Theory of Solution Thermodynamics

188

Chapter 3: Solution Thermodynamics: Theory

3.0 The Notations for Solution Thermodynamics

M = the properties of the solution

(h, s, v)

iM = partial properties (e.g., ih , is )

of species i in the solution

iM = pure-species properties

(e.g., ig , is )

189

3.1 Fundamental Property Relations

We can re-write the following

relationship (Eq. 2.13)

dg vdP sdT= − (2.13)

for a system with a total number of

moles of n, as follows

( ) ( ) ( )d ng nv dP ns dT= − (3.1)

Since ‘ng’ is f(P, T) (do you know

WHY?), we obtain the following

mathematical implication

( ) ( ) ( ), ,T n P n

ng ngd ng dP dT

P T∂ ∂⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ (3.2)

190

By comparing Eq. 3.1 with Eq. 3.2,

we obtain the following relationships

( ),T n

ngnv

P∂⎛ ⎞

=⎜ ⎟∂⎝ ⎠ (3.3)

and ( ),P n

ngns

T∂⎛ ⎞

= −⎜ ⎟∂⎝ ⎠ (3.4)

where

n = total # of moles or # of moles

of all chemical species in the

solution

= 1 2 3 ... nn n n n+ + + +

(for n species)

Accordingly, ( )= 1 2 3f , , , , , ....ng T P n n n

191

Thus, we can re-write Eq. 3.2 as

follows

( ) ( ) ( )

( ), ,

1 , , j

T n P n

n

ii i P T n

ng ngd ng dP dT

P T

ngdn

n=

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∂⎛ ⎞+ ⎜ ⎟∂⎝ ⎠∑

(3.5)

where jn is all species EXCEPT species i

The term ( ), , ji P T n

ngn

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

is defined as

“CHEMICAL POTENTIAL, μi”

( ), , j

ii P T n

ngn

μ∂⎛ ⎞

= ⎜ ⎟∂⎝ ⎠ (3.6)

192

Combining Eqs. 3.2-3.6 together yields

( ) ( ) ( )1

n

i ii

d ng nv dP ns dT dnμ=

= − +∑ (3.7)

In the case that n = 1 i in x= , Eq. 3.7

can be re-written as follows

1

n

i ii

dg vdP sdT dxμ=

= − +∑ (3.8)

From Eq. 3.8, it implies that

( )= 1 2 3f , , , , , ....g T P x x x

Accordingly, it can be implied

mathematically that

,T x

g vP∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

(3.9)

and ,P x

g sT

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (3.10)

193

This confirms that Gibbs free

energy still serves as a generating

function for other thermodynamic

properties, even in the system where

compositions of all species vary with

time

194

3.2 The Chemical Potential & Phase Equilibria

Consider a closed system where

2 phases co-exist and are in equilibrium

with each other

We can write Eq. 3.7 for each phase

(phases &α β ), as follows

( ) ( ) ( )1

n

i ii

d ng nv dP ns dT dnα α α α αμ=

= − +∑ (3.11)

( ) ( ) ( )1

n

i ii

d ng nv dP ns dT dnβ β β β βμ=

= − +∑ (3.12)

195

A total property of the system are

obtained when the property of each

phase is combined together, as

shown in the following equation

( ) ( )nM nM nMα β= + (3.13)

where M = any system property

Hence, by combining Eq. 3.11 with

Eq. 3.12, we obtain

( ) ( ) ( )1 1

n n

i i i ii i

d ng nv dP ns dT dn dnα α β βμ μ= =

= − + +∑ ∑

(3.14) From Eq. 3.1, we knew that

( ) ( ) ( )d ng nv dP ns dT= − (3.1)

196

Thus, Eq. 3.14 becomes

( ) ( )1 1

n n

i i i ii i

d ng d ng dn dnα α β βμ μ= =

= + +∑ ∑

1 1

0n n

i i i ii i

dn dnα α β βμ μ= =

+ =∑ ∑ (3.15)

Since this is a closed system, a

decrease in the number of moles of

Phase α will result in an increase in

the same number of moles of Phase

β , or vice versa, or it means that

i idn dnα β= − (3.16)

Combining Eq. 3.15 with Eq. 3.16

yields

197

( )α α β αμ μ= =

+ − =∑ ∑1 1

0n n

i i i ii i

dn dn

( )α β αμ μ=

− =∑1

0n

i i ii

dn

and, eventually

i iα βμ μ= (3.17)

This principle can also be

extended to any phases, as follows

....i i i iα β χ πμ μ μ μ= = = =

“Multiple phases at the same T &

P are in equilibrium with one

another when the chemical

potential of each species is the

same for all phases”

198

3.3 Partial Properties

We can extend the principle of

chemical potential (Eq. 3.6) to any other

thermodynamic properties, as follows

( ), , j

ii P T n

nMM

n∂⎛ ⎞

= ⎜ ⎟∂⎝ ⎠ (3.18)

where M = any system property

(e.g., h, s)

This is called “PARTIAL PROPERTY” or

“RESPONSE FUNCTION”, which represents

the change in the total property, nM, due

to the addition of a small (differential)

amount of (moles of) species i, at constant

T & P, to the solution

199

Combining Eqs. 3.6 & 3.18 together

results in

( ), , j

i ii P T n

ngg

nμ

∂⎛ ⎞= = ⎜ ⎟∂⎝ ⎠

(3.19)

This implies that the ‘chemical

potential’ is, in fact, ‘partial Gibbs

free energy’

200

3.3.1 Properties & Partial Properties

We can extend Eq. 3.5 to any other

thermodynamic properties (M) as

follows

( ) ( ) ( )

( ), ,

1 , , j

T n P n

n

ii i P T n

nM nMd nM dP dT

P T

nMdn

n=

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∂⎛ ⎞+ ⎜ ⎟∂⎝ ⎠∑

(3.20)

and when combining Eq. 3.20 with Eq.

