Phase Equilibrium: Two Components in Piston-Cylinders Contributions by: John L. Falconer & Will Medlin Department of Chemical and Biological Engineering University of Colorado Boulder, CO 80309-0424 Supported by the National Science Foundation

A 2-componet system is at 98oC and 90 kPa A 2-componet system is at 98oC and 90 kPa. How many different equilibrium phase conditions are possible for this system (for example: liquid, vapor, liquid and vapor)? 1 2 3 1 or 2 1 or 3 ANSWER: E. 1 or 3

A piston-cylinder system contains components A and B in vapor-liquid equilibrium. Assume ideal solution and ideal gas, with xA = 0.25, yA = 0.50. A small weight was added to the piston, while temperature was held constant. What happens? Vapor yA=0.50 Liquid xA=0.25 xA increases, yA decreases xA increases, yA increases xA decreases, yA decreases xA decreases, yA increases ANSWER: B. xA increases, yA increases. Look at PXY diagram.

What are the final contents of the system? One mole of pure hexane is in vapor-liquid equilibrium at 1 bar and 70°C in a piston-cylinder. After 0.2 mole of heptane liquid is injected, the system returns to equilibrium at the same temperature and pressure. What are the final contents of the system? Psat(hexane) > Psat(heptane) Hexane Vapor Liquid 0.2 mole Heptane (l) All liquid All vapor Liquid and vapor with yhexane > xhexane Liquid and vapor with yhexane < xhexane ANSWER: A. All liquid. Look at PXY diagram. ALTERNATIVES questions: 1) Reverse the system (hexane is added to heptane)

What are the final contents of the system? One mole of pure hexane is in vapor-liquid equilibrium at 1 bar and 70°C in a piston-cylinder. After 0.2 mole of heptane vapor is injected, the system returns to equilibrium at the same temperature and pressure. What are the final contents of the system? Psat(hexane) > Psat(heptane) Hexane Vapor Liquid 0.2 mole Heptane (v) All liquid All vapor Liquid and vapor with yhexane > xhexane D. Liquid and vapor with yhexane < xhexane ANSWER: A. All liquid. Look at PXY diagram to visualize solution.

Psat(hexane) > Psat(heptane) Hexane liquid is in equilibrium with N2 gas at 50°C in a piston-cylinder. Some heptane liquid is added to the system at constant temperature and pressure. What happens? Psat(hexane) > Psat(heptane) N2 Gas Hexane Liquid Heptane (l) Some hexane vapor condenses Some hexane liquid vaporizes All the hexane vapor condenses All the hexane liquid vaporizes No change in the hexane in the 2 phases ANSWER: A. Some hexane vapor condenses because the liquid phase fugacity is lowered by adding the less volatile heptane, which will decrease the volume of the vapor (increase P of nitrogen).

Psat(hexane) > Psat(heptane) Hexane liquid is in equilibrium with N2 gas at 50°C in a piston-cylinder. Some heptane vapor is added to the system at constant temperature and pressure. What happens? Psat(hexane) > Psat(heptane) N2 Gas Hexane Liquid Heptane (v) Some hexane vapor condenses Some hexane liquid vaporizes All the hexane vapor condenses All the hexane liquid vaporizes No change in the hexane in the 2 phases ANSWER: A. Some hexane vapor condenses. Some hexane vapor condenses because the liquid phase fugacity is lowered by adding the less volatile heptane, which will decrease the volume of the vapor (increase P of nitrogen).

Psat(hexane) > Psat(heptane) Hexane liquid is in equilibrium with N2 gas at 50°C in a piston-cylinder. Some heptane vapor is added to the system at constant temperature and pressure. What happens? Psat(hexane) > Psat(heptane) N2 Gas Hexane Liquid Heptane (v) The heptane remains in the vapor phase Some heptane condenses All the heptane condenses All the hexane liquid vaporizes ANSWER: B. Some heptane condenses. The fugacity in the liquid phase is xPsat(heptane), which will cause some of the vapor to condense when it is injected (but not all, since x will be greater than zero and P < Psat).

