Completely Miscible Mixtures

1. Completely Miscible Mixtures

a. Liquid–Vapor Mixtures This case is illustrated through the following example.

c. Example 3

A solution consists of 40% methanol (species 1) and 60% water (species 2). Assume that methanol is completely miscible in water and that the solution behaves according to Raoult’s Law. The saturation pressure correlations for the pure components are:

ln p sat 1 (mm of Hg) = 20.61 – 4719.2/T K, and (A) ln p sat 2 (mm of Hg) = 20.60 – 5205.2/T K.

(B)

If the temperature is maintained at 20ºC, at what pressure will a vapor bubble begin to form? At this pressure and at 20ºC, determine the quality W, i.e., the ratio of moles of vapor in the vapor–liquid mixture, and the vapor–phase composition. At 20ºC, what is the pressure at which virtually the entire liquid has vaporized? De- termine the liquid composition and the quality at this pressure.

Solution Water is a low volatility substance and methanol is highly volatile. At 20ºC a vapor bubble first appears in pure water at sat P

2 = 17.03 mm of (0.0224 bar), while for methanol it appears at a higher pressure sat P

1 = 90.33 mm of mercury (0.119 bar). However we are interested in determining the pressure at which bubble is formed from a liquid mixture. A two-phase mixture forms when a bubble embryo first ap- pears in the continuous mother phase of liquid, and Raoult’s Law can be applied at this point. Applying Raoult’s Law for each component in the mixture,

(C) p 2 =X 2(l) P sat 2

p =X P 1 sat 1(l) 1 ,

(D) Using Eq. (A), at T = 20ºC, sat P

1 = 90.33 mm of Hg, (0.119 bar) i.e., from Eq. (C) p 1 = 0.4 × 90.33 = 36.13 mm of Hg(0.048 bar).

(E) Similarly from Eq.(D)

p 2 = 0.6 × 17.03 = 10.22 mm of mercury (0.013 bar). (F) Adding Eqs. (E) and (F), the pressure at which a vapor bubble forms

(G) Generalizing the result for a multicomponent mixture P= ΣX k(l) sat P k (T)

P=p 1 +p 2 = 46.35 mm of mercury (0.061 bar).

Since this state (0.0612 bars ,20ºC) lies on the saturated liquid line, at this pressure, the quality W = 0.

p 1 =X 1 P, (H) where X 1 is the vapor phase mole fraction of species 1. From Eqs, (E), (G), and (H),

X 1 =p 1 /P = 36.13 ÷ 46.35 = 0.78. (I) Similarly,

X 2 = 10.22 ÷ 46.35 = 0.22. (J) The vapor phase has a different composition as compared to the liquid composition.

Generalizing the result for a multicomponent mixture

(K) Applying Raoult’s law,

X k =p k / ΣX k(l) P sat k = X k(l) P sat / ΣX k(l) P sat k k .

X 1 = (X 1(l) P sat

sat

sat

1 /(X 1(l) P 1 +X 2(l) P 2 ), and

2 /(X 1 P 1 +X 2 P 2 ), i.e.,

(M)

Bubble Line

0.08 P vs X

D Wet Mixture

0.06 Dew Curve

V P vs Y

L 0.04

D Vapor

0.02 F S

X 2(l) ,X 2 , Z 2

Figure 3: Illustration of P–X k(l) or P-X k(l) -X k -Z diagram at 20ºC for a methanol (1)–water (2) solution.

When almost the entire liquid is vaporized X 1 = 0.4 and X 2 = 0.6, and from Eqs. (L) and (M) we have

X 1(l) = 0.112, X 2(l) = 0.888, and P = (X 1(l) sat

1 +X P 2 ) = 0.112 × 90.33 + 0.888× 17.03 = 25.24 mm of Hg. The last liquid drop contains 11% methanol and 89% water. One can alter the liquid composition and obtain plots of P vs X HO 2 (l) as shown by curve MFLKH in Figure 3 . Generalizing for K components, one can solve for X k( Ρ) using the K linear relations

P sat

2(l)

(Eqs. (K)) and the known vapor phase mole fractions, X k . At the saturated vapor state the quality W= 1.

