Balances on Multiphase Systems A B, B C A feedstock, B desired product, C byproduct (waste) B(v) to purification and packaging Distillation Column A(g) A(g) A(g),B(g),C(g) B(l), C(l) 80 o F 900 o F 700 o F some S(l) Heater Reactor C(l), some S(l) to disposal Condenser N 2 To stack Cooler A(g), 150 o F to stack B(v), C(v), N 2 some S(v) S(l) Absorber Stripper S(l) B & C (dissolved) S(l) to sewer A(g),B(g),C(g) N 2 250 o F This process is expensive, wasteful, & inefficient how would you modify it? 4-1 Copyright
A B, B C A feedstock, B desired product, C byproduct (waste) B(v) to purification and packaging Distillation Column A(g) A(g) A,B,C(g) A(g) B(l), C(l) 80 o F 900 o F 700 o F 550 o F Some S(l) Heater Reactor Heat Exchanger A(g) 150 o F C(l), some S(l) to disposal S(l) Fresh S(l) Condenser B(v), C(v), N 2 some S(v) Absorber Stripper Q: What types of information do we need to know to design this system? A: S(l), B(soln), S(l) C(soln) A(g),B(g),C(g) N 2 250 o F The function of most chemical process units is to separate mixtures into their components, as in the process just shown. Read text, pp. 237 238; try Test Yourself on p. 238. 4-2 Copyright
Most separation processes work by getting different components into different phases and separating the phases. To design them, we need to know how species distribute themselves between phases at equilibrium. In this section we ll consider liquid-vapor systems and liquidsolid systems. Chapter 6 also considers systems with two liquid phases, but we won t in this class. Phase Diagrams (Section 6.1a) Major features of phase diagram (pure species): Three equilibrium curves: solid-vapor, solid-liquid, vapor-liquid P P c S L V V-L equilibrium curve ends at the critical temperature (T c ) and critical pressure (P c ). Below T c and P c, vapor and liquid can coexist as separate phases: at and above them, can t have separate phases. Look up T c and P c in Table B.1. Diagram shows regions in which species exists as solid, liquid, vapor (below T c ), gas (above T c and below P c ), & supercritical fluid (above T c and P c ). T T c Form pairs designate one pair member as A, other as B. Look at phase diagrams on p. 241. Note that the solid-liquid equilibrium curve for CO 2 slopes to the right, unlike the one for water. (The slopes are highly exaggerated the curves are actually almost vertical.) Consider the cylinder filled with water at the top of p. 242. T & P of the cylinder contents are varied to follow path ABCDE on phase diagram for water. On p. 241, the state of the cylinder is shown for each point on that path. A: Explain the top row to your partner and state how you would calculate the water volume at each point assuming that you know the mass of water in the cylinder. B: Now you do the same for the bottom row. A point on the V-L equilibrium curve for water is (50 o C, 92.5 torr) P(torr) L 50 o C = boiling point of water at P = 92.5 torr 92.5 torr = vapor pressure of water at T = 50 o C 92.5 V 50 T( o C) 4-3 Copyright
CHE 205 Chemical Process Principles Normal boiling point = T b at 1 atm (100 o C for water) Look up normal boiling points of different species in Table B-1 Example: T nbp (toluene) = o C T b (toluene, P = ) = o C; p*(toluene, o C) = A point on the S-L equilibrium curve for water is (0 o C, 760 torr) 0 o C = melting point (or freezing point) of water at 1 atm (= normal melting point) Usually neglect effect of pressure on melting point A point on the S-V equilibrium curve for water is ( 5 o C, 3 torr) 5 o C = sublimation point of water (ice) at 3 torr 3 torr = vapor pressure of ice at 5 o C The equilibrium curves intersect at 0.01 o C and 4.58 torr triple point of water (only point where all three phases can coexist) Note: The statements we have made about the states of water at given conditions apply only to pure water. At normal ambient temperatures and pressures, for example, pure water must be liquid, but water vapor can exist at those conditions in an air-water vapor mixture. (In fact, you re breathing it now.) Table B-3 provides vapor pressure (p * ) of water at any T, boiling point (T b ) of water at any P A: Find p * (37C) [ torr] ; B: Find T b at 32.747 mm Hg [ o C] Work through Test Yourself on p. 243. Estimating Vapor Pressures (Section 6.1b) Volatility: Tendency to go from liquid (or solid) to vapor