1

Lecture 26: First Order Phase Transitions

A first order phase transition (in a single componentsystem) is characterized by a discontinuity in one ormore state variables.

Example of water at 373.15 K and 1atm:

Property Liquid GasMolarvolume, Vm

0.018 L 30.62 L

Molarenthalpy offormation,DHf,m

-285.83+75x0.0753=-280.18 kJ/mol

-241.82+75x0.0336=-239.30 kJ/mol

Molar entropyof formation,DSf,m

69.91+75.291ln(373/298)=86.81J/mol/K

188.83+33.58ln(373/298)=196.36 J/mol/K

Molar heatcapacity,

CP,m

77.06 J/mol/K 33.58 J/mol/K

Enthalpy of vaporization = -239.30+280.18 = 40.88

2

Table 4.2: 40.7 kJ/mol

Entropy of vaporization = 196.36 - 86.81 = 109.6Table 4.2: 109.1 kJ/mol/K

3

DGvap = DHvap - TbpDSvap = 0 (why?)

fl DSvap = DHvap/Tbp

= 40,700/373.15 = 109.1

Q. What is the cause of a phase transition?

A. Multiple minima in mFigure 31.Source of a phasetransition. The slope isan intensive variable,such as the pressure,which provides arestoring force thatdrives the systemtowards ahomogeneousequilibrium state.

Explanation: This might be, for example, a plot of Gm vs Vm forwater at constant T > TBP. The minimum point at B represents themolar volume of liquid water, and that at A is Vm of water vapor(which has a lower density). The only point on the graph thatactually exists in nature at this temperature is the minimum at A.The difference between the two minima is DGm,vap. If somedroplets happen to form at B, they vaporize spontaneously becauseDGm,vap is negative.

4

Minimum A is stable, minimum B is metastable.A random fluctuation at A might cause the system totemporarily reach B.

Q. How can the phase transition A Ø B be made tooccur spontaneously?A. One way is to reduce the temperature untilminimum B is lower than A.

Note: If we cool slowly enough, the system mightremain trapped in A for a while.

Example: crystallization out of a super-cooled liquid.See Figure 2, where X = Vm.

Figure 32. Plot ofchemical potential vs. molarvolume, illustrating a phasetransition, as the temperatureis reduced from T4 to T1. AtT4, the vapor phase , A, ismore stable than the liquid,B. At T2, A and B are inequilibrium, and at T1, B ismore stable.

5

6

Q. What happens if the barrier between the twominima is very large?

A. Phase A becomes metastable.Example: diamond Ø graphite.

Q. How does m vary with T?

dG = -SdT + VdP

A. For a single component system,

dm = -SmdT + VmdP

mP

ST

−=

∂∂µ

This slope is necessarily negative.

It follows that m always decreases with temperature.

Ssolid < S liquid < Sgas fl as the temperature increases amaterial first melts and then boils. (Depending onwhere the curves in Figure 32 cross, it possible forthe solid to go directly to the vapor phase.)

7

Figure 33 shows the value of the minimum of m vs T.(At each T, the three plotted points correspond to adifferent isotherm in Figure 32.)

A phase transition occurs when Dm § 0.

The equals sign denotes equilibrium.

Figure 33 Phase transition from solid to liquid to gas.

8

Q. How does m vary with P?

mT

VP

=

∂∂µ

A. It follows that m always increases with pressure.

Usually, Vm,solid < Vm,liquid

Always, Vm,liquid < Vm,gas

It follows that compressing a gas liquefies it.

Explanation: Because Vm,gas is greater than Vm,liquid,mgas increases with pressure more than does mliquid. IfP is large enough, mgas will exceed mliquid. When thishappens, DGm for condensation becomes negative,and the process becomes spontaneous.

Compressing a liquid usually freezes it. (Exceptionis ice.)

9

Lecture 27. “Topography” of a phasetransition: the phase diagram.

The phase diagram is a map showing the regions of Tand P (or any other suitable pair of intensivevariables) where different phases are the most stable.The boundaries between “countries” denoteequilibrium between phases.

We will prove later that for a single componentsystem, no more than three phases can be inequilibrium.

Figure 34. A“typical” phasediagram and thecorrespondingchemicalpotentials. Gdecreases with Tand increaseswith P.

Note the boundaries of co-existence of two phases,The triple point of co-existence of three phases, andthe critical point where two phases merge into one.

10

Along curve ABCD, the liquid and vapor are inequilibrium, but the barrier falls until the two phasesmerge at the critical point, D.

We can think of this map in two different, butequivalent ways.

