Thermal Physics

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c©2004-2020 G. Anderson Thermal Physics – slide 1 / 72

Thermal PhysicsFree Energy and Chemical Thermodynamics

Greg Anderson

Department of Physics & AstronomyNortheastern Illinois University

Spring 2020

http://physics.neiu.edu/~anderson/

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Outline


Potentials

Legendre Transforms

Phase Transformations

van der Waals

Mixtures

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James Clerk Maxell (1831-1879)


Scottish physicist, James Clerk Maxwell wasone of the 19th centuries greatest scientist.In addition to formulating the equations thatunified electricity and magnetism, and con-tributing to the fields of optics and photogra-phy, Maxwell made important contributionsto thermal physics.

• Kinetic theory of gasses

• Maxwell’s demon paradox

• Maxwell relations, example:

(

∂T

∂V

)

S

= −

(

∂P

∂S

)

V

=

(

∂2U

∂V ∂S

)

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Potentials

Maxwell

PotentialsThermodynamicPotentialsThermodynamicPotentials

Enthalpy

Enthalpy inChemicalProcessesStandardConditionsEnthalpy ofFormationStandardThermodynamicQuantities

Electrolysis I

Electrolysis II

Lead AcidBatteriesLead-AcidBattery

Lead AcidBattery

LegendreTransforms

PhaseTransformations

van der Waals

Mixturesc©2004-2020 G. Anderson Thermal Physics – slide 4 / 72

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Thermodynamic Potentials


These slides covers the application of thermodynamics to chemicalreactions, phase transitions, and other transformations of matter.

Constant temperature: The Helmholtz free energy, F is theenergy that must be provided as work to create the system out ofnothing in a constant temperature environment.

Constant temperature and pressure: The Gibbs free energy, Gis the energy that must be provided as work to create the systemin an environment with constant temperature and pressure.

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Thermodynamic Potentials


U is the energy required to create an isolated system. Creating asystem from nothing in a thermal environment:

• Q = T∆S of “free energy” is available from the heat bath.

• P∆V of expansion work is required to make room for the system.

Enthalpy

H = U + PV

Helmholtz free energy

F = U − TS

Gibbs free energy

G = U − TS + PV

U

F G

H

+PV

−TS

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Enthalpy


Enthalpy, H, the total energy required to create a system U and tomake room for it PV , at constant pressure.

H = U + PV

In a system at constant pressure:

dH = d(U + PV ) = dU + d(PV ) = dU + PdV

Enthalpy can increase by adding energy to or expanding the system.Using the first law: dU = d Q+ dW :

dH = [d Q+ dW ] + PdV

= [d Q− PdV + dWother] + PdV

= d Q+ dWother1

1Eventually Wother ends up as heat added to the system, but unlike Q it is nottaken from a reservoir. It increases the entropy of the universe.

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Enthalpy in Chemical Processes


Enthalpy is the heat transferred during a constant pressure process.

CP =

(

d Q

dT

)

P

=

(

∂H

∂T

)

P

Chemical reactions in open containers and in biological systems oftentake place at constant pressure. For a chemical process:reactants −→ products, the change in enthalpy is:

∆H = Hproducts −Hreactants

• ∆H > 0: endothermic reaction, heat absorbed.

• ∆H < 0: exothermic reaction, heat is released.

• ∆H = 0: no heat is lost or gained.

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Standard Conditions


Notation you will find in the chemistry reference material:

• Enthalpy change in a chemical reaction:

∆Hr = ∆H = Hproducts −Hreactants

In the standard enthalpy change of reaction ∆H⊖

r often the r isleft off (it is assumed).

• A Plimsoll symbol indicates that the process is carried out atstandard conditions

p⊖ = 1 barr = 105 PaT⊖ = 25◦ C = 298 K

• ∆H⊖

f (Standard enthalpy of formation): the enthalpy required toform one mole of substance from its constituent elements in theirstandard states.