3.18, we obtain

( ) ( ) ( )

, ,

1

T n P n

n

i ii

nM nMd nM dP dT

P T

Mdn=

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

+∑

(3.21)

201

Since n = total # of moles of all species

in the solution, it must be constant; thus,

( )

, ,

1

T n P n

n

i ii

M Md nM n dP n dTP T

Mdn=

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

+∑ (3.22)

For a special case, when n = 1, we

obtain the following relationships

, ,T n T x

M MP P

∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

and , ,P n P x

M MT T

∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Thus, Eq. 3.22 can be re-written as

follows

( )

, ,

1

T x P x

n

i ii

M Md nM n dP n dTP T

Mdn=

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

+∑ (3.23)

202

Consider the term 1

n

i ii

Mdn=∑ in Eq. 3.23

Since ii

nxn

= i in x n=

Thus,

( )i i i idn d x n x dn ndx= = +

Accordingly,

( )1 1

n n

i i i i ii i

Mdn M x dn ndx= =

= +∑ ∑ (3.24)

Consider the term ( )d nM in Eq. 3.23

( )d nM ndM Mdn= + (3.25)

Combining Eqs. 3.23-3.25 together

yields

203

( )

, ,

1

T x P x

n

i i ii

M MndM Mdn n dP n dTP T

M x dn ndx=

∂ ∂⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

+ +∑

Rearranging the above equation

yields

, ,

0

i iT x P x

i i

M MdM dP dT Mdx nP T

M x M dn

⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦+ − =⎡ ⎤⎣ ⎦

∑

∑

(3.26) Eq. 3.26 will be true only if

, ,

0i iT x P x

M MdM dP dT MdxP T

⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− − − =⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦∑

and

0i iM x M− =⎡ ⎤⎣ ⎦∑

204

Hence,

, ,

i iT x P x

M MdM dP dT MdxP T

∂ ∂⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∑ (3.27)

i iM x M= ∑ (3.28)

Note that Eq. 3.27 is a special case of

Eq. 3.23 when 1n = and, thus, i in x=

Multiplying Eq. 3.28 with n yields

( )i inM x n M= ∑

i inM n M= ∑ (3.29)

Eq. 3.29 is called “summability relations”

Differentiating Eq. 3.28 gives

i i i idM x dM Mdx= +∑ ∑ (3.30)

205

Combining Eq. 3.27 with Eq. 3.30 yields

, ,

i i i i

i iT x P x

x dM Mdx

M MdP dT MdxP T

+ =

∂ ∂⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑ ∑

∑

Rearranging the above equation gives

, ,

0i iT x P x

M MdP dT x dMP T

∂ ∂⎛ ⎞ ⎛ ⎞+ − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∑

(3.31)

Note that Eq. 3.31 is called a Gibbs/

Duhem equation In the case where T & P are constant, a

Gibbs/Duhem equation (Eq. 3.31) becomes

0i ix dM =∑ (3.32)

206

3.3.2 Partial Properties in Binary Solutions

Consider a solution with 2 components

(in other words, a binary solution) Writing Eq. 3.28,

i iM x M= ∑ (3.28)

for a binary solution yields

1 1 2 2M x M x M= + (3.33)

Differentiating Eq. 3.33 results in

1 1 1 1 2 2 2 2dM x dM Mdx x dM M dx= + + +

(3.34)

207

Writing the Gibbs/Duhem equation

for a system where T & P are constant

results in

0i ix dM =∑ (3.32)

Eq. 3.32 can be written for a binary

solution, as follows

1 1 2 2 0x dM x dM+ = (3.35)


yields

( )1 1 2 2 1 1 2 2dM Mdx M dx x dM x dM= + + +

1 1 2 2dM Mdx M dx= + (3.36)

0

208

Since, for a binary solution, 1 2 1x x+ =

1 21x x= − ; hence, 1 2dx dx= − or 2 1dx dx= −

Eq. 3.36 can then be re-written as

follows:

( )

( )

1 1 2 1

1 1 2 1

1 2 1

dM Mdx M dxMdx M dxM M dx

= + −

= −

= −

1 21

dM M Mdx

= −


results in

1 21

dMM Mdx

= + (3.37)

209

Rearranging Eq. 3.33 gives

2 2 1 1x M M x M= −

12 1

2 2

xMM Mx x

= − (3.38)

Substituting Eq. 3.38 into Eq. 3.37 and

rearranging the resulting equation yield

1

1 12 2 1

11 1

2 2 1

xM dMM Mx x dxx M dMM Mx x dx

⎛ ⎞= − +⎜ ⎟⎝ ⎠

+ = +

2 1 1 1

2 2 1

x M x M M dMx x dx+

= + (3.39)

