PS 8 solutions

PS 8 solutions - Problem 5.30. The slope of a graph of G...

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Unformatted text preview: Problem 5.30. The slope of a graph of G vs. T is —-S, so the slope of each graph should be negative, and become more negative with increasing temperature. Furthermore, at any given temperature, the stable phase should have the lowast value of G. For H20 at atmospheric pressure, the stable phase is ice below 0°C, water between 0°C and 100°C, and steam above 100°C. The plot at left below shows these features qualitatively. Water 0°C 100°C T 0°C 100°C As the pressure is reduced, the relation (BG/BP)T : V tells us that the Gibbs free energy of each phase will decrease. However, that of the gas phase will decrease the fastest by far, since its volume is so much greater. At pressures below the triple point (0.006 bar), there is no temperature at which liquid water is stable. So at 0.001 bar, the graphs of G vs. T would look something like the plot above right. T Problem 5.32. (The water—ice phase boundary.) (a) (b) (C) As ice melts into water the change in entropy (or the latent heat) is positive, while the change in volume is negative (since ice is less dense), so the slope of the phase boundary, AS/AV, must be negative. In more fundamental terms, converting ice to water lets the entropy of the environment increase (by making more volume available), and this effect is more important at high pressure since P = T(BS/8V). So high pressures tend to push the equilibrium in the direction of the phase that takes up less volume. Instead of considering a mole of ice/water, let’s just consider one gram. Then the latent heat is 333 J, the volume of the ice is (917,000)"1 In3 a 1.091 X 10‘“3 m3, and the volume of the water is 1.000 x 10—6 m3. Therefore the slope of the phase boundary is dP L 333 J 7 2a“ ~ r517 ~ “ ‘1‘35 “0 Pa/K " “135 bar/K- So if the temperature decreases by one degree (from 0 to —1°C), the pressure must increase by 135 bars to remain on the phase boundary. In other words, ice will melt at —1°C if the pressure is above 135 bars (or 133 atmospheres). Treating the glacier ice as a fluid, the increase in pressure at depth 2 is simply pgz, where p is the density. (To derive this formula, consider a column of ice extending down to depth z. The weight of the column per unit area is pgz, and this must be balanced by the pressure from below.) In our case, to reach a pressure of 135 bars, P 2:— P9 135 x 105 N/m2 2 (917 kg/m3)(9.8 N/kg) = 1500 m' (01) That’s pretty deep, just to lower the melting temperature by one degree. Apparently the flow of glaciers is not caused primarily by lowering of the melting point under pressure. The blade of an ice skate measures a few millimeters across by perhaps 25 cm long, so the total area is perhaps 10 cm2, Even if you’re leaning on the “corner” of the blade, the total area in contact with the ice is probably more than 1 cm2 = 10‘4 m2. If your mass is 50 kg, then your weight is about 500 N so the pressure on the blade is roughly (500 N) / (10”4 m2) = 5 X 106 Pa = 50 bars. Under this pressure the melting temperature drops by only 50/ 135 w 4°C. This mechanism of friction reduction would work only if the ice temperature is already within less than half a degree of melting, and even then, only when you’re minimizing the area of the blade in contact with the ice. In practice, the ability to glide doesn’t depend so critically on the ice temperature or on how.r the blade touches the ice, so I don’t think this mechanism can be very important. Problem 5.35. (Vapor pressure equation.) Neglecting the volume of the condensed phase and using the ideal gas law for the volume of the gas, the difference in volume between the two phases is approximately RT Ale/Qm? J assuming one mole. Therefore the Clausius-Clapeyron equation becomes d_P_LP EHEdT dT‘ RTz’ or P "RT? Integrating both sides then gives L = __ = —L/RT lnP RT + (const), or P (const)e . Problem 5.36. (Effect of altitude on boiling water.) (a) I’ll use the data given at 100°C, or 373 K. Then the vapor pressure is just 1.013 bar, or 1 atm. Plugging in L = 4.066 x 104 J/mol and R = 8.315 J/mol—K, I find L/R = 4890 K, so I can solve for the constant (which I’ll call P0) in the vapor pressure equation: 1 atm = Poe‘4390/373 —r P0 = 4.91 X 105 atm. (There’s not much point in trying to interpret this constant physically. According to the equation, it’s the pressure at which the boiling temperature would go to infinity. But the critical point is reached long before, and the value of L changes significantly even befOre that.) Here’s a plot of part of the phase boundary: Pressure (atm) 50 6O 70 80 90 100 Temperature (°C) At T = 50°C the predicted pressure is .132 atm, about 7% higher than the actual value. (b) To find the boiling temperature at a given pressure one can either read the value off the graph or solve the vapor pressure equation for T: _ L _ 4890 K _ R ln(P0/P) _ ln(491,000 awn/P). For the elevations of Problem 1.13, I find: T Ogden (4700 ft) .84 atm 95°C