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Enthalpy of Formation


Formation of nitrogen dioxide from elements in their standard states:

1

2N2(g) + O2(g) −→ NO2(g)

Standard enthalpy of formation:

∆H⊖

f = 33.2 kJ/mol− 0 kJ/mol = 34 kJ/mol

Production of ammonia:

1

2N2(g) +

3

2H2(g) −→ NH3(g)

Standard enthalpy of formation:

∆H⊖

f = −45.9 kJ/mol− 0 kJ/mol = −45.9 kJ/mol

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Standard Thermodynamic Quantities


Compound ∆H⊖

f (kJ/mol) ∆G⊖

f (kJ/mol) S⊖ (J/mol K)

C(s, graphite) 0 0 5.7C(s, diamond) 1.9 2.9 2.4

CO(g) -110.5 -137.2 197.9CO2(g) -393.5 -394.4 213.8H+(aq) 0 0 0H2(g) 0 0 130.7H2O(l) -285.8 -237.1 70.0H2O(g) -241.8 -228.6 188.8N(g) 472.7 344.4 153.3N2(g) 0 0 191.6NH3(g) -45.9 -16.4 192.8O(g) 249.2 231.7 161.1O2(g) 0 0 205.2Pb(s) 0 0 64.8

PbO2(s) -277.4 -217.3 68.6PbSO4(s) -920 -813 148.5

SO2−4 (aq) -909.3 -744.5 20.1

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Electrolysis I


Consider the electrolysis of liquid water intohydrogen and oxygen gas:

H2O(l) −→ H2(g) +1

2O2(g)

The change in enthalpy gives the heat tran-ferred during a constant pressure process.∆H for this reacion is 286 kJ.

For the reverse process, if you burned a mole of molecular hydrogen,the heat energy released (enthalpy of combustion for hydrogen) is:

∆H =∑

H(products)−∑

H(reactants)

∆H = HH20(l) −HH2(g) −12HO2(g) = −286 kJ

How much of that heat can be used to do work? Do we need tosupply 286 kJ of work for the electrolysis of a mole of liquid water?

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Electrolysis II


Change in enthalpy:

∆H = U + P∆V = 286 kJ

Entropy per mole:

SH2O = 70 J/K, SH2= 131 J/K, SO2

= 205 J/K

∆U = 282

H2O(l) −→ H2(g) +12O2(g)

System

P∆V = 4 kJpushing atmosphere

∆G = 237 kJ

electrical work

T∆S = 49 kJ

heat

Change in entropy:

∆S = SH2 +1

2SO2 −SH2O = 131 J/K+

1

2205 J/K− 70 J/K = 163 J/K

Change in Gibbs free energy:

∆G = ∆H − T∆S

237 kJ = 286 kJ − (298 K)(163 J/K)

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Lead Acid Batteries


Lead-acid batteries, aka lead storage batteries.

• Invented by French physicist Gaston Plante in 1859.

• The world’s oldest rechargable battery

– The products of reactants at both the anode and cathode(lead sulfate) are insoluble. Thus, available to participate inthe reverse reaction.

• Theoretical cell voltage 2V.

– Typical 6 cell car battery: 12V.

• Growing, global lead-acid battery market: $46.6 billion (2015).

– Car batteries

– Storage for backup power supplies (UPS)

https://www.grandviewresearch.com/industry-analysis/lead-acid-battery-market

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Lead-Acid Battery


PbO2(s)+ Dilute

H2SO4(aq)Pb(s)

−

Fully charged

Positive electrode, discharging: E0 = 1.685 V.

PbO2(s) + HSO−

4 (aq) + 3H+(aq) + 2e− −→ PbSO4(s) + 2H2O(l)

Negative electrode, discharging: E0 = −0.356 V.

