Euler and Gibbs–Duhem Equations

3. Euler and Gibbs–Duhem Equations

The total differential of Eq. (6a) is

dB = ( B / T ) ∂ ∂ P,N dT + ( B / P ) ∂ ∂ T,N dP + ˆ ∑ k b dN k , i.e.,

(7a)

k=1

Since B = N b ,

d(Nb ) = Ndb bdN + = N( b / T ) ∂∂ P,N dT + N( b / P ) ∂∂ T,N dP + ˆ ∑ k b dN k

(7b)

k=1

Furthermore, since N k =X k N, dN k =X k dN + NdX k , further simplification of Eq. (7b) results in the expression

db db K )

N db { −

dT −

dP − ∑ k b dX k } + dN b { − ∑ bX k k } = 0 (8)

dT

dP

Since N is arbitrary and the value of dN can vary, it is apparent from Eq. (8) that the coeffi- cients of N and dN must vanish. Equating the coefficient of dN to zero, we obtain

b = ∑ bX k k , i.e., BbN = = bN k k

(9a)

where B= S, U, H, V, etc. Likewise,

db db K )

db = dT +

(9b) dT

dP − ∑ k b dX k .

dP

The relation in Eq. (9a) is known as the Euler equation. Since all the differentials are exact, we infer from Eq. (9b) that

b = b (T, P, X 1 ,X 2 , ... , X K–1 ).

Note that X 1 +X 2 + ... + X K = 1. The number of independent variables in Eq. (10) is ((K–1)+2) = K+1, while in Eq. (6) it is K+2. At constant temperature and pressure, Eq. (9b) assumes the form K )

db = ∑ k b dX k .

Differentiating Eq. (9a) , K )

db = k b dX k +

∑ (12a)

k db X k .

Equating Eq. (12a) with Eq. (9b), we obtain the relation

db db K

dT +

dP −

∑ k X db k = 0 .

(12b)

dT dP

Upon multiplying this expression by N,  ∂ B 

 ∂ dT B 

+ K dP −

∑ k = 1 N db k k = 0 .

(12c)

 ∂ T 

 ∂ P 

PN ,

TN ,

At a specified temperature and pressure

(13) The expressions in Eqs. (12a)–(12c) and (13) are various forms of the Gibbs–Duhem (GD)

K X db K = 0 .

equations, and apply to liquid, solid, and gas mixtures. Combining Eqs. (12a) and (13) we obtain

db = ∑ k b dX k .

a. Characteristics of Partial Molal Properties Since

ˆb k =( ∂B/∂N k ) T, P = ˆb k (T, P,N 1 ,N 2 , ..., N K ),

applying Euler’s theorem for species k =1 for a partly homogeneous function of order zero,

N 1 ∂ ˆb 1 / ∂N 1 +N 2 ∂ ˆb 1 / ∂N 2 +N 3 ∂ ˆb 1 / ∂N 3 +.... = 0, i.e.,

or, more generally, Σ k (N k ∂ ˆb j / ∂N k ) = 0, j=1, 2,...K.

(15a) Dividing Eq. (15) by N Σ k (X k ∂ ˆb j / ∂N k ) = 0, j=1, 2,...K.

(15b) Consider partial molal property of species 1. Differentiating the partial molal property

ˆb 1 = ∂B/∂N 1 with respect to N 2 , we obtain the relation

∂ 2 ˆb

1 / ∂N 2 = ∂ B/ ∂N 2 ∂N 1 = ∂/∂N 1 ( ∂B/∂N 2 )= ∂ ˆb 2 / ∂N 1 ,

(15c) (15c)

Σ j (N k ∂ ˆb k / ∂N j ) = 0, j=1, 2,...K. (15d) Dividing by N

0, j=1,2…K.

With j=1 in Eq. (15d) ,

N 1 ∂ ˆb 1 / ∂N 1 +N 2 ∂ ˆb 2 / ∂N 1 +N 3 ∂ ˆb 3 / ∂N 1 + ... = 0.

(15e) Note that the partial derivatives ∂ ˆb 1 / ∂N 1 imply that N 2 ,N 3 , etc., are constant Since N 1 =X 1

N, then dN 1 = dX 1 N+X 1 dN. If only N 1 is altered, then the values of all N j≠1 are constant. In that case

dN 1 = dX 1 N+X 1 dN 1 , or (1 – X 1 ) dN 1 = dX 1 N, i.e., dN 1 = dX 1 N/(1 – X 1 ). Using this result in Eq. (15e) ∂ ˆb 1 / ∂N 1 = (1-X 1 )( ∂ ˆb 1 /N ∂X 1 ), ∂ ˆb 2 / ∂N 1 = (1-X 1 )( ∂ ˆb 1 /N ∂X 1 ),…. . Hence, Eq. (15e) with j=1 simplifies to the form

X 1 ∂ ˆb 1 / ∂X 1 +X 2 ∂ ˆb 2 / ∂X 1 +X 3 ∂ ˆb 3 / ∂X 1 +... = 0, i.e., Σ k X k ( ∂ ˆb k / ∂X 1 ) = 0, Generalizing for any “j”

Σ k X k ( ∂ ˆb k / ∂X j ) = 0, j=1,2…K (16) Gibbs function is extensively used in phase and chemical equilibrium calculations. Thus it is

useful to summarize the relations for B=G. i.e.,

= G(T, P, N 1 ,N 2 , ..., N K ), then

g m = Σ k ˆg k X k , and g = Σ k ˆg k Y k . (17) where the second summation is on mass basis. The partial molal Gibbs function ˆg k is the

chemical potential µ k of a species, i.e., ˆg k = (dG/dN k ) T, P, N ji ≠ .

