CHM 201 Problems Solution for Chap 5 Exercises, Fall 2013

104 CHAPTER 5 GASES Pressure 21. a. 4.8 atm × atm 760mm Hg = 3.6 × 103 mm Hg; b. 3.6 × 103 mm Hg × Hg mm 1torr = 3.6 × 103 torr c. 4.8 atm × atm 10 Pa .1 013 5  = 4.9 × 105 Pa; d. 4.8 atm × atm 14 7. psi = 71 psi 22. If the levels of mercury in each arm of the manometer are equal, then the pressure in the flask is equal to atmospheric pressure. When they are unequal, the difference in height in millimeters will be equal to the difference in pressure in millimeters of mercury between the flask and the atmosphere. Which level is higher will tell us whether the pressure in the flask is less than or greater than atmospheric. a. Pflask < Patm; Pflask = 760.  118 = 642 torr 642 torr × torr 760 1atm = 0.845 atm 0.845 atm × atm 10 Pa .1 013 5  = 8.56 × 104 Pa b. Pflask > Patm; Pflask = 760. torr + 215 torr = 975 torr 975 torr × torr 760 1atm = 1.28 atm 1.28 tm × atm 10 Pa .1 013 5  = 1.30 × 105 Pa c. Pflask = 635  118 = 517 torr; Pflask = 635 + 215 = 850. torr CHAPTER 5 GASES 105 23. 4.75 cm × cm 10 mm = 47.5 mm Hg or 47.5 torr; 47.5 torr × torr 760 1atm = 6.25 × 102 atm 6.25 × 102 atm × atm 10 Pa .1 013 5  = 6.33 × 103 Pa 24. a. The pressure is proportional to the mass of the fluid. The mass is proportional to the volume of the column of fluid (or to the height of the column, assuming the area of the column of fluid is constant). d = density = volume mass ; the volume of silicon oil is the same as the volume of mercury in Exercise 22. V = d m ; VHg = Voil; oil oil Hg Hg d m d m  , moil = Hg Hg oil d d m Because P is proportional to the mass of liquid: Poil = PHg       Hg oil d d = PHg       6. 13 .1 30 = (0.0956)PHg This conversion applies only to the column of silicon oil. a. Pflask = 760. torr  (118 × 0.0956) torr = 760.  11.3 = 749 torr 749 torr × torr 760 1atm = 0.986 atm; 0.986 atm × atm 10 Pa .1 013 5  = 9.99 × 104 Pa b. Pflask = 760. torr + (215 × 0.0956) torr = 760. + 20.6 = 781 torr 781 torr × torr 760 1atm = 1.03 atm; 1.03 atm × atm 10 Pa .1 013 5  = 1.04 × 105 Pa b. If we are measuring the same pressure, the height of the silicon oil column would be 13.6/1.30 = 10.5 times the height of a mercury column. The advantage of using a less dense fluid than mercury is in measuring small pressures. The height difference measured will be larger for the less dense fluid. Thus the measurement will be more precise. 25. Suppose we have a column of mercury 1.00 cm × 1.00 cm × 76.0 cm = V = 76.0 cm3: mass = 76.0 cm3 × 13.59 g/cm3 = 1.03 × 103 g × g 1000 1kg = 1.03 kg CHAPTER 5 GASES 106 F = mg = 1.03 kg × 9.81 m/s2 = 10.1 kg m/s2 = 10.1 N 2 2 m cm 100 cm 1. N 10 Area Force         = 1.01 × 105 m2 N or 1.01 × 105 Pa (Note: 76.0 cm Hg = 1 atm = 1.01 × 105 Pa.) To exert the same pressure, a column of water will have to contain the same mass as the 76.0- cm column of mercury. Thus the column of water will have to be 13.59 times taller or 76.0 cm × 13.59 = 1.03 × 103 cm = 10.3 m. Gas Laws 26. T PV = nR = constant, 2 2 2 1 1 1 T V P T P V  P2 = 1 2 2 1 1 T V P V T = 710 torr × 30.) K 273 ( 820) K 273 ( mL 25 10 mL 0.5 2     = 5.1 × 104 torr 27. nT PV = R; for a gas at two conditions: 1 1 1 1 T n P V = 2 2 2 2 T n P V ; because n and V are constant: 2 2 1 1 T P T P  T2 = 1 2 1 P P T = torr 758 294.2K 2500torr  = 970 K = 7.0 × 102 oC For two-condition problems, units for P and V just need to be the same units for both con- ditions, not necessarily atm and L. The unit conversions from other P or V units would cancel when applied to both conditions. However, temperature always must be converted to the Kelvin scale. The temperature conversions between other units and Kelvin will not cancel each other. 