solution thermo part 1

CHE553 Chemical Engineering Thermodynamics 3/30/2015

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OVERVIEW

The purpose of this chapter is to express the theoretical foundation for applications 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 chemical reaction. Thus compositions become essential variables, along with temperature and pressure.

This chapter introduce new property, i.e. chemical potential which facilitate treatment of phase and chemical reaction equilibria; partial properties which are properties of individual species as they exist in solution; and fugacity which provide treatment for real gas mixtures through mathematical formulation.

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

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FUNDAMENTAL PROPERTY RELATION

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 Tds relation 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 PV (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)


2

Equation (6.6) relates the total Gibbs energy of any closed system to

temperature and pressure.

An appropriate application is to a single phase fluid in a closed system

wherein no chemical reactions occur. For such a system the composition is

necessarily constant, and therefore

The subscript n indicates that the numbers of moles of all chemical species are

held constant.

For more general case of a single phase, open system, material may pass

into and out of the system, and nG becomes a function of the numbers of

moles of the chemical species present, and still a function of T and P.

where ni is the number of moles of species i.

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, ,T n P n

nG nGnV and nS

P T

(A)

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

d nG nV dP nS dT (6.6) Total differential of nG is

The summation is over all species present, and subscript nj indicates that all mole numbers except the ith are held constant.

The derivative in the final term is called the chemical potential of species i in 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 fluid systems 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 energy to T, P and {xi}.

For special case of a constant composition solution:

Although the mole numbers ni of eq. (11.2) are independent variables, the mole fractions xi in eq. (11.3) are not, because ixi = 1. Eq. (11.3) does imply

Other solution properties come from definitions; e.g., the enthalpy, from H = 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

THE CHEMICAL POTENTIAL AND PHASE EQUILIBRIA

For a closed system consisting of two phases in equilibrium, each individual phase is open to 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 equilibrium implies 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 nS dT dn

i ii

d nG nV dP nS dT dn

i i i ii i

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


3

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

the sum is

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

The changes dni and dni

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

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nM nM nM

i i i ii i

d nG nV dP nS dT dn dn

0i i i ii i

dn dn

0i i i i idn dn and dn

d nG nV dP nS dT (6.6)

i i i ii i

d nG d nG nV nV dP nS nS dT dn dn Quantities dni

are independent; therefore the only way the left side of the second equation can in general be zero is for each term in parentheses separately 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 chemical potential of ice is larger than water, the ice melts. When chemical potential of water is larger than ice, the water freezes. Water and ice are in equilibrium when their chemical potential are the same.

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

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

Multiple phases at the same T and P are in equilibrium when the chemical

potential of each species is the same in 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 it, i.e. pure component exhibits pure properties.

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

Partial molar property of a species i in a solution is define as

It is the change of total property nM to the addition of a differential amount of species i to a finite amount of solution at constant T and 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

PARTIAL PROPERTIES

iM

11

_

, ,

i

i P T nj

nMM

n

(11.7)

iM

, , , ,i i i i iV U H S G

Example

When one mole of water is added to a large volume of water at 25C, the volume increases by 18 cm3.

However, addition of one mole of water to a large volume of pure ethanol results in an increase in volume of only 14 cm3.

* 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.

In general, the partial molar volume of a substance i in a mixture is the

change in volume per mole of i added to the mixture.

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_

, ,

i

i P T nj

nVV

n


4

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 property M:

The total differential of nM is

where subscript n indicates that all mole numbers are held constant, and subscript

nj that all mole numbers except ni are held constant. Because the first two partial derivatives on the right are evaluated at constant n and because the partial derivative of the last term is given by eq. (11.7), this equation has the simpler form:

where subscript x denotes differentiation at constant composition.

EQUATIONS RELATING MOLAR AND PARTIAL MOLAR PROPERTIES

, ,

i i

iT x P x

M Md nM n dP n dT Mdn

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)

Because ni = xin,

Moreover,

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

The terms containing n are collected and separated from those containing dn to yield

The left side of this equation can be zero if each term in brackets be zero too. 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 iT 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 Mdx

P T

(11.10) i i

i

M x M

(11.11)

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

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

Differentiate eq. (11.11) yields:

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

For changes at constant T and P,

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i i

i

nM n M

(11.12)

, ,

0i iT x P x i

M MdP dT x d M

P T

0i ii

x dM

ii i i

i i

dM x dM M dx

(11.13)

(11.14)

A RATIONALE FOR PARTIAL PROPERTIES

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

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

In the limit as a solution becomes pure in species i, both M and approach the pure species property Mi.