Remarks Consider a 40% methanol and 60% water mixture at 20 °C in a PCW assembly so that pressure is at 0.132 bar (State C). As we start slowly removing smaller weights one

after another, the first vapor bubble (embryo phase) appears in the liquid (mother phase) at a pressure of 46.35 mm of mercury (or 0.061 bar, Figure 3 point F). This pressure is the vapor pressure at 20ºC and the pressure and temperature specify the saturated liquid state of the mixture. At phase equilibrium, inside that bubble, the mole fraction of water vapor is 0.22 and mole fraction of methanol is 0.78 (point D). The vapor in the embryo is still at 46.35 mm of Hg. Although water constitutes 60 % of the liquid phase (point F), it constitutes only 22 % of the vapor phase (Point D), since it is less volatile due to a lower vapor pressure. By altering the composition, the

P-X k( Ρ) diagram can be obtained as M DVGH. Now consider an 88.8% water and 11.2% methanol solution at 20ºC. The first vapor

bubble appears at a pressure of 24.36 mm of mercury (point K, 0.0325 bar) and the vapor composition is 60% water and 40% methanol (point G). Thus, for X 2( Ρ) = 0.6, P

= 0.061 bar, and X 2 = 0.22 while at X 2(l) = 0.888, P = 0.0322 bar, and X 2 = 0.6. As

X 2(l) → 1.0, P → P sat 2 t (i.e., 17.03 mm of mercury or 0.0225 bar, point H). As X 2(l) →

0 (and, consequently, X 1(l) → 1) P = P sat 2 t (90.33 mm of mercury or 0.119 bar, point M). A line connecting the pressures through the bubble points is called the bubble line

(saturated liquid line, MFKH) at which the first bubble appears. Consider a methanol–water mixture with X 2(l) = 0.6 at 20ºC. The mixture is in com- pressed liquid state at a pressure of 100 mm of mercury (point C, 0.132 bar). As the pressure is reduced below 0.061 bar (46.4 mm of Hg or the bubble pressure), say, to

0.0422 bar (32 mm Hg, point W), phase equilibrium requires that X 2(l) = 0.8 and X 2 =

0.43, Representing the initial molal fraction of species 2 as Z 2 , then at “W” (cf. Figure 3 ),

(K) Rewriting,

Z 2 = (N 2 ( l) + N 2 (g))/N, i.e.,

(L) where W = N g /N denotes the molal quality. Consequently,

Z 2 = (X 2(l) N l +X 2 N g )/N = X 2(l)( 1–W) + X 2 W,

W = (X 2(l) –Z 2 )/(X 2(l) –X 2 ).

(M) If Z 2 = 0.6, P = 32 mm of mercury (i.e., 0.042 bar), X 2(l) = 0.8, X 2 = 0.43, and, hence,

W = 0.54. Therefore, 54 % of the original mixture exists as vapor, while 46% is in the liquid state. Water (species 2) constitutes 43% of the vapor. We can apply the ideal gas equation for gas phase to compute the vapor volume. In order to specify the state at 20ºC for a two-phase mixture, we require additional information, such the overall

Vapor 90 Dew

C Liquid

X 2(l) ,X 2 , Z 2

Figure 4: Illustration of a T–X–Y diagram at a pressure of 760 mm of mercury (i.e., 100 kPa) for a methanol (species 1)–water (species 2) solution.

mixture composition Z. Different pressures then lead to different vapor phase compo- sitions. At same T, for a two component mixture, we can have two phase mixture within MFLKH (saturated liquid line for a mixture or bubble line) and MDVGH

(saturated vapor or dew line) with different pressures. For example, at 20ºC, Z 2 or

X 2(l) = 0.6, the pressure can be varied from 46.5 mm of mercury (0.0614 bar) at satu- rated liquid line to 24.36 mm of mercury (0.0322 bar) at saturated vapor line (see the vertical line FWG) but still maintain two phase mixture. Thus, we can have two de- grees of freedom (say T and P) for existence of two phases at given composition. On the other hand, for a pure component such as water if T = 100 C, two phase mixture (vapor and liquid) exists only at a single pressure of P =760 mm of mercury or 1 bar. If the pressure is raised, liquid is formed. Decreasing the pressure produces super- heated vapor.

