Ideal Solution and Raoult’s Law

2. Ideal Solution and Raoult’s Law

a. Vapor as Real Gas Mixture If α represents the liquid and β the vapor phase, then for an ideal solution in the liquid

phase and an ideal mixture of real gases,

(7) where f k( α) denotes the fugacity of component k in its pure state in the α phase (i.e., the same

X k( α) f k( α)( T,P) = X

k( β) f k( β) ,

phase as the mixture) and at the same temperature and pressure and X k mole fraction. The fu- gacity of the liquid phase at any temperature and pressure

f k( α) (T,P) = f k( α) (T,P sat ) POY ( α) , where POY = exp( ∫ ( v () α (T,P)/RT) dP). (8)

P sat

where POY is the Poynting correction factor (Chapter 7). Likewise, in the vapor state where POY is the Poynting correction factor (Chapter 7). Likewise, in the vapor state

P sat

where v ( β) can be obtained as a function of T and P using any real gas equation of state (Ap- pendix B).

b. Vapor as Ideal Gas Mixture In general, the RHS of Eq. (7) is related to the partial pressure of the k–th component if β is the gas phase and Eq. (7) assumes the form

X k( α) f k( α) (T,P) = X k P (10) Typically, v f has a small value and we can neglect the term POY so that

f sat k( α) (T,P) = f k( α) (T,P )=f k( β)( T,P )= P k , and (11) p sat

sat

sat

k =X k( α) P k (12) Note that the capital lettered P sat k represents the saturation pressure of pure species k at given

T. The relation in Eq. (12) is known as Raoult’s law, which can be used to solve problems pertaining to multicomponent phase equilibrium

Consider two components (namely, 1 and 2) at a specified temperature and pressure in two phases α (liquid) and β (vapor). In that case

f 1( α) X 1( α) =X 1 β f 1( β) ,f 2( α) X 2( α) =X 2, β f 2( β) , X 1( α) +X 2( α) = 1, and X 1( β) +X 2( β) =1 (13) Unknowns, X 1( α) ,X 2( α) ,X 1( β) , and X 2( β) . Knowing f 1( α) ,f 1( β) ,f 2( α) ,f 2( β) , and four equations (cf.

Eq. 13), we can solve for the four unknowns at specified T and P.

i. Remarks The ideal solution model allows us to express partial molal properties in terms of pure properties and molal concentrations. For a miscible liquid mixture under phase equilibrium

ˆf 1(f) = ˆf 1(g) . (14) In the future, subscript (g) for gas phase will be omitted

For an ideal liquid mixture, Eq.(14) becomes

(15) For a mixture immiscible in the liquid phase, but which is miscible in the gas phase,

ˆf 1(f)( T,P,X 1 )=X 1(l) f 1(f)( T,P).

the phase equilibrium condition implies that ˆf 1 (g) e., fugacity of component 1 in the vapor mixture = f 1 (T, P), i.e., the fugacity of the pure component in the liquid phase since the component 1 is immiscble.

a. Example 1 Sea water consists of salt (the solute) and water (the solvent). Determine the boiling temperature of solution with an 8% salt concentration at 1 bar. Use both Raoult’s

Law, and the equality ˆ g HO 2 () l = ˆ g HOg 2 () .

Solution The partial pressure of water

sat

HO 2 =X HO 2 1(l) P HO 2 (T).

At 100 kPa, since X HO 2 (l) = 0.92,

100 = 0.92 sat P

sat

HO 2 (T), i.e., P HO 2 (T) = 108.7 kPa.

At this pressure, using the tables T = 102.1ºC, which is the boiling point of the salt water. At phase equilibrium

(A) Using the ideal solution model for the liquid phase in context of Eq. (A),

g ˆ HO 2 () l (T,P,X HO 2 (l) )= ˆ g HOg 2 () (T,P,X HO 2 )

(B) Similarly, for the gas phase,

g ˆ HO 2 () l (T,P,X HO 2 )= g HO 2 () l + R T ln X HO 2 .

g ˆ HOg 2 () (T,P,X HO 2 )= g HOg 2 () (T,P) + R T ln X HO 2 .

(C) Assuming the gas phase to consist of only water vapor, i.e., X HO 2 , l = 1, using Eqs.

(A)–(C),

(D) If we assume the boiling point to be 100ºC,

g HO 2 () l + R T ln X HO 2 (l) = ˆ g HOg 2 () (T,P,1).

g ˆ HOg 2 () (373 K, 100 kPa, 1) = g HOg 2 () (373 K, 100 kPa)

= h HOg 2 () (373 K, 100 kPa) – 373 × s HOg 2 () (373 K, 100 kPa) + 0 = 2676.1 × 18.02 – 373 × 7.3549 × 18.02 = –1212 kJ kmole –1 .