28. PV = nRT, n is constant. T PV = nR = constant, 1 2 2 1 1 2 2 2 1 1 1 T P V P T V , T V P T V P 2   V2 = 1.00 L × 23) K 273 ( 31) K 273 ( . torr 220 . torr 760    = 2.82 L; ΔV = 2.82  1.00 = 1.82 L CHAPTER 5 GASES 107 29. Treat each gas separately, and use the relationship P1V1 = P2V2 (n and T are constant). For H2: P2 = 2 1 1 V P V = 475 torr × 00 L .3 .2 00 L = 317 torr For N2: P2 = 0.200 atm × 00 L .3 .1 00 L = 0.0667 atm; 0.0667 atm × atm 760torr = 50.7 torr Ptotal = 2 2 N H P P  = 317 + 50.7 = 368 torr 30. For H2: P2 = 2 1 1 V P V = 360. torr × 00 L .3 .2 00 L = 240. torr Ptotal = 2 2 2 2 H total N N H P P P , P P    = 320. torr  240. torr = 80. torr For N2: P1 = 1 2 2 V P V = 80. torr × 00 L .1 .3 00 L = 240 torr 31. As NO2 is converted completely into N2O4, the moles of gas present will decrease by a factor of one-half (from the 2 : 1 mol ratio in the balanced equation). Using Avogadro’s law: 2 1 25 0. mL n n V V n , V n V 1 2 1 2 2 2 1 1      = 12.5 mL N2O4(g) will occupy one-half the original volume of NO2(g). 32. a. PV = nRT b. PV = nRT c. PV = nRT PV = constant P =       V nR × T = const × T T =       nR P × V = const × V T V T P V PV CHAPTER 5 GASES 108 d. PV = nRT e. P = V constant V nR  f. PV = nRT PV = constant P = constant × V 1 T PV = nR = constant Note: The equation for a straight line is y = mx + b where y is the y axis and x is the x axis. Any equation that has this form will produce a straight line with slope equal to m and a y intercept equal to b. Plots b, c, and e have this straight-line form. 33. PV = nRT, P is constant. R P V nT  = constant, 2 2 2 1 1 1 V T n V n T  3 3 3 3 1 2 2 1 1 2 10 m 00 .4 10 m 20 .4 K 335 K 294 V T V T n n      = 0.921 34. Because the container is flexible, P is assumed constant. The moles of gas present are also constant. 2 2 1 1 2 2 2 2 1 1 1 1 T V T T , V n V P T n P V   ; Vsphere = 4/3 r3 V2 = . K 280 361K .1( 00cm) 4/3π , 4/3π r( ) T V T 3 3 2 1 1 2   3 2r = . K 280 361K = 1.29, r2 = (1.29)1/3 = 1.09 cm = radius of sphere after heating 35. The decrease in temperature causes the balloon to contract (V and T are directly related). Because weather balloons do expand, the effect of the decrease in pressure must be dominant. 36. For a gas at two conditions: 2 2 2 2 1 1 1 1 T n V P T n P V  Because V is constant: 2 2 2 1 1 1 T n P n T P  , n2 = 2 1 1 2 1 T P n P T n2 = 323K . torr 400 298K 800. torr 50 mol .1    = 2.77 mol Moles of gas added = n2 – n1 = 2.77 – 1.50 = 1.27 mol P V 1/V P PV T P CHAPTER 5 GASES 109 37. n = 24) K (273 mol K 08206L atm .0 2000. L atm 135 RT PV     = 1.11 × 103 mol For He: 1.11 × 103 mol × mol .4 003g He = 4.44 × 103 g He For H2: 1.11 × 103 mol × mol .2 016g H 2 = 2.24 × 103 g H2 38. For the first diagram, there is a total volume of 3X after the stopcock is open. The six total gas particles will be equally distributed (on average) over the entire volume (3X). So per X volume, there will be two gas particles. Your first drawing should have four gas particles in the 2X volume flask and two gas particles in the X volume flask. Applying Boyle’s law, the pressure in the two flasks after the stopcock is opened is: P1V1 = P2V2, P2 = 2 1 1 V P V = X 3 P1  2X = 3 1P 2 The final pressure in both flasks will be two-thirds that of the initial pressure in the left flask. For the second diagram, there is a total volume of 2X after the stopcock is opened. The gas particles will be equally distributed (on average) so that your drawing should have three gas particles in each flask. The final pressure is: P2 = 2 1 1 V P V = X 2 P1  X = 2 P1 The final pressure in both flasks will be one-half that of the initial pressure in the left flask. 39. P = 00 L .4 300. K mol K 08206L atm .0 .01g 44 1mol 0. g 22 V RT n P 2 2 CO CO            = 3.08 atm With air present, the partial pressure of CO2 will still be 3.08 atm. The total pressure will be the sum of the partial pressures. Ptotal = air CO P P 2  = 3.08 atm +        torr 760 1atm 740. torr = 3.08 + 0.974 = 4.05 atm 40. PV = nRT, n is constant. T PV = nR = constant, 2 2 2 1 1 1 T V P T P V  V2 = (1.040)V1, .1 040 000 .1 V V 2 1  CHAPTER 5 GASES 110 P2 = 1 2 2 1 1 T V P V T = 75 psi × 273 19) K ( 58) K 273 ( .1 040 000 .1    = 82 psi 41. PV = nRT, R V P nT  = constant, 2 2 2 1 1 1 P T n P n T  ; moles × molar mass = mass 2 2 2 1 1 1 2 2 2 1 1 1 P T mass P T , mass P (molarmass)T n P n (molarmass)T     mass2 = 2050. psi K 299 650. psi 291K 10 g 00 .1 P T T P mass 3 1 2 1 2 1       = 309 g 42. If we had 100.0 g of the gas, we would have 50.0 g He and 50.0 g Xe. Xe He He He n n n χ   g/mol 131.3 g 50.0 g/mol 4.003 g 50.0 g/mol 4.003 g 50.0   .0 970 .0 381molXe 5. molHe 12 12 5. molHe    PHe = χHePtotal = 0.970 × 600. torr = 582 torr; PXe = 600.  582 = 18 torr 43. H O He P 2 P  = 1.00 atm = 760. torr = PHe + 23.8 torr, PHe = 736 torr nHe = 0.586 g × .4 003g 1mol = 0.146 mol He V = torr 760 1atm torr 736 298K mol K .0 08206L atm 146mol .0 P RT n He He     = 3.69 L 44. The container has 5 He atoms, 3 Ne atoms, and 2 Ar atoms for a total of 10 atoms. The mole fractions of the various gases will be equal to the molecule fractions. He = totalatoms 10 5 He atoms = 0.50; Ne = totalatoms 10 3 Ne atoms = 0.30 Ar = 1.00 – 0.50 – 0.30 = 0.20 PHe = He  Ptotal = 0.50(1.00 atm) = 0.50 atm PNe = Ne  PTotal = 0.30(1.00atm) = 0.30 atm PAr = 1.00 atm – 0.50 atm – 0.30 atm = 0.20 atm CHAPTER 5 GASES 111 45. a. Mole fraction CH4 = .0 250atm 175atm .0 175atm .0 P P χ total CH CH 4 4    = 0.412 χO2 = 1.000 – 0.412 = 0.588 b. PV = nRT, ntotal = 338K mol K 08206L atm .0 10 5. L 425atm .0 RT V Ptotal     = 0.161 mol c. total CH CH total CH CH n χ , n n n χ 4 4 4 4    = 0.412 × 0.161 mol = 6.63 × 102 mol CH4 6.63 × 102 mol CH4 × 4 4 CH mol 16.04g CH = 1.06 g CH4 nO2 = 0.588 × 0.161 mol = 9.47 × 102 mol O2; 9.47 × 102 mol O2 × 2 2 O mol 32.00g O = 3.03 g O2 46. Ptotal = 1.00 atm = 760. torr = 2 2 2 N H O N P P P   + 17.5 torr, PN2 = 743 torr 293K mol K 08206L atm .0 mL) 1000 1L 10 mL ( .2 50 torr) 760 1atm 743torr ( RT V P n 2 N N 2 2         = 1.02 × 102 mol N2 1.02 × 102 mol N2 × 2 2 N mol 28.02g N = 0.286 g N2 47. We can use the ideal gas law to calculate the partial pressure of each gas or to calculate the total pressure. There will be less math if we calculate the total pressure from the ideal gas law. nO2 = 1.5 × 102 mg O2 2 2 .00g O 32 molO 1 mg 1000 1g   = 4.7 × 103 mol O2 nNH3 = 5.0 × 1021 molecules NH3 × 3 23 3 moleculesNH 10 022 .6 molNH 1  = 8.3 × 10−3 mol NH3 ntotal = 3 2 2 NH O N n n n   = 5.0 × 102 + 4.7 × 103 + 8.3 × 103 = 6.3 × 102 mol total atm 4.1 L 0.1 