Solution Thermo(1)

8/10/2019 Solution Thermo(1)

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CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 3/29/20

SOLUTION THERMODYNAMICS:THEORY

The purpose of this chapter is to express the theoretical foundation forapplications of thermodynamics to gas mixtures and liquid solutions.

Separation processes of multicomponent gases and liquids in chemical,petroleum and pharmaceutical industries commonly undergo composition

changes, transfer of species from one phase to another and chemicalreaction. Thus compositions become essential variables, along withtemperature and pressure.

This chapter introduce new property, i.e. chemical potential which facilitatetreatment of phase and chemical reaction equilibria; partial properties whichare properties of individual species as they exist i n solution; and fugacitywhich provide treatment for real gas mixtures through mathematicalformulation.

Another solution properties known as excess properties, is the deviationfrom ideal solution property.

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FUNDAMENTAL PROPERTY RELATION

THE CHEMICAL POTENTIAL AND PHASE

EQUILIBRIA

PARTIAL PROPERTIES

3

The most important property relation is that of Gibbs free energy change.

Gibbs energy:

Multiplied by n and differentiated eq. (6.3):

Enthalpy:

Multiplied by n, differentiated

The first Tdsrelation or Gibbs equation:

Combine eq. (2.11a) and (6.1):

Combine eq. (6.3a) and (6.4) to yield:

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d nU Td nS Pd nV (6.1)

G H TS (6.3)

d nG d nH Td nS nS dT

H U P V (2.11)

d nH Td nS nV dP (6.4)

d nG nV dP nS dT (6.6)

d nH d nU Pd nV nV dP

(6.3a)

(2.11a)


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Equation (6.6) relates the total Gibbs energy of any closed systemtotemperatureand pressure.

An appropriate application is to a single phase fluidin a closed systemwherein no chemical reactions occur. For such a system the compositionis necessarily constant, and therefore

The subscript nindicates that the numbers of moles of all chemical speciesare held constant.

For more general case of a single phase, open system, material may passinto and out of the system, and nGbecomes a function of the numbers ofmoles of the chemical species present, and still a function of Tand P.

where niis the number of moles of species i.

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, ,T n P n

nG nG nV and nSP T

(A)

1 2, , , , .. ., , .. .inG g P T n n n

Total differential of nGis

The summation is over all species present, and subscript njindicates thatall mole numbers except the ith are held constant.

The derivative in the final term is called the chemical potential of species iin the mixture. It is define as

With this definition and with the first two partial derivatives [eqn. (A)]replaced by (nV) and (nS), the preceding equation [eqn. (B)] becomes

Equation (11.2) is the fundamental property relation for single phase fluidsystems of variable mass and composition.

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, ,

i

i P T nj

nG

n

(11.1)

, , , ,

i

i iT n P n P T nj

nG nG nGd nG dP dT dn

P T n

(B)

i ii

d nG nV dP nS dT dn (11.2)

For special case of one mole of solution, n = 1 and ni= xi:

This equation relates molar Gibbs energyto T, P and {xi}.

For special case of a constant composition solution:

Although the mole numbers niof eq. (11.2) are independent variables, themole fractionsxiin eq. (11.3) are not, because ixi= 1. Eq. (11.3) does

imply

Other solution properties come from definitions; e.g., the enthalpy, fromH = G + TS. Thus, by eq. (11.5),

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i i

i

dG VdP SdT dx (11.3)

dG VdP SdT (6.10)

1 2, , , , .. ., , .. .iG G P T x x x

,P x

GS

T

,T x

GV

P

(11.5)(11.4)

,P x

GH G T

T

For a closed system consisting of two phases in equilibrium, each individualphase is opento the other, and mass transfer between phases may occur.Equation (11.2) applies separately to each phase:

where superscripts and identify the phases. The presumption here is that equilibriumimplies uniformity of T and P

throughout the entire system. The change in the total Gibbs energy of the two phase system is the sum

of these equations.

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i ii

d nG nV dP n S dT dn

i ii

d nG nV dP nS dT dn

i i i i i i

d nG d nG nV nV dP nS nS dT dn dn


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When each total system property is expressed by an equation of the form,

the sumis

Because the two phase system is closed, eq. (6.6)is also valid. Comparisonof the two equations shows that at equilibrium,

The changes dniand dni

result from mass transfer between the phases;mass conservation therefore requires

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nM nM nM

i i i i i i

d nG nV dP nS dT dn dn

0i i i i i i

dn dn

0i i i i i dn dn and dn

d nG nV dP nS dT (6.6)

Quantities dniare independent; therefore the only way the left side of the

second equation can in general bezerois for each term in parenthesesseparately to be zero. Hence,

where N is the number of species present in the system.

