CHE553 Chemical Engineering Thermodynamics

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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.

FUNDAMENTAL PROPERTY RELATION

The most important property relation is that of Gibbs free energy change.
Gibbs energy:
G H TS
(6.3)
Multiplied by n and differentiated eq. (6.3):
(6.3a)
d nG d nH Td nS nS dT
Enthalpy:
(2.11)
H U PV
Multiplied by n, differentiated
d nH d nU Pd nV nV dP
The first Tds relation or Gibbs equation:
d nU Td nS Pd nV

(2.11a)

(6.1)

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

d nH Td nS nV dP

(6.4)

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

d nG nV dP nS dT

(6.6)

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(6.6)
d nG nV dP nS dT
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
nG
P nV

T ,n

nG
T nS

P ,n

and

Total differential of nG is
nG
nG
nG
d nG
dni
dP T dT n
i
P T ,n

P ,n
i P ,T ,nj

(B)

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

(A)

nG

ni P ,T ,nj

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.
nG g P , T , n1, n2 ,..., ni ,...

(11.1)

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

d nG nV dP nS dT i dni

(11.2)

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

where ni is the number of moles of species i.

5

THE CHEMICAL POTENTIAL AND PHASE

EQUILIBRIA

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

dG VdP SdT i dx i
i

(11.3)

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:

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

G G P , T , x1, x 2 ,..., x i ,...

d nG nV dP nS dT i dni

For special case of a constant composition solution:

dG VdP SdT

d nG nV dP nS dT i dni

(6.10)

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
G
G
S
V
(11.5)

(11.4)
T P , x
P T , x
Other solution properties come from definitions; e.g., the enthalpy, from
H = G + TS. Thus, by eq. (11.5),
G
H G T

T P , x

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.

d nG d nG nV nV dP nS nS dT i dni i dni
i

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d nG d nG nV nV dP nS nS dT i dni i dni
i

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,

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

nM nM nM

i i

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

For multiple phases ( phases):

the sum is

d nG nV dP nS dT i dni i dni
i

i i ... i

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

i dni i dni 0
i

d nG nV dP nS dT

and

(6.6)

(11.6)

A chemical species is transported from a phase of larger potential to a

phase of lower potential.

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.

dn 0
i

i 1,2,..., N

Multiple phases at the same T and P are in equilibrium when the chemical
potential of each species is the same in all phases.

The changes dni and dni result from mass transfer between the phases; mass
conservation therefore requires

dni dni

i 1,2,..., N

PARTIAL PROPERTIES

Example

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 M i of a species i in a solution is define as

_
nM
Mi

ni P ,T ,nj

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 cm 3.
* 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.

(11.7)

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.

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
Partial properties
Pure species properties

10

_
nV
Vi

ni P ,T ,nj

M,
for example: V, U, H, S, G

M i , for example: Vi , U i , Hi , Si , G i
Mi , for example: Vi , Ui , Hi , Si , Gi
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EQUATIONS RELATING MOLAR AND

PARTIAL MOLAR PROPERTIES

Because ni = xin,
Moreover,

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:

d nM ndM Mdn

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

nM T , P , n1, n2 ,..., ni ,...

M
M dT M x dn ndx
ndM Mdn n
dP n

i i i
i
P T , x
T P , x

The total differential of nM is

nM
nM
nM
d nM
dP T dT n dni
P T ,n

P ,n
i
i
P ,T ,nj

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

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:

M
M
d nM n
dP n T dT Mi dni
P T , x

P , x
i

dni x i dn ndx i

M dP M dT M dx n M x M dn 0

i i i i i i
dM
P T , x
T P , x

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

(11.9)

M
M dT M dx
dM
dP

i i i

T , x
T P , x

where subscript x denotes differentiation at constant composition.

(11.10) M x i Mi
i

(11.11)

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14

A RATIONALE FOR PARTIAL PROPERTIES

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

nM ni Mi

(11.12)

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 M i approach

the pure species property Mi.

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:

dM x i d Mi Mi dx i
i

lim M lim Mi Mi

xi 1

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

equation:
M dP M dT x d M 0

i i i
P T , x
T P , x

x i 1

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,

(11.13)

lim Mi Mi

For changes at constant T and P,

x i 0

x dM 0
i

(11.14)
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PARTIAL PROPERTIES IN BINARY SOLUTIONS

Equations for partial properties can be summarized as follows:
Definition:

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

_
nM
Mi

ni P ,T ,nj

Summability:
M x i Mi

(11.11)

which yields total properties from partial properties.

x1d M1 x 2 d M2 0

Gibbs/Duhem:

M
M dT

dP

P T , x
T P , x

(B)

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

dM x1d M1 M1 dx1 x 2 d M2 M2 dx 2

(A)

Differentiation of eq. (A) becomes

which yields partial properties from total properties.

xdM

M x1 M1 x 2 M2

(11.7)

(C)

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

(11.13)

dM
M1 M2
dx1

which shows that the partial properties of species making up solution

are not independent of one another.

(D)

17

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

Two equivalent forms of eq. (A) result from elimination separately of x 1 and
x 2:
x1 1 x 2
x 2 1 x1

M 1 x 2 M1 x 2 M2

M M1 x 2 M1 M2

x1

M x1 M1 1 x1 M2

M M1 x 2 M1 x 2 M2

18

M x1 M1 M2 x1 M2
M x1 M1 M2 M2

dM
dx1

(11.15)

dM2
x dM1
1
dx1
x2 dx1

dM
dx1

(11.16)

In combination with eq. (D) becomes

M2 M x1

(E)

dM1
x dM2
2
dx1
x1 dx1

and

M1 M x2

d M1
d M2
x2
0
dx1
dx1

(F)

(G)

When M1 and M2 are plotted vs. x1, the slopes must be of opposite sign.

