Lec 4-Solution Thermodynamics-Theory-part...

Chemical Engineering Department | University of Jordan | Amman 11942, Jordan

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Dr.-Eng. Zayed Al-Hamamre

Thermodynamics II

Solution Thermodynamics: Theory


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Content

Introduction

Thermodynamics of gas mixtures and liquid solutions

Chemical potential

Partial properties of mixtures

The Ideal-Gas Mixture Model


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The purpose in this chapter is to develop the theoretical foundation for applications of

thermodynamics to gas mixtures and liquid solutions.

In the chemical, petroleum, and pharmaceutical industries multicomponent gases or liquids

commonly undergo composition changes as the result of mixing and separation processes, the

transfer of species from one phase to another, or chemical reaction.

The property of such systems depend strongly on composition as well as on temperature and

pressure

Introduction


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Fundamental property relation

What is the most important property ?

……G……….

For pure component; G = G (T, P)

For a homogeneous mixture e.g. containing i components mixture;

G = G (T, P, n1, n2, …, ni)

Also, for closed system: no mass transfer across boundary or in a single-phase fluid in a

closed system wherein no chemical reactions occur

dTndPnnd )()()( SVG

G = G (T, P) Since n is the total number of moles of the system (= constant);

dTdPd SVG


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Open system, single phase

,,,,,, 21 innnTPgnG G

ni is the number of moles of species i

i

i

nPTinPnT

dnn

ndT

T

ndP

P

nnd

j,,,,

)()()()(

GGGG

all mole numbers held constant all mole numbers except ni held constant

The fundamental property relation for single phase fluid systems of variable mass and

composition

ii

i dndTndPnnd )()()( SVG


SG

VG

nPnT TP ,,


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nPnT TP ,,

G

SG

V

The chemical potential of species i

jnTPii n

n

,,

)(

G


This equation forms the basis for the definition of partial properties

For pure species

Gi molar Gibbs energy,

inn

0

,

i

ii

TPi

ii nG

n

GnG

n

Gn

G is intensive property independent of the system size ni


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The Chemical Potential and Phase Equilibria

Consider the following:

o A closed system and multicomponent

o Containing two phases in equilibrium.

o Mass transfer occurs if the equilibrium is disturbed

o Each individual phase is an open system, free to transfer

mass to the other

ii


ii


The total Gibbs energy of the two-phase system )()( GGG nnn

ii

iii

i dndndTndPnnd )()()( SVG


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compare with for the whole closed system


dTndPnnd )()()( SVG

0i

iii

ii dndn

Mass conservation requires

ii nn constant 0 ii dndn

0i

iii dn

Quantities idn are independent and arbitrary (never be zero).

),,2,1( Niii Hence


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Chemical Potential

Is an extensive property,

Provides a measure of the work of a system is capable when a change in mole numbers occurs

e.g. chemical reaction or a transfer of mass.

)(

For π phases at equilibrium, and N is the number of species, generalization to multiple phases

in equilibrium

Niiii ,,2,1 N ( – 1) equations

In addition to thermal and mechanical equilibrium criteria

TTT PPP


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Partial (molar) properties

Partial molar properties are defined as partial derivatives with respect to moles

jnTPii n

M,,

M

M denotes for any extensive properties

It is a response function, i.e., a measure of the response of total property nMto the addition at constant T and P of a differential amount of species I to a finite amount of solution.


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GASHUVM ,,,,,

iiiiiii GASHUVM ,,,,,

iiiiiii GASHUVM ,,,,,

Solution properties Pure-species properties

Partial molar properties

NotationsPartial (molar) properties

jnTPiii n

G,,

G Re-call


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),....x,xM(T,P,xnM i21

In general, for a homogenous mixture

The total differential of M is

ii

nTPiP,nT,n

dnn

MdT

T

(nM)dP

P

(nM)d(nM)

j,,

Could also be written as

(11.9) dnMdTT

(M)ndP

P

(M)nd(nM) ii

P,xT,x

differentiation at constant composition


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Since ni = xin

And

Substitute these terms to Eq. (11.9), and then rearrange:

Rearrange,

iii ndxdnxdn

MdnndMd(nM)

=0.0 =0.0


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(11.12)

(11.11)

0

(11.10)

0

iii

iii

iii

iii

P,xT,x

iii

P,xT,x

MnnM

MxM

MxM

dxMdTT

MdP

P

MdM

dxMdT-T

MdP-

P

MdM-


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Differentiating Eq 11.11 i

iii

ii dxMMdxdM

Comparison of this equation with Eq. (11.10) (Subtraction gives) yields

0,,

iii

xPxT

MdxdTT

MdP

P

M

Gibbs-Duhem equation at constant T and P

As a special case at constant T and P: 0i

ii Mdx

This equation must be satisfied for all changes in P, T, and the Mi caused by changes of state in a homogeneous phase


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Partial Properties in Binary Solutions

2211 MxMxM

22221111 dxMMdxdxMMdxdM

For a binary solution

AB

Gibbs-Duhem equation is

02211 MdxMdx C

i

ii MxM

0i

ii Mdx

For a binary solution

Dividing by dx1, we have the Gibbs-Duhem equation in derivative forms

01

22

1

11

dx

Mdx

dx

Mdx


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121 xx

21 dxdx 21

1

MMdx

dM

SinceDEq. B becomes

From Eq A and D

121 dx

dMxMM

112 dx

dMxMM

These equations can be used to obtain partial

molar properties from solution property.