Psat(hexane) > Psat(heptane) Hexane liquid is in equilibrium with N2 gas at 50°C in a piston-cylinder. Some heptane liquid is added to the system at constant temperature and pressure. What happens? Psat(hexane) > Psat(heptane) N2 Gas Hexane Liquid Heptane (l) The heptane remains in the liquid phase Some heptane vaporizes All the heptane vaporizes All the hexane liquid vaporizes ANSWER: B. Some heptane vaporizes. The relation P = xPsat (hex) + (1-x)Psat (hep) + P (N2) will hold because P (N2) can change as the volume of the container changes, so some of each component will be in each phase (VLE).

Two components (A & B) are in VLE in a piston-cylinder container Two components (A & B) are in VLE in a piston-cylinder container. Initially, 1.0 mol of liquid (xA = 0.4) and 0.1 mole of vapor (yA = 0.7) are present. If 0.5 mole of liquid A is added to the system and it returns to equilibrium at the same temperature and pressure as before: Vapor yA=0.7 Liquid xA=0.4 the amount of liquid increases the amount of liquid decreases the concentration of A in the gas phase increases the concentration of A in the liquid phase increases ANSWER: B. the amount of liquid decreases. As you increase the amount of the more volatile component at contant T,P you move toward the vapor line on a Pxy plot; by the lever rule, the fraction of vapor increases.

Two components (A & B) are in VLE in a fixed volume container Two components (A & B) are in VLE in a fixed volume container. Initially, xA = 0.3 and yA = 0.6. 1.0 cm3 of air is injected into the system. The temperature and pressure are constant. The liquid solution is ideal. If liquid and vapor phases are still present, there is _______ before the injection of air. Vapor yA=0.6 Liquid xA=0.3 1 cm3 air more liquid than less liquid than the same amount of liquid as ANSWER: B. less liquid than before the injection of air because the partial pressures, and thus fugacities, in the vapor phase have gone down, providing a driving force for evaporation. ALTERNATIVES: Keep just temperature constant Keep just pressure constant

What phase(s) are present at equilibrium? Two components (A & B) are in VLE in a piston-cylinder. One mole of liquid (xA=0.2) is in equilibrium with one mole of vapor (yA =0.6). The liquid phase does not possess an azeotrope.10 mol liquid A are added at constant temperature and pressure. What phase(s) are present at equilibrium? Vapor yA = 0.6 Liquid xA = 0.2 10 mol A liquid All liquid All vapor More vapor, less liquid, same compositions More liquid, less vapor, same composition More vapor, less liquid, different compositions ANSWER: B. All vapor. Look at Pxy diagram. The final mixture is mostly A such that the bulk composition of A is greater than yA, meaning it’s all in the vapor phase.

Two components (A & B) are in VLE at 60oC and 1 bar in a piston-cylinder. One mole of liquid (xA=0.2) is in equilibrium with one mole of vapor (yA =0.6). The liquid phase does not possess an azeotrope. When 0.3 mol liquid B is added at 60oC and 1 bar, what phase(s) are present at equilibrium? Vapor yA = 0.6 Liquid xA = 0.2 0.3 mol B liquid All liquid All vapor More vapor, less liquid, same compositions More liquid, less vapor, same composition More vapor, less liquid, different compositions ANSWER: D. More liquid, less vapor, same composition Use Pxy diagram to visualize.

Two components (A & B) are in VLE in a fixed volume container Two components (A & B) are in VLE in a fixed volume container. Initially, xA = 0.3 and yA = 0.6. 1.0 cm3 of air is injected into the system. The temperature and pressure are constant. The liquid solution is ideal. If liquid and vapor phases are still present, xA _______. Vapor yA=0.6 Liquid xA=0.3 1 cm3 air increases decreases remains the same ANSWER: B. Decreases. Assuming Raoult’s law, P = xA*PsatA + xB*PsatB + Pair. If Pair goes up, the sum of the other terms must go down. Since PsatB < Psat A, this means that xA must go down and xB must go up.