1. Graph of vapor pressure vs. temperature.

2. Graph of boiling (or sublimation) temperature vspressure.

Slopes of the boundaries on a phase diagram

The boundary between a gas and a condensed phase:

1. S is always positive fl m vs T always has anegative slope.

2. Sm,gas > Sm,condensed fl the slope of m vs T isalways steeper for the gas.

3. Vm,gas > Vm,condensed fl the temperature ofvaporization always increases with pressure, givinga positive slope of P vs T

11

Figure 35. Slopesof the boundaries ina phase diagram.

(a) Vm,condensed < Vm,gas

The figure also appliesfor Vm,solid < Vm,liquid

The result is that thetransition temperatureincreases with P, makingthe phase boundaryslope to the right.

(b) Vm,liquid < Vm,solid

The result is that thetransition temperaturedecreases with P,making the phaseboundary slope to theleft.

12

The boundary between a liquid and solid phase:

1. S is always positive fl m vs T always has anegative slope.

2. Sm,liquid > Sm,solid fl the slope of m vs T is alwayssteeper for the liquid than for the solid.

3. Usually, Vm,liquid > Vm,solid fl the temperature ofmelting usually increases with pressure, giving apositive slope of P vs T. But, if Vm,liquid < Vm,solid (asfor water), the P-T slope will be negative.

13

Quantitative properties of a phase diagram

Figure 35. Calculationof the slope of theboundary in a phasediagram. States A andB are in phase 1; statesA’ and B’ are in phase2.

Consider any two points in equilibrium, A and B.Equilibrium implies that DG = 0.

AA µµ ′=

BB µµ ′=

ABAB µµµµ −=′−′∴Suppose that A and B are very close to each other.

dPVdTS mmAB ,2,1 +−=− µµdPVdTS mmAB ,2,2 +−=′−′ µµ

14

dPVdTSdPVdTS mmmm ,2,2,2,1 +−=+−∴Collecting terms:

( ) ( )dTSSdPVV mmmm ,2,1,2,1 −=−

The Clapeyron Equation:

vapm

vapm

vapm

vapm

mm

mm

VT

H

V

S

VV

SS

dT

dP

,

,

,

,

,1,2

,1,2

∆∆

=∆∆

=−−

=

Boundary between a gas and a condensed phase:

DSm > 0, DVm > 0

P-T slope is therefore positive.

DVm @ Vm,gas = ZRT/P

( )PZRTT

H

dT

dP m

/

∆=

2T

dT

ZR

H

P

dP m∆=

15

ZR

H

Td

Pd m∆−=

)/1(

ln

Slope of ln P vs 1/T gives -DHm/Z.

For constant DHm and Z=1, we get the Clapeyron-Clasusius equation:

−

∆−=

121

2 11ln

TTR

H

P

P m

If we choose T1 to be the boiling point at P1 = 1 bar,then

RT

H

RT

H m

bp

m

eeP∆−∆

=

Note: ∆Hsub = ∆Hfus + ∆Hvap

16

The solid-liquid boundary:

Sm,liquid > Sm,solid

If Vm,liquid > Vm,solid, the P-T slope is positive.

If Vm,liquid < Vm,solid, the P-T slope is negative.

Treating both DHm and DVm as constant,

∫∫ ∆∆=

2

1

2

1

T

Tm

m

P

P T

dT

V

HdP

1

212 ln

T

T

V

HPP

m

m

∆∆

=−

17

Lecture 28: Numerical Examples of PhaseTransitions

Problem: A sled is equipped with two runners 1.5 mlong and 4 mm wide. What is the minimum weightof the sled (plus cargo) if it is to run on a liquid layerabove the ice at −40C?

Answer:m

m

V

S

dT

dP

∆∆=

TV

SP

m

m ∆∆∆

=∆

But we also know that ∆P = mg/A, where A is thecontact area of the runners.

A = 2x1.5x0.004 = 0.012 m2

DVm = 16.5x10-6- 18.0x10-6 = -1.5 x10-6 m3/mol

DSm = 6,008/273.15 = 22.0 J/mol K

DT = -4K

m = 71.8 kg

18

Problem: The boiling point of water on the top of acertain mountain is 900 C. How tall is the mountain?(Ignore variations in the atmospheric temperature.)

Answer: First calculate the vapor pressure of waterat 900 C, using the Claperon-Clausius equation.

atmP 697.015.373

1

15.363

1

31452.8

700,40exp =

−−=

Next, use the barometric formula, with T=273K, andM = 0.018 Kg.

RTMgheP /−=

kmT

T

T

T

Mg

Hh

bpbp

vap 65.40

=

−

∆=

19

The phase diagram for benzene:

The following are the physical properties of benzene,collected from various tables.

