ADVANCED THERMODYNAMICS ENGINEERING phần 7 ppt

In air X N 2 /X O 2 = 3.76. Furthermore, (X O 2( )l /X N 2( )l ) = (X O 2 /X N 2 )/( P N sat 2 (T)/ P O sat 2 (T)) (C) Since P N sat 2 (T) ≈ P O sat 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 k sat 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 cuss the effects with salt addition at 111.4ºC. Solution P sat (111.4ºC) = 1.5 bar. The total volume V = m g v g + m f v f = m (x v g + (1–x) v f ) (A) = m (0.2× 1.159 +0.8× 0.001053) = m (0.233). 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) V f = (m – P o V/RT)/(1/v f – P o /RT) ≈ v f (m – P o V/RT). (C) The pressure increases as additional gas is injected, thereby increasing the Gibbs en- ergy of the liquid and vapor phases. In case of liquid water, g l (T,P) = g l (T,P sat ) + v l (P – P sat ). For an ideal gas mixture in the vapor phase, ˆ g HO 2 (T,P,X H2O ) = g HO 2 (T,p HO 2 ) = g H2O (T,P sat ) + ∫ v H2O(g) dP = g g (T,P sat ) + R T ln(p HO 2 /P sat ). (D) Equating Eq. ( C) with Eq. (D) v (P – P sat ) = RT ln (p HO 2 /P sat ), or ln(p HO 2 /P sat ) = (v l (P – P sat ))/(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 sat = 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, p HO 2 = 1.500445 bar. a value close to saturation pressure at T= 384.6K since v f is small. Further 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- 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 f (P – P sat ))/(RT), in Eq. (A) this implies that p H2O = P sat 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 < ) l g HO 2 () (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- 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 300 350 400 450 500 550 T, K X 2 0 0.2 0.4 0.6 0.8 1 1.2 X 1 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. ergy increase must then cause the entropy of vapor to decrease which corresponds to an increase in the partial pressure of vapor. Consider a component k of a liquid mixture that exists in equilibrium with a vapor phase that also contains a mixture of insoluble inert gases. In this case, µ k(l) (T,P) = µ k(g) (T, P). If the vapor phase is isothermally pressurized, then µ k(l) (T,P) + dµ k(l) = µ k(g) (T,P) + dµ k(g( , v k(l) dP l = v k(g) dP g and dP l /dP g = v k g /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. 2. Immiscible Mixture a. Immiscible Liquids and Miscible Gas Phase This case is illustrated through the following example. k. Example 11 pressure and temperature. You may assume that ln P sat 2 = 13.97 – 5205.2/T (K), and (A) ln P sat 1 = 13.98 – 4719.2/T (K). (B) Solution Employing Eqs. (A) and (B), the normal boiling points of species 1 (methanol) and 2 (water) are, respectively, 64.4 and 100ºC. We will employ Raoult’s Law, in which the liquid mole fractions for water and methanol must be set to unity, since they are immiscible. Therefore, P sat 2 (T) = X 2 P, and (C) P sat 1 (T) = X 1 P. (D) Upon adding Eqs. (C) and (D), we obtain the expression P sat 2 (T) + P sat 1 (T) = P. (E) Figure 8 shows the T- X k( l ) -X k diagram. Remarks In case of immiscible mixtures, partial pressures are only a function of temperature alone. Irrespective of the liquid phase composition, at a specified temperature, P sat 2 can be obtained from Eq. (A), while P sat 1 can be, likewise, obtained using Eq. (B). Using Eqs. (C) and (D), we obtain the values of X 1 and X 2 for a specified pressure, and plots of temperature can be plotted with respect to composition, as shown in Figure 8. The lines BME and EJGA in that figure are called the dew lines for