chapter 5: free energy and chemical thermodynamics - part...

Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction

Chapter 5: Free Energy and ChemicalThermodynamics

Part-2: Phase Transformation

X. Bai

SDSMT, Physics

Fall Semester, 2014

X. Bai Chapter 5: Free Energy and Chemical Thermodynamics


1 Phase Transformation of Pure SubstancesShort reviewThe conceptThe Coexistence CurvesThe Clausius-Claypeyron Relation

2 The van der Waals Modelvan der Waals Model and interaction between particlesvan der Waals isotherms and critical point

3 Maxwell constructionGibbs free energy for vdW fluidMaxwell construction



Free energies

The enthalpy of a system: H = U + PV . The total energy neededto create the system out of nothing and put it in an environmentat pressure P.

The Helmholtz Free Energy: F = U − TS . The total energyneeded to create the system, minus the heat you can get for freefrom an environment at temperature T . It is the energy must beprovided as all work, if you are creating the system out of nothing.All work: including the work done by the the surroundingenvironment.

The Gibbs Free Energy: G = U − TS + PV .The work (energy)you need to do to create a system in an environment withconstant pressure P and temperature P.



The concept

Up to now we have dealt almost exclusively with systemsconsisting of a single phase, solid, liquid, or gas. In this lecture, wewill learn how more complicated, multi-phase systems can betreated by the methods of thermodynamics. The guiding principleis the minimization of the Gibbs free energy in equilibrium for allsystems, including the multi-phase ones.

G = U − TS + PV and dStotal = − 1T dG

To explain what phase transformation is, in terms ofthermodynamic language, let me show you the generic phasediagram of a substance in the P-T coordinates



The concept: P-T plot

The P-T plot of phase transformation.

every point of this diagramis an equilibrium state;

different states of the systemin equilibrium are called phases.

lines dividing different phasesare called the coexistence curves.

along these curves, different phasescoexist in equilibrium,

along these curves, the systemis macroscopically inhomogeneous.

all three coexistence curvescan meet at the triple point -all three phases coexist.

critical point: Where no discontinuous change from phase-A to phase-B.

T

P



The concept: Water

For most normal substances, the slope of the melting curve is positive. Thephase diagram for water shows the characteristic negative slope of thesolid-liquid equilibrium curve.

The ice is less dense than water

As ice melts into water, thechange in entropy (the latentheat) is positive, while thechange in volume is negative,hence the negative slope.

The negative slope of thesolid-liquid coexistence curvemakes ice skating possible:ice melts under the pressureexerted by the skate blade.

Ice I

The Clausius-Clapeyron equation ( dPdT

) provides the connection between ice

skating and the observation that ice floats on water.



The concept: Helium

The P − T plot of Helium phase transformation: Helium has two isotopes 4He(Boson) and 3He (Fermion). Helium is the only element that remains a liquidat T = 0 K and P = 1 bar because the zero-point oscillations of light atomsare large.

Structure: hexagonal closepacked (hcp), face centeredcubic (fcc), body-centeredcubic (bcc) for solid.

The molar volume of 4He (3He)is more than a factor of 2 (3)larger than classical liquid.

The slope of the phase boundarysolid helium superfluid liquidhelium is essentially 0 at T ∼ 0 K .

Superfluid: zero viscosity, very highconductivity.

4He

3He

(Helium-‐I)

(Helium-‐II)



More about phase transformation

Phase transformation caused by other condition changes: Besides the verydifferent details in phase transformations among different substances, phasetransformation caused by other variables:(1) composition: Ice melts when salt is added.(2) external magnetic filed: Superconductor (super conducting and normalmatter), ferromagnet (Magnetized up and down)Classification of phase transformations: according to the abruptness of thechange:

”1st-order” transition: Solid-liquidand liquid-gas transitions. (S and Vare discontinuous at phase boundary,and S and V are the 1st order derivativesof G: dG = −SdT + VdP + µdNEq.(5.23).

”2nd-order” transition: Less abrupt,helium I - helium II transition.

higher orders: derivatives beforeget a discontinuous quantity.

