1. Phase Equilibrium: Two Components
Contributions by:
John L. Falconer & Will Medlin Department of Chemical and Biological Engineering University of Colorado Boulder, CO
Supported by the National Science Foundation

2. The pressure increases for a 50/50 vapor mixture of A and B that is initially at a low pressure. The liquid is ideal. Which plot corresponds to how xA and yA change with pressure at constant temperature if PAsat > PBsat ? A B C D E
0.5
1.0 P yA xA
ANSWER: B. Composition has to be equal to the bulk composition at the endpoints, and satisfy a mass balance such that y and x straddle the bulk composition.

3. The pressure increases for a 50/50 vapor mixture of A and B that is initially at a low pressure. The liquid is ideal. Which plot corresponds to how xB and yB change with pressure if PAsat > PBsat ?
0.5
1.0 P yB xB yB xB
0.5
1.0 P
0.5
1.0 P yB xB A B C yB xB
0.5
1.0 P yB xB
0.5
1.0 P
ANSWER: C. Composition has to be equal to the bulk composition at the endpoints, and satisfy a mass balance such that y and x straddle the bulk composition. D E

4. Isotherm for 60-40 benzene/hexane mixture
Which is the VLE line that most likely connects liquid and vapor phases in equilibrium for a benzene-hexane mixture? x
Isotherm for benzene/hexane mixture A B C
None of these A P B C
ANSWER: D. Since this is a two component mixture, the liquid and vapor compositions are different thus a liquid phase with this composition (and on the line on the graph) must be in equilibrium with a vapor with a different composition – another graph. V

5. Which is the VLE line that most likely connects liquid and vapor phases in equilibrium for a pentene-octene mixture?
Isotherm for pentene/octene mixture A B C
None of these A x P x B x x x x C x x
ANSWER: B. B. The line has to be at constant pressure. Also, pentene will be enriched in the vapor phase so the red line would be the vapor and the black the liquid so line B D
Isotherm for pentene/octene mixture V

6. Which of these diagrams are not possible?
xA, yA P
P-x
P-y
xA, yA P
P-y
P-x 1 2
2 & 6
1, 4, & 6
2 & 5
2, 3, & 5
3 & 5
xA, yA P
P-x
P-y
xA, yA P
P-x
P-y 3 4
ANSWER: D. 2, 3, and because higher pressure causes formation of vapor, 3 because it implies separation into two vapor phases, 5 because high P at compositions greater than 40% A or so leads to formation of vapor.
xA, yA P
P-x
P-y
xA, yA P
P-x
P-y 5 6

7. Mg and Si form the compound Mg2Si
Mg and Si form the compound Mg2Si. At constant pressure, what is the composition of the system when xSi= 0.4 and T = 1000 K?
T (K)
1600
1200
800
1.0
0.5
mole fraction of Si
liquid Mg Si
Mg2Si
One solid phase
Pure Si and Mg
Si and Mg2Si
ANSWER: C. Si and Mg2Si

8. Mg and Si form the compound Mg2Si
Mg and Si form the compound Mg2Si. At constant pressure, what is the composition of the system when xSi= 0.6 and T = 1000 K?
T (K)
1600
1200
800
1.0
0.5
mole fraction of Si
liquid Mg Si
One solid phase containing Mg and Si
Two solid phases: 40% Si, 60% Mg2Si
Two solid phases: 60% Si, 40% Mg2Si
Two solid phases: each contain Si & Mg
Mg2Si
ANSWER: B. Two solid phases: 40% Si, 60% Mg2Si. Apply lever rule.

9. This figure shows saturated liquid and saturated vapor lines for a 2-component system. Which of these are equilibrium compositions and temperature?
x1= 0.6, y1=0.9, 245°C
x1= 0.6, y1=0.3, 210°C
x1= 0.9, y1=0.6, 200°C
x1= 0.6, y1=0.3, 250°C
ANSWER: B. x1= 0.6, y1=0.3, 210°C. Equilibrium must be at the same P and T so needs to be a point when vapor line of one compostion intersects liquid line of another compostion. Also, the saturation pressure of component 1 (x1=1 line ) is lower than the saturation pressure of component 2 (x1 = 0 line), so y1,x1
At 210C, the solid line for x1=0.6 ,which is the liquid, interests the dashed line for y1-0.3, which is the vapor.

