An Introduction to Thermodynamics phần 7

f = p(c - 1) + 2 - c(p - 1) = c - p + 2 (36) Figure 5. Each area corresponds to a single phase. The lines are the common borders, and thus represent two phases in equilibrium (or three phases at a point).. which is Gibbs’ phase rule, for two external degrees of freedom. You will recall from experience with drawing graphs that a point (or any finite number of points) corresponds to no degrees of freedom. A line (or a finite number of lines) represents one degree of freedom; setting one variable determines the other. Two degrees of freedom yield an area (or more than one area). ONE-COMPONENT PHASE DIAGRAMS. Water is a single component (c = 1), so the degrees of freedom may vary from zero to two. (c = 1) f = c - p + 2 = 3 - p (37) A single phase (vapor, liquid, or solid) is therefore represented by an area. Two phases (solid-liquid, solid-vapor, or vapor-liquid) are restricted to a line. Three phases yield a single point, called the triple point (solid + liquid + vapor). The phase diagram (with T on the vertical axis, to match following diagrams) is shown in Figure 5. You can calculate the slopes of the three lines from the Clapeyron and Clausius-Clapeyron equations, knowing the heats of fusion and of vaporization or sublimation and the densities of the phases. The upper end of the liquid-vapor line simply stops — above the critical point the distinction between liquid and vapor disappears. TWO-COMPONENT PHASE DIAGRAMS AT CONSTANT PRESSURE. We could not represent a single phase with two components on a two-dimensional page if we kept both temperature and pressure as external variables. We consider here the restricted example where pressure is maintained constant (e.g., at atmospheric pressure). Then Gibbs’ phase rule tells us the degrees of freedom are, again, with c = 2, (c = 2; const P) f = c - p + 1 = 3 - p (38) It follows that wherever there is a single phase present, that phase must be represented by an area (two degrees of freedom). If there are two phases present, the compositions are restricted to a line (or sometimes more than one line, but still one degree of freedom). Three phases can coexist 7/10/07 3- 79 only at a point, or a set of points, with no remaining degrees of freedom. Accordingly, an area can represent only a region in which there is a single phase, of variable composition. There are two apparent exceptions. First, the solubility of one solid in another is typically so Figure 3.6. Benzene-toluene equilibrium (liquid-vapor, at constant pressure). a) Calculated equilibrium states. b) Phase diagram. The area between the curved lines is not part of the equilibrium phase diagram. There can be no phase of such temperature and composition. small that the area of a “pure” solid will look like a single, vertical line. Second, when two phases, solid or liquid, are in equilibrium, those two phases will have different compositions and therefore different areas, bordered by lines that are separated from each other, as in the benzene-toluene diagram of Figure 6. Each line is simply the border of its respective area. If the overall composition falls between those lines, or areas, the material will split into variable amounts of the left and right areas (phases). The region between phases has been given the unfortunate label of a “two phase” region; a better description is a forbidden area. There can be no phase present with a composition falling in such a forbidden area. The relative amounts of the left phase and right phase are given by the lever-arm rule. If CL and CR are the compositions (as percentages of either phase — say the left-hand phase) and C is the overall composition of the mix at some point, the relative amounts (masses) of the left-hand and right-hand phases are given by m1 (C - CL ) = m2 (CR - C) (39) Similar behavior is shown by two immiscible solids or two immiscible liquids. Carbon tetrachloride and water liquids split into two layers, with very little solubility of either liquid in the other. At higher temperatures, however, the vapors mix freely. Thus the basic phase diagram is as shown in Figure 7a. There are three possible phases: the vapor, at higher temperatures, liquid CCl4 at room temperature and below, and liquid water. The two liquid phases can co-exist, but there is no liquid phase with composition falling in the general range of 95% CCl /5% water to 95% water/5% CCl4 (where the numbers are illustrative only, and depend somewhat on temperature). Figure 7a is somewhat difficult to interpret, so it is customary to add a horizontal line as a marker, to dramatize the lowest vapor temperature of the mixture, as in Figure 7b. (There can be no equilibrium across a horizontal line, which would connect points at different temperatures.) Also, if the diagram represents solid-liquid equilibrium, as for silver and copper in Figure 7c, it was recognized long before Gibbs’ work of the late 19th century that when the liquid 7/10/07 3- 80 solution is cooled, either silver or copper will precipitate out first (depending on whether the composition lies to the left or right of the low point, called the eutectic). The composition of the remaining liquid accordingly moves toward the eutectic composition. In the last stage of precipitation, or freezing, the mixture that comes down is a fine mix of two phases, silver and copper, with the overall composition of the eutectic. Accordingly, a vertical line was drawn below the eutectic point to represent this eutectic mixture — two distinct phases, but with a distinctive appearance. Over decades, the vertical line was changed to a dotted line (to a b c Figure 3.7. a) Bare-bones phase diagram for two immiscible phases with miscible high-temperature phase (e.g., CCl4 and H2O, liquid-vapor, or Silver-Copper, solid-liquid). b) The horizontal line guides the eye. c) Early diagrams showed the eutectic mixture as if it were a separate phase, or (later) as a dotted line. indicate there is no phase with composition falling in this broad forbidden region) and more recently the vertical line is largely omitted. Examining a large number of such phase diagrams, the two patterns shown here will be recognized as appearing in combination. Often a solid compound forms, so the diagram can conceptually be divided in two, element one plus compound on the left and compound plus element two on the right. There may be regions of solubility, resembling the benzene-toluene diagram, within a larger phase diagram. Some compounds may spontaneously decompose as the temperature rises, yielding separate phases. The variety is seemingly endless. But almost all can be recognized as relatively simple combinations of these two basic patterns. Problems 1. For the transformation of graphite to diamond at 25oC, ΔG = 2.9 kJ/mole. The densities are 2.25 and 3.51 g/cm3. What is the minimum pressure required to make diamond thermodynamically stable at this temperature? 