T

S T

G

T

Cp ΔS=Q/T



The Coexistence Curves

Along the coexistence curves, two different phases 1 and 2 coexist inequilibrium. The system undergoes phase separation each time we cross theequilibrium curve (the system is spatially inhomogeneous along the equilibriumcurves).Any system in contact with the thermal bath is governed by the minimum freeenergy principle. We summarized in our last class:

Table: Quantities that govern the thermal processes

Process What governs the process

constant E and V Entropy Sconstant T and V Helmholtz Free Energy F = U − TSconstant T and P Gibbs Free Energy G = U − TS + PV

The shape of coexistence curves on the P-T phase diagram is determined bythe condition: G1(P,T ) = G2(P,T ), ∵ G = µN, so we have

µ1(P,T ) = µ2(P,T ) (1)



The Coexistence Curves

Several remarks: dG = −SdT + VdP + µdN

1 The system would be able to decrease its Gibbs free energy bytransforming the phase with a higher µ into the phase with lower µ:crossing the Coexistence Curves.

2 Two phases are in a state of diffusive equilibrium: there are as manymolecules migrating from 1 to 2 as the molecules migrating from 2 to 1=⇒ The equilibrium is dynamic.

3 For equilibrium between the phases: T1 = T2 (for any two sub-systems inequilibrium) and P1 = P2 (the phase boundary does not move.)

4 Though G is continuous changing across the transition, enthalpy H has astep-like behavior:G = Nµ = (U + PV )− TS = H − TS → 0 = δH − TδS(1) different phases have different values of the enthalpy ← heatexchange always EXISTS during phase transformation!(2) µ =

(∂G∂N

)T ,P

.



Two Phases of Carbon

Carbon has two phases: Diamond and Graphite. See the Gibbs potential in thefigure:

both are solid

graphite is more stable atnormal P: Molar Gibbs FreeEnergy: different by 2.9kJ

pressure dependence of G :(∂G∂P

)T ,N

= V

ρG < ρD , Gg grows fasterthan Gd and dS = − 1

TdG

crossing at about 17 kilo-bars!(1 bar = 105 Pa)

So, natural diamonds must befrom deep underground.

And do not forget the temperature: The right bottom plot.

P (MPa)

G

1 2

diamond

2.9 kJ

G

T (K) 800 1300

diamond 2.9 kJ

300

T = 300K

P = 1 bar



Clausius-Claypeyron Relation

We have seen how Gibbs Free Energy is correlated with P , T , and V :G = U − TS + PV ; and

(∂G∂P

)T ,N

= V ,(∂G∂T

)P,N

= −S . What determines the

phase boundary? - The Clausius-Clapeyron Relation.

(1799 – 1864) French engineer and physicist, one of the founders of thermodynamics.

(1822 – 1888) German physicist & mathemaBcian, one of the central founders of the science of thermodynamics.

ΔP1 ΔP2

ΔT1

ΔT2 On the red line: ΔP1 = ΔP2 ΔT1 = ΔT2



Clausius-Claypeyron Relation - cnt.

(∂G∂P

)T ,N

= V ,(∂G∂T

)P,N

= −S . At the phase boundary G1 = G2. For a small

change in P and T , we would have dG1 = dG2 on the phase boundary line.Therefore,

∵ dG = −SdT + VdP + µdN (2)

− S1dT1 + V1dP1 + µ1dN1 = −S2dT2 + V2dP2 + µ2dN2 (3)

− S1dT1 + V1dP1 = −S2dT2 + V2dP2 (4)

Because along the phase boundary line: dT1 = dT2, dP1 = dP2

S2dT − S1dT = V2dP − V1dP (5)

∴dP

dT=δS

δV(6)



Clausius-Claypeyron Relation - cnt.

If we use latent heat L = Qphase−transition = TδS , Eq. (6) takes the form:

dP

dT=

L

TδV(7)

This is the Clausius-Clapeyron relation, which gives the slope of phaseboundary line on P − T diagram.

Let’s see an examples to help us understand better.

In-class Exercise Ch5-04: Consider the diamond-graphite system. When a mole ofdiamond converted to graphite, its volume increases by 1.9× 10−6 m3. What is theslope of the diamond-graphite phase boundary?Answer: Look at dP

dT= δS

δV, we need δS and δV .