High vapor pressure high volatility Low boiling point high volatility Example: Consider two beakers, one containing n-butane, the other n-octane, at room temperature. n-c 4 H 10 T b = 0.6 o C n-c 8 H 18 T b = 125.5 o C What would happen if the lids were removed? The butane would The octane would Many separation processes rely on the high volatility of some species in a mixture and nonvolatility (evaporation, drying) or much lower volatility (flash vaporization, distillation, condensation) of other species in the mixture. To design separation processes based on relative volatility, we need to know or estimate vapor pressures, p i *(T) for all mixture components (i). 4-4
4 Ways To Estimate Vapor Pressure (p * ) from a given T: 1. Measure it. Reliable but time-consuming and costly. 2. Look it up Table B.3 of F&R for water, Perry s Chemical Engineers Handbook, Handbook of Chemistry & Physics, etc. 3. Use a graphical correlation. The Cox chart, Fig. 6.1-4, p. 247, has nearly linear correlations of vapor pressure with temperature (see pp. 245 246 to know how it is constructed and why the plots are nearly linear): Given T, find vapor pressure p *. Example: (p*) propane (55 o F) = psi Given p *, find boiling point T. Example: (T b ) acetone (2 psi) = o F 4. Fit a function to the data. Use it to estimate p* for any T in range of experimental data. (a) Clausius-Clapeyron equation (Eq. 6.1-3, p. 244). H ln p* RT v B (6.1-3) where H v (kj/mol) = heat of vaporization of the species R [kj/(mol-k)] = 8.314x10 3 (gas constant) T(K) = absolute temperature Given p* vs. T data, plot ln p* vs. 1/T (rectangular) or p* vs. 1/T (semilog). Find H v and B using methods of Ch. 2, then use Eq. (6.1-3) to estimate p* for any given T or vice versa. (See Ex. 6.1-1, p. 244). Note, this expression was derived assuming that the heat of vaporization is constant and independent of temperature. In reality, it varies slightly with T. The approximation is OK over a small temperature range. (b) Antoine equation (Eq. 6.1-4, p. 246 & Table B.4). More accurate than Clausius-Clapeyron but harder to estimate parameters (3 instead of 2). B log 10 p* A (6.1-4) T C Given T( o C), look up A, B, C in Table B.4, calculate log 10 p* from Eq. (6.1-4), then p log 10 p* *(torr) = 10 10 B A T C Example: For water at T = 30 o C log 10 p* 1.5028 1.5028 p* 10 torr = 31.827 torr (From Table B.3, p* = 31.824 torr, so Antoine equation is extremely accurate.) 4-5
Extensive and Intensive Variables and the Gibbs Phase Rule (Section 6.2) Consider the following two closed systems at equilibrium. Think about where each one is on the phase diagram. P(torr) S L H 2 O(v) T( o C), P(torr) V ˆv (L/mol) V H 2 O(l) T( o C), P(torr) T( o C) I II l (g/ml) Q: How many of the variables T, P, V ˆv, and (for System II) l would you have to specify to be able to determine the others for each of these two systems? (Suggestion: Refer to the phase diagram.) A: For System I, you need to specify both T and P (i.e., 2 variables) to know where you are in the vapor region of the phase diagram. Once you know T and P, you can calculate V ˆv = RT/P (or use a real gas equation of state). For System II, you need to specify Once you know, you can calculate Next consider the following closed system with components A, B, C. We can identify a number of variables describing the physical state of the system in both the liquid and vapor phases: Vapor: T, P, V vapor, m v (kg), v (kg/l) n Av (mol A), n Bv, n Cv y A [mol A(v)/mol], y B, y C H 2 O(v) T( o C), P(torr) V ˆv (L/mol) Liquid: T, P, V liq, m l (kg), l (kg/l) n Al (mol A), n Bl, n Cv x A [mol A(l)/mol], x B, x C Say closed system is at equilibrium nothing changes with time. Suppose we double the size of the system but keep system conditions constant: Extensive variables those that would double. (V, m, n i in each phase) Intensive variables those that would remain the same. (T, P, v, l, x i, y i ) Degrees of freedom of an equilibrium system: DF = Number of intensive variables that must be specified to determine all the others. It is different than DF we used for mass balance calculations! Gibbs Phase Rule: DF = 2 + c r # chemical species # phases # independent reactions among the chemical species Apply the phase rule to Systems I and II on the previous page to confirm your previous results. 4-6