(E) In the liquid state, X HO 2 , l = 0.92, and using Eq. (B), (373 K,100 kPa,0.92)= g HO 2 () l (373 K,100 kPa)+8.314 ×373×ln 0.92, and

(F)

g HO 2 () l (373 K,100 kPa) = h HO 2 () l –T s HO 2 () l

419.04 × 18.02– 373 × 1.3069 × 18.02 = –1212 kJ kmole –1 . (G) Using Eqs. (E)–(G),

g ˆ HO 2 () l (373 K, 100 kPa, 0.92) = –1212–8.314 ×373 ln 0.92 = –1407.6 kJ kmole –1 , i.e.,

g ˆ HO 2 () l – ˆ g HOg 2 () = –1470.6 –(–1212) = 258.6 kJ kmole –1 . The partial molal Gibbs free energy of the liquid phase water is lower than in the gas

phase, since the water mole fraction in the liquid phase is lower than unity. The liquid temperature must be increased in order to raise the value of ˆ g HO 2 () l so that it equals

that of the vapor phase water. In this context assume that the boiling point T BP equals 110ºC or 383 K. In that case, for the liquid phase water, Eq. (E) yields

g ˆ HO 2 () l (383 K,100 kPa,0.92)= g HO 2 () l (383 K,100 kPa)+8.314*383*ln 0.92. (H) The Gibbs energy of liquid water at 383 K and 100 kPa is unavailable from the tables,

since this is a superheated vapor state for pure water. The Gibbs energy of saturated liquid water at 383 K is available at a pressure of 143 kPa, i.e., since this is a superheated vapor state for pure water. The Gibbs energy of saturated liquid water at 383 K is available at a pressure of 143 kPa, i.e.,

= 461.3 – 383 × 1.4185 = –81.99 kJ kg –1 = – 1477.5 kJ kmole –1 . (J) We will use the relation

dg T = v dP. (K) Assuming the liquid to be incompressible, and integrating Eq. (K) between the limits

(T,P sat ) and (T,P), for the hypothetical liquid water state,

g HO 2 () l (T,P) – g f (T,P sat )= v f (P – P sat ). where for water v f = 0.018 m 3 kmole –1 at 373 K. Thus

g HO 2 () l (383 K, 100 kPa) = g f (383 K, 143 kPa) + 0.018 ×(100 – 143) = –1477.5 + 0.018 ×(100 – 143) = –1478.2 kJ kmole –1 , i.e.,

(L) It is seen that v sat

f (P-P ) term is negligible. Hence

(K) Using Eq. (B), the partial molal Gibbs function of liquid water can be determined, i.e.,

g –1

HO 2 () l (T,P) ≈ g f (T,Psat) = –1477.5 kJ kmole .

g ˆ HO 2 () l (383 K,100 kPa, 0.92) = –1476.7 + R T ln (0.92) = –1742 kJ kmole –1 . (M) For the gaseous water,

g ˆ HOg 2 () (383 K, 100 kPa) = g HOg 2 () (383 K, 100 kPa) + R T ln (1). In the gaseous state, water is superheated, and from the tables

HOg 2 () (383 K,100 kPa)=2696.4 ×18.02–383×7.41×18.02= –2580.5 kJ kmole . (N) Therefore,

HO 2 () l (383 K, 100 kPa, 0.92) – ˆ g HOg 2 () (383 K, 100 kPa) = 838 kJ kmole . Using Eqs. (H) and (N) we can interpolate for the temperature at which

(ˆ g HO 2 () l –ˆ g HOg 2 () ) = 0. That temperature is 102.2ºC. Remarks

We have assumed the ideal solution model applies in the liquid phase. This model as- sumes that the attractive forces between water–water molecules equal those between water–salt molecules. Therefore, the change in the boiling temperature of water in the mixture compared to that of pure water is uninfluenced by attractive forces. When additional salt is introduced into the solution the mole fraction of liquid water is reduced. At phase equilibrium, the random condensation rate of a component k of a mixture must equal the random evaporation of that component from the liquid phase. The surface of a mass of pure water exposed to the gas phase contains only liquid water molecules, and the condensation and evaporation rates equal one another at a gas–phase pressure of 100 kPa at a temperature of 373 K. In the case of salt water the same area will accommodate a smaller number of liquid water molecules (due to the presence of salt molecules), which reduces the evaporation rate. Phase equilibrium considerations imply that a smaller number of vapor molecules consequently con- dense, and the vapor pressure at 373 K must be lower than 100 kPa. In order to achieve the same vapor pressure as is generated by the pure component, the liquid We have assumed the ideal solution model applies in the liquid phase. This model as- sumes that the attractive forces between water–water molecules equal those between water–salt molecules. Therefore, the change in the boiling temperature of water in the mixture compared to that of pure water is uninfluenced by attractive forces. When additional salt is introduced into the solution the mole fraction of liquid water is reduced. At phase equilibrium, the random condensation rate of a component k of a mixture must equal the random evaporation of that component from the liquid phase. The surface of a mass of pure water exposed to the gas phase contains only liquid water molecules, and the condensation and evaporation rates equal one another at a gas–phase pressure of 100 kPa at a temperature of 373 K. In the case of salt water the same area will accommodate a smaller number of liquid water molecules (due to the presence of salt molecules), which reduces the evaporation rate. Phase equilibrium considerations imply that a smaller number of vapor molecules consequently con- dense, and the vapor pressure at 373 K must be lower than 100 kPa. In order to achieve the same vapor pressure as is generated by the pure component, the liquid