For multiple phases (phases):

Example: A glass of liquid water with ice cubes in it. When the chemicalpotential of ice is larger than water, the ice melts (i.e. above 0 oC). Whenchemical potential of water is larger than ice, the water freezes (i.e. below0oC ). Water and ice are in equilibrium when their chemical potential are the

same (i.e. at 0oC).

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1,2,...,i i i N

... 1, 2,...,i i i i N (11.6)

Multiple phasesat the same T and P are inequilibriumwhen the

chemical potential of each species is the samein all phases.

A chemical species is transported from a phase of larger potential

to a phase of lower potential.

A species exhibits its pure property when no other species exist with i t,i.e. pure component exhibits pure properties.

Species exhibits its partial property when it co-exists with other species ina mixture or solution.

Partial molar property of a species iin a solution is define as

It is the change of total property nMto the addition of a differential

amount of species ito a finite amount of solution at constant Tand P. Three kinds of properties used in solution thermodynamics are

distinguished by the following symbolism: Solution properties M, for example: V, U, H, S, G

Partial properties , for example:

Pure species properties Mi , for example: Vi, Ui, Hi, Si, Gi

iM

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_

, ,

i

i P T nj

nMM

n

(11.7)

iM

, , , ,i i i i i

V U H S G

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When one mole of water is added to a large volume of water at 25C, thevolume increases by 18cm3. The molar volume of pure water would thusbe reported as 18cm3mol-1.

However, addition of one mole of water to a large volume of pure ethanolresults in an increase in volume of only 14cm3. The value 14cm3is said tobe the partial molar volume of water in ethanol.

In general, the partial molar volume of a substance iin a mixture is thechange in volume per mole of iadded to the mixture.

* The increase in volume is different due to intermolecular forces between molecules, size

and shape of molecules are different in mixture rather than pure species.

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_

, ,

i

i P T nj

nVV

n


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Total thermodynamic properties of a homogeneous phase are functions of T, P,and the numbers of moles of the individual species which comprise the phase.Thus, for propertyM:

The total differential of nMis

where subscript nindicates that all mole numbers are h eld constant, andsubscript njthat all mole numbers except niare held constant.

Because the first two partial derivatives on the right are evaluated at constantnand because the partial derivative of the last term is given by eq. (11.7), thisequation has the simpler form:

where subscriptxdenotes differentiation at constant composition.

, ,

i i

iT x P x

M Md nM n dP n dT M dn

P T

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1 2, , , , .. ., , .. .inM T P n n n

, , , ,

i

i iT n P n P T nj

nM nM nMd nM dP dT dn

P T n

(11.9)

13 Because ni= xin,

Moreover,

When dniand d(nM)are replaced in Eq. (11.9), it becomes

The terms containing nare collected and separated from those containingdnto yield

The left side of this equation can be zero if each term in brackets be zerotoo. Therefore,

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i i idn x dn ndx

d nM ndM Mdn

, ,

i i i

T x P x i

M MndM Mdn n dP n dT M x dn ndx

P T

, ,

0i i i i T x P x i i

M MdM dP dT M dx n M x M dn

P T

, ,

i i

T x P x i

M MdM dP dT M dx

P T

(11.10) i ii

M x M

(11.11)

Multiplication of eq.(11.11) by nyields the alternative expression:

Equations (11.11) and (11.12) are known as summability relations, theyallow calculation of mixture properties from partial properties.

Differentiate eq. (11.11) yields:

Comparison of this equation with eq. (11.10), yields the Gibbs/Duhemequation:

For changes at constant T and P,

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i i

i

nM n M

(11.12)

, ,0i i

T x P x i

M MdP dT x d MP T

0i ii

x d M

ii i i

i i

dM x d M M dx

(11.13)

(11.14)

Partial properties have all characteristics of properties of individual speciesas they exist in solution. Thus, they may be assigned as property values tothe individual species.

Partial properties, like solution properties, are functions of composition.

In the limit as a solution becomes pure in species i, bothMandapproach the pure species propertyMi.

For a species that approaches its infinite dilution limit, i.e., the values as itsmole fraction approaches zero, no general statements can be made. Valuescome from experiment or from models of solution behavior. By definition,

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iM

1 1lim lim

i ii i

x xM M M

0lim

ii i

xM M


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Equations for partial properties can be summarized as follows:

Definition:

which yields partial properties from total properties.

Summability:

which yields total properties from partial properties.

Gibbs/Duhem:

which shows that the partial properties of species making up solution

are not independent of one another.