Go through Example 6.2-1 on p. 248 and Test Yourself on p. 249. Don t confuse the degrees of freedom for an arbitrary process system (which you have been calculating since Chapter 4) with the degrees of freedom calculated using the Gibbs Phase Rule. The first one tells you how many process variables (extensive and intensive) must be specified to calculate the rest. The Gibbs Phase Rule tells you the number of intensive variables that must be specified for a system at equilibrium in order to calculate all the other intensive variables. Vapor-liquid equilibrium (VLE) for a single species Evacuate a container, charge with pure liquid water, seal. Water evaporates (molecules of liquid enter vapor phase). As water vapor builds up, reverse flow (condensation) begins. Eventually rate of condensation equals rate of evaporation, & system reaches equilibrium at 25.3 o C. (At equilibrium, no measurable variable changes with time, although molecules are moving constantly within and between phases.) Vacuum H 2 O(v) @ 25.3 o C, P H 2 O Evaporation H 2 O H 2 O H 2 O(l) t H 2 O(l) @ 25.3 o C Q: What is P in the gas phase in the container at equilibrium? (Hint: liquid and vapor are coexisting at the same temperature. Think about the phase diagram.) A: P =. Q: This is also the pressure at the upper surface of the liquid. What happens to P as you move into the liquid (say, to a depth h below the surface)? A: Vapor-liquid equilibrium for a multicomponent gas with a single condensable species (Section 6.3). Apply to evaporation, condensation, drying, air conditioning, humidification. Add liquid water to an open container, seal. Come to equilibrium at 25.3 o C and 800 torr. Neglect dissolution of air in liquid water (it happens, but only to a very slight extent.) Dry air (DA) P atm H 2 O(l) Evaporation H 2 O H 2 O(v), air @ 25.3 o C, 800 torr y w [mol H 2 O(v)/mol] (1 y w ) [mol DA/mol] H 2 O(l) @ 25.3 o C 4-7
Q: What is the condition of the water vapor at equilibrium? Note: If a liquid and its vapor coexist at equilibrium, the vapor must be saturated. (If the gas phase could hold more of the vapor, more liquid would evaporate until the gas phase becomes saturated.) Corollary: You can only condense a saturated vapor. Q: Can you calculate the vapor-phase composition at equilibrium from the data given? Since we know T and P, then all others set (e.g., ρ w and y w can be calculated from an equation of state or another constitutive relation; see Raoult s Law below). Q: What is the vapor-phase composition in the container at equilibrium? Raoult s law for a single condensable species. Suppose Species i is a component of a gas at temperature T and pressure P. If i is (a) the only condensable species in the gas (i.e. the only species that would condense if the temperature were lowered by a moderate amount at the system pressure), and (b) saturated, then to a good approximation pi yip p * i ( T) (6.3 1) Raoult s Law, single condensable species where p i is the partial pressure of Species i and p i * is the vapor pressure at the system temperature. Raoult s Law is an additional equation relating unknown variables that we can count in the DOF analysis. It applies only to saturated vapors. For the system on the preceding page, * o HO 2 p (25.3 C) torr yw 0.0302 mol H2O(v)/mol P torr y 1 y 0.970 mol DA/mol DA w In a chemical process with a single condensable species, if you are told that either (a) the vapor of that species is saturated, or (b) the vapor of that species is coexisting with the liquid at equilibrium (which means it must be saturated), then you can apply Raoult s law [Eq. (6.3-1)] and a relationship for the vapor pressure as a function of temperature (such as the Antoine equation) to relate the mole fraction y i, the gas temperature T, and the total gas pressure P. 4-8
Exercise: Dry Air P = 1 atm What is p N2 (the partial pressure of nitrogen)? Water vapor Liq. water T= 25C At equilibrium, what is the pressure of the vapor in the container? Water vapor and nitrogen T= 25C At equilibrium, what is p H2O (the partial pressure of water)? Liq. water Example: Material balances on an equilibrium condensation process 100 mol/s 0.350 mol C6H 6(v)/mol 0.650 mol N (g)/mol o 2 175 C, 820 torr Gas Liquid n (mol/s), V (L/s) 1 1 y1[mol C6H 6(v)/mol] (1 y )[mol N (g)/mol] n [mol C H (l)/s], V (L/s) 2 6 6 2 o 1 2 o Condenser 15 C, 760 torr 15 C, 760 torr 4-9
