Chapter 8dehoff Thermo

7/26/2019 Chapter 8dehoff Thermo

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Chapter 8

Multicomponent Homogeneous

Nonreacting Systems: Solutions

Notes on

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

by

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


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Extensive Quantities of the

State Functions: F/

, G/

, H/

, S/

, U/

, V/

Using G/as example:

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

/


k

c

k nPTknTnP

dnn

GdP

P

GdT

T

GdG

kjkk

1 ,,

/

,

/

,

//

c

k

k

nPTk

P

T dnn

GdG

kj1 ,,

/

,

/

and at constant T & P:


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Partial Molal Quantities of theState Functions: .

Using as example:

kj nnPTk

k

n

GG

,,

/

KG

kkkkkk VUSHGF ,,,,,


c

k

kk

nTnP

dnGdP

P

GdT

T

GdG

kk 1,

/

,

//

c

k

kkPT dnGdG

1

,

/

and at constant T & P:

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


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Gibbs-Duhem Equation

The contributions of the components sum to thewhole:

c

k

kknGG1

/

Differentiating the products on the right yields:


k k

kkkk GdndnGdG1 1

/

n dGk kk

c

1

0

1

2

12 Gd

n

nGd

For a binary system:

Inspection yields the Gibbs-Duhem equation:


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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 com are values use the same reference state.


Superscript O refers to the reference state.

c

kk

O

k

O

nGG 1

/

Preferably, for a solution use the pure components in

the same phase as the solution as the reference state.

Rule of

mixtures.


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Rule of Mixtures2211

2

1

/ nGnGnGG OO

k

k

O

k

O

21

22

21

11

21

/

nn

nGnn

nGnn

G OOO

2211 XGXGG OOO

O

For binary

For binary

Extensive

Molar


OG1

X

2OG

2X1 2


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Mixing Values for Solutions

F/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


O

kkk GGG

k

c

k

Okk

c

k

kmix nGnGG 11/

c

k

kkmix nGG1

/and

Contributions of the components add to the whole:

given solution.


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Mixing Values

O

G2

X X00

O

somix GGG//

ln

/


OG1

X

lnsoG

2X1 2

mix

2X1 2


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Mixing values for Solutions

F/mix, G/

mix, H/

mix, S/

mix, U/

mix, V/

mix

Differential form:

c

/

c

k

O

kkk

O

kkkkkmix dGndnGGdndnGGd1

/

0 0Gibbs-Duhem


km x

1

c

kkkkkmix

GdndnGGd1

/

n d Gk kk

c

01

Gibbs-Duhem for mixing:

Total derivative:


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Graphical Evaluation ofPartial Molal Values

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

121

XX dX dX

1 2

d G G dX G dX mix 1 1 2 2

Note:


2

22 1dX

GdXGG mixmix

1

11 1dX

GdXGG mixmix

d G

dX

d G

dX

G Gmix mix 2 1

2 1

1

G2

and

Substitute & rearrange:

X2

Gmix

0 1

X1 01


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Derivation: Graphical Evaluationof Partial Molal Values

2211 XGXGGmix

1

d G G dX G dX mix 1 1 2 2 2

121 XX dX dX 1 2 b3a3

G


2

12dX

earrange

1

22

1

1

X

XG

X

GG mix

5Rearrange (1)

211

212

dX

Gd

X

G

X

XXG mixmix

Insert

(5) in (4)

222

1dX

GdXGG mix

mix

Yielding


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Integration of theGibbs-Duhem Equation(s)

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

d GX

X

d G 12

1

2 and


d G G X G X G X

X

X

10 1 2 1 1 1 22

2

1

0

2

2 0

2

2

2

1

21

X

X

dXdX

Gd

X

XG

Now integrate right side:

Integrating the left side from X2 = 0 to X2:


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Molar Values of the State Functions

