Graphite - Diamond

298 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 - S

ogr = 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 -T!So + P!Vo

!G = 0 = 453 (cal/mole) - -0.804 (cal/mole K) T - -

0.045 (cal/mole bar) P

P = (!Ho -T!So) / -!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 - T!So + P!Vo

dP/dT = !S/ !V