Q: The benzene vapor in the product gas stream must be saturated. Explain why. A: Q: In terms of labeled variables, what is the percentage condensation of benzene? A: % C 6 H 6 condensed = Problem: Calculate the volumetric flow rates of the product streams and the percentage of the entering benzene that is condensed. DOF Analysis: unknowns ( n 1, ) balances ( ) _1 _1 _1 = 0 DF How do you know you can use Raoult s Law? System equations. Write all of the equations you need to determine all of the requested quantities from the given information. Solution: V 1 L/s, V 2 L/s, % C6H 6 condensed = 4-10
Saturated and superheated vapors A(v), non-condensable gases at T( o C), P(atm) Suppose A is the only condensable species in the gas mixture. Then Case 1: y A P = p A *(T) vapor is saturated: gas mixture holds as much vapor as possible at T,P. Liquid i may or may not be present. Cooling or compression condensation. Case 2: y A P < p A *(T) vapor is superheated: gas contains less vapor than it can hold at saturation. Cooling or compression no condensation until < becomes =, at which point vapor is saturated Case 3: y i P > p A *(T) equilibrium not physically possible some vapor must condense. Cool a superheated vapor at constant pressure: What happens? Initially, y i P < p i *(T). Lower T at constant P y i & P stay constant (why?) Left-hand-side (LHS) constant, p i * decreases (why?) RHS decreases Eventually at some T (= T dp, dew point temperature), LHS = RHS saturation. Further cooling then leads to condensation. Thus we can define the dew point temperature at that pressure, T dp, by y i P = p i *(T dp ) Degrees of superheat: DS = T T dp Compress a superheated vapor at constant temperature: What happens? Initially, y i P < p i *(T). Raise P at constant T y i (up, down, constant), P LHS p i * RHS Eventually at some P (= saturation pressure), LHS = RHS saturation. Further compression then leads to condensation. Thus we can define the dew point temperature T dp y i P sat n = p i *(T) If a vapor at T is saturated, it is at its dew point. (Note: For a single species, dew point and boiling point are the same temperature (point on the VLE phase diagram. For a mixture, the dew point and the boiling point (bubble point) are different. Stay tuned, Section 6.5). Work through Example 6.3 2. 4-11
Two mechanisms for transferring a liquid to a gas phase (top of p. 253) If p i *(T) of liquid < P, liquid evaporates. Molecules of liquid transfer from liquid surface. If p i *(T) of liquid > P, liquid boils. Bubbles of vapor form in liquid (usually at vessel wall), erupt from liquid. How many ways can we find to give you a mole fraction of a condensable vapor? 1.00 mole% H 2 O(v), 99.0% dry air @ 25 o C, 1000 torr 1. Mole fraction: y w = 0.0100 mol H 2 O/mol 2. Dew point: T dp = 11.2 o C (= T at which condensation would begin if the gas were cooled at constant pressure). Table or Antoine equation * o w * p (11.2 C) torr ywp pw( Tdp) yw 0.0100 P torr 3. Degrees of superheat: D.S. = 13.8 o C (=difference between the actual temperature and the dew point). o o o o dp DS.. TT 13.8 C T 25 C 13.8 C 11.2 C (Proceed as above) dp D.S. = 0 means 4. Relative saturation, or relative humidity for air-water system: h r = 42.1% (= ratio of the partial pressure of the vapor to its saturation partial pressure at the same temperature) given by Eq. (6.3-4): ywp 0.421( torr) hr 100% 42.1% y 0.0100 * o w pw(25 C) torr For a saturated vapor, s r (or h r for air-water) = % 5. Molal saturation (molal humidity): Eq. (6.3-5) 6. Absolute saturation (absolute humidity): Eq. (6.3-6) 7. Percentage saturation (percentage humidity): Eq. (6.3-7) 8. Or, you could be told that the vapor is saturated or that it is in equilibrium with a pure liquid of the same species, in which case * w Vapor is saturated y p ( T) / P w In a material balance problem, if you are given any of the quantities in Items 2 8, label the mole fraction on the flow chart and count Raoult s law or the defining equation of the given quantity as a relation in the degree-of-freedom analysis. Do Test Yourselves on pp. 253 & 254 and Example 6.3-3. 4-12