If Z 2 = 0.6 then at X 2 = 0.6, Eq. (M) suggests that W = 1 regardless of the value of

X 2(l) . Since W = 1, all of the liquid has evaporated at this condition, which is called the saturated vapor state (that starts along (line MDVGH – Figure 4 ). At this state, the vapor concentrations equal the original liquid molal concentrations. Further reduction of pressure causes the vapor mixture to become superheated (state S).

Consider point S, which is at the superheated state, say at X 2 = 0.6 If the pressure of the superheated vapor mixture (the mother phase) is increased, at some pressure a liq- uid drop (the embryo phase) appears (Point G). In this manner, varying the mole frac-

tion of a component X 2 , one can identify the dew point ( Figure 3 ) curve or the satu- rated vapor curve. In context of Figure 3 , the region below the dew point curve MDVGH (cf. Figure 3 ) is the superheated region, while the region within it and the saturated liquid line MFLKH is called the wet region. The region above the saturated liquid line MFLKH is called the compressed (or subcooled) liquid region. P–X k(l) –T Diagram: The methodology of Example 3 can be repeated at various tem- peratures to obtain a three–dimensional space diagram. The maximum temperature at which a species exists in the liquid phase is its critical temperature (647 K in the case of water and 313 K for methanol. At this temperature the intermolecular separation distances are the same for the saturated liquid and the saturated vapor. When T c,1 <T <T c,2 , in that case the diagram does not extend to X 1(l) =1, since T > T c,1 . T–Xk Diagram: This case is illustrated through Example 4 and Figure 4 .

d. Example 4

A solution consists of 40% of methanol (species 1) and 60% water (species 2). As- sume that it follows Raoult’s Law and the following relations apply

ln P sat 1 (in mm of Hg) = 20.61 – 4719.2/T, T in K and (A) ln P sat 2 (in mm of Hg) = 20.60 – 5205.2/T, Tin K .

(B) The normal boiling point is defined when P sat = 100 kPa. From the above relations,

the normal boiling point of species 1 is 64.7 o

C and that of species 2 is 100 o

C. Draw a

graph of T with respect to X 2(l), X 2 , and Z 2 .

Solution Applying Raoult’s Law to each component,

p =X

P 1 sat 1(l) 1 , and

(C)

(D) Using Eq. (A) at 20ºC, sat P

p 2 =X 2(l) P sat 2 .

1 = 90.33 mm of Hg. Then

1 +p 2 = (1 – X 2(l) ) P 1 (T) + X 2(l) P 2 (T) = 760 mm of Hg. (E) From Eqs. (A), (B), and (E),

P = 760 = (1–X 2(l) ) exp (20.61 – 4719.2/T) + X 2(l) exp (20.60 – 5205.2/T). (F) Eq.(F) is non-linear in T at specified X 2(l) ; however it is linear in X 2(l) at specified T.

The graphs can be generated as follows. If T = 75ºC, P = 760 mm, then one solves for X

from Eq.(F): X 2(l) = 0.451, X 1(l) =1–X 2(l) = 0.549, P 2 = 128.41 mm, then using Eq. (D)

sat 2(l)

p 2 =X 2(l) P sat 2 (T)= 0.451 × 128.41 = 57.9 mm = X 2 P=X 2 × 760, i.e.,

X 2 = 0.169 so that X 1 = 0.831. Figure 4 illustrates T-X 2(l) -X 2 diagram at specified pressure while Figure 5 qualitatively illustrates T-X 2(l) -X 2 at various pressures.

b. Relative Volatility The relative volatility is defined as

(16) If X 2 =X 2(l) and X 1 =X 1(l) , then R v,21 =1. If R v,21 > 1, this implies that component 2 is highly

R v,21 = (X 2 /X 2(l) )/(X 1 /X 1(l) )

volatile compared to 1 and vice versa.