Pb(s) + HSO−

4 (aq) −→ PbSO4(s) + H+(aq) + 2e−

PbSO4Dilute

H2SO4

PbSO4

Fully discharged

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Lead Acid Battery


Total reaction:

Pb(s) + PbO2(s) + 2H2SO4(aq) −→ 2PbSO4(s) + 2H2O(l)

Change in Gibbs free energy (source):

∆G⊖ = 2∆G⊖

f (PbSO4(s)) + 2∆G⊖

f (H2O(l))

− ∆G⊖

f (Pb(s))−∆G⊖

f (PbO2(s))− 2∆G⊖

f (H2SO4(aq))

= {2(−813) + 2(−237.1)− 0− (−217.3)− 2(−744.5)} kJ/mol

= −393.9 kJ/mol

Electrical work per electron:

We =∆G⊖

2NA

=393.9 kJ

2(6.022× 1023 molecules/mol)= 3.27×10−19 J = 2.04 eV

Battery voltage at zero current: E0cell = 1.685 V − (−0.356 V) = 2.041 V.

https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Reference_Tables/Thermodynamics_Tables/T1%3A_Standard_Thermodynamic_Quantities

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Legendre Transforms

Maxwell

Potentials

LegendreTransforms

Legendre Trans.

ThermodynamicIdentity

Potentials &Transforms

Helmholtz I

Helmholtz II

Helmholtz

F Decreases

Gibbs IGibbs FreeEnergy

Gibbs II

Gibbs III

Gibbs IV

Gibbs V

G & µ

Grand Potenial

Summary


van der Waals

Mixtures


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Legendre Transformation


Given a function of one varibles F (x),

dF =∂F

∂xdx ≡ sdx

a Legendre transformation of F changes a differential i.e. fromdx to ds with the transformation:

G(s) = F (x(s))− sx(s)

where

dG = dF − sdx− xds= sdx− sdx− xds= −xds

Thenx ≡ −∂G/∂s

x

y

y = F (x)

y=

sx+

G(s

)

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Thermodynamic Identity


Infinitesimal change in entropy:

dS =

(

∂S

∂U

)

V,N

dU +

(

∂S

∂V

)

U,N

dV +

(

∂S

∂N

)

U,V

dN

The thermodynamic identity:

dS =1

TdU +

P

TdV −

µ

TdN

where

1

T≡

(

∂S

∂U

)

N,V

,µ

T≡ −

(

∂S

∂N

)

U,V

,P

T≡

(

∂S

∂V

)

U,N

In terms of dUdU = TdS − PdV + µdN

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Potentials & Transforms


U (S, V,N)

F (T, V,N) G(T, P,N)

H (S, P,N)

+PV−TS

dU(S, V,N) = TdS − PdV + µdN

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Helmholtz Free Energy


Helmholtz free energy, F: energy required to create a system U , minusthe heat energy TS that can be acquired from the environment.

F ≡ U − TS

For a system in contact with a heat bath at temperature T :

dF = d(U − TS) = dU − TdS

= [d Q+ dW ]− TdS

= dW + (d Q− TdS)

Given dS ≥ d Q/T , d Q ≤ TdS, d Q− TdS ≤ 0 , and

dF ≤ dW when T = const.

F is the work-energy required to create the system from nothing, it is the

maximum useful work obtainable from a closed system at constant

temperature.

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Helmholtz Free Energy II


Thermodynamic identity for U(S, V,N)

dU = TdS − PdV + µdN

Definition of F:F ≡ U − TS

Thermodynamic identity for F (T, V,N):

dF = d(U − TS)

= dU − TdS − SdT

= (TdS − PdV + µdN)− TdS − SdT

= −SdT − PdV + µdN

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Helmholtz Function: Summary


Definition:F = U − TS


dF = −SdT − PdV + µdN

Differential relations:

S = −

(

∂F

∂T

)

V,N

, P = −

(

∂F

∂V

)

T,N

, µ = +

(

∂F

∂N

)

T,V

Maxwell relation:(

∂S

∂V

)

T

=

(

∂P

∂T

)

V

= −

(

∂2F

∂V ∂T

)

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F Decreases


For a sytem with fixed T, V, N,the free energy always decreases:

dStotal = dS + dSR

ReservoirSR, UR, T

System

S,U, T

dU = −dUR

dV = 0

dN = 0

Recall

dS =1

TdU +

P

TdV −

µ

TdN

Thus, dSR = dUR/T = −dU/T , and:

dStotal = dS − 1T dU

= − 1T (dU − TdS)