273K mol K .0 08206L atm mol 10 3.6 V RT n P 2 total total         CHAPTER 5 GASES 112 atm 1.1 atm 4.1 mol 10 3.6 mol 10 0.5 ; P n n , P P 2 2 N total N N total N N 2 2 2 2 2             .0 18atm atm 4.1 10 3.6 10 3.8 .0 10atm; P atm 4.1 10 3.6 10 7.4 P 2 3 NH 2 3 O 3 2               48. For O2, n and T are constant, so P1V1 = P2V2. P1 = 1 2 2 V P V = 785 torr × 00 L .2 .1 94 L = 761 torr = PO2 Ptotal = H O H O O 2 2 2 , P P P  = 785 – 761 = 24 torr 49. a. There are 6 He atoms and 4 Ne atoms, and each flask has the same volume. The He flask has 1.5 times as many atoms of gas present as the Ne flask, so the pressure in the He flask will be 1.5 times greater (assuming a constant temperature). b. Because the flask volumes are the same, your drawing should have the various atoms equally distributed between the two flasks. So each flask should have 3 He atoms and 2 Ne atoms. c. After the stopcock is opened, each flask will have 5 total atoms and the pressures will be equal. If six atoms of He gave an initial pressure of PHe, initial, then 5 total atoms will have a pressure of 5/6  PHe, initial. Using similar reasoning, 4 atoms of Ne gave an initial pressure of PNe, initial, so 5 total atoms will have a pressure of 5/4  PNe, initial. Summarizing: Pfinal = Ne, initial He, initial 4 P 5 6 P 5  d. For the partial pressures, treat each gas separately. For helium, when the stopcock is opened, the six atoms of gas are now distributed over a larger volume. To solve for the final partial pressures, use Boyle’s law for each gas. For He: P2 = 2 1 1 V P V = 2 P 2X X P , initial He He, initial   The partial pressure of helium is exactly halved. The same result occurs with neon so that when the volume is doubled, the partial pressure is halved. Summarizing: 2 P ; P 2 P P , initial Ne , final Ne , initial He He, final   CHAPTER 5 GASES 113 Gas Density, Molar Mass, and Reaction Stoichiometry 50. Molar mass = P dRT where d = density of gas in units of g/L. Molar mass = atm 1.000 273.2K mol K 0.08206L atm 3.164g/L   = 70.93 g/mol The gas is diatomic, so the average atomic mass = 70.93/2 = 35.47 amu. From the periodic table, this is chlorine, and the identity of the gas is Cl2. 51. If Be3+, the formula is Be(C5H7O2)3 and molar mass ≈ 13.5 + 15(12) + 21(1) + 6(16) = 311 g/mol. If Be2+, the formula is Be(C5H7O2)2 and molar mass ≈ 9.0 + 10(12) + 14(1) + 4(16) = 207 g/mol. Data set I (molar mass = dRT/P and d = mass/V): molar mass = L) 10 (22 6. torr) 760 1atm 765 2. torr ( 286K mol K .0 08206L atm 2022g .0 PV RT mass 3        = 209 g/mol Data set II: molar mass = L) 10 (26 0. torr) 760 1atm 764 6. torr ( 290. K mol K .0 08206L atm 2224g .0 PV RT mass 3        = 202 g/mol These results are close to the expected value of 207 g/mol for Be(C5H7O2)2. Thus we conclude from these data that beryllium is a divalent element with an atomic weight (mass) of 9.0 g/mol. 52. d = P × (molar mass)/RT; we need to determine the average molar mass of air. We get this by using the mole fraction information to determine the weighted value for the molar mass. If we have 1.000 mol of air: average molar mass = 0.78 mol N2 × 2 2 N mol 28.02 g N + 0.21 mol O2 × 2 2 O mol 32.00g O + 0.010 mol Ar × Ar mol 39.95g Ar = 28.98 = 29 g dair = 273K mol K 08206L atm .0 29g/mol 00atm .1   = 1.3 g/L CHAPTER 5 GASES 114 53. Rigid container: As temperature is increased, the gas molecules move with a faster average velocity. This