Leadville (10,500 ft) .69 atm 90°C Mt. Whitney (14,500 ft) .59 atm 86°C Mt. Everest (29,000 ft) .35 atm 73°C If you’re camping in the mountains at 10,000 ft above sea level, the boiling point is about 90°C. At this temperature, noodles cook much slower than at 100°C. (I’ve heard a rough rule of thumb that cooking time doubles for every drop of 5°C in the boiling temperature. If this is correct, then noodles that are supposed to cook in 5 minutes will actually take 10 minutes in Ogden, 20 minutes in Leadville, and nearly 40 minutes on the summit of Mt. Whitney. I suspect that the actual Cooking times don’t increase quite this dramatically.) (c) From the values computed above, you can see that the boiling temperature decreases by roughly 1°C for every 1000 ft of elevation gain. Alternatively, recall the general result of Problem 1.16: 13(2) = (1 atm)e‘m92/kT : (1 atm)€—z/28,DDO a. Plugging this into the vapor pressure equation yields the boiling temperature as a function of z: L /R L / R : ln(P0/P) P0 Z Mi atm) + 28,000 a TI: The logarithm in the denominator now evaluates to 13.1; dividing through by this constant and plugging in the value of L/ R computed previously yields 373 K 2 1 + 367, 000 ft Tb = For any reasonable value of 2: we can expand this using the binomial theorem to obtain Tb as (373 K) (1 ) = 373 K — (.00102 K/ft)z. z " 367, 000 3: Thus the boiling temperature drops by about 1.02 degrees Celsius per thousand feet of elevation gain, or about 3.35 degrees Celsius per kilometer (assuming as in Problem 1.16 that the atmospheric temperature is constant). Problem 5.42. (Relative humidity and dew point.) (a) The result of Problem 5.35 for the shape of the vapor pressure curve was P = Poe—“RT. Using the data for T = 25°C from Figure 5.11, we have L 43,990 J/mol _ ' E _ 8.315 J/mol-K _ 5290 K’ solving for the constant P0 therefore gives P0 = (0.0317 bar) exp (5290 K/298 K) = 1.626 x 106 bar. I then plotted the vapor pressure curve using the Mathematica instruction Plot [1626000*Exp [—5290/ (t+273)] , {t , O , 40} ,Frame—>True] which produced the graph on the following page. (b) Supersaturated Unsaturated H20 partial pressure (bar) 0 0 10 20 Temperature (° C) 30 40 Indeed, the equilibrium vapor pressure of water approximately doubles for every 10°C increase in temperature: 0.006 bar at 0°C; 0.012 bar at 10°C; 0.023 bar at 20°C 0.043 bar at 30°C; and 0.074 bar at 40°C. At 30°C, a relative humidity of 100% would imply a partial pressure of water vapor equal to 0.043 bar, so a relative humidity of 90% means 90% of this, or 0.038 bar. To find the dew point, just find the temperature at which the equilibrium vapor pressure of water is 0.038 bar; I estimate about 28°C from the graph, and calculate 282°C from the formula. A relative humidity of 40%, on the other hand, implies a partial pressure of water vapor equal to only 0.017 bar. The dew point at this humidity is 149°C. I Problem 5.46. (Nucleation of cloud droplets.) (a) (b) For any pure system we can write the Gibbs free energy as G' = N a For a liquid droplet surrounded by vapor, the total Gibbs free energy would be simply the sum of contributions from the two subsystems: G = Nxm + quv = N1M1+(N — Nam = Np“ + Nam: — run), where the subscript l is for liquid and v is for vapor. If v; is the volume per molecule in the liquid, then M is the ratio of the total volume of the liquid to 1);. Assuming a spherical droplet, we therefore have 47r7'3 3U; G=Nm+ (Ma—Mu)- The Gibbs free energy of the droplet’s surface is simply the surface area times the surface tension, 0. So if we include this contribution, the total Gibbs free energy is 47Tr3 3’0; G : New + (a; — p”) + 47r'r'20'. (C) (d) The r dependence of G comes from the second and third terms in the preceding expression. The third term is always positive and is quadratic in r, so its graph is an upward—opening parabola. The second term is cubic in 7", but can be positive or negative depending on which is larger, in or #1,. If ,u; > pm, indicating that molecules would tend to diffuse from the liquid to the vapor (even neglecting surface tension), then this term is also positive so the graph of G looks like that shown below, left. The only equilibrium point is the minimum at r 2 0, so the tendency of G to decrease will cause any water droplet to evaporate and disappear. On the other hand, if in, > m, then the cubic term is negative so the graph of G vs. T looks like that shown below, right. Now there is a nontrivial equilibrium radius n, where G reaches a maximum. This equilibrium is unstable, however. Droplets smaller than we will evaporate, while droplets larger than T‘c will grow until the vapor becomes depleted of water molecules and [Lu is no longer greater than in. G G #1; <iu'i (LU >3”! 1r ——-| F1” Tc To find the critical radius, set the derivative of 0' equal to Zero: 0'10 411'?"2 0 = w = — '3 1, — 8 c . 