210

Multiplying Eq. 3.39 with x2 yields

2 1 1 1 21

dMx M x M M xdx

+ = +

( )2 1 1 21

dMx x M M xdx

+ = + (3.40)

Since 1 2 1x x+ = (for a binary system),

Eq. 3.40 can be written as follows:

1 21

dMM M xdx

= + (3.41)

By performing the same derivation

for 2M , we obtain the following equation

2 11

dMM M xdx

= − (3.42)

211

Eqs. 3.41 & 3.42 illustrate that any

partial property ( iM) can be calculated

from the property of the solution (M),

when the composition of each species

is known, which can be illustrated

graphically as follows

Let’s consider a plot between M

(M can be either h, s, or v, or etc. –

any system property) and x1 (Figure

3.1)

212

Figure 3.1: A Plot between M (any thermodynamic property) and x1

The slope of the line I1–I2 can be

calculated as follows

2 2

1 1 10M I M IdM

dx x x− −

= =−

(3.43)

and 1 21 2

1 1 0I IdM I I

dx−

= = −−

(3.44)

M

x1

I2

I1

213


2 11

dMI M xdx

= − (3.45)

and rearranging Eq. 3.44 yields

1 21

dMI Idx

= + (3.46)


and rearranging result in

( )

1 11 1

11

1

dM dMI M xdx dx

dMx Mdx

⎛ ⎞= + −⎜ ⎟

⎝ ⎠

= − +

but 1 2 1x x+ = 2 11x x= −

214

Hence,

1 21

dMI M xdx

= + (3.47)

Comparing Eq. 3.47 with Eq. 3.41

and Eq. 3.45 with Eq. 3.42 yields

1 1I M= and 2 2I M=

215

Let’s consider the value of 1

dMdx

when

x1 0 (or it is called “infinite dilution of

species 1”) from Figure 3.2

Figure 3.2: A Plot between M (any thermodynamic property) and x1 when x1 0

x1

M

M2

M1

1M∞

216

Comparing Figure 3.2 with Figure

3.1 gives

1 1 1I M M∞= = and 2 2 2I M M= =

This indicates that when x1 0 (i.e.

pure x2), 2 2M M=

Let’s consider the value of 1

dMdx

when

x1 1 or x2 0 (i.e. infinite dilution of

species 2), in Figure 3.3

217

Figure 3.3: A Plot between M (any thermodynamic property) and x1 when x1 1 (or x2 0)

Comparing Figure 3.3 with Figure 3.1

yields

1 1 1I M M= = and 2 2 2I M M∞= =

meaning that when x1 1 (i.e. pure

species 1): 1 1M M=

x1

M

M2

M1 2M∞

218

Example The enthalpy of a binary

liquid system of species 1 & 2 at

constant T & P is presented by the

following equation

( )1 2 1 2 1 2400 600 40 20h x x x x x x= + + +

(3.48)

Determine the values of 1h , 2h , 1h∞ , &

2h∞

219

Rearranging Eq. 3.48 to be a

function of x1 only (note that, for a

binary system, 2 11x x= − ) gives

31 1600 180 20h x x= − − (3.49)

Differentiating Eq. 3.49 with respect

to x1 yields

21

1

180 60dh xdx

= − − (3.50)

Writing Eqs. 3.41 & 3.42

1 21

dMM M xdx

= + (3.41)

2 11

dMM M xdx

= − (3.42)

for this case yields

220

( )1 11

1 dhh h xdx

= + − (3.51)

2 11

dhh h xdx

= − (3.52)

Combining Eqs. 3.49 & 3.50 with

Eqs. 3.51 & 3.52 and rearranging give

( )( )

31 1 1

21 1

600 180 20

1 180 60

h x x

x x

⎡ ⎤= − −⎣ ⎦+ − − −

2 31 1 1420 60 40h x x= − + (3.53)

32 1600 40h x= + (3.54)

221

From Figures 3.2 & 3.3, we knew

that, when x1 0:

2 2M M= and 1 1M M∞ = and when x1 1:

1 1M M= and 2 2M M∞ =

Hence, by substituting corresponding

numerical values into Eqs. 3.53 & 3.54, we

obtain:

( ) ( ) ( )2 31 1 1 1 420 60 1 40 1 400h h x= = = − + =

( ) ( )32 2 1 0 600 40 0 600h h x= = = + =

( ) ( ) ( )2 31 1 1 0 420 60 0 40 0 420h h x∞ = = = − + =

( )32 2 1( 1) 600 40 1 640h h x∞ = = = + =

222

The values of 1h , 2h , 1h∞ , and 2h∞ can

be shown graphically as follows

0

100

200

300

400

500

600

700

0 0.2 0.4 0.6 0.8 1

x 1

h

h2

h1

2h∞

1h∞

223

3.3.3 Relations among Partial Properties

Combining Eq. 3.7

( ) ( ) ( ) i id ng nv dP ns dT dnμ= − +∑ (3.7)

with Eq. 3.19

( ), , j

i ii P T n

ngg

nμ

∂⎛ ⎞= = ⎜ ⎟∂⎝ ⎠

(3.19)

results in

( ) ( ) ( ) i id ng nv dP ns dT g dn= − +∑ (3.55)

