Immiscible Mixture

2. Immiscible Mixture

a. Immiscible Liquids and Miscible Gas Phase This case is illustrated through the following example.

k. Example 11 Consider binary vapor mixture of methanol (species 1) and water (species 2) that are assumed to be immiscible in the liquid phase. Illustrate their behavior with respect to pressure and temperature. You may assume that

ln sat P 2 = 13.97 – 5205.2/T (K), and (A) ln sat P 1 = 13.98 – 4719.2/T (K).

(B) Solution

Employing Eqs. (A) and (B), the normal boiling points of species 1 (methanol) and 2 (water) are, respectively, 64.4 and 100ºC. We will employ Raoult’s Law, in which the liquid mole fractions for water and methanol must be set to unity, since they are immiscible. Therefore,

(D) Upon adding Eqs. (C) and (D), we obtain the expression

1 (T) = X 1 P.

(E) Figure 8 shows the T- X k(l) -X k diagram.

P sat 2 (T) + P sat 1 (T) = P.

Remarks In case of immiscible mixtures, partial pressures are only a function of temperature alone. Irrespective of the liquid phase composition, at a specified temperature, P sat 2 can be obtained from Eq. (A), while sat P

1 can be, likewise, obtained using Eq. (B). Using Eqs. (C) and (D), we obtain the values of X 1 and X 2 for a specified pressure, and plots of temperature can be plotted with respect to composition, as shown in Figure 8 . The lines BME and EJGA in that figure are called the dew lines for species

1 and 2, respectively. The region above the curve BMEJGA is the superheated vapor mixture region while that below the curve CELD is the compressed liquid region. Consider the following scenario. A vapor mixture is contained in a pis-

ton–cylinder–weight assembly, such that P = 1 bar, X 2 = 0.6, and T = 100ºC (cf. point S). Species 2 exists in the form of superheated vapor, since p 2 = 0.6 bar at T = 100ºC. The cylinder is now cooled. The saturation temperatures T sat 2 = 86.5ºC at p 2 = 0.6 bar, and sat T

1 = 43.87ºC at p 1 = 0.4 bar. The assembly contains a vapor mixture only, as long as T>86.5ºC. As the vapor mixture is cooled, a liquid drop appears at T = 86.5ºC

100 K

90 Vapor Mixture

T vs X

2 Vapor Mixture+Liquid 2

B Species 1

D 60 C

Vapor Mixture+Liquid 1

T dew of 1 50

Separate Liquid Species 1 and 2

X 2 ,Z 2

Figure 8: A T–X l –X diagram for an immiscible liquid solution. (point G). (If the gas phase composition is changed to X 2 = 0.2, in that case the first

liquid drop appears at 61ºC (point E)). If the mixture is cooled to 70ºC (cf. point H), phase equilibrium – that is manifested in the form of Eq. (C) – implies that vapor

phase mole fraction must reduce to X 2 = 0.3 (cf. point J), i.e., more of species 2 must condense. It also implies that X 1 must increase to 0.7 from the initial mole fraction of

0.4. Eqs (D) and (B) dictate that T sat 1 = 52ºC, so that species 1 at 70ºC exists in the form of a superheated vapor. Upon further cooling to 60ºC, phase equilibrium re-

quires that X sat

2 = 0.19 (cf. point E), and T 1 increases to 60ºC. Any further cooling causes both species 1 and 2 to condense, where the condensate phase is an immiscible binary mixture. Within the region EJGADE (i.e., for X 2 > 0.19, 60ºC <T < 100ºC),

liquid species 2 and vapor mixture must coexist. In region BMEC (X 2 < 0.19, 60ºC < T <64.7ºC), liquid species 1 and vapor mixture must coexist. At 60ºC, X 2 = 0.19, and both the liquid and vapor mixture coexist. At point E there are two liquid phases and one vapor phase. According to Gibbs phase rule, F = K + 2 – π = 2 + 2 – 3 = 1. Therefore, there is one independent variable in the