0,, Tc vi (a ml+ WU Solving for Tc gives 2012: Tc 2 . #1: _ #1 Now the chemical potential of the vapor can be written in terms of its partial pressure P using equation 5.40: My = M: + len(P/P°), where ° refers to any convenient reference pressure. If we take the reference pressure to be the nominal vapor pressure, that is, the pressure of vapor that would be in equilibrium with a flat surface of the liquid, then P/P° is just the relative humidity (which I’ll call RH) and ,LL: is equal to M. The difference pm A My can therefore be written as len(RH), and our expression for the critical radius becomes 1" — 20m or RH—ex 202” -ex 26V! C‘krinmii) _ p kTrc _ p arr, ’ where H is the volume of one mole of the liquid. (This result is known as the Kelvin equation.) For water at 20°C, 20V; f 2(0.073 J/m2)(18 x 10‘6 m3) RT “ (8.315 J/K)(293 K) = 1.08 x 10—9 m = 1.08 nm, so the Kelvin equation becomes __ 1.08 nm _ ln(RH) are or RH = (20'08 “mm”. The critical radius goes to zero as the relative humidity goes to infinity, and goes to infinity as the relative humidity goes to 100% (from above). A plot of re vs. relative humidity is shown at right. We know from expe— rience that the relative humidity in our 10 atmosphere is never much greater than 100%; therefore, according to this analy— 8 sis, droplets smaller than about 10 nm in E 5 radius should never be stable. But a 10— 5, nm droplet contains more than 105 mole— 53 4 cules, so it could never form spontaneously out of a random density fluctuation We 2 are forced to conclude that clouds droplets in our atmosphere must nucleate through 100% 20%“ 30;}? $0??— {05% some other process. Rfilative humidity Problem 5.48. From the van der Waals equation (5.52) we can differentiate to obtain 8P NkT + 2aN2. 82F 2NkT EinN2 av _ (V M Nb)2 V3’ swim? V4' At the critical point, both of these expressions should equal zero, so Arch _ 2aN2 and 2Nch _ 6aN2 (1) (Va—~Nb)2 _ V3 (Va —Nb)3 “ V4 ‘ C C Dividing the first of these equations by the second gives 1 1 5(W—Nblflgvca 0r chsNb‘ To obtain the critical temperature, plug this result into the first equation in- (1): JVch _ EaN2 => JVch _ 2aN2 z} kT __ it; (3N1) — Nb)2 — (3Nb)3 4N252 — 27N3b3 '2 — 271). Finally, to obtain the critical pressure, plug both these results back into the van der Waals equation itself: Vc—Nb V3 _ 2N5 "9N2b2 P _ Nch GN2 _ 8Na/27b £th2 (4 1):; 1 a. C" 27 9 ‘ Problem 5.51. Substituting T = tTc, P 2 pH, and V = 111/; into the van der Waals equation gives ' N ktTa aN2 W, s Nb " Lei/3' Plugging in the results of Problem 5.48, we then obtain 1 a N t 8 a aN?‘ a ( 8t 3 ) 'EE "“ 3Nbv — Nb 'EE _ 9sz202 2752 3'0 — 1 — E laps: P that is, St 3 31) w- 1 — The Constants a and b have conveniently disappeared. This means that when we use reduced variables, our calculations apply to all van der Waals fluids, regardless of their a and 6 Values. p: Problem 5.61. One method of purifying oxygen from air is shown on the phase diagram below. Lower the temperature of air until a portion of it (perhaps about a third) condenses, resulting in a liquid with composition A. Then discard the remaining gas, let the liquid evaporate, and partially condense it a second time to obtain a liquid with composition B. Again discard the gas and repeat the process to obtain compositions C, D, etc., until the condensed liquid has the required purity. For the case shown in the diagram, the four steps indicated plus a fifth step should suffice to produce at least a small amount of oxygen that is 95% pure. Of course, this liquid is only a tiny fraction, less than 1%, of the original air. 92—I—ir—g—r-z1—r—I-I—r-g-t—I—I 90 Gas 88 I f 86 : I Q 84 ! V I E. 82 ' 80 78 76 0 0.2 0.4 0.6 0.8 1.0 N2 33 _*.’ 02 Alternatively, and a bit more simply, you could start by liquefying all the air, then raise its temperature until a nitrogen—rich gas boils off. Instead of keeping this gas in the same container, pump it away continuously and discard it. Then the composition of the remaining liquid will move upward and to the right along the lower phase boundary line, all the way to the end point of 100% oxygen, although again, only a tiny fraction of the original fluid will remain by the end. Problem 5.65. The illustration at right ShOWS the free energy graphs of the gas and liquid at one particular temperature. Because the liquid’s free energy curve is less concave than that of the gas, the curves can intersect in two places, as shown. Therefore there are two com— position ranges at which a combination of gas and liquid is most stable (at this temperature). At higher temper— atures, the gas’s free energy curve moves down relative to that of the liquid (BC/6T = —S), so the intersections move farther apart. At lower temperatures, the gas’s free energy curve moves up relative to that of the liquid, so T the intersections move closer together until finally the T two curves just kiss each other at a single point. The composition at that point is the azeotrope; a mixture of this composition condenses abruptly at a well—defined temperature, just as a pure substance would. ...
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PS 8 solutions - Problem 5.30. The slope of a graph of G...

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