species 1 and 2, respectively. The region above the curve BMEJGA is the superheated vapor mixture region while that below the curve CELD is the compressed liquid region. Consider the following scenario. A vapor mixture is contained in a pis- ton–cylinder–weight assembly, such that P = 1 bar, X 2 = 0.6, and T = 100ºC (cf. point S). Species 2 exists in the form of superheated vapor, since p 2 = 0.6 bar at T = 100ºC. The cylinder is now cooled. The saturation temperatures T sat 2 = 86.5ºC at p 2 = 0.6 bar, and T sat 1 = 43.87ºC at p 1 = 0.4 bar. The assembly contains a vapor mixture only, as long as T>86.5ºC. As the vapor mixture is cooled, a liquid drop appears at T = 86.5ºC Consider binary vapor mixture of methanol (species 1) and water (species 2) that are assumed to be immiscible in the liquid phase. Illustrate their behavior with respect to (point G). (If the gas phase composition is changed to X 2 = 0.2, in that case the first liquid drop appears at 61ºC (point E)). If the mixture is cooled to 70ºC (cf. point H), phase equilibrium – that is manifested in the form of Eq. (C) – implies that vapor phase mole fraction must reduce to X 2 = 0.3 (cf. point J), i.e., more of species 2 must condense. It also implies that X 1 must increase to 0.7 from the initial mole fraction of 0.4. Eqs (D) and (B) dictate that T sat 1 = 52ºC, so that species 1 at 70ºC exists in the form of a superheated vapor. Upon further cooling to 60ºC, phase equilibrium re- quires that X 2 = 0.19 (cf. point E), and T sat 1 increases to 60ºC. Any further cooling causes both species 1 and 2 to condense, where the condensate phase is an immiscible binary mixture. Within the region EJGADE (i.e., for X 2 > 0.19, 60ºC <T < 100ºC), liquid species 2 and vapor mixture must coexist. In region BMEC (X 2 < 0.19, 60ºC < T <64.7ºC), liquid species 1 and vapor mixture must coexist. At 60ºC, X 2 = 0.19, and both the liquid and vapor mixture coexist. At point E there are two liquid phases and one vapor phase. According to Gibbs phase rule, F = K + 2 – π = 2 + 2 – 3 = 1. Therefore, there is one independent variable in the set (P, T, X 2 ). In case the pressure is fixed, then the temperature and X 2 are fixed (i.e., 60ºC, and X 2 = 0.19) for coexistence the three phases to coexist. If the mixture is cooled from 100ºC (cf. point K), species 1 condenses, increasing the mole fraction of species 2 until X 2 = 0.19. Now assume that the liquid mixture is heated at the condition X 2( l ) = 0.6 and P = 1 bar in a piston–cylinder–weight assembly. At low temperatures, the sum of the saturation pressures (cf. Eq. (E)) is insufficient to create the imposed 1 bar pressure. Therefore, at T < 60ºC (point Q), the fluid exists as a compressed liquid. At ≈60ºC (cf. point L), the sum of the saturation pressures is roughly 1 bar. The temperature at this condition can be predicted using Eqs. (A), (B), and (E) as 60ºC. Consequently, the values of X 1 and X 2 can be determined as 0.81 and 0.19 using Eqs. (C) and (D). Thus first vapor bubble at a 1 bar pressure appears at 60ºC. At this point there are three phases (two 40 50 60 70 80 90 100 0 0.2 0.4 0.6 0.8 1 X 2 , Z 2 T, C T dew of 1 Vapor Mixture Separate Liquid Species 1 and 2 Vapor Mixture+Liquid 2 Vapor Mixture+Liquid 1 Species 1 Species 2 T vs X 2 S G H L Q B M E J AK D C Figure 8: A T–X l –X diagram for an immiscible liquid solution. immiscible liquid phases, since they are immiscible, and a vapor phase). As more heat is isobarically added, the temperature cannot rise according to Eq. (E), but the vapor bubble can grow. If