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

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iM

1 1lim limi i

i ix xM M M

0limi

i ixM M


5

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 xi

M Mx dM dP dT

P T

(11.13)

(11.7)

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

Differentiation of eq. (A) becomes

When M is known as a function of x1 at constant T and P, the appropriate form of the Gibbs/Duhem equation is eq. (11.14), expressed as

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

PARTIAL PROPERTIES IN BINARY SOLUTIONS

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)

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

Two equivalent forms of eq. (A) result from elimination separately of x1 and x2:

In combination with eq. (D) becomes

Thus for binary systems, the partial properties are calculated directly from an expression for the solution property as a function of composition at constant T and P.

2 1

1

dMM M x

dx

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1 2

1

dMM M x

dx

(11.15)

(11.16)

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

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

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1 21 2

1 1

0dM dMx 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 2 and M M


6

Moreover,

Similarly,

Thus, plot of vs. x1 become horizontal as each species approaches purity.

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1 1

1 2

1 11 1

lim 0 Provided lim is finitex x

d M dM

dx dx

2 2

2 1

1 11 1

lim 0 Provided lim is finitex x

d M dM

dx dx

1 2 and M M

EXAMPLE 11.2

Figure 11.1 (a) shows a representative plot of M vs. x1 for a binary system. The tangent line shown extend across the figure, intersecting the edges (at x1 = 1 and x1 = 0) at points label I1 and I2. Two equivalent expressions can be written for the slope of this tangent line: The first equation is solved for I2; it combines with the second to give I1. Comparison of these expression with eqs. (11.16) and (11.15) show that The tangent intercepts give directly the values of the two partial properties.

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 xdx dx

1 21 2andI M I M

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

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

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

1 1M M

2 2M M

2 2M M

1 1M M

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EXAMPLE 11.3

The need arises in a laboratory for 2000 cm3 of an antifreeze solution consisting of 30 mole % methanol in water. What volumes of pure methanol and of pure water at 25oC (298.15K) must be mixed to form the 2000 cm3 of antifreeze, also at 25oC (298.15K)? Partial molar volumes for methanol and water in a 30 mole % methanol solution and their pure species 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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i i

i

V x V

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 number of moles required is

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

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

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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

tVnV

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 1053 cm

t

t

V

V

i in x n

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

intercept,

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

intercept,

1 2and V V

Values of for the binary solution methanol(1)/water(2) at 25oC (298.15K) are plotted in Fig. 11.2 as functions of x1.

The line drawn tangent to the V vs x1 curve at x1 = 0.3 illustrates the graphical procedure by which values of

may be obtained.

The curve becomes horizontal at x1 = 1 and the curve for becomes horizontal at x1= 0 or x2 = 1.

The curves for appear to be horizontal at both ends.

1 2, and V V V

1 2 and V V

1V

1 2 and V V

2V

1 1V V

2 2V V

2 2V V

1 1V V

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EXAMPLE 11.4

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

where H is in Jmol-1. Determine expressions for as a functions of x1, numerical values for the pure species enthalpies H1 and H2, and numerical 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 2 and H H

Solution:

Replace x2 by 1 x1 in the given equation for H and simplify:

By equation (11.15),

Then,

Replace x2 by 1 x1 and simplify:

By eq. (11.16),

or

2 3

1 1 1420 60 40H x x

3 3

2 1 1 1 1 1

1

600 180 20 180 60dH

H H x x x x xdx

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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)

3

2 1600 40H x

(C)


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1 2 and H H

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

-1.

H2 is 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 = 0 in 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 of the Gibbs/Duhem equation.

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--1 -1

1 2420 Jmol and H = 640 JmolH

-

=

RELATIONS AMONG PARTIAL PROPERTIES

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)

dz Mdx Ndy

Plus the two additional equations:

where subscript n indicates constancy of all ni, and subscript nj indicates that 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 of eqs. (11.4) and (11.5).

,

ii

T x

GV

P

31

, ,, j

i

i P T nT n

nVG

P n

(11.18) (11.19)

,

ii

P x

GS

T

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.

, ,, 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 ni at constant T, P, and nj yields

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 a constant-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 nVP

n n n

i i iH U PV

iG

, ,

i ii

T x P x

G GdG dP dT

P T

i i idG V dP S dT

Similar to eq. (2.11)

HU+PV

Similar to eq. (6.10)

dG=VdP-SdT


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REFERENCES

Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to Chemical Engineering 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] 03-55436333/019-2368303

solution thermo part 1

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