δV is given. We need the change in entropy when a mole of diamond is convertedto graphite.The entropy increases is (p.404): Sg − Sd = 5.74− 2.38 = 3.36 J/KdPdT

= δSδV

= 3.4 J/K

1.9×10−6 m3

dPdT

= 1.8× 106 J/K

m3 = 1.8× 106 N·mK ·m3

dPdT

= 1.8× 106Pa/K = 18 bar/K



Model for phase transition

To understand phase transformations, can we use Ideal Gas Model?PV = kNT (k is Boltzmann’s constant 1.381× 10−23 J/K)

1 point-like particles

2 zero interaction between particles except elastic collisions with zerocross-section

But in phase transitions, what govern the process are the potentials. So, theanswer is NO.How about the other model we have introduced - the van der Waals Model?(P + aN2

V 2

)(V − Nb) = kNT .

Possible because it takes into account the interactions between particles. Thetwo corrections:

1 point-like particles ← each particle has size by (V − Nb)

2 zero interaction between particles ← short range attraction force whenthey are NOT touching



van der Waals Model-1

U(r)

short-distance repulsion

long-distance attraction

-3

-2

-1

0

1

2

3

4

1.5 2.0 2.5 3.0 3.5 4.0distance

Energy

r

Johannes van der Waals, 1873



van der Waals Model-2

Why the corrections are V − Nb and P + aN2

V 2 ? Because of the interactions:

1 The strong short-range repulsion: the molecules are ”rigid” as soon asthe molecules touch each other. ”b”: the excluded volume per particle

2 The weak long-range attraction forces between the molecules tend tokeep them closer together: Total potential energy due to attraction:

−αN NV

= −αN2

Vwhere N and V are the total number of particles and

the volume of the solid/liquid/gas.

3 pressure associated with the potential energy: dE = −Pf dV ,

Pf = − dEdV

= −αN2

V 2

The model can help explain phase transformation behavior, but far from preciseenough!

1 inhomogeneous on the microscopic scale when gas becomes denser, withdifferent phases coexist.

2 corrections to the interaction: non-isotropic, may depend on other stateparameters.

Good for QUALITATIVE discussions.



van der Waals Model parameters

The constant a and b in the van der Waals equation for typicalsubstances:

Substance

a’ (J. m3/mol2)

b’ (10-5 m3/mol)

Pc (MPa)

Tc (K)

Air

.1358

3.64

3.77

133 K

Carbon Dioxide (CO2)

.3643

4.27

7.39

304.2 K

Nitrogen (N2)

.1361

3.85

3.39

126.2 K

Hydrogen (H2)

.0247

2.65

1.30

33.2 K

Water (H2O)

.5507

3.04

22.09

647.3 K

Ammonia (NH3)

.4233

3.73

11.28

406 K

Helium (He)

.00341

2.34

0.23

5.2 K

Freon (CCl2F2)

1.078

9.98

4.12

385 K



van der Waals isotherms

1 The isotherms on P − V plane: Lines with fixed temperatures.

2 The black isotherm is at T = Tc , corresponding to critical behavior.

3 Real isotherms can NOT have negative slope dP/dn or positive slopedP/dV (shaded area!).

4 Critical isotherm: boundary between isotherms along which there are/notphase transformations.

5 Critical point C : The point on isotherm where ∂P/∂V = ∂2P/∂V 2 = 0

0

N·b



The critical point

Critical point: The point on critical isotherm where ∂P/∂V = ∂2P/∂V 2 = 0.So, what is the critical point for van der Waals gas?

(P +

aN2

V 2

)(V − Nb) = kNT

P =kNT

V − Nb− aN2

V 2

∂P

∂V= − kNT

(V − Nb)2+

2aN2

V 3

∂2P

∂V 2=

2kNT

(V − Nb)3− 6aN2

V 4

At critical point, we have



The critical point-cnt.

kNTc

(Vc − Nb)2=

2aN2

V 3c

2kNTc

(Vc − Nb)3=

6aN2

V 4c

Combine to give:(Vc − Nb)

2=

Vc

3⇒ Vc = 3Nb.

and, kTc =8a

27b, Pc =

a

27b2.