the heating process begins with 0.4 kmole of species 1 and 0.6 kmole of species 2 and vaporization occurs until the vapor phase is at state E (i.e., T sat 1 = 60ºC), since the vapor phase mole fraction of species 1 is 0.81, the ratio of the moles of species 2 that are vaporized to those of species 1 is 0.19÷0.81. Therefore, for every 0.4 kmole of species 1 that are vaporized, the moles of species 2 that are va- porized equal 0.4×0.19÷0.81 = 0.094 kmole. Hence, the vapor mixture contains 0.4 kmole of species 1, 0.094 kmole of species 2, and 0.6–0.094 = 0.506 kmole of species 2 remain in the liquid phase. Now the species 1 from liquid have been completely va- porized. Once T>60ºC, further vaporization of species 2 occurs, thereby increasing the mole fraction of species 2 in the vapor state, and it is possible to determine the value of X 2 along the curve EJGA. As the temperature reaches 86.5ºC (cf. point G), all of the initial 0.6 kmole of species 2 in the liquid phase vaporize so that X 2 = 0.6. b. Miscible Liquids and Immiscible Solid Phase Oftentimes two species 1 and 2 are miscible in the liquid phase, but are immiscible in the solid phase and each species forms its own aggregate in the solid phase (i.e., upon cooling of the liquid mixture, the two species form two separate solid phases). In this case, at phase equilibrium, ˆ f 1( l ) = ˆ f 1(s) , and ˆ f 2( l ) = ˆ f 2(s) (17a) Under the ideal solution assumption and since X 1(s) = 1 due to immiscibility X 1, l f 1( l )( T,P) = f 1(s)( T,P). (17b) For example, pure H 2 O at a temperature of –5ºC and a pressure of 1 bar should exist as ice. However, if the water is a component of a binary solution (e.g. salt addition), then ˆ f HO 2 ()l = X HO 2 ( l ) f H2O( l ) (–5ºC, 1 bar) = X HO 2 ( l ) f H2O( l ) (–5ºC, P sat ) POY ( l ) , where POY = exp (v ( l ) (P – P sat )/RT). Generalizing, X 1( l ) f 1( l )( T,P) = X 1( l ) f 1( l ) (T,P 1 sat ) POY 1 ( l ) = f 1,s (T,P 1 sat ) POY 1 (s) . (18) Since f 1( l ) (T,P 1 sat ) = f 1(s) (T,P 1 sat ), X 1( l ) POY 1 ( l ) = POY 1(s) , and X 2( l ) POY 2 ( l ) = POY 2( s) . (19) However, X 1( l ) + X 2( l ) = 1, so that POY 1(s) /POY 1 ( l ) + POY 2(s) /POY 2 ( l ) = 1 (20) Following example 11, the pressure can be determined from Eq. (20) at a specified tempera- ture. The mole fractions X 1( l ) or X 2( l ) at that pressure can be obtained using Eq. (19). 3. Partially Miscible Liquids a. Liquid and Gas Mixtures Many liquids are miscible within a certain range of concentrations. The solubility of liquids with one another generally increases with the temperature. The corresponding pres- sure–temperature relationships are a combination of the corresponding relationships for misci- ble and immiscible liquids. Figure 9 illustrates the T-X k( l ) -X k diagram for a partially miscible liquid. In the context of Figure 9 assume that methanol and water are partially miscible. Let X 1( l ) denote the methanol mole fraction and X 2, l the water mole fraction in the liquid. Suppose that water (species 2) is soluble in methanol (species 1) up to 10% by mole fraction at 40ºC. As the temperature is increased, the increased “ve” can overcome the attractive forces between methanol molecules and hence its solubility increases as the temperature approaches 60ºC. Line FLC represents the boundary between miscibility and immiscibility. When solubility re- mains constant, the line is vertically oriented (cf. line VF). Water is insoluble from X 2( l ) = 0.1 to X 2( l ) = 0.8 say at temperatures less than 40ºC, but at 60ºC it is immiscible