10. the mole fraction of A in the liquid phase __________.
A tank contains a 50/50 mixture of A and B (yA = 0.6). As the vapor is removed and the liquid boils away,
the mole fraction of A in the liquid phase __________.
increases
decreases
remains the same
ANSWER: B. decreases because A is enriched in the vapor phase

11. A liquid mixture (x1 = 0.4) evaporates at constant pressure by increasing the temperature. The first bubble has a composition of y1 = 0.7 because component 1 is more volatile and preferentially evaporates. As the last drop of liquid is evaporating, which is correct about the mole fraction of component 1 in the vapor phase?
y1 = 0.7
y1 = 0.4
y1 > 0.7
0.4 < y1 < 0.7
ANSWER: B. Since liquid phase starts at mol fraction x1 =0.4, then when it is all vapor at the end, the vapor composition will be the same, y1=0.4.

12. A liquid mixture (x1 = 0.4) evaporates at constant pressure by increasing the temperature. The first bubble has a composition of y1 = 0.7 because component 1 is more volatile and preferentially evaporates. As the last drop of liquid is evaporating, which is correct about the mole fraction of component 1 in the liquid phase?
x1 = 0.4
x1 > 0.4
x1 < 0.7
x1 <<< 0.4
ANSWER: D. The vapor has a composition of 0.4 when the last drop of liquid evaporates and since the vapor is enriched in component 1, the liquid morel fraction must be much less than 0.4

13. (Note: Tsat(methanol) = 65°C, Tsat(ethanol) = 78°C)
A closed system has methanol and ethanol in vapor-liquid equilibrium at 1 bar and 65°C. The temperature of the system is raised to 70°C at 1 bar. If two phases are still present, how do the methanol mole fractions change for each phase?
(Note: Tsat(methanol) = 65°C, Tsat(ethanol) = 78°C)
xm increases, ym decreases
xm increases, ym increases
xm and ym do not change
xm decreases, ym decreases
xm decreases, ym increases
ANSWER: D. xm increases, ym increases. Refer to a Txy diagram

14. We want to separate 50/50 mixtures of C6 isomers in the gas phase
We want to separate 50/50 mixtures of C6 isomers in the gas phase. As the temperature was lowered, your technician observed that n-hexane (n-C6) condensed before 2,2, dimethylbutane (DMB). Can this happen?
Yes, if n-C6 has a lower vapor pressure than DMB
Yes, if n-C6 has a higher vapor pressure than DMB
No, because both species have to condense
It depends on the system pressure as to whether one or two species condense.
ANSWER: C. No, because both species have to condense

15. A mixture of n-hexane and acetone were in the liquid phase in a piston-cylinder at 1 bar pressure. As the temperature increased, your technician observed that acetone evaporated first. Can this happen?
Yes, if acetone has the higher Psat
Yes, if n-hexane has the higher Psat
No, both species evaporate
It depends on the system pressure as to whether one or two species evaporate
ANSWER: C. No, both species have to evaporate

16. A mixture of n-hexane and acetone were in the liquid phase in a piston-cylinder at 1 bar pressure. As the pressure decreased, your technician observed that acetone evaporated first. Can this happen?
Yes, if acetone has the higher Psat
Yes, if n-hexane has the higher Psat
No, both species evaporate
It depends on the system pressure as to whether one or two species evaporate
ANSWER: C. No, both species evaporate

17. A mixture of water and acetone was placed in an open beaker in a room
A mixture of water and acetone was placed in an open beaker in a room. The next morning, 90% of the acetone had evaporated, but none of the water had evaporated. Can this happen?
Yes, if acetone has the higher Psat
Yes, if n-hexane has the higher Psat
No, both species evaporate
It depends on the system pressure as to whether one or two species evaporate
ANSWER: C. No, both species evaporate

18. Will this system exhibit an azeotrope at 100°C?
At 100°C, PAsat = PBsat.
Will this system exhibit an azeotrope at 100°C?
Yes No
Maybe…need more information.
ANSWER: A. Yes. Not possible to have vapor and liquid compostion the same for the entire compostion range so the pressure above the liquid must be above or below horizontal

19. Which statement is not correct?
A binary mixture that has a minimum boiling point azeotrope ____________.
has a maximum pressure azeotrope
has one composition where the boiling point does not change as liquid evaporates
cannot be modeled by a modified Raoult’s Law
allows for either component 1 or 2 to be enriched in the vapor phase, depending on the liquid composition
has activity coefficients greater than one
ANSWER: C. cannot be modeled by a modified Raoult’s Law.