= − 1T dF

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Gibbs Function


Gibbs Free Energy, G: the energy required to create asystem U , minus the heat energy TS that can be acquiredfrom the environment plus the work done to make room

G ≡ U − TS + PV = F + PV = H − TS

For a system in contact with a heat bath T , at constantpressure P

dG = d(U − TS + PV ) = dU − TdS + PdV

= [d Q+ dW ]− TdS + PdV

Given d Q− TdS ≤ 0 , and dW = −PdV + dWother:

dG ≤ dWother when T, P = const.

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Gibbs Free Energy


∆G is an indicator of whether a reaction willspontaneously occur at contant temperature and pressuresince, in the absence of applied, non-mechanical work:

∆G ≤ 0

• A system at constant pressure in contact with a heatbath evolves to a state which minimizes G.

• Spontaneous processs lower the Gibbs function.

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Gibbs Free Energy II


Thermodynamic identity for F (T, V,N)


Definition of G:

G ≡ U − TS + PV = F + PV = H − TS

Thermodynamic identity for G(T, P,N):

dG = d(U − TS + PV ) = d(F + PV )

= dF + PdV + V dP

= (−SdT − PdV + µdN) + PdV + V dP

= −SdT + V dP + µdN

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Gibbs Free Energy III


DefinitionG = U − TS + PV

Thermodynamic identity for G(T, P,N):

dG = −SdT + V dP + µdN


S = −

(

∂G

∂T

)

P,N

, V = +

(

∂G

∂P

)

T,N

, µ = +

(

∂G

∂N

)

T,P

Maxwell relation:

−

(

∂S

∂P

)

T

=

(

∂V

∂T

)

P

=

(

∂2G

∂P∂T

)

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Gibbs Free Energy IV


For a system with fixed T, P, N,the Gibbs function always decreases:

dStotal = dS + dSR

ReservoirSR, UR, T

System

S,U, T

dU = −dUR

dP = 0

dN = 0

Recall

dS =1

TdU +

P

TdV −

µ

TdN

Thus, dSR = (dUR + PdVR)/T = −(dU + PdV )/T , and:

dStotal = dS − 1T (dU + PdV )

= − 1T (dU − TdS + PdV )

= − 1T dG

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Gibbs Free Energy


Partial derivative relation for chemical potential:

µ = +

(

∂G

∂N

)

T,P

For fixed T, P , µ = const. Thus,

G = Nµ

System

dG = µ

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The Gibbs Function and µ


RecallG = U − TS + PV = µN

For an ideal gas, PV = NkT :

∂µ

∂P=

1

N

∂G

∂P=

V

N=

kT

P

Multiply by dP and integrate:

µ(T, P )− µ(T, P⊖) =

∫ P

P⊖

(

∂µ

∂P

)

dP = kT

∫ P

P⊖

dP

P

By convention, the reference pressure P⊖ = 1 bar.

µ(T, P ) = µ(T, P⊖) + kT ln

(

P

P⊖

)

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Grand Potenial


Grand free energy or grand potential, Φ:

Φ ≡ U − TS − µN = F − µN = −PV



Thermodynamic identity for Φ(T, V, µ):

dΦ = dF − µdN −Ndµ= −SdT − PdV −Ndµ


S = −

(

∂Φ

∂T

)

V,µ

, P = −

(

∂Φ

∂V

)

T,µ

, N = −

(

∂Φ

∂µ

)

T,V

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Summary


U (S, V,N)

F (T, V,N)G(T, P,N)

H(S, P,N)