for values of X 2( l ) < 0.7. Region I is a miscible liquid mixture but rich in species 2, while region II is a miscible liquid mixture but rich in species 1. The boundary DQG represents the variation of miscibility with temperature in the region richer in X 2, l . The region above line CED is similar to the im- miscible case we have just discussed. Consider the vapor mixture at a 90% water vapor concentration (point K). As we cool the vapor from state K to M, first a liquid drop appears containing both species that has a com- position corresponding to point R, while the vapor has a composition corresponding to point M as discussed for miscible liquids. As the temperature is decreased to point N, the last liquid will have composition at N (at the bubble line) while the vapor is at state T. If temperature is further decreased, a liquid mixture fixed at a composition N forms. If we start at point S, then we obtain the first drop at point T with drop composition corresponding to N, which is in the miscible region. As we cool further to point U, the liquid composition is at D (miscible limit at 60ºC) while the vapor is at E. However there is still wa- ter and methanol vapor left in the mixture. Condensation will occur at a constant vapor com- position with the liquid-I composition at D (rich in species 2) and liquid-II composition C (rich in species 1). If the temperature drops below 60ºC, there are two separate liquid phases I (composition rich in species 2 along DQG) and II (composition lean in species 2 along FLC). However, the fraction of species 2 in liquid–II will increase since the solubility of species 2 increases (DQG) while the fraction of species 2 in liquid–II will decrease (FLC). b. Liquid and Solid Mixtures When a solid (a solute, such as salt) is dissolved in a liquid (a solvent, e.g., water), the dissolved solid can be considered as a liquid in the liquid solution. It is pertinent to know the maximum amount of solute that can be dissolved in a solvent. We will denote the salt in solid phase as s(s) and that in the liquid as s(l). At the equilibrium state of a saturated liquid solution with a solid salt, ˆ f s( l ) = X s l f s( l )( T,P) = f s(s) (T,P), where (21) f s(s)( T,P) = f s(s) (T,P sub ) POY s(s) , and (22) POY s(s) = exp [v s(s) (P–P sub )/RT]. (23) The P sub denotes the saturation pressure for the sublimation of a salt at a specified temperature. Since f s(s) (T,P) = f s(g) (T,P sub ), f s(s) (T,P) = f s(g) (T, P sub ) POY s(s) . (24) If the vapor phase behaves as an ideal gas, f s(s) (T,P) = P sub POY s(s) . (25) Similarly, f s( l )( T,P) = φ s( l )( T,P) P. (26) Employing Eqs. (21) and (26) X s( l ) φ s( l )( T,P) P = P sub (T) POY s(s) , and (27) X s( l ) = (P sub (T) POY s(s) )/(P φ s( l ) ). (28) At low pressures, an increase in the pressure causes the solubility to decrease while at higher pressures the value of φ s( l ) may decrease and, consequently, the solubility may also increase. D. DISSOLVED GASES IN LIQUIDS Gases dissolve in liquid solutions through a process called absorption. (This should not be confused with adsorption, which is a process during which molecules are attached to a the surface of a solid material due to strong intermolecular forces.) The solubility of a compo- nent in a mixture is expressed as a ratio of the maximum amount of solute that can be present in a specified amount of solvent. In case of gases, the solubility is typically expressed in units of ppm. We will treat dissolved gaseous species within a liquid as though they behave like liquids. → 60 C 