20. A liquid mixture of 50 mol% n-pentane (Psat = 5 bar) and 50 mol% n-heptane (Psat = 1 bar) is at high pressure. The mixture is partially vaporized by isothermally lowering the pressure to just below 3 bar. Which statement is correct?
n-heptane is enriched in the gas phase
n-pentane is enriched in the gas phase
The gas phase has a 50/50 composition
No vapor forms
ANSWER: B. n-pentane is enriched in the gas phase.

21. Often, wine or beer is added to soup that is then
simmered. When the soup is ready to be served,
what has happened to the alcohol?
It all evaporates
It all remains in the soup
Some of it evaporates
ANSWER: C. VLE for a liquid of water and alcohol would have the vapor enriched in alcohol but it would not all evaporate.

22. ________ is the driving force for component A to move from liquid to vapor in order to reach equilibrium.
Pressure
Entropy
Enthalpy
Concentration
Gibbs free energy
ANSWER: E. Gibbs free energy. The fugactiy or chemial potential or the Gibbs free energy represent driving forces for mass transfer

23. 1. 0 mol of CH4 at 1 bar and 50°C is mixed with 1
1.0 mol of CH4 at 1 bar and 50°C is mixed with 1.0 mol of O2 at 1 bar and 50°C. The final mixture is at 2.0 bar and 50°C. Assume ideal gases. The Gibbs free energy change is _______.
positive
negative
zero
ANSWER: C. The enthalpy does not change since ideal gases. (G = H-TS)
The entropy does not change because the partial pressure of CH4 does not change and the partial pressure of O2 does not change and the entropy change is related to the ln of of the ratio of partial pressures.

24. An n-butane/n-heptane mixture is in vapor-liquid equilibrium
An n-butane/n-heptane mixture is in vapor-liquid equilibrium. The liquid composition is 30% n-butane. Which of these vapor compositions (% n-butane) are not possible?
20%
40%
60%
80%
All of these compositions are possible
ANSWER: A. 20%. The vapor pressure of n-butane is lower than that of n-heptane so the vapor must be enriched in n-butane

25. Consider a system at equilibrium that contains n-hexane and water, which are immiscible in the liquid phase. Which set of phases could not exist?
Two Liquids (hexane + H2O)
Hexane liquid and two vapors (hexane + H2O)
Two Liquids (hexane + H2O) and hexane vapor
H2O liquid and two vapors (hexane + H2O)
None of the above
ANSWER: C. Not possible to have liquid water and vapor space above it and no water vapor.

26. Consider a 1 L constant-volume container with n-hexane in VLE
Consider a 1 L constant-volume container with n-hexane in VLE. If half the volume is liquid and 0.1 L of liquid H2O is added to the container, what happens? Assume constant temperature and that water is immiscible with n-hexane.
Some hexane evaporates
Some hexane condenses
All the H2O evaporates
Some water evaporates and some hexane condenses
All the hexane evaporates
ANSWER: D. The two components do not interact in liquid phase so they each exert their own vapor pressure. Some n-hexane will condense because less vapor volume available. Some water will evaporate to get to VLE but not all because 0.1 L of liquid water would occupymuch more than 0.4 L of vapor volume.
ALTERNATIVE:
1) another answer: some hexane evaporates and some water evaporates

27. PAsat > PBsat > PCsat
An ideal liquid solution that is 30% A, 30% B, and 40% C is heated at a constant temperature until 80% of the original liquid has evaporated. Which component would you expect to have completely evaporated at that point?
PAsat > PBsat > PCsat
Component A
Component B
Component C
None of them
ANSWER: D. None of them. Must have each component in each phase for equilibrium.