Multicomponent Gas-Liquid Systems (Sect. 6.4) Earlier in Chapter 6 we discussed systems with one condensable species. Now let s extend that to systems with several species that can condense. We want to know how components are distributed in the liquid and the vapor phases, so we can design and analyze several common separation processes. Vaporization/Condensation: Partially vaporize a liquid mixture of volatile species or partially condense a vapor mixture. The vapor product will be richer in the more volatile species, the liquid product richer in the less volatile species. 71.4 mol/s 0.600 mol C 5 H 12 (v)/mol 0.400 mol C 6 H 14 (v)/mol 100 mol/s 0.500 mol C 5 H 12 (l)/mol 0.500 mol C 6 H 14 (l)/mol 28.6 mol/s 0.250 mol C 5 H 12 (l)/mol 0.750 mol C 6 H 14 (l)/mol Distillation: Do a series of vaporizations and condensations to improve the separation of components you can get in a single stage operation. Schematics of multistage distillation are shown on p. 296 of the text and in the Visual Encyclopedia of Chemical Engineering Equipment (Chemical separations Distillation columns Plate distillation columns) Absorption or (environmental) scrubbing: Bubble (sparge) a non-condensable gas through a liquid solvent or spray a liquid solvent mist into a stream of gas, generally at low T and high P, getting some or essentially all of the gas into solution. (Examples: SO 2 in stack gas dissolves in solvent in a scrubbing tower; HCl dissolves in water to make hydrochloric acid; CO 2 dissolves in liquid to make soda). The higher the solubility of the gas at equilibrium, the more concentrated the product solution can be. Schematics of absorbers are shown on p. 296 of the text and in the Visual Encyclopedia of Chemical Engineering Equipment (Chemical separations Absorbers Spray columns, bubble columns, wet scrubbers) Desorption or stripping: Bring a dissolved species out of solution into the gas phase at high T and/or low P (desorption), possibly by bubbling an insoluble gas through the solution (stripping). Note that absorption and stripping operations may be linked to transfer a species from one gas mixture to another one. A(g) B(g), C(g), N 2 (g) S(l), B(soln), C(soln) A(g), B(g), C(g) Absorber (or Scrubber) Low T and/or high P S(l) N 2 (stripping gas) Stripper High T and/or low P 4-13
Example SO 2 scrubbing: A = stack gases, B = SO 2, no C, S = alkaline solvent or slurry Why is this an important system? The stack gas comes from a coal-fired furnace or boiler. Trace amounts of sulfur in the fuel result in SO 2 formed during combustion: 2SO 2 + O 2 2SO 3. SO3, sulfur trioxide, can react to form sulfuric acid (H2SO4), a component of acid rain (SO 3 + H 2 O H 2 SO 4 ) Objective: to transfer most of the SO 2 from the gas phase to a liquid phase before it can get into the atmosphere. What information do we need to know to design this system? o o o o How soluble is SO 2 in water? How much SO 2 is coming in? What are the regulatory requirements for how much SO 2 can leave in the emissions? How much solvent will I lose in the gas stream? This is a concern especially if the solvent is expensive if it s water, we may not care. Other Examples: Here are some familiar phenomena. See if you know what s going on. 1. A cold can of soda is opened and bubbles slowly form and emerge. What are they, and why is that happening? 2. A warm can of soda is opened and bubbles rapidly form and emerge. Why is this process different from the previous process? 3. A pot is partially filled with tap water at 20 o C and heated on a stove. You first see a lot of very small bubbles coming out of the water (the water is only lukewarm at the time), then the flow of small bubbles stops. Eventually the water boils large bubbles form below the water surface (mostly at the bottom surface of the pot) and burst out. What do you think is going on? In Chapter 6, we ll find out these answers and answers to other questions about familiar phenomena. Vapor-liquid equilibrium data: Read through Section 6.4a and make sure you understand Example 6.4-1 on p. 255. (Uses tabulated data for partial pressures for SO 2 -H 2 O system.) Raoult s law and Henry s law: Simple equilibrium relations for multicomponent gas-liquid systems. Consider a 2-component condensable gas-liquid system at equilibrium. n v (mol vapor) y A [mol A(v)/mol] y B [mol B(v)/mol] (= 1 y A ) T( o C), P(torr) n l (mol liquid) x A [mol A(l)/mol] x B [mol B(l)/mol] (= 1 x A ) T( o C), P(torr) 4-14