+PV

−TS

• U, V = constant: dS ≥ 0

• T, V = constant: dF ≤ 0

• T, P = constant: dG ≤ 0

Extensive Variables : V, N, S, U, H, F, GIntensive Variables : P, µ, T

U = TS + µN − PV

F = U − TS = µN − PV

G = U − TS + PV = F + PV = H − TS = µN

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Phase Transformations

Maxwell

Potentials

LegendreTransforms


Phases of MatterPhaseTransitionsEhrenfest’sClassificationClausius-Clapeyron

Phase Diagramfor H2O

WaterPhase Diagramfor CO2

Carbon Dioxide

Helium

Helium 4

G of C

PT CarbonCalciumCarbonateFree energy offormationCalciumCarbonate PhaseDiagram

van der Waalsc©2004-2020 G. Anderson Thermal Physics – slide 34 / 72

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Phases of Matter


Four elements:Earth

Air

Water Fir

e

Four phases of matter:

Solid

Gas

Liquid

Plasm

a

Greek Philosopher Empedocles, Agrigentum, Sicily, 5th Century BCE

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Phase Transitions


Plasma

Gas

Liquid SolidMelting

Vaporization D

eposition

Ionization

Condensation Sublim

ation

Freezing

Recombination

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Ehrenfest’s Classification


The order of a phase transition is the order of the lowestdifferential of the Gibbs functions which shows adiscontinuity.


First order: Involve latent heat. Examples: solid-liquid,solid-vapour, liquid-vapour, superconducting transitions inmagnetic field.

Second order: Superconducting transitions in zero field.

Third order: Curie point of ferromagnets.

Modern usage is to call all higher-order phase transitions continuousphase transitions.

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Clausius-Clapeyron Relation


Recall that thermodynamic identity for the Gibbs function:


Along a phase boundary:

dGα = dGβ

−SαdT + VαdP = −SβdT + VβdP

(Vα − Vβ)dP = (Sα − Sβ)dT

dPdT

=Sα−Sβ

Vα−Vβ

With ∆S = L/T :

dP

dT=

L

T∆VdT

dP

P

T

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Phase Diagram for H2O


−100 0 100 200 300 40010−4

10−3

10−2

10−1

100

101

102

103

Triple Point

(0.01,0.006)

Critical Point(374,221)

Solid (Ice)

Gas (Steam)

Liquid

Temperature (◦C)

Pressure

(bar)

b

T (◦C) Pv (bar) L (kJ/mol)

−40 0.00013 51.16−20 0.00103 51.13

0 0.00611 51.07

0.01 0.00612 51.07

25 0.0317 43.9950 0.1234 40.92100 1.013 40.66150 4.757 38.09200 15.54 34.96250 39.74 30.90300 85.84 25.30350 165.2 16.09374 220.6 0.00

Water

http://en.wikipedia.org/wiki/File:Phase_diagram_of_water.svg

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Phase Diagram for CO2


−120 −80 −40 0 400

10

20

30

40

50

60

70

80

Triple Point(-56.6,5.18)

Critical Point

(31,75)

Solid

Gas

Liquid

Temperature (◦C)

Pressure

(bar)

T (◦C) Pv (bar)

−120 0.0124−100 0.135−80 0.889

−78.6 1.0−60 4.11

−56.6 5.18

−40 10.07−20 19.72

0 34.8520 57.231 73.8

Carbon Dioxide

http://en.wikipedia.org/wiki/File:Carbon_dioxide_pressure-temperature_phase_diagram.svg

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Helium


• Abundance

– 4He: 99.9998%

– 3He: 0.0002%

• Helium is the only element that re-mains a liquid at T = 0. The solidphase requires very high pressures.

• 4He has two distinct liquid phases, anormal phase called Helium I, and asuperfluid phase called Helium II. Su-perfluids have zero viscosity.