28. PAsat = 1.5; PBsat = 1.0 bar ; T = 70°C
Assume ideal gas and ideal solution behavior for the A-B system:
PAsat = 1.5; PBsat = 1.0 bar ; T = 70°C
What is the total pressure at equilibrium above a liquid that is 60% A and 40% B?
2.5 bar
1.5 bar
1.2 bar
1.0 bar
None of the above
ANSWER: C bar.

29. Which of the following statements is NOT true at an azeotrope?
xA = yA
xA = xB
None of the above
ANSWER: B. xA = xB

30. Which of the following statements is NOT true at an azeotrope?
xA = yA
None of the above
ANSWER: E. All are correct.

31. At VLE,
𝑓𝑖 𝐿 = 𝑓𝑖 𝑉
Which is the correct equation to use when solving for the liquid phase fugacity of component 2 using an equation of state?
x2*g2*P
x2*f2L*P
x2*f2L*Psat
x2*g2*Psat
None of the above
ANSWER: B.

32. One mole of component A is in VLE at 1 bar and 75°C
One mole of component A is in VLE at 1 bar and 75°C. One mole of B is added to form a mixture, and the pressure is isothermally increased to 2 bar. What is the state of the system? The vapor pressure of B at 75°C is 2 bar. Assume A and B form an ideal solution.
VLE
All liquid
All vapor
ANSWER: B. All liquid.

33. Which of the following is true for a binary azeotrope at low pressure?
The activity coefficients of the two components are the same
The vapor pressures of the two components are the same
The value of the vapor pressures multiplied by the activity coefficients are the same
None of these
ANSWER: C. The value of the vapor pressures multiplied by the activity coefficients are the same

34. Are there conditions in which liquid water (or solid ice) at barospheric pressure will not have a vapor phase partial pressure when in contact with nitrogen? (Assume the system is below the normal boiling point of nitrogen).
No.
Yes, if the temperature is low enough
Yes, if the temperature is high enough
ANSWER: B. Yes, if the temperature is low enough

35. At 105°C: P1sat = 1.8 bar, P2sat = 1.2 bar
You want to completely condense a vapor containing
50% methanol(1) and 50% ethanol(2). What is the minimum pressure at which the condenser must operate at 105°C?
 At 105°C: P1sat = 1.8 bar, P2sat = 1.2 bar
3.0 bar
1.8 bar
1.5 bar
1.2 bar
ANSWER: C. 1.5 bar

36. At what temperature is 59 mol% ethane liquid not in equilibrium with a vapor?
T (°F) P
700
100
300
500 M
89 mol% C2
77 mol% C2
27 mol% C2
CC2H6
CC7H16
59 mol% C2
335 °F
275 °F
315 °F
400 °F
ANSWER: D. Both phases are not present at 400F.

37. 59 mol% ethane liquid is in equilibrium with what ethane mol% vapor composition?
T (°F) P
700
100
300
500 M
89 mol% C2
77 mol% C2
27 mol% C2
CC2H6
CC7H16
59 mol% C2 77 89 27
All of those shown
A & B
ANSWER: E. The liquid phase intersects with the other two compositions.

38. A vapor with 59% ethane is in equilibrium with which phase?
T (°F) P
700
100
300
500 M
89 mol% C2
77 mol% C2
27 mol% C2
CC2H6
CC7H16
59 mol% C2
89%
77%
59%
27%
None of the above
ANSWER: D. Equilibrium must have the two phases at the same temperature and pressure. There are an infinite number of equilibrium compositions for liquid phases that are in equilibrium with a gas phase of 59%. For the constant composition lines in the figure, only the intersection with the liquid at 27% at 400C.
Posted in Double VLE part 2

39. When the interaction parameter, kij, increases, for a fixed liquid phase, fixed temperature system, the equilibrium pressure ________.
𝒂 𝒊𝒋 = 𝒂 𝒊𝒊 𝒂 𝒋𝒋 𝟏− 𝒌 𝒊𝒋
increases
decreases
remains the same
ANSWER: A. As kij increases, the aij term decreases so the interactions between molecules I and j are weakened, so the pressure increases
In the Peng-Robinson equation of state, the parameter a represents molecular interactions in the fluid. The parameter aij represents the interactions between I and j molecules in the mixture.