Raoult s law (Eq. 6.4 1) p y Px p * ( T) i i i i where p i * = vapor pressure of Component i. Raoult s Law is an approximation that applies to vapor and liquid phases in equilibrium. Note: If x i = 1, (6.4-1) reduces to (6.3-1) for a single condensable species. Raoult s law most accurate when applied to mixtures of structurally similar liquids (straight-chain alcohols, aromatic hydrocarbons,...) pentane, hexane, heptane methanol, ethanol, propanol benzene, toluene, xylene a component of a liquid mixture for which x i 1 (the solvent in a very dilute solution) Apply with care to dissimilar liquids, never to immiscible liquids (e.g., hydrocarbons & water). Example: A liquid mixture contains 40.0 mole% C 6 H 6 (l) (benzene) and 60.0 mole% C 7 H 8 (l) (toluene) at 90 o C. Find vapor phase pressure and composition. Solution: B = C 6 H 6, T = C 7 H 8 y B (mol B(v)/mol) (1 y B ) (mol T(v)/mol) 90 o C, P(torr) 0.400 mol B(l)/mol 0.600 mol T(l)/mol 90 o C Gibbs Phase Rule: DF = 2 + n species n phases = 2 + 2 2 = 2. Since two intensive variables have been specified for the system (x B = 0.400, T = 90 o C), all other intensive variables (in this case, P and y B ) are fixed. Both components are aromatic hydrocarbons apply Raoult s law for each one (2 eqs. in 2 unknowns), using the Antoine equation (Table B.4) for the vapor pressures: p B * = 10^(6.89272 1203.531/(90+219.888)) p T * = y B P = Antoine equation for benzene Antoine equation for toluene Raoult s law for benzene Raoult s law for toluene 4-15
Solutions: p * B (90 o C) = 1021 torr p * T (90 o C) = 407 torr P = 652 mm Hg, yb = 0.626 Q: Which is more volatile benzene or toluene? A: Observe: (a) Pressure is weighted average of component vapor pressures at 90 o C. (b) Vapor is enriched in more volatile component: x B = 0.400 mol B(l)/mol, y B = 0.626 mol B(v)/mol. s Henry s law (Eq. 6.4 2): pi yip xihi( T) where H i (atm/mole fraction) = Henry s law constant for Component i. Note that H i (T) denotes a function of T, not times T. It is specific to a pair of species (e.g. SO 2 in H 2 O). Most accurate when applied to a nondissociating, nonionizing, nonreactive component of a liquid mixture for which x i 0 (e.g., the solute in a very dilute solution, or absorbed gas with a low solubility). Look up H i (T) in Perry s Chemical Engineers Handbook & other standard references. Q: The (higher, lower) the value of H, the greater the solubility of a gas in a liquid. A: ( ) Component (in water @ 25 C) H (units of atm) O 2 4.259 x 10 4 H 2 7.099 x 10 4 N 2 9.770 x 10 4 * Values from http://www.rolf-sander.net/henry/ Ideal solution: VLE relationships for all components can be described by either Raoult s or Henry s Law over the entire composition range. If a solution is not ideal, need to use more complex phase equilibrium relations (a topic treated in CHE 316). When you use Raoult s law or Henry s law for a solution component and are asked to justify doing so, you can say any of four things: (1) x i 1 (Raoult s law); (2) x i 0 (Henry s law, nondissociating nonionizing nonreactive species); (3) mixture of structurally similar compounds (Raoult s law); and if all else fails, (4) we assume ideal solution behavior. 4-16
Example: A system at equilibrium at 20 o C and pressure P(atm) contains water and CO 2 in liquid and gas phases. The gas phase is 10.0 mole% CO 2, and CO 2 is only slightly soluble in water. We wish to determine P and the composition of the liquid phase. 0.100 mol CO 2 (g)/mol 0.900 mol H 2 O(v)/mol 20 o C, P(atm) x C [mol CO 2 (dissolved)/mol] (1 x C ) [mol H 2 O(l)/mol] 20 o C (a) Use Gibbs Phase Rule to demonstrate that all unknown intensive variables can be determined (at least in principle) from the given information. (b) Which VLE correlations (laws) would you use to express the vapor-liquid equilibrium relationship for CO 2 : s law, because H 2 O: s law, because (c) The Henry s law constant for carbon dioxide in water at 20 o C is 1.38x10 4 atm/mole fraction. Calculate P and x C. 4-17