From an “exact differential expression”,

we obtain the following relationships

224

( ) ( ), ,P n T n

nv nsT P

∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

, ,P n T n

v sT P∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(3.56)

( ), , , j

i

P n i P T n

nsgT n

∂⎛ ⎞∂⎛ ⎞ = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ (3.57)

and ( ), , , j

i

T n i P T n

nvgP n

∂⎛ ⎞∂⎛ ⎞ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ (3.58)

By employing Eq. 3.18

( ), , j

ii P T n

nMM

n∂⎛ ⎞

= ⎜ ⎟∂⎝ ⎠ (3.18)

we can re-write Eqs. 3.57 & 3.58 as follows

,

ii

P n

g sT

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (3.59)

,

ii

T n

g vP

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠ (3.60)

225

When comparing with Eqs. 3.10 & 3.9

,P n

g sT

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (3.10)

,T n

g vP∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

(3.9)

we can conclude that

“Every equation that provides a linear

relation among thermodynamic

properties of a constant-composition

solution has its counterpart as an

equation connecting the corresponding

partial properties of each species in the

solution”

226

Example In the case of a constant-

composition solution ( ). . 0ii e dn = , we

know that ( ),ig f T P=

Hence,

, ,

i ii

P n T n

g gdg dT dPT P

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Combining the above equation

with Eqs. 3.59 & 3.60 results in

i i idg sdT v dP= − + (3.61)

which is in accord with Eq. 2.13

dg sdT vdP= − + (2.13)

227

3.4 Ideal-Gas Mixtures

If n moles of a mixture of ideal

gases contain in a container with a

volume of tV at the temperature of T,

the total pressure of the container

can be calculated using the

following equation:

tnRTPV

= (3.62)

228

and if in moles of species i contain in the

same container, the partial pressure of

species i ( )iP at the same temperature,

T, is as follows:

ii t

nRTPV

= (3.63)

(3.63)/(3.62) gives

i

ti i

i

t

nRTP nV xnRTP n

V

= = =

or i iP x P=

229

Partial volume of species i (ideal gas)

can be written in the form of equation

as follows:

, ,

, ,

, ,

j

j

j

igig

ii T P n

i

T P n

i T P n

nvvn

RTnP

n

RT nP n

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

⎛ ⎞⎛ ⎞∂ ⎜ ⎟⎜ ⎟⎝ ⎠= ⎜ ⎟∂⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

Since i jn n n= +∑ , but jn is constant,

we obtain

( ), , , ,j j

i j

i iT P n T P n

n nnn n

⎛ ⎞∂ +⎛ ⎞∂= ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑

( ), , , ,

1j j

ji

i iT P n T P n

nnn n

⎛ ⎞∂⎛ ⎞∂= + =⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑ 0

230

Thus,

ig igi i

RTv vP

= =

which illustrates that, in the case of an

ideal-gas mixture

ig igi iv v= (3.64)

The above relation can also be

extended to other thermodynamic

properties:

“A partial molar property 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 same

temperature, but at a pressure equal to its

partial pressure in the mixture”

231

which can be written in the form of

equation as follows:

( ) ( ), ,ig igi i iM T P M T P= (3.65)

3.4.1 For Enthalpy: ( ) ( )=, ,ig ig

i i ih T P h T P

However, we have known (from

Chapter 2) that an enthalpy of an ideal

gas is independent of P

Hence,

( ) ( ) ( ), , ,ig ig igi i i ih T P h T P h T P= =

meaning that

ig igi ih h= (3.66)

232

Applying Eq. 3.28

i iM x M= ∑ (3.28)

to enthalpy, yields

ig ig igi i i ih x h x h= =∑ ∑ (3.67)

This indicates that, for an ideal-gas

solution, the summation of an enthalpy

of each “pure” species is equal to the

enthalpy of the solution

233

3.4.2 For Entropy: Since an entropy of an ideal gas

depends on both T & P (see Chapter 2),

by considering Eq. 2.31 when T is

constant, we obtain

igi p

dT dPds c RT P

= −

ln

igi

dPds RP

Rd P

= −

= −

Integrating the above equation

from P = Pi to P = P yields

0

234

ln

ln

i i

i

P Pig

iP P

P

P

ds Rd P

R d P

= −

= −

∫ ∫

∫

( ) ( ) [ ], , ln lnig igi i i is T P s T P R P P− = − −

( ) ( ), , lnig igi i i

i

Ps T P s T P RP

− = −

but ii

PxP

= i iP x P=

Accordingly,

( ) ( ) 1, , ln lnig igi i i

i i

Ps T P s T P R Rx P x

− = − = −

( ) ( ), , lnig igi i i is T P s T P R x− = (3.68)

235

Applying Eq. 3.65 to entropy yields

( ) ( ), ,ig igi i is T P s T P= (3.69)

Combining Eq. 3.68 with Eq. 3.69 results in

( ) ( ), , lnig igi i is T P s T P R x− = (3.70)


( ) ( ), , lnig igi i is T P s T P R x= − (3.71)

Multiplying Eq. 3.71 with xi gives

( ) ( ), , lnig igi i i i i ix s T P x s T P Rx x= − (3.72)

Applying Eq. 3.28 ( )i iM x M= ∑ to Eq. 3.72

yields

( ) ( ), , lnig igi i i i i ix s T P x s T P R x x= −∑ ∑ ∑

( ), lnig igi i i is x s T P R x x= −∑ ∑ (3.73)

236


( ) 1, lnig igi i i

i

s x s T P R xx

− =∑ ∑ (3.74)

Note that ( ),ig igi is x s T P−∑ = entropy

change of mixing for ideal gas

Since 1 ix > 1, ( ),ig igi is x s T P−∑ > 0

(positive, +, sign) in agreement

with the 2nd law of thermodynamics

(what does the 2nd law say?)