40. A liquid mixture containing species A & B is boiled by increasing the temperature at constant pressure. The saturation pressure is greater for A than for B. What happens?
xA increases and yA increases
xA increases and yA decreases
xA decreases and yA decreases
xA decreases and yA increases
ANSWER: C. Observing a Txy diagram for this system, as temperature is increased from liquid phase to eventual vapor phase, the more volatile component A will decrease its molar composition in the liquid and vapor phases as T is increased.

41. What is the fraction of this mixture that is liquid?
1/3
2/3 70 80 90
100
110
120
130
140
0.2
0.4
0.6
0.8 1 X
, Y
Temperature
T-x
T-y
ANSWER: A. Lever rule applies at equilibrium composition.

42. In a bubble pressure calculation using a EOS, the mole fractions of liquid are known, and the mole fractions of vapor must be calculated by iteration by guessing the pressure. After the first iteration, Syi > 1. For the next iteration, _______________.
raise the pressure
lower the pressure
keep the pressure the same but change the composition
ANSWER: A. Raise the pressure, because the yi’s adding to number larger than one means that the calculation indicates the liquid wants to be in the vapor phase more than it is possible. A higher pressure will decrease the drive to be in the vapor phase (will raise the vapor phase fugacities)

43. A EOS spreadsheet is used to calculate VLE for a binary mixture (x1= 0
A EOS spreadsheet is used to calculate VLE for a binary mixture (x1= 0.9) with a non-ideal liquid phase and a non-ideal gas. Which statement about the fugacity coefficients is most likely to be correct?
φ 1 L = φ 1 V
φ 1 L = φ 2 V
φ 1 L < φ 1 V
φ 1 L > φ 2 V
ANSWER: C. The liquid phase is further away from an ideal gas than the vapor and the fugacity coefficient (fi/xiP or fi/yiP) is closer to one for gas phase so the fugacity coefficient is smaller for the liquid phase
Posted in Double VLE part 2

44. Which phases are present?
6 mol A and 4 mol B are in equilibrium at 100°C and 3.0 bar. A and B are completely immiscible in the liquid phase. Their vapor pressure at 100°C are:
Which phases are present?
Liquid B and vapor of A + B
Two liquids
Two liquids in equilibrium with vapor
All vapor
Liquid A and vapor of A + B
ANSWER: B. If both liquids are present an in equilibrium with the vapor, pressure = = PAsat + PBsat = 2.5 bar. That is the highest vapor phase possible. Since P=3.0 bar, only liquid is present. When liquids are completely immiscible, they exert their own vapor pressures independent of the other liquid.

45. Which phases are present?
6 mol A and 4 mol B are in equilibrium at 100°C and 2.0 bar. A and B are completely immiscible in the liquid phase. Their vapor pressure at 100°C are:
Which phases are present?
Liquid B and vapor of A + B
Two liquids
Two liquids in equilibrium with vapor
All vapor
Liquid A and vapor of A + B
ANSWER: A.
P = 2.0 = PA + PB.
B) If both liquids are present an in equilibrium with the vapor, pressure = = PAsat + PBsat = 2.5 bar (WRONG).
E) PA = 2.0 bar and so PB = 0. (WRONG)
A) PB = 0.5 bar so PA = 1.5 bar. yA = 0.75 but overall mole fraction is So liquid B will make overall mole fraction less than So liquid B and A & B in vapor is correct. When liquids are completely immiscible, they exert their own vapor pressures independent of the other liquid.

46. A gas phase mixture at 100°C is 75 % A and 25 % B
A gas phase mixture at 100°C is 75 % A and 25 % B. As pressure increases isothermally, liquid A starts to condense at 1.6 bar. At 2.4 bar liquid B starts to condense. What are the vapor pressures of A and B?
PAsat = 1.6 bar, PBsat = 2.4 bar
PAsat = 1.2 bar, PBsat = 1.6 bar
PAsat = 1.6 bar, PBsat = 1.2 bar
PAsat = 1.8 bar, PBsat = 0.6 bar
PAsat = 1.2 bar, PBsat = 1.2 bar
ANSWER: 1.6 bar, PA = 0.75(1.6) = 1.2 bar so PB = 0.4 bar. So PAsat = 1.2 bar.
@ 2.4 bar PA = 1.2 bar since saturated so PB = 1.2 bar (so total is 2.4). So PBsat =1.2 bar.