results in more frequent and more forceful collisions, resulting in an increase in pressure. Density = mass/volume; the moles of gas are constant, and the volume of the container is constant, so density in this case must be temperature-independent (density is constant). Flexible container: The flexible container is a constant-pressure container. Therefore, the final internal pressure will be unaffected by an increase in temperature. The density of the gas, however, will be affected because the container volume is affected. As T increases, there is an immediate increase in P inside the container. The container expands its volume to reduce the internal pressure back to the external pressure. We have the same mass of gas in a larger volume. Gas density will decrease in the flexible container as T increases. 54. We assume that 28.01 g/mol is the true value for the molar mass of N2. The value of 28.15 g/mol is the average molar mass of the amount of N2 and Ar in air. Assume 100.00 mol of total gas present, and let x = the number of moles that are N2 molecules. Then 100.00  x = the number of moles that are Ar atoms. Solving: 28.15 = .00 100 )(39.95) (100.00 (28.01) x x   2815 = (28.01)x + 3995  (39.95)x, (11.94)x = 1180. x = 98.83% N2; % Ar = 100.00  x = 1.17% Ar Ratio of moles of Ar to moles of N2 = .83 98 .1 17 = 1.18 × 102. 55. Out of 100.0 g of compound, there are: 87.4 g N × .01 g N 14 1molN = 6.24 mol N; .6 24 .6 24 = 1.00 12.6 g H × 008 g H .1 1molH = 12.5 mol H; 24 .6 12 5. = 2.00 Empirical formula is NH2. P × (molar mass) = dRT, where d = density. Molar mass = torr 760 1atm . torr 710 373K mol K 08206L atm .0 L 977g .0 P dRT     = 32.0 g/mol Empirical formula mass of NH2 = 16.0 g. Therefore, the molecular formula is N2H4. 56. P × (molar mass) = dRT, d = volume, mass P × (molar mass) = V mass × RT CHAPTER 5 GASES 115 Molar mass = .0 256L torr) 760 1atm 750. torr ( 373K mol K .0 08206L atm 800g .0 PV RT mass       = 96.9 g/mol Mass of CHCl  12.0 + 1.0 + 35.5 = 48.5; 5. 48 96 9. = 2.00; molecular formula is C2H2Cl2. 57. 2 NaClO3(s)  2 NaCl(s) + 3 O2(g) Ptotal = H O total O H O O 2 2 2 2 P P , P P P    = 734 torr – 19.8 torr = 714 torr 22) K (273 mol K 08206L atm .0 .0 0572L torr 760 1atm torr 714 RT V P n 2 2 O O              = 2.22 × 103 mol O2 Mass NaClO3 decomposed = 2.22 × 103 mol O2 × 3 3 2 3 NaClO mol .44g NaClO 106 molO 3 2 molNaClO  = 0.158 g NaClO3 Mass % NaClO3 = 8765g .0 .0 158g × 100 = 18.0% 58. For ammonia (in 1 minute): 496K mol K 08206L atm .0 500. L .atm 90 RT V P n 3 3 3 NH NH NH      = 1.1 × 103 mol NH3 NH3 flows into the reactor at a rate of 1.1 × 103 mol/min. For CO2 (in 1 minute): 496K mol K 08206L atm .0 600. L atm 45 RT V P n 2 2 2 CO CO CO      = 6.6 × 102 mol CO2 CO2 flows into the reactor at 6.6 × 102 mol/min. To react completely with 1.1 × 103 mol NH3/min, we need: 3 2 3 3 molNH 2 molCO 1 min 10 molNH 1.1   = 5.5 × 102 mol CO2/min CHAPTER 5 GASES 116 Because 660 mol CO2/min are present, ammonia is the limiting reagent. urea mol .06g urea 60 molNH 2 molurea 1 min 10 molNH 1.1 3 3 3    = 3.3 × 104 g urea/min 59. 