237

Applying the “relations among

partial properties” to the Gibbs free

energy for an ideal gas

ig ig igg h Ts= − (3.75)

gives

ig ig igi i ig h Ts= − (3.76)

Combining Eq. 3.76 with Eqs. 3.66

& 3.71 yields

( )lnig ig igi i i ig h T s R x= − − (3.77)

Rearranging Eq. 3.77 results in

( ) lnig ig igi i i ig h Ts RT x= − +

lnig igi i i ig g RT xμ= = + (3.78)

238

For an ideal gas when T is constant,

Eq. 2.13

dg vdP sdT= − (2.13)

becomes

igidg vdP

RT dPP

dPRTP

=

=

=

lnigidg RTd P= (3.79)

Integrating Eq. 3.79 yields

( )lnigi ig RT P T= + Γ (3.80)

where ( )i TΓ = integration constant

= f(T)

239


results in

( )( ) ( )

lnln ln

ln ln

ig igi i i

i i

i i

g g RT xRT P T RT x

T RT x P

= +

= + Γ +⎡ ⎤⎣ ⎦= Γ + +

( ) lnigi i ig T RT x P= Γ + (3.81)

Applying Eq. 3.28 to Eq. 3.81 yields

( ) lnig igi i i i i ig x g x T RT x x P= = Γ +∑ ∑ ∑

(3.82)

240

3.5 Fugacity and Fugacity Coefficient: Pure Species

Re-writing Eq. 3.80

( )lnigi ig RT P T= + Γ (3.80)

for pure & real fluids at constant T

gives

( )lni i ig RT f T= + Γ (3.83)

The property if in Eq. 3.83 is called

a “FUGACITY”

Accordingly, fugacity has the

same unit as pressure

241

(3.83) – (3.80) yields

( ) ( )ln ln

ln

igi i i i i

i

g g RT f T RT P TfRTP

− = + Γ − + Γ⎡ ⎤⎣ ⎦

=

but ig Ri i ig g g− = (see Chapter 2); thus,

lnR ii

fg RTP

=

Note that ii

fP

φ= = FUGACITY

COEFFICIENT

Hence,

lnRi ig RT φ= (3.84)

242

In the case of an ideal gas: if P=

1ii

fP

φ = = ( )ln 1 0Rig RT= =


lnRi

igRT

φ = (3.85)


(see Chapter 2)

( )0

1PRg dPZ

RT P= −∫ (2.67)

yields

( )0

ln 1PR

ii i

g dPZRT P

φ = = −∫ (3.86)

243

In the case of cubic EoS (e.g.,

Redlich-Kwong, Soave-Redlich-Kwong,

& Peng-Robinson EoS), the property Rg

RT

can be written for species i, as follows

( ) ( )ρ= − − − + −ln 1 1Ri

i ig b Z qI ZRT

(2.85a)

Combining Eq. 2.85a with Eq. 3.85

lnRi

igRT

φ = (3.85)

gives

( ) ( )ln ln 1 1R

ii i i

g b Z qI ZRT

φ ρ= = − − − + − (3.87)

244

but, from Eq. 2.95 (See Chapter 2)

B bZ

ρ= (2.95)

or, for species i

i

i

B bZ

ρ= (2.95a)

Eq. 3.87 can, thus, be re-written as

follows:

( ) ( )ln ln 1R

ii i i i

g Z B qI ZRT

φ = = − − − + − (3.88)

245

3.6 Vapour/Liquid Equilibrium (VLE) for Pure Species

When a system consists of liquid (l)

and vapour (v), we can write Eq. 3.83

for each phase, as follows:

( )lnv vi i ig RT f T= + Γ (3.89)

( )lnl li i ig RT f T= + Γ (3.90)

(3.89) – (3.90) yields

ln ln

ln

v l v li i i i

vil

i

g g RT f RT ffRTf

− = −

=

246

At equilibrium,

v li ig g=

or 0v li ig g− =

Thus, this means that

ln 0v

il

i

ff=

or v l sati i if f f= = (3.92)

“For a pure species coexisting

liquid and vapour phases are in

equilibrium when they have the

same T, P, and FUGACITY”

247

Dividing Eq. 3.92 with P results in

v l sat

i i if f fP P P= =

v l sati i iφ φ φ= = (3.93)

which implies that, when two phases

are in equilibrium with each other at

constant T & P, in addition to the fact

that the fugacity of each phase is

equal to each other, their fugacity

coefficients are also identical

248

Example For H2O at a T of 300 oC and for

P up to 10,000 kPa (100 bar), calculate

values of if and iφ using data from steam

tables

Writing Eq. 3.83 for a constant-P system

at P = P gives

( )lni i ig RT f T= + Γ (3.83)

and at P = low P yields

( )* *lni i ig RT f T= + Γ (3.94)