P sat HO 2 = (P/X k(l) ) (O) Recall from Chapter 7 (Clausius Clapeyron relation) that we can approximately ex-

press P sat HO 2 in the form ln sat P

HO 2 = A – B/T. (P) Employing Eq. (O) in (P) and solving for T,

T = B/(A – ln(P/X k(l) )). (Q) For the pure component X k, (l) =1, and T pure = B/(A – ln (P)).

(R) Using Eqs. (Q) and (R), (1/T pure – 1/T) = (1/B) ln X k(l) .

(S) Since T = T pure + δT, and δT/T pure « 1, the LHS of the above relation can be expanded

in a binomial series. Retaining only the first order terms, Eq. (S) yields δT = –( T 2 pure /B) ln X k(l)

(T) Recall from the Clausius Clapeyron relation, Chapter 7 that B = h fg /R. Therefore,

δT = –( 2 T pure R/h fg ) ln X k(l) . (U) In case of water B = 5205.2 K and T pure = 373 K. For the case X k(l) = 0.92, δT = 2.23

K, i.e., the boiling temperature T BP = 102.3 C. At a specified pressure, the boiling temperature is increased by the amount δT. In the current example, the salt concentration is much less compared to water. Thus

X H2O(l) = 1– X salt ,X salt << 1, then ln X H2O(l) = ln (1– X salt ) . –X salt . Using this result in Eq. (U), the boiling point elevation is given as δT = (T 2 pure R/h fg )X salt, (V)

For H 2 O, one can show that δT = 28.441 X salt . Sometimes we write δT = k b Mo solute , where Mo solute is the molality of the solution (Eq.(1e), Chapter 8). Comparing this

empirical expression with Eq. (V) and replacing the molal fraction with molality (see Example 1 in Chapter 8), we can show that

b =RT pure /(1000 ×h fg ) = (8.314/18.02) ×373 /(1000 ×2257) = 0.51 (W) Since gas phase consists of only H 2 O, the salt water is purified by distillation process.

The anti–freeze in your car consists of glycol and water solution. The glycol is almost non–volatile while water is volatile. Thus we can use Eq. (U) to determine the rise in boiling temperature with addition of glycol. The freezing point depression with addi- tion of salt can also be determined using similar derivation.

b. Example 2 In general, when hydrocarbon droplets burn in air, the droplets first vaporize, follow- ing which the vapor is transported away from the droplet surface before burning. Phase equilibrium is often assumed between the drop and the surrounding air. In this problem we will determine the partial pressure of a vapor at the surface of a droplet. Consider a water droplet at a uniform temperature of 90ºC that is vaporizing in air at a pressure of 1 bar. The air consists of a mixture of oxygen and nitrogen, while the

droplet consists of a pure water. Assume ideal gas behavior to apply in the gaseous state. Determine: The water vapor pressure. The mole fraction of the vapor at the droplet surface. The mole fraction of water vapor in case the pressure is raised to 10 bar, but the pres- sure is held constant (e.g., injection of a liquid fuel spray in gas turbines). The Gibbs energy of water in the liquid and vapor phases at the elevated pressure.

Solution

The pressure in the liquid droplet equals that of the ambient, while the water vapor

pressure p v is lower. In the droplet X HO 2 (l) = 1, and applying Raoult’s Law at the drop- let surface

p sat

v =X HO 2 (l) P HO 2 (T).

The saturation pressure of water sat P HO 2 = 0.7014 bar at 90ºC. Hence p H2O = 0.7014bar The gas phase consists of multiple components so that

p v =X HO 2 (l) P, i.e., X HO 2 (l) = 0.7014.

If total pressure is 10 bar, the at T = 90 C,

X HO 2 (l) =p v /P = 0.7014 ÷10 = 0.07014, i.e., X Air = 0.92986. Note that increased P at same T reduces the mole % .

The Gibbs free energy

g HO 2 ( Ρ) (90ºC, 10 bar) = h HO 2 ( Ρ) (90ºC, 10 bar) – Ts HO 2 Ρ) ( (90ºC, 10 bar)

≈ sat

h sat

HO 2 () l (90ºC) – T s HO 2 () l (90ºC) × 1.1925 = –55.96 kJ kg = 376.92 – 363 –1 .

The vapor adjusts its mole fraction in the gas phase such that

g HO 2 Ρ)( ( 90ºC, 10 bar) = g HO 2 (g)( 90ºC, 10 bar, X H2O ).

In an ideal gas mixture, pH2O(g) = 0.07014 * 10 = 0.7014 bar

g HOl 2 () (90ºC, 10 bar) = ˆ g HOg 2 () (90ºC, 0.7014 bar).