40 C 20 C Immiscible L Q V X G I II Vapor Liq S T U N M K R Rich 1::V Rich2:V 0.10 0.20 0.70 0.80 0.90 X 2(l) , X 2 F Figure 9: T–X k,, l –X k diagram for partially immiscible liquid mixture, where I denotes the miscible vapor, II the miscible liquid, AE and BMTE the dew lines; AC and BRND the bubble lines, and CLF and DQG the boundaries separating the miscible and immiscible regions. (From Smith and Van Ness, Introduction to Chemical Engineering Thermody- namics, 4th Edition, McGraw Hill Book Company, 1987, p. 455. With permission.) 1. Single Component Gas As carbon dioxide is dissolved in water, at some concentration a vapor bubble con- taining pure CO 2 will start to form. At that saturated condition. ˆ f CO 2 ( l )( T,P,X CO 2 , l ) = f CO 2 (T,P). (29) Employing the ideal solution model, X CO 2 ( l ) f CO 2 ( l ) (T,P) = f CO 2 (T,P). If the gas phase behaves as an ideal gas X CO 2 , l f CO 2 ( l ) (T,P) = P, else f CO 2 ( l )( T,P) = f CO 2 ( l )( T,P sat )POY CO 2 ( l ) , where POY CO 2 ( l ) = exp {v CO 2 ( l ) (P – P sat )/RT}, and f CO 2 ( l )( T,P sat ) = f CO2(g) (T,P sat ) = P sat . Therefore, X CO 2 ( l ) = P/ {P CO2 sat POY CO 2 ( l ) }. (30) In general, POY CO 2 ( l ) ≈ 1, and X CO 2 ( l ) = P/P CO2 sat . (31) Generalizing for solute k dissolved in a solvent, X k ( l ) = P/P k sat . This methodology works for values of P sat (T) > P. In the case of carbon dioxide dissolved in soda water at 25ºC, P sat = 66.7 bar. Consequently, at P=1 bar, X CO 2 ( l ) = 1÷67 = 0.015, imply- ing a solubility of 1.5%. As the temperature increases, P sat increases and the solubility of gases decreases. This is an opposite trend to the solubility of liquid components in liquids or of sol- ids in liquid. Recall that chemical potentials of solute in vapor and liquid phases determine whether k is absorbed in or distilled from the solvent. For example, when the pure distilled water is exposed to pure carbon dioxide, if µ CO 2 ,g > µ CO 2 ( l ) , the carbon dioxide is transferred (absorbed) from the gas into the liquid phase. If µ CO 2 ,g < µ CO 2 ( l ) , carbon dioxide is transferred (distilled) from the liquid to the gas phase. The relation shown in Eq. (31) presumes Raoult’s law or ideal solution behavior. At low carbon dioxide mole fractions, since a relatively large number of water molecules sur- round the molecules of carbon dioxide, the liquid water molecules dominate the intermolecular attraction forces. If the attractive forces between water molecules significantly differ from those between the CO 2 molecules, the ideal solution model breaks down. The ideal solution model is also not applicable at higher pressures, since the dioxide no longer behaves as an ideal gas. 2. Mixture of Gases Consider a gaseous mixture (e.g., of carbon dioxide and oxygen) above a liquid sur- face. In that case ˆ f CO 2 ( l ) = ˆ f CO 2 (g) , and using the ideal solution model X CO 2 ( l ) f CO 2 ( l )( T,P) = X CO 2 f CO 2 (g) (T,P) Treating the gases as ideal, X CO 2 ( l ) f CO 2 ( l ) (T,P) = X CO 2 P = p CO 2 , and proceeding as before X CO 2 ( l ) POY CO 2 ( l ) P CO sat 2 = X CO 2 P = p CO 2 , i.e., X CO 2 = p CO 2 /( P CO sat 2 POY CO 2 ( l ) ), or (32) p CO 2 = X CO 2 ( l ) P CO sat 2 POY CO 2 ( l ) . (33) In this case, the total pressure that appears in Eq. (31) is replaced by the partial pres- sure. In power plants, water exists under large pressures and hence air may be dissolved in it in the boiler drums. Since solubility decreases at low pressures, the air is released in the con- denser sections (Eq. (31)). Oxygen is