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_

, ,

i

i P T nj

nMM

n

i i

i

M x M (11.11)

, ,

i i

T x P x i

M Mx d M dP dT

P T

(11.13)

(11.7) For binary solution, the summability relation, eq.(11.11) becomes

Differentiation of eq. (A)becomes

When Mis known as a function of x1at constant T and P, the appropriateform of the Gibbs/Duhem equation is eq. (11.14), expressed as

Because x1+ x2= 1, dx1 + dx2 = 0dx1= - dx2. Substitute eq. (C) into eq.(B) to eliminate dx2gives

1 1 2 2 0x d M x d M

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1 1 2 2M x M x M

(A)

1 1 1 1 2 2 2 2dM x d M M dx x d M M dx

(B)

(C)

1 2

1

dMM M

dx

(D)

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Two equivalent forms of eq. (A) result from elimination separately of x1and x2:

In combination with eq. (D) becomes

Thus for binary systems, the partial properties are calculated directly froman expression for the solution property as a function of composition atconstant T and P.

2 1

1

dMM M x

dx

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1 2 2 1

2 1 2 2 1 1 1 2

1 2 1 2 2 1 1 2 1 2

1 2 1 2 1 1 2 2

1 1

1 1

and

x x x x

M x M x M M x M x M

M M x M x M M x M M x M

M M x M M M x M M M

1 2

1

dMM M x

dx

(11.15)

(11.16)

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Eq. (C), the Gibbs/Duhem equation, may be written in derivative forms:

When are plotted vs. x1, theslopes must be of opposite sign.

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1 21 2

1 1

0d M d M

x xdx dx

1 2 2

1 1 1

dM x dM

dx x dx

(E)

(F)

2 1 1

1 2 1

dM x dM

dx x dx

(G)

1 2andM M


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Moreover,

Similarly,

Thus, plot of vs. x1become horizontalas each speciesapproaches purity.

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1 1

1 2

1 11 1

lim 0 Provided lim is finitex x

d M d M

dx dx

2 2

2 1

1 11 1

lim 0 Provided lim is finitex x

d M d M

dx dx

1 2andM M

Figure 11.1 (a) shows a representative plot ofMvs. x1for a binary system.

The tangent line shown extend across the figure,intersecting the edges (at x1= 1 and x1 = 0) atpoints label I1and I2.

Two equivalent expressions can be written forthe slopeof this tangent line:

The first equation is solved for I2; it combineswith the second to give I1.

Comparisonof these expression with eqs.(11.16) and (11.15)show that

The tangent intercepts give directly the values ofthe two partial properties. 22

Describe a graphical interpretation of eqs. (11.15) and (11.16).

Solution:

21 2

1 1 1

anddM M I dM

I Idx x dx

2 1 1 11 1

and 1dM dM

I M x I M x dx dx

1 21 2andI M I M

The limiting values are indicated by Fig.11.1 (b).

For the tangent line drawn at x1= 0(pure species 2), and at theopposite intercept,

For the tangent line drawn at x1= 1(pure species 1), and at the

opposite intercept,

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1 1M M

2 2M M

2 2M M

1 1M M

The need arises in a laboratory for 2000 cm3of an antifreeze solutionconsisting of 30 mole % methanol in water. What volumes of puremethanol and of pure water at 25oC (298.15K) must be mixed to form the2000 cm3of antifreeze, also at 25oC (298.15K)? Partial molar volumes formethanol and water in a 30 mole % methanol solution and their purespecies molar volumes, both at 25oC (298.15K), are

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3 -1 3 -1

1 1

3 -1 3 -1

2 2

Methanol 1 : 38.632 cm mol 40.727 cm mol

Water 2 : 17.765 cm mol 18.068 cm mol

V V

V V


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Solution:

Equation (11.11) is written for the molar volume of the binary antifreeze solution,and known values are substituted for the mole fractions and partial volumes:

Because the required total volume of solution is Vt= 2000 cm3, the total numberof moles required is

Of this, 30% is methanol, and 70% is water:

The total volume of each pure species is Vit= niVi; thus,

i ii

V x V

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1 1 2 2

3 -1

0.3 38.632 0.7 17.765

24.025 cm mol

V x V x V

200083.246 mol

24.025

tV

nV

25

1

2

0.3 83.246 24.974 mol

0.7 83.246 58.272 mol

n

n

3

1

3

2

24.974 40.727 1017 cm

58.272 18.068 1053cm

t

t

V

V

i in x n

1 2andV V

Values of for thebinary solutionmethanol(1)/water(2) at 25oC(298.15K) are plotted in Fig.11.2 as functions of x1.