(3.83) – (3.94) results in

**ln i

i ii

fg g RTf

− =

*

*ln i i i

i

f g gRTf−

= (3.95)

249

Writing the following equation

i i ig h Ts= − (3.96)

for a low-P system gives

* * *i i ig h Ts= − (3.97)

(3.96) – (3.97) results in

( ) ( )* * *i i i i i ig g h h T s s− = − − − (3.98)

Dividing Eq. 3.98 with RT yields

( ) ( )* **i i i ii i

h h s sg gRT RT R

− −−= −

Substituting the above Eq. into Eq. 3.95

results in

( ) ( )* *

*ln i i i ii

i

h h s sfRT Rf− −

= − (3.99)

250

Let low P = 6 kPa, and when T = 300 oC

(note that, at low P (& high T as T = 300 oC),

water vapour (H2O) can be assumed to be

an ideal gas; hence, * * 6 kPaif P= = )

h = 3,076.8 kJ/kg & s = 9.5162 kJ/kg-K (by reading from a superheated steam table)

At P = 4,000 kPa (4 MPa) & T = 300 oC

(superheated vapour), we obtain the

following values

h = 2,958.6 kJ/kg & s = 6.3531 kJ/kg-K Substituting corresponding numerical

values into Eq. 3.99 results in

251

[ ]( )

[ ]( )

[ ]( )

*

kJ18.02 2,958.6 3,076.8 kmolln

kJ8.314 300 273 Kkmol-K

kJ18.02 6.3561 9.5162 kmol-K

kJ8.314 kmol-K

6.395

i

i

ff

−=⎛ ⎞ +⎜ ⎟⎝ ⎠

−−

⎛ ⎞⎜ ⎟⎝ ⎠

=

(note that we have to multiply the values of h

& s obtained from the steam table with MW of water

(= 18.02) to convert the units of h & s to be kJ/kmol

& kJ/kmol-K, respectively)

Thus,

* exp(6.395) 598.84i

i

ff= =

( )*598.84 598.84 6 kPa 3,593 kPai if f= = =

and

3,593 kPa 0.8984,000 kPa

ii

fP

φ = = =

252

When P = 10,000 kPa & T = 300 oC

(the system (H2O) is now in the form of

“compressed liquid”), we obtain the

following

h = 1,347.3 kJ/kg & s = 3.2519 kJ/kg-K

Substituting corresponding numerical

values into Eq. 3.99 yields

[ ]( )

[ ]( )

[ ]( )

−=⎛ ⎞ +⎜ ⎟⎝ ⎠

−−

⎛ ⎞⎜ ⎟⎝ ⎠

=

*

kJ18.02 1,347.3 3076.8 kmolln

kJ8.314 300 273 Kkmol-K

kJ18.02 3.2519 9.5162 kmol-K

kJ8.314 kmol-K

7.057

i

i

ff

253

Hence,

= =* exp(7.057) 1,160.96i

i

ff

( )*1,160.96 1,160.96 6 kPa 6,966 kPai if f= = =

and

6,966 kPa 0.69710,000 kPa

ii

fP

φ = = =

It can be observed, from this

Example, that when P is getting higher,

the discrepancy between the values of

if and P is getting larger Also, when P is getting higher, the

value of φi is lower and getting more

and more deviated from unity (i.e. 1)

254

3.7 Fugacity and Fugacity Coefficient: Species in Solutions

Combining Eq. 3.78

lnig ig igi i i ig g RT xμ = = + (3.78)

with Eq. 3.80

( ) lniig

ig T RT P= Γ + (3.80)

gives the following equation:

( )( )ln lnig igi i i ig T RT P RT xμ = = Γ + +


gives

( ) ( )lnig igi i i ig T RT x Pμ = = Γ + (3.100)

which is the same as Eq. 3.81

255

If this is a “real” solution (can be

either gaseous or liquid solution), Eq.

3.100 can be re-written as follows:

( ) ˆlni i i ig T RT fμ = = Γ + (3.101)

where

if = fugacity of species i in a solution

We have just learned that, if there

are 2 phases (e.g., phases α & β ) co-

existing in equilibrium at given T & P, it

results in the fact that

i if fα β=

256

The above equation can also be

applied to any species in a solution

as follows

ˆ ˆi if fα β=

“Multiple phases at the same

T & P are in equilibrium when

the fugacity of each constituent

species is the same in all phases”

257

We have learned, from Chapter 2,

that

R igM M M= − (2.55)

Hence, (3.101) – (3.100) results in

( )ˆln ln

R ig igi i i i i

i i

g g g

RT f RT x P

μ μ= − = −

= −

ˆ

lnR ii

i

fg RTx P

⎛ ⎞= ⎜ ⎟

⎝ ⎠ (3.102)

Let ˆ ˆ

ˆ i ii

i i

f fx P P

φ = =

Thus,

ˆlnRi ig RT φ= (3.103)

where iφ = fugacity coefficient of

species i in a solution

258

We do the same as did previously

for iφ (see Page 242) and, thus,

obtain

( )0

ˆln 1P

i idPZP

φ = −∫ (3.104)