Boiling point = Tbp = 80.1 C = 353.25 K

Melting point = Tmp = 5.5 C = 278.65 K

Density of liquid = rliq = 0.879 g/cm3

Density of solid = rsol = 0.891 g/cm3

Triple point: Tt = 5.50 C = 278.65 KPt = 36 Torr = 4800 Pa

Critical Point: Tc = 562.7 KPc = 48.6 atmVm,c = 260 cm3/molZc = PcVm,c/RTc = 0.274

DHm,fus = 10,600 JDHm,vap = 30,800 J

Surface tension = g = 28.88 mN/m

20

Let’s sketch out the phase diagram.

Vm,liq = 78/0.879 cm3/mol = 8.874 x 10-5 m3/mol

Vm,sol = 78/0.891 cm3/mol = 8.754 x 10-5 m3/mol

DVm = 1.195 x 10-6 m3/mol

Figure 37.Phase diagram ofbenzene.

21

Equation of solid-liquid boundary:

T

x

VT

H

V

S

dT

dP

m

fusm

m

fusm9

,, 1091.8=∆

∆=

∆∆

= pa K−1

tt T

TxPP ln1091.8 9=−

e.g., T=6.0 C = 279.15 K fl P = 1.6 x 107 Pa = 157.6 atm

22

Solid-vapor boundary:

−

∆=

TTR

H

P

P

t

subm

t

11ln ,

DHm,sub = DHm,fus + DHm,vap = 41,400 J

TP

P

t

/979,48692.17ln −=

e.g., T = 5.0 C = 278.15 Kfl P = 0.9692 Pt = 34.9 Torr

Liquid-vapor boundary:

TTTR

H

P

P

t

vapm

t

/370429.1311

ln , −=

−

∆=

e.g., T = 6.0 C = 279.15 Kfl P = 1.0254 Pt = 36.9 Torr

Estimate of the normal boiling point:

Set P = 760 Torr and solve for T: Find Tbp @ 361.7 K

23

Question: What is the effect of external pressure onthe vapor pressure?

Answer: At equilibrium, mliq = mvap. At constant T,

vap

vapvapvapmvap P

RTdPdPVd == ,µ

extliqmliqliqmliq dPVdPVd ,, ==µ

extliqm

vap

vap PVP

PRT ∆= ,*

ln

RT

PV

vapvap

extliqm

ePP∆

=,

*

P* is the vapor pressure of the liquid without anyexternal pressure.

24

Example: What is the vapor pressure of benzene in anatmosphere of 100 atm of helium at its normalmelting point?

388.065.27831451.8

1001325.110874.8ln

75,

*===

−

x

xxx

RT

PV

P

P extliqm

vap

vap

Pvap = 1.47Pvap* = 53 Torr

What is the effect of atmospheric pressure on thevapor pressure of water?

455

,

*104.7

29831451.81001325.1108.1

ln −−

=== xx

xxx

RT

PV

P

P extliqm

vap

vap

The effect is negligible.

25

Lecture 29: Physical Source of the PhaseTransition

How can we calculate the temperature and pressureof a phase transition, given only the equation ofstate?

We have a stability criterion that tells us that certainconditions are unstable:

01 >

∂∂−=

T

m

mT P

V

Vκ

Figure 38.Isotherm of a non-ideal substance.Note that theinverse is multi-valued.

The unstable region lies between F and M. Whereexactly does the phase transition occur? What is thevalue of Vm?

We calculate µ(Vm) for a fixed T, using point A as areference:

26

∫+=B

A

P

P

mAB dPPV )(µµ

Figure 39. Calculation ofthe chemical potential fora van der Waals gas.

We find that m(P) increases between PA and PF,decreases from PF to PM, and then increases again.The equilibrium state is always on the lowermostbranch of m(P). The phase transition occurs at pointsD Ø O, where PD = PO and Vm has a discontinuity.

Figure 40Location of thetransition point.The phase transitionoccurs along lineDKO, with areas Iand II equal. PointD is a vapor, andpoint O is a liquid.Point K doesn’tactually exist.

27

What is the value of the transition pressure?

m(PO) = m(PD)

∫→

=OD

m dPPV 0)(

∫∫∫∫ =+++O

M

M

K

K

F

F

D

P

P

m

P

P

m

P

P

m

P

P

m dPPVdPPVdPPVdPPV 0)()()()(

∫∫∫∫ =−=−O

M

K

M

F

K

F

D

P

P

m

P

P

m

P

P

m

P

P

m dPPVdPPVdPPVdPPV 0)()()()(

This equation shows that the two areas are equal.

28

The Lever Rule

The points along the “tie line” OKD exist only in anaverage sense.

<Vm> = xDVm,D + xOVm,O

xD = mole fraction of the vapor = xliq

xO = mole fraction of the liquid = xvap

xliq + xvap = 1

(xliq + xvap) <Vm> = xliqVm,liq + xvapVm,vap

[ ] [ ]mvapmvapliqmmliq VVxVVx −=− ,,

liqmm

mvapm

vap

liq

VV

VV

x

x

,

,

−−

=

Vm is discontinuous; <Vm> is continuous.