Exercise. Use Henry s Law pi yip xihi( T) to explain some familiar phenomena: (a) : A cold can of soda (CO 2 dissolved in water and nonvolatile additives) is opened and bubbles slowly form and emerge. Explain why, using Henry s law in your explanation. (b) A warm can of soda is opened and bubbles rapidly form and emerge. Explain why this process is different from the previous process, again using Henry s law. What is the effect of T on H CO2? (c) A pot is partially filled with tap water at 20 o C and heated on a stove. You first see a lot of very small bubbles coming out of the water (the water is only lukewarm at the time), then the flow of small bubbles stops. Eventually the water boils large bubbles form below the water surface (mostly at the bottom surface of the pot) and burst out. What are the small bubbles? (Hint: They re not water.) Why are they forming? What are the large bubbles? Why is the temperature at which they form slightly greater than 100 o C? Is any vaporization occurring between the emission of the small bubbles and boiling? Explain. 4-18
Bubble and Dew Point Calculations for Ideal Solutions (Section 6.4c) T n bp (benzene) = 80.1 o C T n bp (toluene) = 110.6 o C Heat 0.40 mol C 6 H 6 (l)/mol 0.60 mol C 7 H 8 (l)/mol 20 o C, 760 torr Heat liquid mixture at constant pressure. At some temperature, the first bubble of vapor forms. Speculate on the temperature at which it happens (the bubble point temperature) and the composition of the vapor in the bubble. A logical guess would be that a bubble of pure benzene would form at 80.1 o C. That would be wrong. In fact, the first bubble would form at 95 o C and would contain 62.1 mole% benzene and 37.9% toluene. Now let s find out how to do those calculations. 4-19
T n bp (benzene) = 80.1 o C T n bp (toluene) = 110.6 o C 0.400 mol C 6 H 6 (l)/mol 0.600 mol C 7 H 8 (l)/mol 20 o C, 760 torr Heat liquid mixture at constant pressure. Find T at which first vapor bubble forms (bubble point temperature) and composition of the bubble. 0.400 mol C 6 H 6 (l)/mol 0.600 mol C 7 H 8 (l)/mol T bp ( o C), 760 torr y B (mol C 6 H 6 (v)/mol) (1 y B ) (mol C 7 H 8 (v)/mol) T bp ( o C), 760 torr Since liquid and vapor (bubble) are in equilibrium, and benzene and toluene are similar in structure, Raoult s law applies to each species: Benzene: (p B =) y B (760) = 0.40p B *(T bp ) Toluene: (p T =) (1 y B )(760) = 0.60p T *(T bp ) Substitute Antoine eqn. for vapor pressures, use Solver with 2 eqns. in 2 unknowns (or add equations to get Eq. (6.4 4) on p. 259 for T bp ): x B = 0.400 ; P = 760 p B * = 10^(6.89272 1203.531/(T bp + 219.888)) p T * = 10^(6.95805 1346.773/(T bp + 219.693)) y B *760 = x B *p B * (1 y B )*760 = (1 x B )*p T * specify xb (mol B(l)/mol) and P (mm Hg) vapor pressure of benzene vapor pressure of toluene Raoult s law for benzene Raoult s law for toluene Solution: T bp p B * p T * y B 95.146 1181.4 479.064 0.621792 Note that T bp is between normal boiling points of benzene (80.1 o C) and toluene (110.6 o C), & the vapor enriched in more volatile species. Can easily repeat the calculation for new values of x B and P, or do a sweep of x B from 0 to 1. Dew point calculations proceed similarly 4-20
0.40 mol C 6 H 6 (v)/mol 0.60 mol C 7 H 8 (v)/mol 150 o C, 760 torr Cool at constant pressure, find T at which first liquid droplet forms (dew point temperature), composition of the droplet. 0.40 mol C 6 H 6 (v)/mol 0.60 mol C 7 H 8 (v)/mol T dp ( o C), 760 torr x B (mol C 6 H 6 (l)/mol) x T (mol C 7 H 8 (l)/mol) [= 1 x B ] T dp ( o C), 760 torr Exercise: Derive the equations for the dew point temperature and initial droplet composition (mole fraction of benzene). Also derive Eq. (6.4 7) for the dew point temperature. Notes: Henry s Law applies only for components that are dilute AND nondissociating, nonionizing, nonreactive. For example, HCl in water would quickly ionize completely so Henry s Law cannot be used. In this course, when you have a piece of equipment and vapor and liquid streams exit as products, you can assume that the streams are in equilibrium and that the vapor is saturated (i.e. can apply Raoult s or Henry s Law), unless told otherwise. 4-21