(18a) Equation (12c) implies that

0=– s dT + v dP + X 1 d ˆg 1 +X 2 d ˆg 2 + ..., and

(18b)

(19a) where ( ∂G/∂T) P,N = –S and ( ∂G/∂P) T,N = V. At a specified temperature and pressure

0 = – S dT + V dP +N 1 d ˆg 1 +N 2 d 2 + ... ,

X 1 d ˆg 1 +X 2 d ˆg 2 + ... = 0.

(19b)

b. Physical Interpretation Consider the extensive property V of a k–component mixture. Each species in the mixture contributes an amount V k towards the total mixture volume, and its partial molal vol- ume

1L

Ideal, 4L

Methanol(j)

(a)

(b) C

Partial V B Molal

D volume A

N H2O

(C )

Figure 1 a. Mixing of two miscible species. b. Mixing of two immiscible species. c. Determination of the partial molal vol- ume from a plot of total mixture volume vs. the number of moles of water in a water/alcohol mixture.

ˆv k =( ∂V/∂N k ) T, P = ˆv k (T, P, N 1 ,N 2 , ..., N K ),

When the same component is considered in its pure state at the same temperature and pressure,

its specific volume v k , and, in general, ˆv k ≠ v k .

If a liter of water is added at standard conditions to three liters of pure alcohol, you will find that both species completely mix at the molecular level. This is an example of a mis- cible mixture. Under standard conditions, the total mixture volume is not four liters, suggesting that the mixture must have contracted due to a change in the intermolecular attractive forces. The volume occupied by 1 kmole of water, i.e., 6 26 ×10 water molecules in the mixture is its

partial molal volume. As more water is added to the mixture the total volume increases as shown by curve ADBC in Figure 1 . The slope of the mixture volume with respect to the num- ber of moles of water provides a measure of the partial molal volume. The point A in the figure represents a condition corresponding to trace amounts of water in the mixture, while point B represents alcohol in trace quantities. An immiscible mixture is formed if two species do not mix at a molecular level and, in that case, the partial molal volume loses meaning.

i. Remarks In Chapters 1 and 6 we have discussed the functional form for the intermolecular at-

traction forces given by the Lennard– Jones empirical potential Φ(l) between a pair of mole- cules. For like pairs of molecules,

12 Φ(l) = 4ε((σ/l) 6 –( σ/l) ). (20a) For an unlike molecular pair consisting of species k and j,

(20b) where ε kj =( ε k ε j ) 1/2 , and the collision diameter σ kj =( σ k + σ j )/2. Note that concentration effects

Φ(l) = 4ε kj (( σ kj /l) 12 –( σ kj /l) 6 ),

on σ kj and ε kj are not included in this model. If F k (= ∂φ/∂l) denotes the intermolecular attrac- tion forces between the molecules of species k, and F kj denotes the corresponding forces be-

tween dissimilar molecules of the two species k and j,

F kk = 4(

6 ε/l)(6(σ/l) 12 – 12( σ/l) ), and (21a)

F 6 kj 12 = 4( ε kj /l) (( σ kj / l) –( σ kj /l) ). (21b) Consider the following scenarios: (1) F kj =F kk . In this case an ideal solution (or ho-

mologous series) is formed, e.g., a mixture of toluene and benzene. (2) F kj >F kk . This is an example of a non ideal solution (e.g., the volume contraction upon mixing of water in alcohol). (3) F kj <F kk . Also a non ideal solution, but in this case there is a volumetric expansion upon mixing).

If F kj »F kk at all concentrations, the mixture is miscible at a molecular level. Alterna- tively, if F kj «F kk , at all concentrations, the mixture is completely immiscible. In some cases, the mixture is miscible up to a certain mole fraction beyond which F kj «F kk . Such mixtures are called partially miscible mixtures.

In miscible mixtures, energy must be initially utilized to overcome the intermolecular attraction forces between like molecules (e.g., k–k). Inserting a molecule of j and forming the j–k pairs alters the intermolecular attraction forces. Consequently, the system may reject en- ergy (during exothermic mixing) or require it (endothermic mixing).

In an ideal solution F kk =F kj , and ˆv k (T, P, X 1 ,X 2 , ...) = v k (T, P). Recall that

V= Σ k ˆv k (T, P,X 1 ,X 2 , ...)N k .

Therefore, for an ideal mixture

V id = Σ k v k (T, P)N k . (22) This relation is called law of additive volumes or the Amagat–Leduc Law for mixtures. It is

particularly valid for gas mixtures at low pressures. Similarly, for any property (other than the entropy),

B id = Σ k b k (T, P)N k . (23) Mixing is always an irreversible process. For adiabatic ideal mixing, i.e., when there is no

volumetric change and heat is neither absorbed nor removed. Since the Second Law states that

dS – δQ/T b = δσ,

the difference in the entropy after and before mixing is given by the expression S final –S initial –0= σ, the difference in the entropy after and before mixing is given by the expression S final –S initial –0= σ,

Even after ideal mixing, the entropy of a species k inside the mixture at T and P is larger than the entropy of the pure species at same temperature and pressure. This is due to the increase in the intermolecular spacing between the k and j molecules, which increases the number of quantum states for each species (Chapter 1).