Relationship Between Molal and Pure Properties

4. Relationship Between Molal and Pure Properties

a. Binary Mixture For a two component system, Eq. (14) suggests that at specified values of T and P )

(24a) Since X 1 +X 2 = 1, Eq. (24a) assumes the form

db = 1 b dX 1 + 2 b dX 2

/ db dX 2 =−+ b 1 b 2 (24b) Since the mixture property is provided by Eq (9a),

db = b 1 { − dX 2 } + 2 b dX 2 , i.e.,.

b = bX 1 1 + bX 2 2

(24c)

Multiplying Eq (24b) by X 2 , rearranging and simplifying,

b 1 X 2 = b 2 X 2 - X 2 d b /dX 2 )

and using Eq. (24c) to eliminate

(24d) Similarly

b 1 = b- X 2 d b /dX 2

(24e) Equations (24d) and (24e) express the partial molal property of the species in terms of the mo-

b 2 = b- X 1 d b /dX 1

lal property of the mixture.

b. Multicomponent Mixture Similarly we can extend the derivation to a multi-component mixture, i.e.,

b ˆ i =− b ∑ X k ( ∂∂ b / X kTPki ) ,, ≠ .

k =≠ 1 , ki

b. Example 2

A small amount of liquid water (species 2) is added to liquid methanol (species 1) in

a constant temperature bath held at 25 o

C and 1 atm. The variation of X is shown in

Figure 2 . Assume that v 1 = 0.04072 m kmole , v 2 = 0.0181 m kmole . Determine the specific volume v id under ideal mixing.

Plot ˆv 1 and ˆv 2 with respect to X 1 .

ˆv 1, x 1 →1 , ˆv 1 , x 1 →0 (species 1 is in infinite dilution or in trace amounts). ˆv 2, x 2 →1 , ˆv 2, x 2 →0 (species 2 is in infinite dilution or in trace amounts)

Determine an approximate expression for v in terms of X 1 when X 1 →0.

Solution The specific volume of any mixture or solution

(A) For an ideal solution,

v = vx ˆ 11 + vx ˆ 2 2 .

(B) v id = v

ˆv id

id

1 = v 1 , ˆv 2 = v 2 , i.e.,

(C) Similarly on mass basis v id = ˆv m m 1 Y 1 + ˆv m 2 Y m 2 . Since v 1 , and v 2 are fixed once

the temperature and pressure are specified, and X 2 = (1–X 1 ), using Eq. (C), we obtain the relation v id = v

(D) Equation (D) indicates that a plot of v id with respect to X 1 is linear. However, meas- urements indicate that this is not so. Using the measured results for v with respect to X 1 shown in Figure 2 , it is possible to obtain ˆv 1 and ˆv 2 using Eqs. (24d) and (24e) with b = v . Since X 1 +X 2 = 1, dX 1 = –dX 2 , and

1 X 1 + v 2 (1– X 1 ).

v 1 = v + (d v /dX 1 )(1– X 1 ),

(E) v ˆ 2 =− v X dv dX 1 / 1 , (F)

Figure 2 shows a plot of specific volume, partial molal volumes of methanol (1) and

PM

Figure 2: Partial and molal volumes of methanol (1) and water (2) at 25 C, 1 atm (From Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, 4th Edi- tion, McGraw Hill Book Company, 1987, p. 428. With permission.)

DQ

Figure 3: Determination of partial molal properties. water (2) at 25ºC, 1 atm while Figure 3 illustrates a graphical method that can be used

to determine ˆv 2 by applying Eq. (F). For instance, a tangent to the v –X 1 curve at point R yields the slope d v /dX 1, and the intercept PS represents ˆv 2 at that value of X 1

(i.e., at R). As X 1 →1, the mixture is virtually pure so that species 1 is mainly sur- rounded by like molecules, and

ˆv

1, x 1 →1 = v 1 = 0.04072 m kmole . (Point D, Figure 2 ). As X 1 →0, Eq. (E) yields

ˆv 1, x 1 →0 = v x 1 →0 + (d v /dX 1 ) x 1 →0 ≈ v 2 + ( dv dX / 1 ) X 1 → 0 .