• The solid-liquid boundary for 3He hasa negative slope below T . 1 K!

http://www.physics.berkeley.edu/research/packard/current_research/nielsweb/intro.html

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Diamond and Graphite


Gm

Pressure (kbar)

5 10 15 20

Diamo

nd(Vm

= 3.34 cm

3 )

Graphite(Vm

=5.30 cm3 )

(

∂G∂T

)

P,N= −S

(

∂G∂P

)

T,N= V

Gm(P ) = G⊖

m + Vm(P − P⊖)

2.9kJ

S(graphite) = 5.75

S(diamond) = 2.38 J/K

http://commons.wikimedia.org/wiki/File:Carbon_basic_phase_diagram.png

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Calcium Carbonate


The limestone, chalk, marble, tufa and travertine rocksare made from calcium carbonate CaCO3. Calciumcarbonate exists in three mineral forms: calcite,aragonite, and vaterite.

Calcite Aragonite

http://en.wikipedia.org/wiki/Aragonite

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Free energy of formation


The enthalpy of formation ∆fH, (Gibbs free energy of formation:∆fG) is the change in H (G) upon forming one mole of material fromits elements. Consider for example calcium carbonate:

Ca(s) + C(s) + 11

2O2(g) −→ CaCO3(s)

where (s) = solid, (l) = liquid, (g) = gas, and (aq) = aqueous.

CaCO3 ∆fH(kJ) ∆fG(kJ) S(J/K) CP (J/K) V (cm3)

Calcite -1206.9 -1128.8 92.9 81.88 36.93Aragonite -1207.1 -1127.8 88.7 81.25 34.15

The Gibbs free energy of calcite is less than the Gibbs free energy ofaragonite at standard pressure:

G◦(aragonite)−G◦(calcite) = ∆fG(aragonite)−∆fG(calcite) = 1 kJ

http://www.geol.ucsb.edu/faculty/hacker/geo102C/lectures/part4.html

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van der Waals Model

Maxwell

Potentials

LegendreTransforms


van der Waals

Model

a & b

dimensionless

Gibbs Function

Parametric Plot

Parametric Plot

Parametric Plot

Parametric Plot

Maxwell Const.Fig. 5.22Isotherm

PV Diagram

PV Diagram

PT Diagram

Mixtures


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The van der Waals Model


Johannes Diderik Van der Waals Ph.D. Thesis 1873:(

P + aN2

V 2

)

(V −Nb) = NkT

or

P =NkT

V −Nb− a

N2

V 2

• Attraction: short-range intermolecular forces.

U/N ⊃ −aN

V, P = −

(

∂U

∂V

)

S

⊃ −aN2

V 2

• Repulsion: b = finite molecular volume.

P → ∞ as V → Nb

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Empirical Constants


b a/bSubstance A3 (eV)

Water 50.7 0.188

Sulfur dioxide 94.7 0.125

Carbon dioxide 71.3 0.0884

Oxygen 52.9 0.0451

Argon 53.8 0.0439

Nitrogen 64.3 0.0367

Hydrogen 44.3 0.00966

Helium-4 39.4 0.00152

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Dimensionless van der Walls equation


1 2 3 40

−1

−2

−3

1

2

3

4

p=

P/P

c

v = V/Vc

t = 1.2, 1.1, . . . , 0.7, 0.6

Equation of State

P =NkT

V −Nb− a

N2

V 2

Dimensionless EoS

p =8t

3v − 1−

3

v2

Dimensionless variablesv ≡ V/Vc, Vc = 3Nb

p ≡ P/Pc, Pc =127

ab2

t ≡ T/Tc, kTc =827

ab

At the critical temperature Tc: P′(Vc) = P ′′(Vc) = 0.

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Dimensionless van der Walls equation


1 2 3 40

−1

−2

−3

1

2

3

4

p=

P/P

c

v = V/Vc

t = 1.2, 1.1, . . . , 0.7, 0.6

Equation of State

P =NkT

V −Nb− a

N2

V 2

Dimensionless EoS

p =8t

3v − 1−

3

v2

Dimensionless variablesv ≡ V/Vc, Vc = 3Nb

p ≡ P/Pc, Pc =127

ab2

t ≡ T/Tc, kTc =827

ab

At the critical temperature Tc: P′(Vc) = P ′′(Vc) = 0.