259

3.8 Generalised Correlations for the Fugacity Coefficient

Combining Eq. 3.86

( )0

ln 1P

i idPZP

φ = −∫ (3.86)

with the following equations

rc

PPP

=

c rP P P= (3.105)

( )c r

c r

dP d P PP dP

=

= (3.106)

results in

260

( )0

ln 1rP

c ri i

c r

P dPZP P

φ = −∫

( )0

ln 1rP

ri i

r

dPZP

φ = −∫ (3.107)

We can write Pitzer et al theorem

of corresponding states

( ) ( )ω= +0 1Z Z Z (1.21)

for species i as follows

( ) ( )ω= +0 1i i iZ Z Z (3.108)

261

Subtracting both sides of Eq. 3.108

with 1 results in

0 11 1i i iZ Z Zω− = − + (3.109)


gives

( )

( )

( )

φ

ω

φ ω

= −

= − +

= − +

∫

∫

∫ ∫

0

0 1

0

0 1

0 0

ln 1

1

ln 1

r

r

r r

Pr

i ir

Pr

i ir

P Pr r

i i ir r

dPZP

dPZ ZP

dP dPZ ZP P

262

Let

( )0 0

0

ln 1rP

ri i

r

dPZP

φ = −∫

and 1 1

0

lnrP

ri i

r

dPZP

φ = ∫

Thus,

0 1ln ln lni i iφ φ ω φ= +


yields

( )

( )( )

0 1

0 1

ln ln ln

ln

i i i

i i

ω

ω

φ φ φ

φ φ

= +

⎡ ⎤=⎣ ⎦

( )( )0 1i i i

ωφ φ φ= (3.110)

263

Note that the values of 0iφ and 1

iφ in table

format (Tables A.3.1 and A.3.2) are

included at the end of this Chapter (on

Pages 284-285 and 286-287, respectively)

Example Determine the values of

fugacity coefficient ( iφ ) and fugacity ( if)

of R-134a at 300 oC, and (a) 80 bar and

(b) 284 bar

For R-134a:

374.26 KcT =

40.59 barcP =

0.326ω =

264

(a)

80 bar 1.9740.59 barr

c

PPP

= = =

[ ]300 273 K1.53

374.26 Krc

TTT

+= = =

By performing double interpolations

of ( )φ 0i and ( )φ 1

i appeared in Tables A.3.1

and A.3.2, we obtain

( )φ =0 0.8495i

( )φ =1 1.1828i

Hence,

( )( )0.3260.8495 1.18280.897

iφ =

=

and ( )( )φ= = =0.897 80 bar 71.8 bari if P

265

(b) 284 bar 7.0

40.59 barrc

PPP

= = =

[ ]300 273 K1.53

374.26 Krc

TTT

+= = =

Thus, also by reading from Tables A.3.1

and A.3.2 and performing double

interpolations, we obtain

( )φ =0 0.6866i

( )φ =1 1.5817i Hence,

( )( )0.3260.6866 1.58170.797

iφ =

=

and

( )( )0.797 284 bar226.3 bar

i if Pφ= =

=

266

3.9 The Ideal Solution In the case that we are dealing

with an ideal liquid solution, Eq. 3.78

lnig igi i ig g RT x= + (3.78)

can be re-written for an ideal solution

as follows:

lnidi i ig g RT x= + (3.111)


to T, while P and composition are kept

constant, yields

, ,,

lnidi i i

P x P xP x

g g RT xT T T

⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

267

Combining the above equation with Eq.

3.59

,

ii

P x

g sT

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (3.59)

results in

,

lnid ii i

P x

gs R xT

∂⎛ ⎞− = +⎜ ⎟∂⎝ ⎠

and ,

lnid ii i

P x

gs R xT

∂⎛ ⎞= − −⎜ ⎟∂⎝ ⎠ (3.112)

Combining Eq. 3.10

,P x

g sT

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (3.10)

re-written for species i

,

ii

P x

g sT

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠ (3.10a)

with Eq. 3.112 gives

268

( ) lnln

idi i i

i i

s s R xs R x

= − − −

= − (3.113)


to P, while T and composition are kept

constant, yields

, ,,

lnidi i i

T x T xT x

g g RT xP P P

⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ (3.114)


,

ii

T x

g vP

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠ (3.60)

results in

,

0id ii

T x

gvP

∂⎛ ⎞= +⎜ ⎟∂⎝ ⎠

Note that ln iRT x is a constant

269

By combining the above equation

with Eq. 3.9 re-written for species i

,

ii

T x

g vP

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠ (3.9a)

we obtain

idi iv v= (3.115)

When we applied the following

equation

g h Ts= −

to an ideal solution, we obtain

id id idi i ig h Ts= −

or

id id idi i ih g Ts= + (3.116)

270

Combining Eq. 3.116 with Eqs. 3.111 &

3.113 gives

[ ] [ ]= + + −

= + =

ln lnidi i i i i

i i i

h g RT x T s R xg Ts h

which indicates that

idi ih h= (3.117)