Txy/Pxy diagrams (Fig. 6.4 1, p. 262) CHE 205 Chemical Process Principles 1. For binary mixtures (A & B) at a specified pressure, assume x (liquid-phase mole fraction of lighter component A), calculate T bp and y (mole fraction of A in vapor phase in equilibrium with liquid) as above. Plot T vs. x and y as shown: Liquid Vapor T bp T x 2. Repeat for full range of x from 0 to 1, generate two curves. y Fixed P T Both phases Vapor T vs. y T vs. x T bp Liquid 0 z 1 Mole fraction of A Start with liquid mixture, x = z. Heat slowly at constant pressure. First bubble forms when T reaches the T-x curve. Go horizontally right to T-y curve, read mole fraction of A in vapor. Keep heating move into two-phase region. More liquid evaporates. At a specified T, go left to T-x curve to read liquid-phase composition, right to T-y curve to read vapor-phase composition. Heat more. Eventually reach T at which only a drop of liquid remains. Mole fraction in vapor is then z; read mole fraction in liquid drop by going left to T-x curve. Questions: 1. What is the physical significance of T where the two curves meet at x = 0? 2. What about T at the intersection at x = 1? 4-22
Exercises: Use the Txy diagram for benzene and toluene to solve the following problems: 1. An equimolar liquid mixture of B & T is slowly heated from 20 o C to 105 o C at 1 atm. (a) At what temperature does the first bubble form? (b) What is the vapor composition of the first bubble? (c) What happens to the liquid and vapor phase amounts and compositions (x and y) as heating proceeds? (d) At what temperature does the last droplet vaporize? (e) What was the liquid composition of the last droplet? (f) From the graph, what is the normal boiling point of benzene? (g) What is the normal boiling point of toluene? (h) What happens if we start out with pure liquid benzene and heat from 20 o C to 105 o C? 4-23
2. A 60% benzene 40% toluene vapor mixture is slowly cooled from 120 o C to 50 o C at 1 atm. (a) What is the dew point temperature? (b) What is the composition of the first liquid droplet? (c) What happens to the liquid and vapor phase amounts and compositions as cooling proceeds? (d) At what temperature does the last bubble condense? (e) What is the vapor composition of the last bubble? 3. An equimolar B-T vapor mixture is cooled from 120 o C to 95 o C at 1 atm. Use the Txy diagram to determine the fraction condensed and compositions of the liquid and vapor streams. n v (mol) y B [mol B(v)/mol] Basis: 100 mol 0.50 mol B(v)/mol 0.50 mol T(v)/mol 120 o C, 1 atm Finish labeling the flowchart. DOF Analysis From the Txy diagram, y B = and x B = 4-24
Briefly state how you would calculate the fraction (moles condensed/mole fed)? Exercise Suppose a test problem begins, A liquid mixture contains 30.0 mole% A and 70.0 mole% B. (Specific species are given in place of A and B.) What questions might follow? What else might be asked if the species are benzene and toluene and the pressure is constant at 1 atm? Boiling vs. Evaporation. Read the two paragraphs on p. 263 ending with Eq. (6.4-9). Be able to (a) describe and distinguish between the mechanisms of boiling and evaporation (vaporization), (b) estimate the boiling point of a liquid mixture of known composition at a specified pressure assuming Raoult s law is valid, (c) explain why the actual boiling point of the mixture would be slightly greater than the calculated value (see footnote on p. 263). Practice Example 6.4-2 on p. 258. Example 6.4-3 on p. 260. Understand the trial-and-error procedures shown in the solution, and also recognize that you don t need to use any of them if you have Excel with Solver, both of which do all the trial-and-error calculations for you. Example 6.4-4 on p. 262. Example 6.4-5 on p. 263. Be able to explain how you would do the calculation if you didn t have the Txy diagram. Test Yourself on p. 264. Problem 6-60 in workbook. 4-25
Colligative Solution Properties (Section 6.5c) Dissolve a nonvolatile (negligible vapor pressure), nonreactive, nondissociative solid, A, in a solvent, S. p s (torr) [no solute in vapor phase] Apply Raoult s law to the solvent. x (mol A/mol) (1 x) [mol S(l)/mol] * * s s s p ( T) (1 x) p ( T) p ( T) (6.5-1) The effective vapor pressure of the solvent is lowered by the presence of the solute. The extent of the lowering is * * s xps p (6.5-3) regardless of what the solute and solvent are. Fig. 6.5-2 Figure 6.5-2 shows that the vapor pressure lowering also has the effect of raising the boiling point and lowering the freezing point of the solvent at a fixed pressure. For dilute solutions (x 0), the boiling point elevation and freezing point depression are given by the approximate relations of Eqs. 6.5-4 and 6.5-5. 2 bo RT Tb Tbs Tbo x Hˆ 2 mo RT Tm Tmo Tms x Hˆ v m (6.5-4) (6.5-5) Work through Test Yourself, Example 6.5-4 on p. 270. 4-26