Northeastern

Illinois

University

Gibbs Function of a van der Waals Gas


Thermodynamic identity for G:


Differential relation:(

∂G∂V

)

N,T= V

(

∂P∂V

)

N,T= V ∂

∂V

(

NkTV−Nb − aN2

V 2

)

N,T

= − (NkT )Nb(V−Nb)2 +

2aN2

V 2

Integrating with respect to V :

G = −NkT ln(V −Nb) +(NkT )Nb

V −Nb−

2aN 2

V+ c(T )

Northeastern

Illinois

University

Parametric van der Waals Plot


Dimensionless equation of state for a van der Waals gas:

p =8t

3v − 1−

3

v2

g =G

PcVc

=8t

3

[

1

3v − 1− ln(3v − 1)

]

−6

v+ c(T )

1 20.4

0.2

0.6

0.8

1.0

PPc

v = V/Vc

t = 0.95

b1

b2b

3b

4b

5

b6

b7

0.7 0.8 0.9−6.7

−6.6

−6.5

−6.4

t = 0.95

g

p = P/Pc

b

1

b3

b4b

2,6b5

b7

gas

liquid

Northeastern

Illinois

University




p =8t

3v − 1−

3

v2

g =G

PcVc

=8t

3

[

1

3v − 1− ln(3v − 1)

]

−6

v+ c(T )

1 2 30.2

0

0.4

0.6

0.8

1.0

PPc

v = V/Vc

t = 0.9

b 1b 2

b3

b4

b

5

b6

b7

0.3 0.5 0.7 0.9−7.0

−6.8

−6.6

−6.4

−6.2

t = 0.9

g

p = P/Pc

b1

b3

b5

b

2,6

b4b

7

gas

liquid

Northeastern

Illinois

University




p =8t

3v − 1−

3

v2

g =G

PcVc

=8t

3

[

1

3v − 1− ln(3v − 1)

]

−6

v+ c(T )

0 1 2 3 40.2

0.4

0.6

0.8

1.0

PPc

v = V/Vc

t = 0.85

b1b

2

b3

b4

b5

b6

b7

−0.1 0.1 0.3 0.5 0.7 0.9−7.0

−6.8

−6.6

−6.4

−6.2

t = 0.85

g

p = P/Pc

b1

b

3

b5

b

2,6

b

4

b

7

gas

liquid

Northeastern

Illinois

University




p =8t

3v − 1−

3

v2

g =G

PcVc

=8t

3

[

1

3v − 1− ln(3v − 1)

]

−6

v+ c(T )

0 1 2 3 4−0.3

−0.1

0.1

0.3

0.5

PPc

v = V/Vc

t = 0.80b1b

2

b3

b4

b5

b6

b7

−0.5 0 0.5−7.0

−6.8

−6.6

−6.4

−6.2

−6.0

t = 0.80

g

p = P/Pc

b 1

b3

b

5

b

2,6

b4

b

7

gas

liquid

Northeastern

Illinois

University

Maxwell Construction


The homogeneous states are unstable.

At fixed temperature and pressure, the equilibrium state of thesystem minimizes G, the minima follow from the thermodynamicidentity for G


For fixed N and T , dG = V dP . The integral around the closed loopcontaining unstable states must vanish.

0 =

∫

loop

dG =

∫

loop

(

∂G

∂P

)

T,N

dP =

∫

loop

V dP

Maxwell construction: For each isotherm T the vapor pressure isthe presure at which the liquid-gas tranformation takes place.

c©Addison-Wesley

http://physics.weber.edu/thermal/

Northeastern

Illinois

University

PV Phase Diagram


1 2 3 40

−1

1

2

3

p=

P/P

c

v = V/Vc

t = 1.2, 1.15, . . . , 0.85, 0.8

Northeastern

Illinois

University

PV Phase Diagram


1 2 3 40

−1

1

2

3

p=

P/P

c

v = V/Vc

t = 1.2, 1.15, 1.1, . . . , 0.8

Northeastern

Illinois

University

Van der Waals Phase Diagram


0 0.2 0.4 0.6 0.80

0.2

0.4

0.6

0.8

1.0

1.2p=

P/P

c

t = T/Vc

b

Liquid

Gas

Critical point

Northeastern

Illinois

University

Mixtures

Maxwell

Potentials

LegendreTransforms


van der Waals

Mixtures

Mixology

Entropy ofMixing

Free Energy &Mixing

Free Energy &Mixing II

Fig. 5.26 Mixing

Free Energy &Mixing III

Fig. 5.27 Mixing

Further Study


Northeastern

Illinois

University

Mixology


Change in Gibbs free energy G = U + PV − TS upon mixing:

∆Gmix = ∆Hmix − T∆Smix

Miscible: can be mixed in all proportions. Example water and

ethanol.

Immiscible: can not undergo mixing or blending. Example: oil

and water.

Ideal Mixing: U and V do not change upon mixing: ∆Hmix = 0

Solubility gap (Miscibiliy gap): a region in a phase diagram for amixture of components where the mixture exits and two or morephases. This occurs when the Gibbs free energy of the separatedsystem is lower than the fully mixed system.

Northeastern

Illinois

University

Entropy of Mixing


Two ideal gas species, N = Nα +Nβ, which are initially separated at

equal P , T . Let x =Nβ

N=

Vβ

V.

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V = Vα + Vβ

Recall that the entropy of an ideal gas satisfies:

S = Nk lnV + · · ·

∆Sα = Sα(V )− Sα(Vα)= Nαk lnV −Nαk lnVα

= (1− x)Nk lnV − (1− x)Nk ln(1− x)V= −(1− x)Nk ln(1− x)

∆Sβ = Nβk lnV −Nβk lnVβ

= xNk lnV − xNk ln xV= −xNk ln x

Northeastern

Illinois

University

Entropy of Mixing


Two ideal gas species, N = Nα +Nβ, which are initially separated at

equal P , T . Let x =Nβ

N=

Vβ

V.

b b

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V = Vα + Vβ

∆Sα = −(1− x)Nk ln(1− x)

∆Sβ = −xNk ln x

∆Smix = −Nk [x ln(x) + (1− x) ln(1− x)]

0 0.2 0.4 0.6 0.8 1.0−1

0

∆SR

Northeastern

Illinois

University

Free Energy & Mixing


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G = U + PV − TS

Unmixed:

G = (1− x)G◦

α + xG◦

β, x =Nβ

Nβ +Nα

0 0.2 0.4 0.6 0.8 1.00

1

G0α

G0β

Northeastern

Illinois

University

Free Energy & Mixing


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G = U + PV − TS

Ideal mixture:

G = (1− x)G◦

α + xG◦

β +RT [x ln(x) + (1− x) ln(1− x)]

no mixing

ideal m

ixing

0 0.2 0.4 0.6 0.8 1.00

1

x

G0α

G0β

Northeastern

Illinois

University

Free Energy & Mixing II


Entropy mixing:

∆Smix = −Nk [x ln(x) + (1− x) ln(1− x)] (5.60)

Ideal mixing:

G = (1− x)G◦

α + xG◦

β +RT [x ln(x) + (1− x) ln(1− x)] (5.61)

Equilibrium phases almost always contain impurities:

limx→0

d(∆Smix)

dx= ∞, lim

x→0

dG

dx= ∞

Everything else being equal, substances tend to mix. Exceptions tothis trend occur when mixing two substances increases the energy.


Northeastern

Illinois

University

Free Energy & Mixing III


G = U + PV − TS

Entropy mixing:

∆Smix = −Nk [x ln(x) + (1− x) ln(1− x)]

• For T 6= 0 there is a competition between U and −TS.

• A concave-down G indicates an unstable mixture.

• Connect any two points on the curve by a straight line: thestraight line gives G of the inhomogeneous, unmixed combination.


Northeastern

Illinois

University

Further Study


• Read page 192–194 on the experimental phase diagram fornitrogen and oxygen at atmospheric pressure.

• Read page 202–204 on osmotic pressure.

Documents

Thermal Physics