Using Vc ,Tc ,Pc to normalize P,V ,T , T = Tc T ,P = Pc P,V = Vc V the vander Waals equation becomes:(

P +3

V 3

)(V − 1

3

)=

8T

3(8)

p =8t

3v − 1− 3

v 2often called the vdW equation in reduced variables (9)



Gibbs free energy for vdW fluid

When T and P are given, the equilibrium state of a system is determined by itsGibbs free energy G . Let’s see if we can obtain G for vdW fluid.dG = −SdT + VdP + µdN which becomes dG = VdP at fixed T and N.Therefore, we have(

∂G

∂V

)N,T

= V

(∂P

∂V

)N,T

(10)

P and V are correlated by vdW equation:(P +

aN2

V 2

)(V − Nb) = kNT (11)

P =kNT

V − Nb− aN2

V 2(12)(

∂P

∂V

)N,T

= − kNT

(V − Nb)2+

2aN2

V 3(13)(

∂G

∂V

)N,T

= − kNTV

(V − Nb)2+

2aN2

V 2(14)



Gibbs free energy for vdW fluid - cnt.

G =

∫ (− kNTV

(V − Nb)2+

2aN2

V 2

)dV (15)

G =

∫ (−kNT (V − Nb) + kNTNb

(V − Nb)2+

2aN2

V 2

)dV (16)

G =

∫ (− kNT

V − Nb− kTN2b

(V − Nb)2+

2aN2

V 2

)dV (17)

G = −kNTln(V − Nb) +kTN2b

(V − Nb)− 2aN2

V+ c(T ) (18)

G in reduced variables (off-class exercise):

G

kNTc= −tln(3v − 1) +

t

(3v − 1)− 9

4v+ c(T ) (19)

where c(T ) is the integration constant, independent from V . Here we see:(1) G is a function of T , V , N, and other parameters.

(2) Plotting G as function of V or P: with V and P associated through the

equation of state.



Gibbs free energy for vdW fluid - cnt.

This is an example for vdW fluid when T = 0.9Tc .

1 States between 2 and 6 (2,3,4,5,6) are unstable: one P has multiple Vvalues.

2 The thermodynamically stable state has minimum G .

3 Point-6 is fluid; point-2 should be gas (by the dP/dV at these points)

4 Point-2 and 6 are equally stable because they have same G .

5 The straight red line connecting point-2 and 6 represents the phasetransformation: V changes while P is fixed.

P/Pc G

V/Vc P/Pc

1 2

3

4 5

6

7

1

2,6

3

4

5 7



About plotting:

How to draw the plots of P/Pc vs V /Vc or G vs P/Pc ?Eq. (9): p = 8t

3v−1− 3

v2

Eq. (19): GkNTc

= −tln(3v − 1) + t(3v−1)

− 94v

+ c(T )

cV/V0 0.5 1 1.5 2 2.5

cP

/P

0.6

0.7

0.8

0.9

1

1.1

1.2

cP/P0.7 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2

G/k

NT

-2.5

-2.48

-2.46

-2.44

-2.42

-2.4

-2.38

-2.36

c++ code available at the teaching website: plot 5 21b.CC



Maxwell construction

The question is how to determine the pressure for phase transition.Two ways:

1 From G -plot: P at points 2,4,6.

2 From the PV diagram - invented by James Clerk Maxwell.

Here is why and how it works:The net change in G as the system changes around the loop 2-3-4-5-6 is zero∫

loop

dG =

∫loop

(∂G

∂P

)N,T

dP (20)

dG = −SdT + VdP + µdN (21)∫loop

dG =

∫loop

VdP (22)∫loop

dG = SA + SB = 0 (23)

The straight line that divides the enclosed area into two parts each with half of

the total area - the Maxwell construction.



Maxwell construction plots

P/Pc

V/Vc

1 2

3

4 5

6

7



Vapor pressure and critical point

(1) vapor pressure Pv for each T: at which the liquid-gas transformation takesplace(2) critical pressure Pc , critical temperature Tc , and critical volume Vc : thecritical point.

Pc

Tc Vc

Pv


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