Two components (A & B) are in VLE in an adiabatic piston-cylinder Two components (A & B) are in VLE in an adiabatic piston-cylinder. Assume ideal solution and ideal gas. If 0.2 mole of liquid A is rapidly injected into the system, the temperature at equilibrium is _______ before the injection of A. xA = 0.2, yA = 0.7 Vapor yA = 0.7 Liquid xA = 0.2 0.2 mole A liquid higher than lower than the same as ANSWER: B. Lower than. The more volatile A will be driven to evaporate by the lower fugacity in the vapor phase. This will require energy, which will lower T. ALTERNATIVES: 1) Add vapor instead of liquid.

Two components (A & B) are in VLE in a fixed volume container with 2 moles vapor and 1 mole liquid. You add 30 moles of liquid B at constant temperature and pressure. This system does not have an azeotrope. What happens? xA = 0.2, yA = 0.7 Vapor yA=0.7 Liquid xA=0.2 30 moles B(l) Some liquid evaporates All the liquid evaporates Some vapor condenses All the vapor condenses ANSWER: D. All the vapor condenses. The bulk mole fraction of A will now be less than 0.2, bringing the system outside of the phase envelope.

Two components (A & B) are in VLE in a fixed volume container with 2 moles vapor and 1 mole liquid. You add 30 moles of vapor A at constant temperature and pressure. This system does not have an azeotrope. What happens? xA = 0.3, yA = 0.8 Vapor yA=0.8 Liquid xA=0.3 30 moles A(v) Some liquid evaporates All the liquid evaporates Some vapor condenses All the vapor condenses ANSWER: B. Adding that much A (the more volatile component) will cause the bulk composition to be greater than 80% A, bringing the system out of the phase envelope.

A fixed container, with components A and B, is in chemical and phase equilibrium (xA = 0.6, yA = 0.2). For the reaction (A  B), the addition of liquid B _______ the mole fraction of A in the liquid. Vapor yA=0.2 Liquid xA=0.6 B(l) increases decreases remains the same ANSWER: C. The equilibrium constant is fixed at fixed T. The fugacity of A is therefore in a fixed proportion to the fugacity of B, therefore the partial pressures and thus compositions must remain the same at fixed T.

A liquid mixture is in equilibrium in a piston-cylinder system with a gas phase. The N2 does not dissolve in the liquid. If N2 is removed completely isothermally using a selective membrane at constant pressure, then xA ________. Vapor yA = 0.5 yN2 = 0.2 Liquid xA = 0.4 xA = 0.4, xB = 0.6 yA = 0.5, yB = 0.3, and yN2 = 0.2 increases decreases remains the same ANSWER: Assuming Raoult’s law, P = xAPAsat + xBPBsat + PN2. As PN2 is decreased, keeping P constant requires the other two terms to collectively increase. Since A is more volatile a PAsat is thus larger, XA must increase. N2

A liquid mixture is in equilibrium in a piston-cylinder system with a gas phase. The N2 does not dissolve in the liquid. If N2 is removed completely isothermally using a selective membrane at constant pressure, then xA ________. Vapor yA = 0.5 yN2 = 0.2 Liquid xA = 0.4 xA = 0.4, xB = 0.6 yA = 0.5, yB = 0.3, and yN2 = 0.2 increases decreases remains the same ANSWER: A. increases. With the N2 removed, the partial pressure of A increases to keep pressure constant, so xa and ya both increase. (See Pressure vs xa, ya graph). N2

Water liquid and vapor are in equilibrium in a piston-cylinder Water liquid and vapor are in equilibrium in a piston-cylinder. An insoluble polymer containing a salt is in the liquid. The salt slowly diffuses out of the polymer and dissolves in the water. Temperature and pressure are constant. What happens? 1 kg All the water condenses All the water evaporates Some water condenses Some water evaporates Vapor ANSWER: A. All the water condenses. As the salt enters the water, it lowers the liquid water fugacity, which creates a driving force for mass transfer. In other words, the fugacity of liquid water becomes less than the pressure/fugacity of water vapor, therefore all the water condenses. Polymer Water

