chapter 8 multicomponent homogeneous...

Chapter 8Multicomponent Homogeneous Nonreacting Systems: Solutions

Notes onThermodynamics in Materials Science

09/19/2001 Notes from R.T. DeHoff, Thermodynamics in Materials Science (McGraw-Hill, 1993) 8-1

Thermodynamics in Materials Scienceby

Robert T. DeHoff(McGraw-Hill, 1993).

Extensive Quantities of theState Functions: F/, G/, H/, S/, U/, V/

Using G/ as example:G G T P n n n nk c( , , , ... ... )1 2

A differential form of G/:


k

c

k nPTknTnP

dnnGdP

PGdT

TGdG

kjkk1 ,,

/

,

/

,

//

c

kk

nPTkPT dn

nGdG

kj1 ,,

/

,/

and at constant T & P:

A differential form of G/:

Partial Molal Quantities of theState Functions: .

Using as example:

kj nnPTkk n

GG,,

/

A differential form of G/ (use definition of ):

KG

kkkkkk VUSHGF ,,,,,

G


c

kkk

nTnP

dnGdPPGdT

TGdG

kk1,

/

,

//

c

kkkPT dnGdG

1,

/

and at constant T & P:

A differential form of G/ (use definition of ):kG

Gibbs-Duhem EquationThe contributions of the components sum to the

whole: c

kkknGG

1

/

c c

Differentiating the products on the right yields:


c

k

c

kkkkk GdndnGdG

1 1

/

n dGk kk

c

10

12

12 Gd

nnGd

For a binary system:

Inspection yields the Gibbs-Duhem equation:

Reference States:F/O, G/O, H/O, S/O, U/O, V/O

• T, S, V, P have absolute values.• F, G, H, U have relative values.• The difference in values between states is unique.• To compare values, use the same reference state.


• To compare values, use the same reference state.• Superscript O refers to the reference state.

c

kk

Ok

O nGG1

/

• Preferably, for a solution use the pure components in the same phase as the solution as the reference state.

Rule of mixtures.

Rule of Mixtures2211

2

1

/ nGnGnGG OO

kk

Ok

O

21

22

21

11

21

/

nnnG

nnnG

nnG OO

O

2211 XGXGG OOO

OG2X

For binary

For binary

Extensive

Molar


OG1 X

G2XOG

2X1 2

Mixing Values for SolutionsF/

mix, G/mix, H/

mix, S/mix, U/

mix, V/mix

For solution: Gibbs free energy of mixing.O

somix GGG //ln

/

For component k: Change experienced when 1 mole of k is transferred from its reference state to the given solution.


Okkk GGG

k

c

k

Okk

c

kkmix nGnGG

11

/

c

kkkmix nGG

1

/and

Contributions of the components add to the whole:given solution.

Mixing Values

OG2XG G

X X00

Osomix GGG //

ln/


OG1 X

XlnsoG

2X1 2

mixG

2X1 2

Mixing values for SolutionsF/

mix, G/mix, H/

mix, S/mix, U/

mix, V/mix

Differential form:

c

kkmix dnGGd /

c

k

Okkk

Okkkkkmix dGndnGGdndnGGd

1

/

0 0Gibbs-Duhem


kkkmix dnGGd

1c

kkkkkmix GdndnGGd

1

/

n d Gk kk

c

01

Gibbs-Duhem for mixing:

Total derivative:

Graphical Evaluation of Partial Molal Values

Consider a binary system (alloy):G G X G Xmix 1 1 2 2

121 XX dX dX1 2

d G G dX G dXmix 1 1 2 2

GSubstitute & rearrange:

Note:


222 1

dXGdXGG mix

mix

111 1

dXGdXGG mix

mix

d GdX

d GdX

G Gmix mix

2 12 1

G1

G2

and

Substitute & rearrange:

X2

Gmix

0 1X1 01

Derivation: Graphical Evaluationof Partial Molal Values

2211 XGXGGmix 1

d G G dX G dXmix 1 1 2 2 2

121 XX dX dX1 2 b3a3GdGG mix 4Rearrange (2)


212 dX

GdGG mix 4Rearrange (2)

1

22

11 X

XGXGG mix 5Rearrange (1)

211

212 dX

GdXG

XXXG mixmixInsert

(5) in (4)

222 1

dXGdXGG mix

mixYielding

Integration of the Gibbs-Duhem Equation(s)

For a binary system (alloy):X d G X d G1 1 2 2 0

d GXX

d G12

12and


1

d G G X G X G XX

X

10

1 2 1 1 1 22

2

10

2

2 02

2

2

1

21

X

X

dXdX

GdXXG

Now integrate right side:

Integrating the left side from X2 = 0 to X2:

Molar Values of the State FunctionsNote: c

kk

kk

n

nX

1

dG G dXk kk

c

1d G G dXmix k k

k

c

1

Then,


k 1

G G Xk kk

c

1

X dGk kk

c

10

k 1

G G Xmix k kk

c

1

X d Gk kk

c

10

Chemical Potential of (Open) Multicomponent Systems

U U T P n n n nk c( , , , . . . . . . )1 2

dU TdS PdV dnk kk

c

1

U c

termsc 2


kk S V n n

Un

j k, ,

W dnk kk 1

kk S V n k S P n k T V n k T P n

Un

Hn

Fn

Gn

j j j j, , , , , , , ,

Chemical Potential of (Open) Multicomponent Systems

kj nnPTkkk n

GG,,

kkk TT

GS kkkkk T

TSTGH


kk nPnPk TT ,,

kk nT

k

nT

kk PP

GV,,

knPkkkk T ,

kk nP

k

nT

kkkkk T

TP

PVPHU,,

knT

kkkkk P

PSTUF,

Activities and Activity Coefficients

Definition of activity, ak (dimensionless):

kkOkk aRTln

O XRTlnXaDefinition of activity coefficient, k (dimensionless):


kkkOkk XRTlnkkk Xa

If k<1 ak<Xk k is less apparent than its mole fraction.k=1 ak=Xk k is as apparent as its mole fraction.k>1 ak>Xk k is more apparent than its mole

fraction.

Ideal SolutionNo heat of mixing. 0mixH0kH

0mixV0kV

0mixU0kUc

Entropy increases.

No change in internal energy.No volume change.


kk XRS ln k

c

kkmix XXRS ln

1

kk XRTF ln k

c

kkmix XXRTF ln

1

kk XRTG ln k

c

kkmix XXRTG ln

1

Gibbs free energy decreases.

Helmholtz free energy decreases.

Ideal Solution

mixGmixS

mixH


T2X 2X

mix00

Ideal Solution

• All plots (e.g. Gmix vs. Xk) are symmetrical with composition.

• Slopes of plots of Smix, Fmix, Gmix are infinite at Xk =0 & Xk =1.


infinite at Xk =0 & Xk =1.• Entropy of mixing is independent of

temperature.

Ideal SolutionActivity is the same as mole fraction.

Activity coefficient is one.1

kk Xa

1


a2 a1

X20

1k

0 1

Slope = 1

Dilute Solutions:Raoult & Henry’s Laws

Raoult’s Law for the solvent in dilute solutions:1111

lim XaX

lim Xa O

Henry’s Law for the solute in dilute solutions:1slope 1slope


2202

lim Xa O

X

20 XsmallforOOslope 2

2a

2X

1a

2X

Oslope 1

1slope 1slope

Real Solutions: Relation of Activity Coefficient to Free Energy

kkk Xa

kkkkkk XRTRTXRTG lnlnlnID

kXS

kkk GGG


Ideal partial molal free energy of mixing:kkkk GGG

kID

k XRTG ln

kXS

k RTG lnExcess partial molal free energy of mixing:

Regular SolutionHeat of mixing is a function of composition, only.

ckmixmix XXXXHH ,...,..., 21

kk XRS ln k

c

kmix XXRS ln

PTHH mixmix ,Entropy is the same as for ideal solution.


kk XRS ln kk

kmix XXRS ln1

kkk XRTUF ln k

c

kkmixmix XXRTUF ln

1

kkk XRTHG ln k

c

kkmixmix XXRTHG ln

1

Gibbs free energy decreases.

Helmholtz free energy decreases.

Regular SolutionmixS

mixGX 2X00


mixH

mixG2X 2X

T

Regular SolutionmixSmixH

mixGX 2X00


mixG2X 2X

T


mixGX 2X00


mixG2X 2X


mixGX 2X00


mixG2X 2X

T

Problem 8.6 DeHoffmoleJXXH CnPnPn

2500,12)(XfHmix

1222 XaXH

Find: Given:Rewrite in general form:

2211222 2 dXXaXdXaXHd

22 2 XaXaXHdDifferentiate:

Substitute dX1= dX2:


2122

2 2 XaXaXdX

121222

2 312 XaXXXaXdX

Hd

Substitute dX1= dX2:

Substitute X1+X2=1:

202

2

1

21

2

2

dXdX

HdXXH

X

X

20 121

21

2

2

31( dXXaXXXH

X

X

Gibbs-Duhem:

chapter 8 multicomponent homogeneous...

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