The line drawn tangent to theV vs x1curve at x1= 0.3illustrates the graphicalprocedure by which values of

may be obtained. The curve becomes

horizontal at x1= 1 and thecurve for becomeshorizontal at x1= 0 or x2= 1.

The curves forappear to be horizontal atboth ends.

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1 2, andV V V

1 2andV V

1V

1 2andV V

2V

For the tangent line drawn at x1= 0 (pure species 2), and at the oppositeintercept,For the tangent line drawn at x1= 1 (pure species 1), and at the oppositeintercept,

1 1V V

2 2V V

2 2V V

1 1V V

The enthalpy of a binary liquid system of species 1 and 2 at fixed T and P isrepresented by the equation

where H is in Jmol-1. Determine expressions for as a functionsof x1, numerical values for the pure species enthalpies H1and H2, andnumerical values for the partial enthalpies at infinite dilution

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1 2 1 2 1 2400 600 40 20H x x x x x x

1 2andH H

1 2andH H

Solution:

Replace x2by 1x1in the given equation for H and simplify:

By equation (11.15),

Then,

Replace x2by 1x1and simplify:

By eq. (11.16),

or

2 31 1 1420 60 40H x x

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3 3

2 1 1 1 1 1

1

600 180 20 180 60dH

H H x x x x x dx

3

1 1600 180 20H x x (A)

2

1

1

180 60dH

xdx

1 2

1

dHH H x

dx

3 2

1 1 1 2 1 2600 180 20 180 60H x x x x x

(B)

32 1600 40H x

(C)

28


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A numerical value for H1results by substitution of x1= 1 in either eq (A) or(B). Both eqn. yield H1= 400 J mol

-1.

H2is found from either eq. (A) or (C) when x1= 0.

The result is H2= 600 J mol-1.

The infinite dilution are found from eq. (B) and (C) when x1= 0in eq. (B) and x1= 1 in eq. (C). The results are:

Exercise: Show that the partial properties as given by eqs. (B) and (C)combine by summability to give eq. (A), and conform to all requirements ofthe Gibbs/Duhem equation.

1 2andH H

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--1 -1

1 2420 Jmol and H = 640 JmolH -

=

29

By eq. (11.8),

and eq. (11.2) may be written as

Application of the criterion of exactness, eq. (6.12),

yields the Maxwell relation,

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ii G

i ii

d nG nV dP nS dT G dn

(11.17)

yx

M N

y x

(6.12)

, ,P n T n

V S

T P

(6.16)

i ii

d nG nV dP nS dT dn

(11.2)

(11.8)

d z Md x Nd y

Plus the two additional equations:

where subscript nindicates constancy of all ni, and subscript njindicatesthat all mole numbers except the ith are held constant.

In view of eq. (11.7), the last two equations are most simply expressed:

These equations allow calculation of the effects of P and T on the partial

Gibbs energy (or chemical potential). They are partial property analogs ofeqs. (11.4) and (11.5).

,

ii

T x

GV

P

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, ,, j

i

i P T nT n

nVG

P n

(11.18) (11.19)

,

ii

P x

GS

T

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Every equation that provides a linearrelation among thermodynamic

properties of a constant-compositionsolution has as its counterpart an

equation connecting the corresponding partial properties of each

species in the solution.

, ,, j

i

i P T nP n

nSG

T n

An example is based on the equation that defines enthalpy:

H = U + PV

For n moles,

Differentiation with respect to niat constant T, P, and njyields

By eq. (11.7) this becomes

In a constant-composition solution, is a function of P and T, and therefore:

By eqs. (11.18) and (11.19),

These examples illustrate the parallelism that exists between equations for aconstant-composition solution and the corresponding equations for partial

properties of the species in solution.

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nH nU P nV

, , , , , ,j j ji i iP T n P T n P T n

nH nU nV P

n n n

i i iH U P V

iG

, ,

i ii

T x P x

G Gd G dP dT P T

i i id G V dP S dT

Similar to eq. (2.11)HU+PV

Similar to eq. (6.10)dG=VdP-SdT


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Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to ChemicalEngineering Thermodynamics. Seventh Edition. Mc Graw-Hill.

http://www.chem1.com/acad/webtext/thermeq/TE4.html

http://mpdc.mae.cornell.edu/Courses/ENGRD221/LECTURES/lec26.pdf

http://science.csustan.edu/perona/4012/partmolvolsalt_lab2010.pdf

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PREPARED BY:MDM. NORASMAH MOHAMMED MANSHOR

FACULTY OF CHEMICAL ENGINEERING,UiTM SHAH ALAM.

[email protected]/019-2368303

Solution Thermo(1)

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