29

At the critical point the maximum and minimum in mcoalesce into a point of inflection.

Figure 41. Temperaturedependence of the phasetransition. The lowermostcurve corresponds to thetriple point.

30

Lecture 30. Surface Tension and CapillaryAction

Surface Tension

In order to lower its energy, a liquid spontaneouslyalters its shape so as to maximize the number ofmolecules in the condensed phase.

The work done do distort the surface area, s, atconstant volume and temperature, is given by

dw = gds

g is the surface tension, having units of J/m2 = N/m

dw is the change in Helmholtz free energy, dA.

The pressure inside a droplet increases as a result ofthe surface distortion from planarity. It is analogousto the increased pressure inside a balloon caused bythe stretched rubber.

31

Mechanical derivation of the pressure differential:

Force inside = Force outside + Force from surface tension

Fin = Fout + Fsurface

Fin = 4pr2Pin

Fout = 4pr2Pout

Fsurfacedr = w = gds = gd(4pr2) = 8pgrdr

Fsurface = 8pgr

DP = Pin − Pout = 2g/r

32

Thermodynamic derivation, at constant T and V.

Consider a slight expansion of the droplet:

dAliq = -PliqdVliq + mliqdnliq + gds

dAvap = -PvapdVvap + mvapdnvap

dAtot = -(Pliq - Pvap)dVliq + (mliq - mvap)dnliq + gds

At equilibrium,0=

∂∂=

∂∂

TliqTliq n

A

V

A

mliq = mvap

(Pliq - Pvap)dVliq = gds

s = 4pr2, ds = 8prdr

Vliq = 4/3 pr3, dVliq = 4pr2dr

ds = (2/r)dVliq

Pliq - Pvap = 2g/r

33

What is the pressure differential for a droplet ofbenzene with r = 0.01 mm?

DP = 2 x 0.02888/10-5 Pa = 5776 Pa = 43.3 Torr

This pressure differential leads to an enhanced vaporpressure above a meniscus (in a capillary, forexample).

rRT

V

RT

PV liqmextliqm

ePePP,, 2

**γ

==∆

In the present case, this effect is very small: P/P* =1.0002

34

Capillary Action

The pressure below a meniscus is lowered by 2g/r ascompared to the external pressure.

This allows the liquid to rise in a capillary until thesurface force is balanced by the hydrostatic force.

rgh = 2g/r

h = 2g/rgr

For a 0.5 mm radius capillary, (assuming the radiusof the meniscus = radius of the capillary):

cmmxxx

xh 34.10134.0

105.081.987902888.02

3=== −

35

Lecture 31. Partial Molar Quantities

We know from previous lectures that by definition,

dU = TdS - pdV + mdn

We will now derive a very remarkable result, knownas the Euler relation,

U = TS - PV + mn

This expression stems from the fact that S, V, and nare extensive variables, and that T, P, and m areintensive. The technical terminology is that U is ahomogeneous first order function of S, V, and n: Forany constant l,

U(lS, lV, ln) = lU(S, V, n)

For example, doubling S, V, and n is equivalent todoubling U.

36

Next, differentiate both sides with respect to l, usingthe chain rule,

λλ

λλλλ

λλ

λλλλ

λλ

λλλλ

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂ )(

)(

),,()(

)(

),,()(

)(

),,( n

n

nVSUV

V

nVSUS

S

nVSU

nn

nVSUV

V

nVSUS

S

nVSU

)(

),,(

)(

),,(

)(

),,(

λλλλ

λλλλ

λλλλ

∂∂+

∂∂+

∂∂=

= U(S, V, n)

Next, set l=1,

Unn

UV

V

US

S

U =∂∂+

∂∂+

∂∂

But we recognize that the derivatives are theintensive variables.

PVnSnV n

U

V

UP

S

UT

,,,

,,

∂∂=

∂∂=−

∂∂= µ

Therefore,

U = TS - PV + mn

37

Next, calculate the perfect differential of U,

dU = TdS +SdT -PdV - VdP + mdn + ndm

But we also know from the First Law that

dU = TdS - PdV + mdn.

Subtracting, we obtain the Gibbs-Duhem relation,

SdT -VdP + ndm = 0.

We can easily generalize to a mixture of two (ormore) substances:

U = TS - PV + mAnA + mBnB

dU = TdS - PdV + mAdnA + mBdnB

SdT -VdP + nAdmA + nBdmB = 0

At constant temperature and pressure,

nAdmA + nBdmB = 0

38

We can apply this method to any extensive,homogeneous first order variable. Consider thevolume of a mixture of liquids,

V = VAnA + VBnB + …

Where

,...,,, CB nnTpAA n

VV

∂∂=

,...,,, CA nnTpBB n

VV

∂∂=

VA is the increase in the total volume when a smallamount dnA of substance A is added to the solution.