Figure 5: P–X–T diagram for a binary mixture, where C denotes the critical point; MFH the bubble line, and MVH the vapor line (from G. J. Wiley, and R. Sonntag, Fundamentals of Classical Thermodynamics, 3 rd Ed., John Wiley & Sons, 1986. With permission.).

c. P–T Diagram for a Binary Mixture The generation of this graph is illustrated through the following example.

e. Example 5

A solution consists of 40% of methanol (species 1) and 60% water (species 2). As- sume that it follows Raoult’s Law and the following relations apply

ln P sat 1 (in mm of Hg) = 20.61 – 4719.2/T, T in K, and (A) ln P sat 2 (in mm of Hg) = 20.60 – 5205.2/T, T in K .

(B) The normal boiling point is defined when P sat = 100 kPa. From above relations, the

normal boiling point of species 1 is 64.7 o

C. Draw a graph of P with respect to T with X 2(l) as a parameter.

C and that of species 2 is 100 o

Solution Applying Raoult’s Law to each component,

p =X

sat

1 1(l) P 1 , and

(C)

p 2 =X 2(l) P sat 2 , so that

(D)

(E) Fix a value for X 2(l) . Vary T, obtain P from Eq. (E). Thus P vs T at given X 2(l) can be

P=p 1 +p 2 = (1 – X 2(l) ) P sat 1 (T) + X 2(l) P 2 sat (T).

plotted.

d. P–X k(l) –T diagram We can repeat Example 3 for various temperatures and qualitatively obtain 3 D plot as shown in Figure 5 . For example, if species 2 is water at 20ºC, the curve MFH represents the bubble line while MVH represents the dew line. Repeating the calculation at T sat 2 = 40ºC, we can obtain similar curves. The line HC 2 descrbes the saturation pressure of pure species 2 while the line MC 1 shows the corresponding curve for species 1. We know that maximum temperature for species 2 is the critical temperature (647 K) while the corresponding tempera- ture is 313 K for methanol. When T

< T< T c, 1 c,2 , then the P-T-X k(l) -X k diagram will not extend to X 1(l) =1 since T > T c,1 and hence it is always vapor at x 1 =1. Similarly if T > T c,2 then it will not extend near X 2(l) =1. But it can have curves for some intermediate values of X 1(l) and X 2(l) .

e. Azeotropic Behavior The term azeotrope is used for situations that have no composition change. During the boiling of a binary mixture, at a certain liquid composition, the vapor composition can be the same as the liquid composition i.e., X 2 = X 2(l) . Therefore, the liquid components cannot be

separated. This state is an azeotrope. At a specified total pressure, as the mixture composition is changed, an azeotropic mixture reaches a minimum ( Q in Figure 6 ) or maximum in the boiling temperature (e.g., water–antifreeze mixtures). In a nonazeotropic mixture the boiling

temperature varies monotonically with composition (e.g., water–NH 3 mixtures, Figure 4 for H2O:methanol mixtures). Figure 6 presents a T–X k(l) diagram that illustrates the azeotropic

behavior of a binary mixture of species A and B. At point Q the azeotropic composition has a corresponding temperature T Q .

f. Example 6 Lake water at 25ºC is exposed to ambient air at 1 bar. Phase equilibrium occurs and Raoult’s Law can be assumed to be applicable. Assume that air is dissolved in the liq- uid water so that X air(l) = 0.019. What is the mole fraction of the water vapor in the gas phase? What will this vapor mole fraction be if no air is dissolved in the liquid water?

Include the POY correction factor to answer the question. If the partial pressure of the water vapor at phase equilibrium is 3.1 kPa, determine the mole fraction of air in the liquid solution.