set (P, T, X 2 ). In case the pressure is fixed, then the temperature and X 2 are fixed (i.e., 60ºC, and X 2 = 0.19) for coexistence the three phases to coexist. If the mixture is cooled from 100ºC (cf. point K), species 1 condenses, increasing the mole fraction of species 2 until X 2 = 0.19. Now assume that the liquid mixture is heated at the condition X 2(l) = 0.6 and P = 1 bar in a piston–cylinder–weight assembly. At low temperatures, the sum of the saturation pressures (cf. Eq. (E)) is insufficient to create the imposed 1 bar pressure. Therefore, at T < 60ºC (point Q), the fluid exists as a compressed liquid. At ≈60ºC (cf. point L),

the sum of the saturation pressures is roughly 1 bar. The temperature at this condition can be predicted using Eqs. (A), (B), and (E) as 60ºC. Consequently, the values of X 1

and X 2 can be determined as 0.81 and 0.19 using Eqs. (C) and (D). Thus first vapor bubble at a 1 bar pressure appears at 60ºC. At this point there are three phases (two and X 2 can be determined as 0.81 and 0.19 using Eqs. (C) and (D). Thus first vapor bubble at a 1 bar pressure appears at 60ºC. At this point there are three phases (two

T sat 1 = 60ºC), since the vapor phase mole fraction of species 1 is 0.81, the ratio of the moles of species 2 that are vaporized to those of species 1 is 0.19 ÷0.81. Therefore, for every 0.4 kmole of species 1 that are vaporized, the moles of species 2 that are va- porized equal 0.4 ×0.19÷0.81 = 0.094 kmole. Hence, the vapor mixture contains 0.4 kmole of species 1, 0.094 kmole of species 2, and 0.6–0.094 = 0.506 kmole of species

2 remain in the liquid phase. Now the species 1 from liquid have been completely va- porized. Once T>60ºC, further vaporization of species 2 occurs, thereby increasing the mole fraction of species 2 in the vapor state, and it is possible to determine the

value of X 2 along the curve EJGA. As the temperature reaches 86.5ºC (cf. point G), all of the initial 0.6 kmole of species 2 in the liquid phase vaporize so that X 2 = 0.6.

b. Miscible Liquids and Immiscible Solid Phase Oftentimes two species 1 and 2 are miscible in the liquid phase, but are immiscible in the solid phase and each species forms its own aggregate in the solid phase (i.e., upon cooling of the liquid mixture, the two species form two separate solid phases). In this case, at phase equilibrium,

ˆf 1(l) = ˆf 1(s) , and ˆf 2(l) = ˆf 2(s) (17a) Under the ideal solution assumption and since X 1(s) = 1 due to immiscibility

X 1,l f 1(l)( T,P) = f 1(s)( T,P). (17b) For example, pure H 2 O at a temperature of –5ºC and a pressure of 1 bar should exist as ice.

However, if the water is a component of a binary solution (e.g. salt addition), then ˆf HO 2 () l =

f H2O(l) (–5ºC, 1 bar) = X HO 2 (l) f H2O(l) (–5ºC, P ) POY (l) , where POY = exp (v (l) (P – P sat )/RT). Generalizing,

sat

HO 2 (l)

f 1(l)( T,P) = X 1(l) f 1(l) (T,P 1 ) POY 1(l) =f 1,s (T,P 1 sat ) POY 1(s) . (18) Since f sat

X sat

1(l)

(T,P 1 )=f 1(s) (T,P 1 sat ),

1(l)

X 1(l) POY 1(l) = POY 1(s) , and X 2(l) POY 2(l) = POY 2(s) . (19) However, X 1(l) +X 2(l) = 1, so that POY 1(s) /POY 1(l) + POY 2(s) /POY 2(l) =1

(20) Following example 11, the pressure can be determined from Eq. (20) at a specified tempera-

ture. The mole fractions X 1(l) or X 2(l) at that pressure can be obtained using Eq. (19).