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Chapter 10 - Section B - Non-Numerical Solutions

10.5 For a binary system, the next equation following Eq. (10.2) shows that P is linear in x1. Thus no

maximum or minimum can exist in this relation. Since such an extremum is required for the existence

of an azeotrope, no azeotrope is possible.

10.6 (a) Because benzene and toluene are chemically similar and the pressure is only 1(atm), this system

can be modeled by Raoults law to a good approximation.

(b) Although n-hexane and n-heptane are chemically similar, a pressure of 25 bar is too high for

modeling this system by Raoults law.

(c) At 200 K, hydrogen is supercritical, and modeling the hydrogen/propane system at this tempera-

ture by Raoults law is out of the question, because no value of P sat for hydrogen is known.

(d) Because isooctane and n-octane are chemically similar and at a temperature (373.15 K) close to

their normal boiling points, this system can be modeled by Raoults law to a good approximation.

(e) Water and n-decane are much too dissimilar to be modeled by Raoults law, and are in fact only

slightly soluble in one another at 300 K.

10.12 For a total volume Vt of an ideal gas, PVt = nRT. Multiply both sides by yi , the mole fraction of

species i in the mixture:

yi PVt= niRT or piV

t=

mi

MiRT

where mi is the mass of species i , Mi is its molar mass, and pi is its partial pressure, defined as

pi yi P. Solve for mi :

mi =Mi piV

t

RT

Applied to moist air, considered a binary mixture of air and water vapor, this gives:

mH2O =MH2OpH2OV

t

RTand mair =

MairpairVt

RT

(a) By definition,

h mH2O

mairor h =

MH2O

Mair

pH2O

pair

Since the partial pressures must sum to the total pressure, pair = P pH2O; whence,

h = MH2OMair

pH2OP pH2O

(b) If air is in equilibrium with liquid water, then the partial pressure of water vapor in the air equals

the vapor pressure of the water, and the preceding equation becomes:

hsat =MH2O

Mair

P satH2O

P P satH2O

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(c) Percentage humidity and relative humidity are defined as follows:

hpc h

hsat=

pH2O

P satH2O

P P satH2O

P pH2O(100) and hrel

pH2O

P satH2O(100)

Combining these two definitions to eliminate pH2O gives:

hpc = hrelP P satH2O

P P satH2O(hrel/100)

10.14 Because the vapor space above the liquid phase is nearly pure gas, Eq. (10.4) becomes P = xiHi .

For the same mole fraction of gas dissolved in the liquid phase, P is then proportional to Hi . Values

given in Table 10.1 indicate that were air used rather than CO2, P would be about 44 times greater,

much too high a pressure to be practical.

10.15 Because Henrys constant for helium is very high, very little of this gas dissolves in the blood streams

of divers at approximately atmospheric pressure.

10.21 By Eq. (10.5) and the given equations for ln 1 and ln 2,

y1P = x1 exp(Ax22)P

sat1 and y2P = x2 exp(Ax

21)P

sat2

These equations sum to give:

P = x1 exp(Ax22)P

sat1 + x2 exp(Ax

21)P

sat2

Dividing the equation for y1P by the preceding equation yields:

y1 =x1 exp(Ax

22)P

sat1

x1 exp(Ax22)P

sat1 + x2 exp(Ax

21)P

sat2

For x1 = x2 this equation obviously reduces to:

P =P sat1

P sat1 + Psat

2

10.23 A little reflection should convince anyone that there is no other way that BOTH the liquid-phase and

vapor-phase mole fractions can sum to unity.

10.24 By the definition of a K-value, y1 = K1x1 and y2 = K2x2. Moreover, y1 + y2 = 1. These equations

combine to yield:

K1x1 + K2x2 = 1 or K1x1 + K2(1 x1) = 1

Solve for x1: x1 =1 K2

K1 K2

Substitute for x1 in the equation y1 = K1x1:

y1 =K1(1 K2)

K1 K2

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Note that when two phases exist both x1 and y1 are independent ofz1.

By a material balance on the basis of 1 mole of feed,

x1L+ y1V= z1 or x1(1 V)+ y1V= z1

Substitute for both x1 and y1 by the equations derived above:

1 K2

K1 K2(1 V)+

K1(1 K2)

K1 K2V= z1

Solve this equation for V: V=z1(K1 K2) (1 K2)

(K1 1)(1 K2)

Note that the relative amounts of liquid and vapor phases do depend on z1.

10.35 Molality Mi =ni

ms

=xi

xsMs

where subscript s denotes the solvent and Ms is the molar mass of the solvent. The given equation

may therefore be written:

xi

xsMs= kiyi P or xi

1

xsMski

= yi P

Comparison with Eq. (10.4) shows that

Hi =1

xsMskior for xi 0 Hi =

1

Mski

For water, Ms = 18.015 g mol1 or 0.018015 kg mol1.

Thus, Hi =1

(0.018015)(0.034)= 1633 bar

This is in comparison with the value of 1670 bar in Table 10.1.

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