(G)

However the slope (d v /dX 1 ) x 1 →0 ≠0. Then (G) yields

( dv dX / 1 ) X 1 → 0 = . 0 0194 . (H) We must, therefore, resort to experiments from which we find that

1 , x 1 →0 = ˆv 1 = 0.0375 m kmole (point G, Figure 2 ) At this condition, species 1 is in trace amounts and is surrounded mainly by unlike

3 ˆv –1

molecules with force fields dominated by molecules of species 2. Likewise,

3 ˆv –1

= v 2 = 0.0181 m kmole , (point A) and ˆv 2,X2 →0 = v

2,X1 →1

2 = 0.015 m 3 kmole (point M).

Applying Eq. (A) as. X 1 →0, and X 2 →1, using Eq.(G), v =X 1 ˆv 1 , X1 →0 +X 2 ˆv 2,X2 →1 =X 1 ˆv 1 , X1 →0 + (1-X 1 )v 2

Simplifying,

(I) An extension to multiple components is given in remarks. Since,

v = v 2 + Xv 1 (ˆ 1 , X 1 → 0 − v 2 )

ˆv 3

1 , x 1 →0 = 0.0375 m kmole –1 , and

= v = 0.0181 m 3 ˆv –1 2 2 kmole , since X 2 →1.

Then for small values of X 1 , Eq. (I) yields v = 0.0181 + 0.0194 X 1 .

Thus the mixture volume increases linearly with X 1 as X 1 → 0. Similarly, as X 2 →0, one can show that

v = 0.0407 (1 – X 2 ) + 0.015 X 2

Remarks The expression v id = v 1 X 1 + v 2 X 2 = v 1 X 1 + v 2 (1–X 1 )= ( v 1 – v 2 )X 1 + v 2 is also known as the Law of Additive Volumes (or the Lewis–Randall rule for volume). The law presumes that the intermolecular attraction forces between unlike molecules are the same as those between like molecules. This is a reasonable assumption for succes- sive homologous series of hydrocarbons (e.g., pentane, hexane, etc). Using a Taylor series expansion,

V(N 1 +dN 1 ,N 2 +dN 2 ,N 3 +dN 3 ,..) = V(N 1 ,N 2 ,N 3 ,..) + ∂V/∂N 1 dN 1 + ∂V/∂N 2 dN 2 +... . Suppose initially there is only species 1 and hence initial amounts of N 2 ,N 3 .. etc. are

equal to zero. Assume that small amounts of species 2, 3, 4, … are being added to species 1. In that case, δN 1 = 0, and δN 2 =N 2 , δN 3 =N 3 , δN 4 =N 4 , … are small.

Therefore, V(N 1 , 0+N 2 , 0+N 3 ) = V(N 1 )+ ∂V/∂N 1 ×(0) + ∂V/∂N 2 N 2 + ∂V/∂N 3 N 3 + ... , i.e.,

V(N 1 ,N 2 ,N 3 , ...) = V(N 1 )+ ˆv 2 N 2 + ˆv 3 N 3 + ... .

Dividing throughout by the total number of moles N (recall that N →N 1 ), v (X 1 →1, X 2 →0) = v 1 + ˆv 2 N 2 /N + ˆv 3 N 3 /N + ... = v 1 + ˆv 2 X 2 + ˆv 3 X 3 + ... . where X 2 ,X 3 , … denote mole fractions of trace species. In a salt (solute – species 2)

and water (solvent – species 1) solution, an upper limit X 2,upper ≈ 0.3 exists at standard conditions, which is called the solubility limit. In a mixing tank, the addition of salt

beyond a 30% salt mole fraction usually results in the settling of solid salt. Therefore,

one may not reach the limit X 2 →1 or X 1 → 0 in a solution.

c. Example 3 The following relation describes the volume change in a water (solvent – species 1) and salt (solute – species 2) solution:

V = 1.001 + 16.625N + 56.092 N 3/2

(A) where V is expressed in units of liters, and N 2 in kmole. Obtain an expression for ˆv 2

2 2 + 119.4 N 2 ,

1 . Determine v ˆ 2 , X 2 → 0 or v ˆ 2 in liter per kmole. Obtain an expression for ˆv 1 in terms of N 2 , and one in terms of X 1 . Solution Differentiating Eq. (A),

in terms of N 2 , and determine the value of ˆv

ˆv 2 = ∂V/∂N 2 = 16.625 + 84.138 N 1/2 2 + 238.8 N 2 in units of l kmole –1 . (B)

As N 2 →0 (or X 2 →0), the solute volume at infinite dilution ˆv 2 = 16.625 l kmole –1 .