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.
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EXAMPLE 11.2
Moreover,

d M1
lim
0
x1 1 dx
1
Similarly,

d M2
0
x 2 1 dx
1
lim

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

d M2
Provided lim
is finite
x1 1 dx

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:
dM M I2
dM

and
I1 I2
dx1
x1
dx1

Solution:

d M1
Provided lim
is finite
x

1
2

dx1

The first equation is solved for I2; it combines with

the second to give I1.

Thus, plot of M1 and M2 vs. x1 become horizontal as each species

approaches purity.

I2 M x1

dM
dx1

and

I1 M 1 x1

dM
dx1

Comparison of these expression with eqs. (11.16)

and (11.15) show that

I1 M1

and

I2 M2

The tangent intercepts give directly the values of the

two partial properties.
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EXAMPLE 11.3
The need arises in a laboratory for 2000 cm 3 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

The limiting values are indicated by

Fig. 11.1 (b).
For the tangent line drawn at x1 = 0
(pure species 2), M2 M2 and at the

opposite intercept, M M
1

Methanol 1 : V1 38.632 cm3 mol-1

For the tangent line drawn at x1 = 1

(pure species 1), M1 M1 and at the

opposite intercept, M M
2

Water 2 : V2 17.765 cm mol

3

-1

V1 40.727 cm3 mol-1

V2 18.068 cm3 mol-1

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V x i Vi
i

V x1 V1 x 2 V2
0.338.632 0.7 17.765
24.025 cm3 mol-1

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

83.246 mol
V 24.025
Of this, 30% is methanol, and 70% is water:

ni x i n

V1 and V2

n1 0.3 83.246 24.974 mol

n2 0.7 83.246 58.272 mol

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

For the tangent line

drawn at x1 = 0 (pure species 2), V2 V2 and at the opposite

intercept, V1 V1

For the tangent

line
drawn at x1 = 1 (pure species 1), V1 V1 and at the opposite

intercept, V2 V2

V1t 24.974 40.727 1017 cm3

V2t 58.272 18.068 1053 cm3
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EXAMPLE 11.4

26

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

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

represented by the equation

H 600 180 x1 20 x13

(A)

dH
180 60 x12
dx1

H 400 x1 600 x 2 x1x 2 40 x1 20 x 2

Values of V1, V2 and V 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

V1 and V2 may be obtained.

The curve V1 becomes

horizontal at
x1 = 1 and the

curve for V2 becomes

horizontal at x1= 0 or x2 = 1.

The curves for V1 and V2

appear to be horizontal at both
ends.

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:

where H is in Jmol-1. Determine expressions for H1 and H2 as a functions of x 1,

numerical values for the pure species enthalpies H 1 and
H2, and
numerical

values for the partial enthalpies at infinite dilution H1 and H2

By equation (11.15),

H1 H x 2

dH
dx1

H1 600 180 x1 20 x13 180 x 2 60 x12 x 2

Then,
Replace x2 by 1 x1 and simplify:

H1 420 60 x12 40 x13

By eq. (11.16),

or
27

H2 H x1

H2 600 40 x13

(B)

dH
600 180 x1 20 x13 180 x1 60 x13
dx1

(C)
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RELATIONS AMONG PARTIAL PROPERTIES

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.

The infinite dilutionH1 and H2 are found from eq. (B) and (C) when x 1 = 0 in
eq. (B) and x1 = 1 in eq. (C). The results are:

-1

H = 420 Jmol

and

By eq. (11.8),

i G i
and eq. (11.2) may be written as

H2 is found from either eq. (A) or (C) when x1 = 0.

The result is H2 = 600 J mol-1.

(11.8)

d nG nV dP nS dT i dni

d nG nV dP nS dT G i dni
i

-1

H = 640 Jmol

(11.2)

(11.17)

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

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.

dz Mdx Ndy

M N

y x x y

(6.12)

V S


T P ,n
P T ,n

(6.16)

yields the Maxwell relation,

29

An example is based on the equation that defines enthalpy:

H = U + PV
For n moles,

Plus the two additional equations:

G
nV
i
P
ni P ,T ,n
j

T ,n

G
nS
i

T
ni P ,T ,n j

P ,n

nH nU P nV

Differentiation with respect to ni at constant T, P, and nj yields

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:
G
i
P

Vi

T , x

(11.18)

G
i
T

Si

P , x

30

nH
nU
nV

ni P ,T ,n j ni P ,T ,n j
ni P ,T ,n j
By eq. (11.7) this becomes

(11.19)

Similar to eq. (2.11)

HU+PV

Hi Ui P Vi

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

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).

d Gi i
P

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.

G
dP i

T
T , x

dT

P , x

By eqs. (11.18) and (11.19),

Similar to eq. (6.10)

dG=VdP-SdT
These examples illustrate the parallelism that exists between equations for a constantcomposition solution and the corresponding equations for partial properties of the
species in solution.

d G i Vi dP Si dT

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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
PREPARED BY:
MDM. NORASMAH MOHAMMED MANSHOR
FACULTY OF CHEMICAL ENGINEERING,
UiTM SHAH ALAM.
norasmah@salam.uitm.edu.my
03-55436333/019-2368303

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