Note:

c

kk

kk

n

nX

1

dG G dX k k

c

d G G dX mix k k

c

Then,


G G Xk kk

c

1

X dGk kk

c

1

0

G G Xmix k k k

c

1

X d Gk kk

c

1

0


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Chemical Potential of (Open)Multicomponent Systems

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

dU TdS PdV dnk kk

c

1

c

termc 2


k k S V n nn j k

, ,

W dnk kk 1

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

U

n

H

n

F

n

G

nj j j j

, , , , , , , ,


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Chemical Potential of (Open)Multicomponent Systems

kj nnPTk

kknGG

,,

kkk

GS

kkkkk TSTGH


kk nPnP ,,

kk nT

k

nT

kk

PP

GV

,,

knP,

kk nP

k

nT

kkkkk

TT

PPVPHU

,,

knT

kkkkk

P

PSTUF

,


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Activities and Activity Coefficients

Definition of activity, ak (dimensionless):

kk

O

kk aRTln Definition of activity coefficient,

k (dimensionless):


kkk

O

kk XRT lnkkk Xa

If

kXk k is more apparent than its mole

fraction.


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Ideal SolutionNo heat of mixing. 0mixH0 kH

0mixV0 kV

0mixU0 kU

c

Entropy increases.

No change in internal energy.No volume change.


kk XRS ln kk

kmix XXRS ln1

kk XRTF ln kc

k

kmix XXRTF ln1

kk XRTG ln kc

k

kmix XXRTG ln1

Gibbs free energy decreases.

Helmholtz free energy decreases.


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Ideal Solution

mixG

mixS

mixH


T2X

2X00


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Ideal Solution

All plots (e.g. Gmix vs. Xk) are symmetricalwith composition.

Slopes of plots of Smix, Fmix, Gmix are= =


k k .

Entropy of mixing is independent of

temperature.


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Ideal SolutionActivity is the same as mole fraction.

Activity coefficient is one.1

kk Xa


a2 a1

X20

1k

0 1

Slope = 1


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Dilute Solutions:Raoult & Henrys Laws

Raoults Law for the solvent in dilute solutions:

1111

lim XaX

O

Henrys Law for the solute in dilute solutions:


2202

m aX

20 XsmallforO

Oslope 2

2a

2X

1a

2X

Oslope 1

ope s ope

R l S l ti R l ti f A ti it


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Real Solutions: Relation of ActivityCoefficient to Free Energy

kkk Xa kkkkkk XRTRTXRTG lnlnln

IDXS GGG


Ideal partial molal free energy of mixing:

k

ID

k

XRTG ln

k

XS

k RTG lnExcess partial molal free energy of mixing:

R l S l i


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

ckmixmix XXXXHH ,...,...,

21

XRS ln

c

XXRS ln

PTHH mixmix ,

Entropy is the same as for ideal solution.


k 1

kkk XRTUF ln k

c

k

kmixmix XXRTUF ln1

kkk XRTHG ln k

c

k

kmixmix XXRTHG ln1

Gibbs free energy decreases.

Helmholtz free energy decreases.


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Regular SolutionmixS

G 00


mixH

2 2

T


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Regular Solution

mixSmix

H

G 00


2 2

T


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Regular Solution

mixSmixH

G 00


2 2


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Regular Solution

mixSmixH

G 00


2 2

T


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Regular Solution

mixSmixH

G 00


2 2

T


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Problem 8.6 DeHoffmoleJXXH CnPnPn

2500,12)(XfHmix

1

2

22 XaXH

Find: Given:

Rewrite in general form:

2211

2

22 2 dXXaXdXaXHd

22Hd

Differentiate:

8 29

212

2dX

12122

2

2 312 XaXXXaXdX

Hd

1

2

Substitute X1+X2=1:

20

2

2

1

21

2

2

dXdX

Hd

X

XH

X

X

20

12

1

21

2

2

31( dXXaX

X

XH

X

X

Gibbs-Duhem:

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Chapter 8dehoff Thermo