This is a pretty amazing result!

The Gibbs-Duhem relation for a binary mixture is

nAdVA + nBdVB = 0

or

AB

AB dV

n

ndV −=

The slopes have opposite sign.

39

Liquid mixtures

Example from text: ethanol, water mixture

Figure 42. Partial molarvolumes of an ethanol,water mixture. Note theopposite signs of theslopes.(Fig. 7.1 from Atkins.)

We will use the notation of MA to designate themolecular weight of A, nA for the number of moles ofA, and mA for the mass of A, so that

mA = nA MA,

and R = mA/mB

40

Molar volumes for pure liquids: Vm,E = ME/rE

The partial molar volume of E is designated by VE.

Vm,W =VW(xE = 0) = 18 cm3/mol

Vm,E = VE(xE = 1) = 58 cm3/mol

(Note: Fig. 7.1 is inconsistent with the tabulated value of rE, whichgives a value of 58.4.)

Partial molar volumes for a solute added to a puresolvent:

VW(xE = 1) = 14 cm3/mol

VE(xE = 0) = 54 cm3/mol

Partial molar volumes for a 50% mixture:

VW(xE = 0.5) = 17 cm3 /mol

VE(xE = 0.5) = 57 cm3/mol

41

Lecture 32. Application of the Euler andGibbs-Duhem equations to solutions

Using the Euler equation

Example: What is the volume of a mixture of oneliter of ethanol and one liter of water?

In general, n = Vr/M

nW = 1000 x 1 /18.015 = 55.51 molesnE = 1000 x 0.789/46.07 = 17.13 molesmole fraction of ethanol = xE = 17.13/72.64 = 0.236

From the figure, VW = 0.0175 L, VE = 0.0560 L fl

V = 0.0175(55.51) + 0.0560(17.31) = 1.94 L

42

Self –test problem, 7.1: Given that a 1:1 mixture (bymass) of ethanol to water has a density of 0.914g/cm3, if the partial molar volume of water is 17.4cm3/mol, find the partial molar volume of ethanol.

From the Euler relation we know that

V = nEVE + nWVW

Given:

r = (mE + mW)/V = 0.914

R = mE/mW = 1

Solve for VE in terms of VW.

EWWE nVnVV /)( −=

whereV = (mE + mW)/r

nW = mW/MW

nE = mE/ME

43

It follows that

( )EE

WWWEW

MmMVmmm

EV /// −+= ρ

Divide through by mW,

EW

E

W

WWE

E

Mm

mM

Vmm

V1

/1 −+

= ρ

E

WW

MR

MVR

/

//)1( −+= ρ

Here, R = 1, r = 0.914, VW = 17.4, MW = 18.015, and

ME = 46.07 fl VE = 56.3 cm3

44

Verifying the Euler and Gibbs-Duhem equations:

Application of the Gibbs – Duhem relation

Suppose

BA nnC21*

BB*AA VnVnV ++=

Doubling nA and nB causes V to double.

To express V in terms of mole fractions, divide by nA

+ nB,

BAmBBAmAm xxCVxVxV2

1,, ++=

Calculate VA and VB:

ABAA nnCVV /4

1* +=

BABB nnCVV /4

1* +=

BABBBAAABBAA nnCVnnnCVnVnVn 41*

41* +++=+

This shows that the Euler equation is satisfied.

45

Next, calculate the derivatives of the partial molarvolumes:

2/12/12/12/3

8

1

8

1 −−− +−= BABAA nCnnCndV

2/32/12/12/1

81

81 −−− −= BABAB nCnnCndV

Use this result to confirm the Gibbs-Duhem equation.

46

Application of the Gibbs-Duhem equation.

It is most useful in its integrated form.

Expressing the partial molar volumes as functions ofnA and nB, we integrate over nA:

A

n

n

A

A

B

A

nV

nV

AB

AmBBAB nd

n

V

n

nVd

n

nVnnV

B

AAA

AA

′

′∂∂

′−=′

−=− ∫∫

= 0

)(

)0(

,),(

where VB,m=VB* is the volume of the pure liquid B.

Look carefully at the integration limits, starting frompure B.

Suppose you are given that

cbVV mAA += ,

where b is the molality of A,

BB

A

B

A

Mn

n

m

nb ==

and c is a constant. Calculate VB and V.