150 g (CH3)2N2H2 × 2 2 2 3 2 2 2 2 3 (CH ) N H mol molN 3 .10g 60 1mol(CH ) N H  = 7.5 mol N2 produced V nRT P 2 N  = L 250 300. K mol K .0 08206L atm mol 5.7   = 0.74 atm We could do a similar calculation for PH2O and PCO2 and then calculate H O N total 2 2 P P ( P   ) P  CO2 . Or we can recognize that 9 total moles of gaseous products form for every mole of (CH3)2N2H2 reacted. This is three times the moles of N2 produced. Therefore, Ptotal will be three times larger than . P 2 N Ptotal = 3 × PN2 = 3 × 0.74 atm = 2.2 atm. 60. Rigid container (constant volume): As reactants are converted to products, the moles of gas particles present decrease by one-half. As n decreases, the pressure will decrease (by one- half). Density is the mass per unit volume. Mass is conserved in a chemical reaction, so the density of the gas will not change because mass and volume do not change. Flexible container (constant pressure): Pressure is constant because the container changes volume in order to keep a constant pressure. As the moles of gas particles decrease by a factor of 2, the volume of the container will decrease (by one-half). We have the same mass of gas in a smaller volume, so the gas density will increase (is doubled). 61. For NH3: P2 = 2 1 1 V P V = 0.500 atm  00 L .3 .2 00 L = 0.333 atm For O2: P2 = 2 1 1 V P V = 1.50 atm × 00 L .3 .1 00 L = 0.500 atm After the stopcock is opened, V and T will be constant, so P n. The balanced equation requires: 4 5 P P n n 3 2 3 2 NH O NH O   = 1.25 The actual ratio present is: 333atm .0 500atm .0 P P 3 2 NH O  = 1.50 CHAPTER 5 GASES 117 The actual ratio is larger than the required ratio, so NH3 in the denominator is limiting. Because equal moles of NO will be produced as NH3 reacted, the partial pressure of NO produced is 0.333 atm (the same as PNH3 reacted). 62. 2 NaN3(s)  2 Na(s) + 3 N2(g) 273K mol K 08206L atm .0 70 0. L 00atm .1 RT PV n 2 N     = 3.12 mol N2 are needed to fill the air bag. Mass NaN3 reacted = 3.12 mol N2 × 3 3 2 3 NaN mol .02g NaN 65 molN 3 2 molNaN  = 135 g NaN3 63. 273K mol K 08206L atm .0 cm 1000 L 1 m 100cm 4800m atm 0.1 RT PV n 3 3 3 H2                     = 2.1 × 105 mol 2.1 × 105 mol H2 are in the balloon. This is 80.% of the total amount of H2 that had to be generated: 0.80(total mol H2) = 2.1 × 105, total mol H2 = 2.6 × 105 mol H2 2.6 × 105 mol H2 × Fe mol .85g Fe 55 H mol molFe 1 2  = 1.5 × 107 g Fe 2.6 × 105 mol H2 × 4 2 4 2 4 2 2 4 2 g H SO 98 g reagent 100 H SO mol .09g H SO 98 H mol 1molH SO   = 2.6 × 107 g of 98% sulfuric acid 64. 0.2766 g CO2 × .009gCO2 44 12.011g C = 7.549 × 102 g C; % C = 1023g .0 g .7 549 10 2  × 100 = 73.79% C 0.0991 g H2O × .02gH O 18 016g H .2 2 = 1.11 × 102 g H; % H = 1023g .0 g 10 .1 11 2  × 100 = 10.9% H PV = nRT, 273K mol K 08206L atm .0 L 10 27 6. 00atm .1 RT PV n 3 N2       = 1.23 × 103 mol N2 1.23 × 103 mol N2 × 2 2 N mol 28.02g N = 3.45 × 102 g nitrogen Mass % N = 4831g .0 g 10 .3 45 2  × 100 = 7.14% N CHAPTER 5 GASES 118 Mass % O = 100.00  (73.79 + 10.9 + 7.14) = 8.2% O Out of 100.00 g of compound, there are: 73.79 g C × .011g 12 1mol = 6.144 mol C; 7.14 g N × .01g 14 1mol = 0.510 mol N 10.9 g H × 008g .1 1mol = 10.8 mol H; 8.2 g O × .00g 16 1mol = 0.51 mol O Dividing all values by 0.51 gives an empirical formula of C12H21NO. Molar mass = torr 760 1atm torr 256 400. K mol K 08206L atm .0 L 02g .4 P dRT     = 392 g/mol Empirical formula mass of C12H21NO ≈ 195 g/mol and 195 392 ≈ 2. Thus the molecular formula is C24H42N2O2. 65. 2 NH3(g)  N2(g) + 3 H2(g); as reactants are converted into products, we go from 2 moles of gaseous reactants to 4 moles of gaseous products (1 mol N2 + 3 mol H2). Because the moles of gas doubles as reactants are converted into products, the volume of the gases will double (at constant P and T). PV = nRT, P = n V RT       = (constant)n; pressure is directly related to n at constant T and V. As the reaction occurs, the moles of gas will double, so the pressure will double. Because 1 mol of N2 is produced for every 2 mol of NH3 reacted, . 