Applying Eq. 3.28

i ii

M x M= ∑ (3.28)

to an ideal solution gives id id

i ii

M x M= ∑

271

Thus,

ln

ln

id idi i i i i i

i i i i

g x g x g x RT x

x g RT x x

= = +

= +∑ ∑ ∑∑ ∑

(3.118)

lnid idi i i i i is x s x s R x x= = −∑ ∑ ∑ (3.119)

id idi i i iv x v x v= =∑ ∑ (3.120)

id idi i i ih x h x h= =∑ ∑ (3.121)

272

3.10 The Lewis/Randall Rule

When applying Eq. 3.101

( ) ˆlni i i ig T RT fμ = = Γ + (3.101)

to an ideal solution, we obtain

( ) ˆlnid id idi i i ig T RT fμ = = Γ + (3.122)

Combining Eq. 3.83

( ) lni i ig T RT f= Γ + (3.83)

with Eq. 3.111

lnidi i ig g RT x= + (3.111)

yields

273

( ) ( )ˆln ln lnidi i i i iT RT f T RT f RT xΓ + = Γ + +⎡ ⎤⎣ ⎦


gives

( )

ˆln ln lnln

idi i i

i i

RT f RT f RT xRT x f

= +

=

idi i if x f= (3.123)

Eq. 3.123 = “Lewis/Randall Rule”

“Fugacity of each species in an

ideal solution is proportional to its

mole fraction, where fugacity of

pure species i is a proportionality

constant”

274

Dividing Eq. 3.123 with ix P yields

id

i i i i

i i

f x f fx P x P P

= =

We have defined (see Page 257) that

ˆ

ˆ ii

i

fx P

φ =

and when applying the above definition to

an ideal solution, we obtain

ˆ

ˆid

id i ii

i

f fx P P

φ = =

275

Since

ii

fP

φ =

(see Page 241)

idi iφ φ= (3.124)

(Note: for IDEAL SOLUTION only)

276

3.11 Excess Properties

As well as a residual property

R igM M M= − (2.55)

an excess property indicates how far

a real solution deviates from an ideal

solution, which can be written in the

form of equation as follows

E idM M M= − (3.125)

(3.125) – (2.55) results in

( ) ( )( )

E R id ig

id ig

M M M M M M

M M

− = − − −

= − − (3.126)

277

Since an ideal-gas mixture is an

ideal-gas solution, when we replace ig

in Eq. 3.118

= +∑ ∑ lnidi i i ig x g RT x x (3.118)

with igig , the resulting equation is as

follows:

= +∑ ∑ lnig igi i i ig x g RT x x (3.127)

(by applying Eq. 3.28 to Eq. 3.78)

Eqs. 3.119-3.121 can also be re-

written as a function of igiM in the same

manner as Eq. 3.127 (please try to do it

yourself)

278

Thus, (3.118) – (3.127) yields − = − =∑ ∑ ∑id ig ig R

i i i i i ig g x g x g x g

When performing the same derivation

for Eqs. 3.119-3.121, we obtain the following

general equation for any property id ig ig R

i i i i i iM M x M x M x M− = − =∑ ∑ ∑

(3.128)

When combining Eq. 3.128 with

Eq. 3.126, we obtain

( )E R Ri iM M x M− = − ∑

E R Ri iM M x M= −∑ (3.129)

279

Note that Eq. 3.125 can be applied

to any partial property, as follows:

E idM M M= − (3.130)

280

3.12 The Excess Gibbs Energy and the Activity Coefficient

We can write Eq. 3.101

( ) ˆlni i i ig T RT fμ = = Γ + (3.101)

for an ideal solution, as follows

( ) ˆlnid idi i ig T RT f= Γ + (3.131)


(Lewis/Radall Rule)

idi i if x f= (3.123)

gives

( ) lnidi i i ig T RT x f= Γ + (3.132)

281

(3.101) – (3.132) results in

ˆ

lnid E ii i i

i i

fg g g RTx f

− = = (3.133)

Let ii

i i

fx f

γ = = activity coefficient of

species i in a solution

Thus,

lnEi ig RT γ= (3.134)

and

lnEi

igRT

γ= (3.135)

282

When applying Eq. 3.28 to Eq. 3.135,

we obtain

lnEEi

i i igg x x

RT RTγ= =∑ ∑ (3.136)

The Nature of Excess Properties

(See Figure 3.4)

• Strongly depend on T

• Weakly depend on P, especially

at moderate T

• Depend on composition, but

the dependence varies from

solution to solution

283

Figure 3.4: Examples of the Nature of Excess Properties

(from Introduction to Chemical Engineering Thermodynamics: 7th ed, by Smith, Van Ness, and Abbott, 2005)

284

Table A.3.1: Values of φ0 as a function of Tr and Pr (from Introduction to Chemical Engineering Thermodynamics:

7th ed, by Smith, Van Ness, and Abbott, 2005)

285

Table A.3.1: Values of φ0 as a function of Tr and Pr (cont.) (from Introduction to Chemical Engineering Thermodynamics:


286

Table A.3.2: Values of φ1 as a function of Tr and Pr (from Introduction to Chemical Engineering Thermodynamics:


287

Table A.3.2: Values of φ1 as a function of Tr and Pr (cont.) (from Introduction to Chemical Engineering Thermodynamics:


Theory of Solution Thermodynamics

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