iixx

MMMii

11

limlim

As a solution becomes pure in species i, both properties approach pure species

property

In the limit of infinite dilution iiix

MMMi

0lim


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1

2222 ,,1,,1,,1,,1

2211

2211

2211

2211

nVTnPTnSVnPS

T,VT,V

T,PT,P

V,SV,S

S,PS,P

n

A

n

G

n

U

n

H

dn) (μdn)(μPdV -SdT dA

dn) (μdn)(μVdP -SdT dG

dn) (μdn)(μPdVTdS dU

dn) (μdn)(μVdPTdS dH

For a binary (For two components) solution

Also,


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The need arise in a laboratory for 2000 cm3 of an antifreeze solution consisting of 30mol % methanol in water. What volumes of pure methanol and of pure water at 25 Cmust be mixed to form the of antifreeze, also at 25 C ? Partial molar volumes formethanol and water in a 30 mol % methanol solution and their pure-species molarvolume, both at 25 C , are:Methanol (1) and water (2):

Example

132

132

131

131

molcm 068.81 molcm 765.17

molcm 40.727 molcm 632.38

VV

VV


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mol 272.58) 246.83)(7.0(

mol 974.24) 246.83)(3.0(

246.83 025.24

2000

025.24

)765.17)(7.0()632.38)(3.0(

22

11

313

3

13

2211

nxn

nxn

cmmolcm

cm

V

Vn

molcmV

VxVxV

t

Example Cont.


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SolutionThe line drawn tangent to the V-x1 curve at x1=0.30, illustrates the values of V1=40.272 cm3 mol-1 and V2=18.068 cm3 mol-1.

32

132

31

131

1053

) 068.18)( 272.58(

1017

) 727.40)( 497.24(

cmV

molcmmolV

cmV

molcmmolV

t

t

t

t

Example Cont.


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We show now how partial properties are related to one another. By Eq. (11.8), μi ≡ Gi, and Eq. (11.20 may be written:

) . (dnG(nS)dT(nV)dPd(nG) ii 1711

Application of the criterion of exactness, Eq. (6.12) , yields the Maxwell relation,

). ( P

S -

T

V

T,nP,n

166

Relations among Partial Properties

jj P,T,niT,n

i

P,T,niP,n

i

n

(nV)

P

G and

n

(nS) -

T

G


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One can write the RHS in the form of partial molar, and change the composition from n to x.


T,x

ii

T,x

i

P,x

ii

P,x

i

PV

P

G

T S

T

G

Every equation that provides a linear relation among thermodynamic properties of a

constant-composition solution has as its counterpart an equation connecting the

corresponding partial properties of each species in the solution.


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P(nV) nU , nH moles n For

PVU H

dTSdPVGd

dTT

GdP

P

GGd

P

nVnUnH

iii

xP

i

xT

ii

iii

,,

iii VUH

)

P))

jjj nT,P,nT,P,nT,P,n

(

n

(

n

(


This may be compared with Eq. (6.10). These examples illustrate the parallelism that exists

between equations for a constant composition solution and the corresponding equations for

the partial properties of the species in solution. We can therefore write simply by analogy

many equations that related partial properties.


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Example

212121 2040600400 xxxxxxH

12 1 xx 3

11 20180600 xxH

21

1

60180 xdx

dH

121 dx

dHxHH

31

211 4060420 xxH

112 dx

dHxHH

312 40600 xH

4201 H 6402 H


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Dalton Law: Every gas has the same V and T.

iit

t

iitii

PP

n

nyPyP where



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Application of partial properties to molar volume

jji

niP,T,niP,T,ni

igiig

i

nnn

n

n

P

RT

n

(nRT)/P

n

)(nVV

jjj

P

RTVVV igig

iig

i

partial molar volume = pure species molar volume= mixture molar volume

Note: Partial pressure of species i (It is not partial molar property)

Molar Volume and Partial Molar Volume


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Properties of each component species are independent of the presence of other species.

A partial molar property (other than volume) of a constituent species in an ideal-gas mixture is equal to the corresponding molar property of the species as a pure ideal gas at the mixture temperature but at a pressure equal to its partial pressure in the mixture.



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Partial molar entropy


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Partial molar Gibbs energy


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Property Change of Mixing


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y

yR SyS

S

/P p y yy -RSyS

HyH

UyU

HyH

) (T,pM(T,P)M

iii

igi

ii

ig

igi

iiii

iigi

ii

ig

igi

ii

ig

igi

ii

ig

igi

ii

ig

iigi

igi

1ln

is mixing gas idealan of changeentropy The

P. and T mixture at the valuespecies-pure theis

(11.25))(where ln

0

(11.23)

(11.23)

(11.21)

Summary


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).( P y yRT (T) ΓyG

). ( P yRT (T) Γμ

(T) Γ

). ( P RT (T) ΓG

PRTd dPP

RTdPV dG

).( yRT GGμ

ii

iii

iig

iiigi

i

iigi

igi

igi

iigi

igi

igi

2911ln

2811ln

constant n integratio theis

2711ln

ln From

2611ln

Summary


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What is the change in entropy when 0.7 m3 of CO2 and 0.3 m3 of N2 each at 1 bar and 25 C blend to form a gas mixture at the same condition? Assume ideal gases.

Example

1-1-

1-1-

iii

igi

ii

igmixing

K mol) (g J

K mol) (g J 8.314

y

yR SySS

is mixing gas ideal an of changeentropy The

079.5

)7.0

1ln7.0

3.0

1ln3.0(

1ln

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