2. At 1114oC the vapor pressure of Ni was observed to be 7.50 x 108 torr and at 1142 C it was 14.33 x 10-8 torr. What is the heat of vaporization of nickel in this range? 3. Is the fugacity of ice increased or decreased by exerting pressure on the ice? Which will show the greatest change in fugacity for a pressure increase of 10 atm: ice, or liquid water? Explain. 4. The vapor pressure of a liquid compound, Y, obeys the equation ln P = a + bT -1, with P the pressure in atm and T in kelvin. 7/10/07 3- 81 a. What is the boiling point of Y? b. What is the heat of vaporization at the normal boiling point? 5. Find the mole fractions of the several components if 24 g of methane, CH , is mixed with 7 g of CO, 14 g of N2, and 8 g of He. 6. A saturated solution of benzoic acid, C6H5COOH, contains 0.32 g in 100 g of water. The density of the solution is 1.3 g/cm3. What is the molarity of the solution? 7. Find the mole fraction, molarity, and molality of iodine when 0.10 g of iodine is dissolved in 100 g of CCl4 (density 1.59 g/cm3). 8. a. What is the molarity of water in pure water? b. What is the molality of water in pure water? c. What is the molarity of CCl4 in pure CCl4 (density 1.59 g/cm3)? d. What is the molality of CCl4 in pure CCl4? e. What is the molarity of an ideal gas at 1 atm and 25oC? 9. A solution contains, by weight, 40% water, 35% ethanol (C2H5OH), and 25% acetone (CH3COCH3). a. Find the mole fraction of each component. b. Find the molality of the ethanol and acetone if water is considered to be the solvent. c. Find the molality of the water and acetone if the ethanol is considered to be the solvent. 10. The solubility of CCl4 in water at room temperature is about 0.9 g/L and the vapor pressure of CCl4 is about 100 torr. What would be the equilibrium vapor pressure of CCl above a beaker containing CCl4 covered with a layer of water? 11. Water in equilibrium with air (20.9% O2, 79.1% N2 by pressure, volume, or mole fraction) at 0oC contains 1.28 x 10-3 mole of air per liter, of which 34.91% is O2. Calculate the Henry’s law constant a. for O2 b. for N2 12. Is the Henry’s law constant for benzene in water large or small? 13. The distribution coefficient for mercuric bromide, HgBr2, between water and benzene is 0.90. That is, with concentrations expressed as molarities, C(HgBr2 /H2O) C(HgBr2 /C6H6 ) An aqueous solution contains 0.010 M HgBr2. 50 ml of this solution is to be extracted with 150 ml of benzene. a. What fraction of the HgBr2 is left in the aqueous phase if the extraction is made with one 150 ml portion of benzene? b. What fraction is left if the extraction is made with three successive 50 ml portions of benzene? 14. Over a certain range of concentrations, the volume of a solution containing m moles of NaCl in 1 kg water has been found to be, in ml, V = 1002.935 + 16.670 m + 1.636 m3/2 + 0.170 m2 Find the partial molal volume of a. NaCl in a 2 m solution b. water in a 2 m NaCl solution c. pure NaCl (density 2.165 g/cm3) 7/10/07 3- 82 15. The vapor pressures of benzene and toluene, at 27oC, are 120 and 40 torr. Assuming an ideal solution, what is the composition of the vapor above a solution containing 100 g of benzene (C6H6) and 150 g of toluene (C6H5CH3)? 16. In your work for Adulterated Chemicals, Inc., you have isolated a new antibiotic through a lengthy series of extractions, biological tests, and so forth. A few milligrams are available, and by the ultra-centrifuge method it has been found that the molar mass is 10,000. It is desired to check this by another method. For a 1% by weight solution of the substance in water, calculate a. the freezing-point depression b. the boiling-point elevation c. the change in vapor pressure at 25oC d. the osmotic pressure, at 25oC, in cm H2O. (The density of mercury is 13.6 g/cm3.) Which method would you recommend? 17. Boiling occurs in a solution when the sum of the vapor pressures of the components is equal to the atmospheric pressure (or the applied pressure), but these vapor pressures are lower than for the pure materials (cf. Raoult’s law, equation 29c). If two liquids are immiscible, each exhibits its own vapor pressure (in each phase), and “steam distillation” occurs when the sum of the vapor pressures is equal to the atmospheric pressure. a. Calculate the vapor pressure of pure water at 99oC. b. What vapor pressure must a compound, immiscible with water, have to steam distill at 99oC when the atmospheric pressure is 745 torr? c. What would be the composition of the vapor in such a steam distillation? 18. A solution of 3.795 g of sulfur in 100 g of CS2 boils at 46.66oC. The boiling point of pure CS2 is 46.30oC and the heat of vaporization of CS is 26.8 kJ/mol. From this experimental result, what is the probable formula of the sulfur molecule in the solution? 19. Show that if two phases (A and B) in equilibrium are both slightly impure, the resultant change in temperature (change in melting point or boiling point, relative to pure phases), ΔT, is given by ΔN1 = ΔHTΔT where ΔN1 is the difference in mole fraction of the major component (ΔN = N A - N B) and ΔH is the enthalpy of transition(ΔH1 = H1A H1 ) . 7/10/07 3- 83 ... - tailieumienphi.vn