Solution We will use the expressions

k (T,P,X k, )= µ k (T,P) + RT ln X k = µ k (T,P ) + RT ln (p k /P (T)), and

(B) Equating Eqs. (A) and (B),

µ (T,P,X )= µ (T,P sat k k(l) k )+v f (P – P sat ) + RT ln X k(l) .

v f (P – P sat ) + RT ln X k(l) = RT ln (p k /P sat (T)), where, p k /(X k(l) P sat (T)) = exp (v

(C) Ignoring the Poynting correction, and applying Raoult’s Law for water,

f (P – P )/RT) = POY.

sat

p HO =X HO 2 ,l P 2 sat HO 2 (25C) = 0.981 × 3.16. When p HO 2 = 3.1 kPa and P = 100 kPa,

X HO 2 = 3.1 ÷100 = 0.031. In the absence of any dissolved air, X sat

HO 2 ,l = 1 so that P HO 2 =P = 3.16 kPa, and

X HO 2 = 3.16 ÷100 = 0.0316. The water vapor mole fraction at 25ºC is reduced if air is dissolved in liquid water.

Figure 6: The T–X k, –X k diagram for a binary azeotropic mixture con- taining two species A and B, where g denotes the dew line (saturated vapor), f the bubble line (saturated liquid), and Q the azeotropic point. (From A. Bejan, Advanced Engineering Thermodynamics, John Wiley and Sons., 1988, p. 271. With permission)

If the RHS of Eq. (C) has a significant value, then If the RHS of Eq. (C) has a significant value, then

g. Example 7 An ideal solution contains alcohol (species 1 present on a 60% mole basis in the liq- uid phase and 87% in the vapor phase) and water (species 2) at a temperature and

pressure of 60ºC and 433 mm of Hg. At 60ºC, P sat = 625 mm of Hg, and P sat 1 2 = 144 mm of Hg. Determine ˆf 1(g) , ˆf 2(g) , ˆf 1(l) , and ˆf 2(l) . Solution

The specified pressure is greater than P sat 2 but less than P sat 1 . Therefore, at the speci- fied temperature and pressure, pure water exists as compressed liquid while pure al- cohol exists as superheated vapor.

ˆf 1(g) =X 1(l) 1(l) f (60ºC, 433 mm of hg) = X 1(l) × 433 mm of mercury = 0.87 × 433 = 377,

since the fugacity of the vapor equals the vapor pressure, assuming ideal gas behav- ior. Similarly,

ˆf 2(l) =X 2(l) f 2(l)( 60ºC, 433 mm of Hg), ≈X 2(l) f 2(g)( 60ºC, 144 mm of Hg), where

f 2(g) ≈ 144 mm of Hg, since the liquid fugacities do not change significantly with pressure. Hence,

ˆf 2(l) = 0.4 × 144 = 57.6 mm of Hg. Likewise.

ˆf 1(l) =X 1(l) 1(l)( f 60ºC, 433 mm of Hg).

In its pure state, the alcohol exists in the form of a vapor. We must determine the fu- gacity at the hypothetical liquid state (T,P = 60ºC and 433 mm of Hg). We will as- sume that

f (60ºC, P 1(l) sat 1 at 60 C) ≈f 1(l)( 60ºC, 625 mm of Hg),

since the term v l

(l)( 433 – 625) « f (60 C, 625 mm of Hg). However,

f 1(l)( 60ºC, 625 mm of Hg) = f 1(g)( 60ºC, 625 mm of Hg) = 625 mm of Hg, since the vapor is assumed to be ideal.

Therefore, ˆf 1(l) = 0.6 × 625 = 375 mm of Hg.

ˆf 2(l) = 0.13 ×f 2 (60ºC, 433 mm of Hg). We will use a hypothetical ideal gas state for f 2 (60ºC, 433 mm of Hg), i.e., ˆf 2(g) = 0.13 × 433 mm of mercury = 56 mm of Hg.

Remarks We see that

ˆf 2(g) ≈ ˆf 2(l) , and ˆf 1(g) ≈ ˆf 1(l) , thereby satisfying the phase equilibrium criterion. We have assumed ideal gas behavior for the hypothetical vapor state. Real gas be-

havior can also be accounted for, as shown below. Recall that f 2 (60ºC, 433 mm of Hg) must be calculated for a hypothetical liquid state. To do so, we can first deter- mine the fugacity of saturated vapor or liquid at the state (60ºC, 144 mm of Hg). Then we can employ the relation d(ln(f)) = ∫vdP/(RT) = ∫(Pv/(RT))d(ln P) = ∫Zd(ln P).