A container with 5 cm3 of benzene liquid and 1 cm3 of benzene vapor is in equilibrium. You inject 1 cm3 of toluene vapor. Which of the following mole fractions of toluene in the vapor composition at equilibrium will be closest to the correct value: 1 cm3 Toluene 90 % 50 % 10 % 1 % Vapor 1 cm3 ANSWER: D. 1%. You’re injecting a small number of moles since it is in the vapor phase, so x and y for toluene should be quite low (well under 1%). ALTERNATIVE: Ask about the fraction of toluene that you expect to condense Benzene 5 cm3

A fixed-volume tank contains two components in VLE. 1 A fixed-volume tank contains two components in VLE. 1.0 mole of N2 is added. Assume N2 is not soluble in the liquid. The vapor phase is an ideal gas. The amount of component A in the liquid phase ____________. Vapor yA = 0.6 Liquid 1 mole N2 increases decreases stays the same ANSWER: C. stays the same. Ideal gas so the partial pressures do not change for the 2 components so the liquid does not change

A fixed-volume tank contains two components in VLE. 1 A fixed-volume tank contains two components in VLE. 1.0 mole of N2 is added. Assume N2 is not soluble in the liquid. The vapor phase is an ideal gas. The amount of component A in the vapor phase ____ Vapor yA = 0.6 Liquid 1 mole N2 increases decreases stays the same ANSWER: C. stays the same. Ideal gas so the partial pressures do not change for the 2 components so the liquid does not change

A fixed-volume tank contains two components in VLE. 1 A fixed-volume tank contains two components in VLE. 1.0 mole of N2 is added. Assume N2 has no solubility in the liquid. The vapor phase is an ideal gas. The partial pressure of component A ________. Vapor yA = 0.6 Liquid 1 mole N2 increases decreases stays the same ANSWER: C. stays the same.

Tb (chloroform) = 60°C @ 0.95 bar Tb (acetone) = 55°C @ 0.95 bar Acetone and chloroform form a maximum temperature azeotrope. A piston-cylinder contains pure chloroform at 55°C and 0.95 bar. It has 0.9 mole vapor, 0.1 mole liquid. What is the final condition when 0.1 mole acetone is added at constant temperature and pressure? Tb (chloroform) = 60°C @ 0.95 bar Tb (acetone) = 55°C @ 0.95 bar Vapor 0.9 mole CHCl3 Liquid 0.1 mole Acetone All liquid All vapor Vapor-liquid mixture ANSWER: A. All liquid. Look at a Txy diagram to visualize the behavior

0.1 mole of A vapor is in VLE with 10 mol liquid A at 70°C in a piston-cylinder. After 0.1 mol of liquid B is injected, the system returns to equilibrium at the same temperature and pressure. PAsat = 2 bar ; PBsat = 1.3 bar The liquid is non-ideal and the system has a maximum pressure azeotrope. What are the final contents of the system? 0.1 mole A 10 mole A All liquid All vapor Liquid and vapor with yA > xA Liquid and vapor with yA < xA ANSWER: B. All vapor. Start by drawing a constant temperature Pxy diagram. 0.2 mole B

1.0 mole of pure hexane is in VLE at 1 bar and 70°C in a piston-cylinder. After 0.2 mole of octane liquid is injected, the system returns to equilibrium at the same temperature and pressure. If octane has a lower vapor pressure than hexane, what are the final contents of the system? Hexane Vapor Liquid All liquid All vapor Liquid and vapor with yhexane > xhexane Liquid and vapor with yhexane < xhexane ANSWER: A. All liquid. Look at phase diagram

A binary mixture is in VLE in a piston-cylinder with 10x more moles in the vapor phase than in the liquid phase with yA = 2xA. You push down on the piston until the volume is half its original value. Temperature is constant. What happens? Pressure increases Pressure remains the same All vapor condenses, pressure is constant All vapor condenses, pressure is higher Nothing changes B. There is far more vapor than liquid, so as the volume is decreased some of the vapor will just condense to form liquid, with fugacity driving forces remaining the same.

A binary mixture is in VLE in a piston-cylinder with 10x more moles in the vapor phase than in the liquid phase with yA = 2xA. You push down on the piston until the volume is half its original value. Temperature is constant. What happens? Vapor and liquid at same pressure Vapor and liquid at higher pressure Some vapor condenses xA increases, yA decreases xA and yA increase