corrosive to metals, and it, therefore, becomes necessary to remove the dissolved air or oxygen from water prior to sending water to the boiler. Deaera- tors are used to remove the dissolved gases from water. They work by heating the water with steam (P sat increases, Eq. (31)), and then allowing it to fall over a series of trays in order to expose the water film so that the gases are removed from the liquid phase as much as possible. Another example pertains to diving in deep water. The human body contains air cavi- ties (e.g., the sinuses and lungs). As a diver proceeds to greater depths, the surrounding pres- sure increases. In order to prevent the air cavities from collapsing at greater depths, the divers must adjust the air pressure they breathe in. They do so by manipulating their diving equip- ment to equalize the cavity pressures with the surrounding water pressure. Consequently, the pressurized air gets dissolved in the blood (Eq. (31)). Upon rapid depressurization, in the process of reaching phase equilibrium, the dissolved air is released into the blood stream in the form of bubbles that can be very harmful to human health. Raoult’s Law may be applied to estimate the concentration of air in blood. Similarly when a person develops high blood pres- sure, the amount of soluble O 2 and CO 2 may increase. If we assume blood to have the same properties as water, we can determine the solu- bility of oxygen at a 310 K temperature and 1 atm pressure as follows. The vapor pressure data of oxygen can be extrapolated from a known or reference condition to 310 K using Clau- sius–Clayperon equation (which is valid if (h fg /Z fg ) is constant), namely, ( P k sat /P ref ) = exp ((h fg,k /(R k Z fg,k ))(1/T ref – 1/T)). (34) The saturation pressure at 310 K can be determined using the relation ln (P sat ) = 9.102 – 821/T (K) bar, i.e., P sat (310 K) = 635 bar. In air, at 1 atm p O 2 = 0.21 bar, and the resulting solubility of O 2 in water is 300 ppm. Another example pertains to hydrocarbon liquid fuels (e.g., fuel injected engines) that are injected into a combustion chamber at high pressures (≈ 30 bar). The gaseous carbon dioxide concentration in these chambers is of the order of 10%. At 25ºC, the solubility of the dioxide in the fuels is ≈0.1×3 MPa÷61MPa = 0.005. This solubility increases as the pressure is increased. 3. Approximate Solution–Henry’s Law Rewrite Eq. (33) as, p k = X k, l H k (T,P), where (35) H k (T,P) = P k sat (POY) k( l ) . (36) Where “k is the solute dissolved in a liquid solvent. The symbol H k denotes Henry’s constant for the k–th gaseous species dissolved in the liquid solution. The units used for H k are typically those of pressure. Since v f has a relatively small value, POY ≈ 1. Therefore, H k (T,P) ≈ H k (T) = p k sat (T), i.e., (37) Hence Eq. (35) is written as, p k = H k (T) p k sat (T). (38) At 25ºC, for molecular oxygen and nitrogen, respectively, H(25ºC) = 4.01×10 4 and 8.65×10 4 bar when p O 2 <1 bar, and p N 2 <1 bar. The ideal solubility of oxygen in water is X k, l = 0.21x10 6 /40100 = 5.2 ppm. This result is only approximate. The solubility of O 2 in water is found to be as high as 170 ppm. Rewriting Eq. (35) X k, l = p k /H k (T,P). (39) Multiplying Eq. (39) by the number of moles per unit volume n l , X k( l ) n l = p k n l /H k (T,P), so that (40) n k( l ) = (X k n l /H k (T,P)) P, i.e., n v,k = H k ´ P, where (41) H k ´ = X k( l ) /H k (T,P) ≈ X k n l /H k (T) = X k n l / p k sat (T). (42) Eq. (41) states that moles of gas dissolved in a unit volume liquid is proportional