47

B

A

BmAA n

n

M

cVV += ,

ABB

A dnnM

cdV =

ABB

AA

B

AB dn

nM

cndV

n

ndV

2−=−=

This is the Gibbs-Duhem equation.Integating it gives the Euler equation:

2

2

, 2 BB

AmBB nM

cnVV −=

B

A

BmBBmAABBAA n

n

M

cVnVnVnVnV

2

,, 2++=+=

48

Lecture 33. Chemical Potential of Liquids

Question: What happens when you step out of theshower?

Answer: Evaporating water removes heat andmakes you feel cold. Let’s calculate how muchheat is removed, and determine whether theprocess occurs spontaneously.

Suppose that room temperature is 298 K, and thatone mole of water evaporates. Let’s calculate DGfor this process. (It would have been better to assume thatthe room and water the temperatures are the same as your bodytemperature, but we will ignore this detail.)

Gibbs-Helmholtz Equation:

2T

H

T

G

TP

∆−=

∆∂∂

Integrated form:

−∆=

∆−

∆

121

1

2

2 11)()(

TTH

T

TG

T

TG

49

−∆+∆=∆

1

21

1

22 1)()(

T

THTG

T

TTG

Choose T1=373 K, T1 = 298 K

DG(373) = 0, DH(298) = 44,014 cal/mol

Conclusion: DG(298) = 8850.

This result is positive! What’s going on?

Note for later reference:

−∆−=∆

bTTR

H

RT

TG 11)(

22

2

50

We could also have reached this result byconstructing a thermodynamic cycle.

What is DHvap(H2O, 298 K)?1

H2O(liq, 298 K) Ø H2O(gas, 298 K)

∞ 2 Æ 4

H2O(liq, 373 K) Ø H2O(gas, 373 K)3

DH1 = DH2 + DH3 + DH4

DH2,m = CP,m(H2O, liq) x (373-298) = 5.647 kJ/mol

DH3,m = DHvap,m(H2O, 373 K) = 40.656 kJ/mol

DH4,m = -CP,m(H2O, gas) x (373-298) = -2.519kJ/mol

\ DHvap,m(H2O, 298 K) = 40.656 + 3.126 = 43.784kJ/mol

(Table 2.3 lists 44.016 kJ/mol. Why?)

This heat of vaporization that is absorbed by thewater accounts for your feeling cold.

51

Now let’s calculate the entropy changes.

DS298,m = DS2 + DS3 + DS4

DS2,m = CP,m(H2O, liq) ln (373/298) = 16.90 Jmol-1K-1

DS3,m = DHvap,m(H2O, 373 K)/373 = 109.0 Jmol-1K-1

DS4,m = CP,m(H2O, gas) ln (298/373) = -7.54 Jmol-1K-1

\ DS298,m = 118.36 J mol-1K-1

Question: Is this process spontaneous?

Answer: Calculate DGm = m = DHm -TDSm to findout.

At 373K, DGm is clearly zero. Is it positive ornegative at 298 K?

DG298,m = DH298,m - 298DS298,m

= 43,784 - 298(118.36) = 8,513 J/mol > 0

52

Resolution of the paradox: The above result is forPo = 1 bar, whereas the water vapor pressure islower. In general,

oomm P

PRTTGPTG ln)(),( +∆=∆

At equilibrium at 298 K,

TorrPP

PRTPG

om 210ln850,8),298( =⇒=

+=∆

Note: You get the identical result using theClapeyron-Calusius equation:

−

∆=

TTR

H

P

P

b

vap 11

*ln

But we recognize from the Gibbs-Helmholtz equationthat the right hand side is just RTG o

m /∆ .

53

Lecture 35. Equilibrium between liquidsolutions and the gas phase

Equilibrium occurs when mliq = mvap

Let’s calculate mvap. Strictly speaking, we should usethe fugacities rather than partial pressures.

The chemical potential of the vapor of pure liquid A:

+== o

A

AovapAvapAliqA f

fRT

*

,,*

, lnµµµ

where “o” refers to the fugacity at 1 bar, and * refersto the pure liquid or pure vapor.

For a liquid solution,

+== o

A

AovapAvapAliqA f

fRT ln,,, µµµ

Subtracting,

+= *

*,, ln

A

AliqAliqA f

fRTµµ

This is a general result.

54

For an ideal solution,

*AAA fxf =

AliqAliqA xRT ln*,, += µµ

This result is called Raoult’s Law, which is derivedfrom the idea that the probability of A escaping froma liquid mixture is proportional to the frequency withwhich it strikes the interface.

Even if fA is not linear over the entire range of xA

(from fA = 0 to **AA Pf ≈ ), it still may be linear at low

concentrations:AAA Kxf = .

This limit is called is called an “ideal dilute solution,”and the equation is called Henry’s Law:

+=

**

,, lnA

AAliqAliqA f

KxRTµµ

Remember, *Af is essentially the vapor pressure of the

pure liquid.

