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6 Chapter Suggested Problems P* 1 1 R = *+ ln this equation results from 6.4 (a) The pertinent equation is, vap H P T T the algebraic manipulation of equation 6.12. Plugging in the values we have from the problem we obtain, 1 1 8.314 J mol-1 K -1 400 torr = + ln T 297.3K 2.87 10 4 J mol-1 500 torr so T = 303 K. 6.10 (b) We need to use two pieces of information. The enthalpy of sublimation is given by the sum of the enthalpy of fusion and the enthalpy of vaporization. Then we can use information from the triple point from Fig 6.4 (p=0.006 atm, T=273.16 K) and we can P* H 1 1 sub * to obtain the vapor pressure of ice at 5 then use the equation, ln = P R T T degrees C. sub H = 40.656 + 6.008 = 46.664 kJ mol-1. Now we can plug in to obtain P, 4.56 torr 4.6664 10 4 J mol-1 1 1 ln = which when solved for P -1 -1 P 8.314 J mol K 268.15 K 273.16 K gives, 3.1 torr. Because the partial pressure of water given in the problem is below this value, the frost will sublime (the air is not saturated). In order for the frost to remain, the partial pressure of water must be at least 3.1 torr. 6.11 (b) (a) the quickest way to estimate the enthalpy of vaporization is to use Trouton s rule (from Chapter 4) which says, vap H = T b 85 J mol-1 K -1 = (85)(342.2) = 29.1 kJ mol-1 Now we can use this value for part (b) (b) P vap H ln 2 = R P 1 1 1 T T 1 2 Solving at 25 oC, 1.00 atm 2.91 10 4 J mol-1 1 1 = ln and P=0.22 atm = 168 torr P -1 -1 8.314 J mol K 298.2 K 342.2 K 1 1.00 atm 2.91 10 4 J mol-1 1 1 = ln and P = 0.76 atm = 576 torr. -1 P -1 8.314 J mol K 333.2 K 342.2 K 1 Solving at 60 oC, 6.14 (b) We first need to know the molar volume, M 18.02 g mol-1 Vm = = = 1.803 10 5 m3 mol-1 5 -3 9.994 10 g m 2 Vm * plugging in Now we can use the Kelvin equation (eq 6.17) which is, P = P exp rRT we have to estimate the surface tension of water at 35 oC. From the table in the appendix, it must be slightly smaller than 72.0. Looking up a value in the Handbook of Chemistry and Physics suggests 70.4 mN m-1) 2(7.04 10 2 N m-1)(1.803 10 5 m3 mol-1) P = 5.623 kPa exp (20.0 10 9 m) (8.314 J mol-1 K -1) (308.2 K) so P = 5.91 kPa. 6.2 This is a problem similar to 6.11 with a practical application to it the equation we need is, P vap H ln 2 = P R 1 1.00 atm 2.025 10 4 J mol-1 1 1 = ln so P1 = 9.1 atm. P -1 -1 8.314 J mol K 313.2 K 243.95 K 1 1 1 T T 1 2 - and plugging in values we have, 6.8 (a) The normal boiling point occurs when the pressure equal 760 torr. We can estimate this from a graph of the data provided 1400 1200 1000 800 600 400 200 0 0 20 40 60 80 100 120 from this data, a reasonable estimate would be 83 C, or 356 K. (b) Using this estimate we can calculate the heat of vaporization using the equation, P vap H 1 1 and any two points we are given, so we have ln 2 = P T T R 1 1 2 467 torr vap H 1 1 ln and vapH = 36.4 kJ = -1 -1 670 torr 8.314 J mol K 353.2 K 343.2 K o mol-1. (note that this number changes slightly based on which pair of points used, so variations in answers from 36 to 39 kJ/mol could be expected.)

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