1/2 P P o NH N 3 2  Owing to the 3 to 2 mole ratio in the balanced equation, o NH H 3 2 3/2 P P  . Note: o NH o NH o NH N H total 3 3 3 2 2 2 P 1/2P 3/2P P P P      . As we said earlier, the total pres- sure doubles as reactants are completely converted into products for this reaction 66. 1.00 × 103 kg Mo × .94g Mo 95 molMo 1 kg 1000g  = 1.04 × 104 mol Mo 1.04 × 104 mol Mo × 3 2 3 MoO mol molO 7/2 Mo mol 1molMoO  = 3.64 × 104 mol O2 CHAPTER 5 GASES 119 00atm .1 290. K mol K .0 08206L atm 10 mol 64 .3 P RT n V 4 O O 2 2      = 8.66 × 105 L of O2 8.66 × 105 L O2 × L O2 21 100L air = 4.1 × 106 L air 1.04 × 104 mol Mo × Mo mol 3 molH 2 = 3.12 × 104 mol H2 00atm .1 290. K mol K .0 08206L atm 10 mol 12 .3 V 4 H2     = 7.42 × 105 L of H2 67. Ptotal = 2 2 2 N H O N , P P P  = 726 torr – 23.8 torr = 702 torr  torr 760 1atm = 0.924 atm 298K mol K 08206L atm .0 L 10 31 8. 924atm .0 RT V P n 3 N N 2 2        = 1.20 × 103 mol N2 Mass of N in compound = 1.20 × 103 mol N2 × mol 28.02g N 2 = 3.36 × 102 g nitrogen Mass % N = 253g .0 g .3 36 10 2  × 100 = 13.3% N 68. 10.10 atm  7.62 atm = 2.48 atm is the pressure of the amount of F2 reacted. PV = nRT, V and T are constant. 2 1 2 1 2 2 1 1 n n P or P n P n constant, P n P    24atm .1 48atm .2 Xe reacted Mol MolF reacted 2  = 2.00; so: Xe + 2 F2  XeF4 69. Because P and T are constant, V and n are directly proportional. The balanced equation requires 2 L of H2 to react with 1 L of CO (2 : 1 volume ratio due to 2 : 1 mole ratio in the balanced equation). The actual volume ratio present in 1 minute is 16.0 L/25.0 L = 0.640 (0.640 : 1). Because the actual volume ratio present is smaller than the required volume ratio, H2 is the limiting reactant. The volume of CH3OH produced at STP will be one-half the volume of H2 reacted due to the 1 : 2 mole ratio in the balanced equation. In 1 minute, 16.0 L/2 = 8.00 L CH3OH are produced (theoretical yield). CHAPTER 5 GASES 120 nCH3OH = RT PV = 273K mol K 08206L atm .0 .8 00 L 00atm .1   = 0.357 mol CH3OH in 1 minute 0.357 mol CH3OH × CH OH mol .04g CH OH 32 3 3 = 11.4 g CH3OH (theoretical yield per minute) Percent yield = l yield theoretica actual yield × 100 = 4. g 11 .5 30 g  100 = 46.5% yield 70. CH3OH + 3/2 O2  CO2 + 2 H2O or 2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g) 50.0 mL × .04g 32 mol 1 mL .0 850g  = 1.33 mol CH3OH(l) available nO2 = RT PV = 300. K mol K 08206L atm .0 22 8. L 00atm .2   = 1.85 mol O2 available 1.33 mol CH3OH × molCH OH 2 molO 3 3 2 = 2.00 mol O2 2.00 mol O2 are required to react completely with all the CH3OH available. We only have 1.85 mol O2, so O2 is limiting. 1.85 mol O2 × 2 2 molO 3 4 molH O = 2.47 mol H2O 71. Molar mass = 000atm .1 273 2. K mol K 08206L atm .0 L 70902g .0 P dRT    = 15.90 g/mol 15.90 g/mol is the average molar mass of the mixture of methane and helium. Assume 100.00 mol of total gas present, and let x = mol of CH4 in the 100.00 mol mixture. This value of x is also equal to the volume percentage of CH4 in 100.00 L of mixture because T and P are constant. 