55

Vapor pressure of a liquid mixture

Suppose you have a liquid mixture of A and B, withmole fractions xA and xB. What is the composition ofthe vapor phase?

For an ideal liquid mixture,

*AAA PxP =

*BBB PxP =

What is the composition of the vapor?

**

*

BBAA

AA

BA

AA

PxPx

Px

PP

Py

+=

+=

**

*

BBAA

BB

BA

BB PxPx

Px

PP

Py

+=

+=

56

Suppose that A is more volatile than B. Which phaseis richer in A?

Our assumption is that .**BA PP >

AABBA

AA x

PPxx

xy >

+=

** /

(Explanation: xA + xB = 1. Therefore the denominator issmaller than 1.)

Conclusion: The gas phase is richer in A.

An ideal liquid solution is not one in which themolecules do not interact. Rather, it is a mixture inwhich the interactions of all species are the same(both like and unlike species).

57

Theorem: If the solute obeys Henry’s law, thesolvent obeys Raoult’s Law.

We will prove it using the Gibbs-Duhem relation form:

nAdmA + nBdmB = 0

Divide through by n = nA + nB:

xAdmA + xBdmB = 0

Differentiate with respect to xA:

0,,

=

∂∂+

∂∂

TPA

BB

TPA

AA x

xx

xµµ

But dxA = -dxB for a binary mixture. Therefore,

0,,

=

∂∂−

∂∂

TPB

BB

TPA

AA x

xx

xµµ

TPB

B

TPA

A

XX,,

lnln

∂∂=

∂∂ µµ

Let A be the solvent and B the solute.

58

Recall Henry’s law,

+=

**

,, lnA

AAliqAliqA f

KxRTµµ

From the above, RTx

TPB

B =

∂∂

,ln

µ

Therefore, for the solvent RTx

TPA

A =

∂∂

,ln

µ

Integrating, CxRT AA += lnµ

This result is valid in the limits of xBØ0, xAØ1.

Therefore, C = mA*, giving Raoult’s Law.

The gas phase composition is therefore

BBAA

AAA KxPx

Pxy

+=

*

*

BBAA

BBB KxPx

Kxy

+=

*

59

Figure 43.LiquidSolutions.Left: IdealmixtureillustratingRaoult’s Law.Right: Non idealmixture,showing idealdilute behavior.

60

Colligative Effects

Suppose component B is totally non-volatile. Also,suppose that it remains behind in the liquid phasewhen the solvent freezes. Then its only effect is tolower mA,liq. The net result is to lower the freezingpoint and raise the boiling point of A. Thephenomenon is conceptually related to vapor pressureelevation by an external pressure.

Figure 44.Colligativeeffects: Boilingpoint elevationand freezingpointdepression.

61

The key concept is that the chemical potential of thesolvent, A, equals that of the vapor, which is pure A.

*,

*,, ln vapAAliqAliqA xRT µµµ =+=

AliqAvapA xRT ln*,

*, =− µµ

Amvap xRTG ln, =∆

Amvapmvap x

RT

STHln,, =

∆−∆

But DSvap,m = DHvap,m/Tb

AB

mvap

xRT

T

TH

ln

1,

=

−∆

Ab

mvap xTTR

Hln

11, =

−

∆

62

After some manipulation, for xB << 1,

The same reasoning holds for freezing. The chemicalpotential of the solid, which is assumed to be pure A,equals that of the solvent:

*,

*,, ln solidAAliqAliqA xRT µµµ =+=

The freezing point is depressed by an amount

BmfusA

ff x

H

RTT

,,

2

∆=∆

In both cases, DT is independent of the properties ofthe solute.

BmAvap

bb x

H

RTT

,,

2

∆=∆

63

Suppose the solvent does not freeze, but instead thesolute, B, precipitates. Then the same reasoning gives

*,

*,, ln solidBBliqBsolutionB xRT µµµ =+=

We can use this relation to calculate the solubility of B:

−

∆−=

f

mfusBB TTR

Hx

11ln ,,

64

Lecture 36. Entropy and Energy of Mixing

Microscopic approach: S = k lnW, where W is thenumber of equivalent configurations.

Suppose you have a crystal with N sites, and youwish to fill it with NA atoms of type A and NB atomsof type B, with N = NA + NB. Assume that theinteraction energies AA, BB, and AB are equivalent.What is the entropy of the crystal?

Assume that the A atoms are indistinguishable, as arethe B atoms. The atoms can be permuted among theN sites N! different ways, but of these NA!permutations of A and NB! permutations of B areindistinguishable. Therefore

!!

!