to total pres- sure. For carbon dioxide that exists in a liquid solution with water, H k ´ = 0.0312 k mole of CO 2 m –3 bar –1 at 25ºC. Since the volume at STP for the gas is 24.5 m 3 , H k ´ = 0.764 m 3 of CO 2 (STP) per m 3 of liquid per bar. The solubility of oxygen in blood is 0.03 mL (STP) per liter of blood per mm of Hg. If partial pressure of oxygen in the lungs is 100 mm of Hg, the dissolved oxygen is 3 mL per liter of blood or 4 mg per liter of blood. Figure 10 presents the variation of H(T) for various dissolved gases in liquid water. E. DEVIATIONS FROM RAOULT’S LAW Consider two species k and j that form a binary mixture. The attraction force between similar molecules of species k is denoted as F kk and between dissimilar molecules as F kj . The following scenarios ensue: (1) F kj = F kk so that the ideal solution model and Raoult’s Law ap- ply, e.g., toluene–benzene mixtures and mixtures of adjacent homologous series; (2) F kj > F kk implying a nonideal solution in which contraction occurs upon mixing, e.g., acetone–water mixtures and other examples of hydrogen bonding; and (3) F kj < F kk , which corresponds to a nonideal solution in which the volume expands upon mixing, e.g., ethanol–hexane and other polar–non polar liquids. In case of the second scenario, since the intermolecular attraction forces are stronger between k–j pairs than between k–k molecular pairs, the vapor pressure of species k can be lower than that predicted using Raoult’s Law, which is referred to as a nega- tive deviation from the Law. In case (3) the attraction forces are lower, and a larger amount of vapor may be produced as compared with the Raoult’s Law prediction, i.e., both the second and third scenarios suggest that we must involve activity coefficients, γ k( l ) . It will now be shown that p k = γ k( l ) X k, l p k sat . (43) where γ k( l ) = ˆ f k( l ) (T,P)/ ˆ f k( l ) id (T,P) = ˆ f k( l ) (T,P)/(X k( l ) f k( l ) (T,P)). 1. Evaluation of the Activity Coefficient We have previously employed the ideal solution model to predict the vapor pressure of a component k in an ideal solution. If the measured component vapor pressure differs from that prediction, then it is apparent that the ideal solution model is not valid. We can determine the activity coefficient (that represents the degree of non-ideality from the measured vapor pressure data) as follows. [...]... C 148 y 1 47. 5 L 1 47 Liquid 146.5 146 0.04 0.05 0.06 0. 07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0. 17 0.18 0.19 0.2 3 v, m / kmole Figure 3: Variation in the entropy vs volume for water modeled as a VW fluid at u = 75 00 kJ kmole–1 entropy of each section is initially SD = (S1 = N1 ¯s 1) = (S2 = N2 ¯s 2) = 2×149.83 =299 .7 kJ (A) The total entropy of the tank is S = 2SD = 299 .7 + 299 .7 = 599.4 kJ... sections 1 and 2 of a rigid container 159 10000 1 57 10000 9500 s, kJ/ kmole K 9000 155 8500 153 8000 151 u = 75 00 kJ/kmole 149 1 47 145 0.02 0. 07 0.12 0. 17 v, m3 / kmole 0.22 0. 27 Figure 5: Variation of s vs v in case of water according to the VW state equation relation δWopt = – dU + T0 δS, with disturbance at a stable state yields δWopt = T0 δS < 0 (7) If the system is initially unstable (cf State... N1 ¯s 1) = (S2 = N2 ¯s 2) = 2×149.83 =299 .7 kJ (A) The total entropy of the tank is S = 2SD = 299 .7 + 299 .7 = 599.4 kJ After the perturbation, v 1 = 0.14÷2 = 0. 