Furthermore, V = ˆv 1 N 1 + ˆv 2 N 2 , i.e.,

(C) Since a liter of water corresponds to a 1 kg mass,

ˆv 1 = (V – ˆv 2 N 2 )/N 1 .

N 1 = 1 kg/18.02 kg kmole –1 = 0.0555 kmole, Therefore, applying Eq. (C),

ˆv 1 = 18.02 – 505.4 N 3/2 2 – 2151.6 N 2 2 in units of l kmole –1 . (D) The mole fraction X 2 = (1 – X 1 )=N 2 /(N 2 +N 1 )=N 2 /(N 2 + 0.055), i.e.,

N 2 = 0.055 (1/X 1 –1), and

1 = 18.04–6.625(1/X 1 –1) –6.567(1/X 1 –1) ,X 1 >0, in units of l kmole . Example 4

2 ˆv –1

A large flexible tank is divided into two sections A and B by a partition. Section A consists of 2 kmole of C 2 H 2 (species 1) at 320 K and 100 bar, and section B consists of 3 kmole of CO 2 (species 2) at the same temperature and pressure. The partition is removed, but the temperature and pressure are maintained constant. Determine V A , and V B using the RK equation of state for each component before mixing and the total mixture volume using an ideal mixture model.

Solution The RK equation has the form

P = RT/( v – b )– a /(T 1/2 v ( v + b )).

Therefore,

v 1 = 0.09 m 3 kmole –1 , and V A =2

× 0.09 = 0.18 m 3 .

2 = 0.10 m kmole , and V B =3 × 0.1 = 0.3 m .

1 1 2 2 1 1 2 2 × 0.09 + 3 × 0.1 = 0.48 m . Remarks

V id = v N=( v X + v X )N = v N + v N =2

This is an illustration of the law of additive volumes. When the partition is removed, the total volume is the same as the combined original volume at the same temperature and pressure.

The actual volume V = id ˆv

1 N 1 + ˆv 2 N 2 can differ from V , i.e., if the partition is re- moved in the rigid system, the final pressure may not be 100 bar.

d. Example 5 Determine V id for a gaseous mixture of 0.3 kmole of H

2 O (species 1) and 9.7 kmole of N 2 (species 2) at 160ºC and 100 kPa, and 25ºC and 100 kPa.

Solution Since the 433 K temperature is high, we expect each species to behave as though it is an ideal gas in its pure state. Therefore,

1 = (433 K,100 kPa) + 9.7 v 2 (433 K,100 kPa), where v (433 K, 100 kPa) = 8.314

V id = 0.3 v

1 × 423 ÷ 100 = 36 m kmole –1 (from the tables for super- heated vapor v

3 v –1

2 (433 K, 100 kPa) = 8.314 × 423 ÷ 100 = 36 m kmole .

V 3 = 0.3 × 36 + 9.7 × 36 = 10 × 36 = 360 m . At 298 K,

id

V id = 0.3 v 1 (298 K, 100 kPa) + 9.7 v 2 (298 K, 100 kPa).

Although water exists as a liquid under these conditions, in the mixture it exists as a vapor. Therefore,

V id = 0.3 v

1 (298 K, 100 kPa, liquid) + 9.7 v 2 (298 K, 100 kPa) = 0.3

× 0.018 + 9.7 × 24.8 = 240.5 m 3 . If we use a hypothetical gaseous state for water at 25ºC and 100 kPa, then using the

ideal gas law

V id = 0.3 v 1 (298 K, 100 kPa) + 9.7 v 2 (298 K, 100 kPa)

= 0.3 × 24.8 + 9.7 × 24.8 = 248 m 3 . Remarks

It is more accurate to determine V id using the specific volumes of pure components in the same state as they exist in the mixtures. We have used a hypothetical state to determine the volume pure water, since the wa- ter changes phase in the mixture.

Suppose we have 0.3 kmole of H 2 O(l) in compartment A at P =100 kPa and 9.7 kmole of N 2 at P= 100 kPa in compartment of B of a PCW assembly, which is immersed in

a bath at 25 °C. If the partition is removed and allowed to equilibrate at 25°C, 100KPa, then the vapor will become H 2 O(g). Since the water changes phase after

mixing, it must be endothermic, i.e., heat must be supplied from the thermal bath to evaporate the water molecules. Therefore, after mixing, F kj «F kk .