BA NN

NW =

Assume that NA, NB >> 1

65

Sterling approximation: NNN −≈ ln!ln

BBBAAA NNNNNNNNNkS +−+−−≈ lnlnln/

BBAABA NNNNNNN lnlnln)( −−+=)ln(ln)ln(ln BBAA NNNNNN −+−=

BA

BB

BA

AAmixing NN

NkN

NN

NkNS

+−

+−=∆ lnln

0lnln >−−= BBAA xkNxkN

Number of moles = (NA + NB)/Navog = nA + nB = n

BBAAmmixing xRxxRxS lnln, −−=∆

66

Macroscopic approach: Suppose you have two idealgases,{nA, VA P, T} and {nB, VB P, T}. Mix themirreversibly to the final equilibrium condition,{nA, nB, V, P, T}. Calculate the entropy change usinga reversible path.

Step 1. Expand each gas separately, reversibly, andisothermally to a final volume, V, and partialpressures PA and PB:

PA = nART/V

PB = nBRT/V

P = PA + PB

V = VA + VB

BB

AA V

VRn

V

VRnS lnln1 +=∆

67

Step 2. Mix the gases reversibly, while maintaining atotal volume V.

Membrane 1 is permeable only to A. The partialpressure PA is the same on both sides of themembrane.

Membrane 2 is permeable only to B. The partialpressure PB is the same on both sides of themembrane.

Membrane 3 is impermeable to both A and B.

Membranes 1 and 3 are connected by a movablepiston having area A. Membrane 2 is stationary.

The force pushing the piston to the right equalsPAA + PBA.

The force pushing the piston to the left equals(PA + PB) A.

68

Net force on the piston is zero fl w = 0.

DU = 0 because T is constant.

The process is reversible. \ qrev = 0 fl DS2 = 0

BB

AAmixing V

VRn

V

VRnSS lnln1 +=∆=∆∴

BBAAmixing xRnxRnS lnln −−=∆∴

BBAAmmixing xRxxRxS lnln, −−=∆∴

General result:

∑−=∆i

iimmixing xxRS ln,

The entropy of a mixture of perfect gases is the sumof the entropies that each pure gas would have if itoccupied the same volume as the mixture at the sametemperature.

69

What is the energy of mixing?

Consider the case of an ideal gas.

mixing

nnP

mixing ST

G

BA

∆−=

∂∆∂

,,

TdTSTGTGT

mixingmixingmixing ′′∆−=∆=∆ ∫0

)()0()(

But we proved that for an ideal gas, DSmixing isindependent of temperature.

\ DGmixing = -TDSmixing

Your textbook attacks the problem in the oppositedirection. First it solves for DGmixing and thendetermines DSmixing.

Before mixing, the chemical potential is just the sumfor the pure materials:

BBAAunmixed nnG µµ +=

++

+=o

oBBo

oAA

P

PRTn

P

PRTn lnln µµ

70

After mixing, the partial pressures have dropped toPA and PB:

++

+=oBo

BBoAo

AAmixed P

PRTn

P

PRTnG lnln µµ

(Note the typo in the book.)

The free energy of mixing is the difference betweenthese two quantities:

+

=∆P

PRTn

P

PRTnG B

BA

Amixing lnln

Assuming Dalton’s Law,

BBAAmixing xRTnxRTnG lnln +=∆

BBAAmmixing xRTxxRTxG lnln, +=∆

What is the entropy of mixing?

)lnln(,,

,, BBAA

nnP

mmixingmmixing xxxxR

T

GS

BA

+−=

∂∆∂

−=∆

71

What is the enthalpy of mixing?

DH = DG + TDS = 0

This result is not surprising because we assumed thatthe gas is ideal. What are the results for real gases?

BBAAunmixed nnG µµ +=

++

+= o

BoBBo

AoAA f

fRTn

f

fRTn

**

lnln µµ

++

+=

oBo

BBoAo

AAmixed f

fRTn

f

fRTnG lnln µµ

+

=∆

**lnln

B

BB

A

AAmixing f

fRTn

f

fRTnG

+

=∆

**, lnlnB

BB

A

AAmmixing f

fRTx

f

fRTxG

72

Because f=f/P depends on temperature, it no longerfollows that DSmixing = -DGmixing/T and that DHmixing=0. However, for gases it is agood approximation (the rule of Randall and Lewis)that

AA

A xf

f≈

*

so that for gases it is still true that .0≈∆ mixingH

What is the story for solutions?

For solvent A, having vapor pressure PA,

AAA

AAA aRT

P

PRTliq lnln)( *

** +=

+= µµµ

where we have defined the activity, aA. For an ideal(Raoult) solvent, aA = xA..

73

For an ideal-dilute solute,

+=

+=

**

** lnln)(

B

BBB

B

BBB P

xKRT

P

PRTliq µµµ

BBB xRTliq ln)( # += µµ

whereB

BB K

Px =

Finally, for a real solute,

BBB aRTliq ln)( # += µµ

whereB

BB K

Pa =