07 m3 kmole–1, u1 = 75 00 kJ kmole–1, and v 2 = 0 26÷2 = 0.13 m3 kmole–1, u2 = 75 00 kJ kmole–1 The corresponding states are represented by points M and N in Figure 3, i.e., SM= S1 = N1 ¯s 1 = 2×149.8 = 299.6 kJ, and SN = S2= N2 ¯s 2 = 2×149.9... bar, cv = 30.09 kJ kmole–1 K, a = 142.64 bar m3 kJ kmole–2, b = 0.02110 m3 kmole–1, (∂T/∂v)u = – 173 .04 kmole K m–3, ∂2s/∂v2 = –31.92 kJ kmole m– 6 K– 1 , ∂2s/∂u2 = –9.4× 10 –8 kmole kJ–1 K–1 , and ∂2s/(∂v∂u) = 0.000492 kmole m–3 K–1 For stability, D2 = S UUS UV > 0 , so that S VUS VV (16) (suusvv – svu2 = = 2 .77 ×10–6 kmole2 m–6 K–2) > 0 When calculations are repeated at v = 0.1 m3 kmole–1, D2 < 0 Remarks... (X2(l) ln γ2(l))/(X1(l) ln γ1(l)))2, (54) B= ln γ2(l) (1+ (X1(l) ln γ1(l))/(X2(l) ln γ2(l)))2 (55) At the azeotropic condition, at which Xk(l) = Xk, Eq ( 47 ) assumes the form pk = Xk(l) P = γk(l) Xk(l) p sat k (56) γk(l)(Xk,azeotropic)= P/ p sat (T) k ( 57) Thus, The Wilson equation for activity coefficient has the form ln γ1(l) = AX2,l2/((A/B) X1(l) +X2(l))2, ln γ2 = B X1(l) 2/(X1(l) +(B/A)X2(l))2, (58)... to a plot of s vs both u and v for a real or ideal gas Using the values ¯a = 5.3 bar m6 kmole–2 , ¯b = 0.0305 m3 kmole–1 , u = 70 00 kJ kmole–1, Tref = 1 K, v ref = 1 m3 kmole–1, c v0 = 28 kJ kmole–1 K–1, u ref,0 = c v0 Tref = 28 kJ kmole–1 in Eq (K), a plot of s vs v at u = 75 00 kJ kmole–1 is presented in Figure 3 Remarks The relation s =s (u,v) is the entropy fundamental equation It is somewhat more... 6RT/(v–b)4 + 24a/v5 (38) At T = Tc, and v = vc, ∂P/∂v = 0, ∂2P/∂v2 = 0 The third derivative w.r.t v, (∂3P/∂v3 = – 6RTc/(vc–b)4 + 24 a/vc5 = – 6RTc/((2/3)vc)4 + 24(9/8)(RTcvc)/vc5 = – (243/8) RTc/vc4 + 27 RTc/vc4 = – ( 27/ 8) RTc/vc4) < 0 (39) This stability condition is satisfied at the critical point Remark The specific heat cv is finite at the critical point, but (cp–cv) = –T(∂P/∂T)2/(∂P/∂v)→∞ at that state... stability analysis can help determine the state at which the fluid becomes a liquid First, let us consider an isolated system with specified values of U, V and m and express S = S(U, V, m) From Chapter 7, s = s (T,v) and u = u(T,v), i.e., s = s(u,v), (1) which is the entropy fundamental equation a Example 1 Consider the Van der Waals equation of state Obtain an expression for s = s(u,v) for water assuming... constant and equal to the ideal gas value cv0 = 28 kJ kmole–1 Select the reference condition such that s = cv0 ln (u/cv0 + a/(v cv0)) + R ln (v–b) Plot the entropy vs volume for an internal energy value of 70 00 kJ kmole–1 Solution Since du = cv dT + (T (∂P/∂T)v – P) dv, duT = (T(∂P/∂T)v – P) dv (A) Using P = RT/(v–b) – a/v2 (B) in Eq (A), duT = T(R/(v–b)) – (RT/(v–b) – a/v2) = (a/v2) dv Integrating at constant... measured vapor pressure at a specified value of Xk, namely, γk(l) = pk/(Xk,l p sat ), or γk,l = pk/pk,Raoult k (49) Recall that the γk’s for any phase are related to ( g E/( R T)) (see Eqs (114) to (1 17) , Chapter 8) If, X1(l) X2(l) R T/ g E = B´ + C´ (X1(l) - X2(l)), (50) then we obtain the Van Laar Equations: ln γ1 = A(T) X2(l)2/((A(T)/B(T)) X1(l) + X2(l))2, ln γ2 = B(T) X1(l)2/((X1(l) + (B(T)/A(T)) . ¯s 2 ) = 2×149.83 =299 .7 kJ. (A) The total entropy of the tank is S = 2S D = 299 .7 + 299 .7 = 599.4 kJ. After the pertur- bation, v 1 = 0.14÷2 = 0. 07 m 3 kmole –1 , u 1 = 75 00 kJ kmole –1 , and v 2 . the figure, and the extensive 146 146.5 1 47 1 47. 5 148 148.5 149 149.5 150 150.5 151 0.04 0.05 0.06 0. 07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0. 17 0.18 0.19 0.2 v, m 3 / kmol e s, kJ/ kmole. Example 11 pressure and temperature. You may assume that ln P sat 2 = 13. 97 – 5205.2/T (K), and (A) ln P sat 1 = 13.98 – 471 9.2/T (K). (B) Solution Employing Eqs. (A) and (B), the normal boiling