Integrating between the limits P sat and P, and assuming that Z = Z sat , we obtain the relation ln(f/f sat ) ≈Z sat ln (P/P sat ), or (ln ( φ))/(ln (f sat /P sat )) = Z sat (60ºC, 144 mm of Hg).

A plot of ln ( φ) with respect to P R at specified T R is approximately linear.

h. Example 8

A fuel droplet contains a binary mixture of 60% n–heptane (species 1) and 40% hexa- decane (species 2). Assume that air (species 3) is insoluble in the liquid phase. The Cox–Antoine relation can be assumed to apply, i.e.,

ln P sat k =E k +F k /(T + G k ), k = 1,2 (A) where E k , F k , and G k are constant for a specified fuel component. If the pressure is

expressed in mm of Hg, E 1 = 15.89, F 1 = –2911.32, and G 1 = –56.4, and E 2 = 24.66,

F 2 = –10660.2, and G 2 = 54.1 when 1 < P< 40 mm of Hg. Determine the partial pres- sures of the two species 297 K in air if the ambient pressure is 100 kPa and the gas–phase composition.

Solution Applying Raoult’s Law for miscible mixtures p k =x k P sat k , k=1,2, since

p 1 +p 2 +p 3 = P, X 1,l sat P +X

sat

1 2,l P 2 +p 3 = P, i.e.,

(C) At 20ºC, P sat 1 = 44.245 mm of Hg, and P sat 2 = 0.00338 (i.e., species 2 is almost non-

vaporizable). Therefore, p 1 =X 1,l P sat 1 = 0.6 × 44.245 = 26.55 mm of Hg, and

p = (1–X

sat

2 1,l ) P 2 = 0.4 × 0.00338 = 0.001352 mm of Hg.

Using Eq. (C), p air = 760 – (26.55 + 0.001335) = 733.45 mm of Hg, i.e.,

X 1 = 26.55 ÷760 = 0.0349, and

X air = 733.45 ÷ 760 = 0.965, (Since hexadecane is virtually nonvaporizing, we neglect its mole fraction.) The mass

fraction of n–heptane Y 1 = 0.0349 × (7 × 12 + 16 × 1) ÷ (0.0349 × 100 + 0.965 × 28.97) = 0.11, i.e.,

Y 3 = 0.89 The variation of the temperature of a droplet containing a binary mixture consisting of n–heptane (60 %) and hexadecane with respect to the n–heptane mole fraction is il- lustrated in Figure 7 at P=100 kPa.

i. Example 9 Dissolved air in water provides the oxygen and nitrogen that are necessary to sustain marine life. Obtain approximate relations that can be used to determine the trace mole

fractions of oxygen and nitrogen in liquid water. Recall that in air (X O 2 /X N 2 ) = 3.76. Solution

We will assume that Raoult’s Law applies, i.e.,

X sat

O 2( ) l

=X O 2 P/ P O 2 (T), and

(A)

X N 2( ) l =X N 2 P/ P sat N 2 (T).

(B)

In air X N 2 /X O 2 = 3.76. Furthermore,

O 2 N 2 )/( P N 2 (T)/ P O 2 (T))

O 2( ) l N 2( ) l

(C)

Since P sat (T) ≈ P sat N 2 O 2 (T), (X O 2( ) l /X N 2( ) l ) = (X O 2 /X N 2 ) = 3.76.

Remarks As the temperature rises, the value of P sat k for a substance increases. Hence X k,l de-

creases. The warming of river water decreases the O 2 and N 2 concentrations in it. j. Example 10

A 20-liter rigid volume consists of 80% liquid and 20% vapor by mass at 111.4ºC and

1.5 bar. A pin is placed on piston to prevent its motion. Gaseous nitrogen is isother- mally injected into the volume until the pressure reaches 2 bar. What is the nitrogen mole fraction in the gas phase? Assume that N 2 does not dissolve in the liquid. What happens if there is no pin during the injection of N 2 . Instead of adding N 2 , dis- cuss the effects with salt addition at 111.4ºC.

Solution P sat (111.4ºC) = 1.5 bar. The total volume

(A) = m (0.2 × 1.159 +0.8× 0.001053) = m (0.233).

V=m g v g +m f v f = m (x v g + (1–x) v f )

Therefore, m = 20 ×0.001/0.233 = 0.0858 kg Using the ideal gas law for the vapor phase

m=V f /v f + (V – V f )/(RT/P o ),

(B)

(C) The pressure increases as additional gas is injected, thereby increasing the Gibbs en-

V f = (m – P o V/RT)/(1/v f –P o /RT) ≈v f (m – P o V/RT).

ergy of the liquid and vapor phases. In case of liquid water,

g sat l (T,P) = g l (T,P )+v l (P – P sat ). For an ideal gas mixture in the vapor phase, ˆg HO 2 (T,P,X H2O )= g HO sat 2 (T,p HO 2 )= g H2O (T,P )+ ∫ v H2O(g) dP

(D) Equating Eq. ( C) with Eq. (D)

g (T,P g sat )+ = R T ln(p /P sat HO 2 ).

v (P – P sat ) = RT ln (p /P sat ), or ln(p /P sat HO 2 HO 2 ) = (v (P – P sat l ))/(RT) (E) This relation is known as the Kelvin–Helmholtz formula which shows the effect of

total pressure on partial pressure of vapor. Note that the partial pressure of H 2 O in the vapor phase is not the same as saturation pressure at T.

For water, v

l = 0.001053 m 3 kmole –1 ,P = 1.5 bar, and for this case P = 2 bar, and T = 384.56 K. Therefore, the partial pressure of H 2 O in vapor phase,

sat

p HO 2 = 1.500445 bar.

a value close to saturation pressure at T= 384.6K since v f is small. Further

Figure 7: The variation of the mole fraction of heptane vapor with droplet temperature for a mixture containing 60 % n–heptane and 40 % hexadecane at 100 kPa.

X HO 2 = 0.75022, and X N2 = 0.24798. The vapor mass

m v =p v V v /RT = p v (V – V f )/RT, and

the liquid mass

m f =V f /v f .

Adding the two masses,

m=p v (V – V f )/RT + V f /v f .

Therefore,

V f = (m – p v V/RT)/(1/v f –p v /RT) ≈v f (m – p v V/RT).

(F) Since P v after N 2 injection is slightly higher than P v before N 2 injection, there should

be more vapor; thus the volume of liquid decreases. According to Le Chatelier, the system counteracts the pressure increase by increasing the volume of the vapor phase. If we ignore the term (v (P – P ))/(RT), in Eq. (A) this implies that p

=P f sat sat H2O and X v = 0.75. The injection of nitrogen implies that X ) H2O <1. Pressure remains constant. Therefore, ) )

g HO 2 = g HO 2 (T,P) + R T ln X H2O . Since X H2O <1, g HO 2 < g HO 2 () l (T,P), as long as the temperature and pressure are maintained, vaporization continues until all of the liquid

vaporizes. Similarly when we add salt in water( or an impurity), the Gibbs function of the liquid

H 2 O decreases which causes the vapor molecules to cross over from the vapor into the liquid phase. Remarks At a specified temperature, an increase in pressure causes the "g" of liquid to increase slightly. The Gibbs free energy of the vapor equals that of the liquid. If the vapor is an ideal gas, the enthalpy of the vapor will remain unchanged. The slight Gibbs en- H 2 O decreases which causes the vapor molecules to cross over from the vapor into the liquid phase. Remarks At a specified temperature, an increase in pressure causes the "g" of liquid to increase slightly. The Gibbs free energy of the vapor equals that of the liquid. If the vapor is an ideal gas, the enthalpy of the vapor will remain unchanged. The slight Gibbs en-

µ k(l) (T,P) = µ k(g) (T, P). If the vapor phase is isothermally pressurized, then

g k(l) (T,P) + d µ k(l) = µ k(g) (T,P) + d µ k(g( ,v k(l) dP l =v k(g) dP g and dP l /dP g =v k /v k l . An increase in the pressure in the vapor phase requires a large change in the liquid

phase pressure to ensure that liquid–vapor equilibrium is maintained.