15.90 = .00 100 )( .4 003) (100.00 (16.04) x x   , 1590. = (16.04)x + 400.3  (4.003)x 1190. = (12.04)x, x = 98.84% CH4 by volume; % He = 100.00  x = 1.16% He by volume CHAPTER 5 GASES 121 Kinetic Molecular Theory and Real Gases 72. a. Containers ii, iv, vi, and viii have volumes twice those of containers i, iii, v, and vii. Containers iii, iv, vii, and viii have twice the number of molecules present than containers i, ii, v, and vi. The container with the lowest pressure will be the one which has the fewest moles of gas present in the largest volume (containers ii and vi both have the lowest P). The smallest container with the most moles of gas present will have the highest pressure (containers iii and vii both have the highest P). All the other containers (i, iv, v and viii) will have the same pressure between the two extremes. The order is ii = vi < i = iv = v = viii < iii = vii. b. All have the same average kinetic energy because the temperature is the same in each container. Only temperature determines the average kinetic energy. c. The least dense gas will be in container ii because it has the fewest of the lighter Ne atoms present in the largest volume. Container vii has the most dense gas because the largest number of the heavier Ar atoms are present in the smallest volume. To figure out the ordering for the other containers, we will calculate the relative density of each. In the table below, m1 equals the mass of Ne in container i, V1 equals the volume of container i, and d1 equals the density of the gas in container i. Container i ii iii iv v vi vii viii mass, volume m1, V1 m1, 2V1 2m1, V1 2m1, 2V1 2m1, V1 2m1, 2V1 4m1, V1 4m1, 2V1 density     volume mass 1 1 V m = d1 1 1 1 d 2V m 2 1  1 1 V m 2 = 2d1 1 1 V 2 m 2 = d1 1 1 V m 2 = 2d1 1 1 V 2 m 2 = d1 1 1 V 4 m = 4d1 1 1 V 2 4 m = 2d1 From the table, the order of gas density is ii < i = iv = vi < iii = v = viii < vii. d. µrms = (3RT/M)1/2; the root mean square velocity only depends on the temperature and the molar mass. Because T is constant, the heavier argon molecules will have a slower root mean square velocity than the neon molecules. The order is v = vi = vii = viii < i = ii = iii = iv. 73. The kinetic molecular theory assumes that gas particles do not exert forces on each other and that gas particles are volumeless. Real gas particles do exert attractive forces for each other, and real gas particles do have volumes. A gas behaves most ideally at low pressures and high temperatures. The effect of attractive forces is minimized at high temperatures because the gas particles are moving very rapidly. At low pressure, the container volume is relatively large (P and V are inversely related), so the volume of the container taken up by the gas particles is negligible. 74. a. At constant temperature, the average kinetic energy of the He gas sample will equal the average kinetic energy of the Cl2 gas sample. In order for the average kinetic energies to be the same, the smaller He atoms must move at a faster average velocity than Cl2 molecules. Therefore, plot A, with the slower average velocity, would be for the Cl2 CHAPTER 5 GASES 122 sample, and plot B would be for the He sample. Note the average velocity in each plot is a little past the top of the peak. b. As temperature increases, the average velocity of a gas will increase. Plot A would be for O2(g) at 273 K, and plot B, with the faster average velocity, would be for O2(g) at 1273 K. Because a gas behaves more ideally at higher temperatures, O2(g) at 1273 K would behave most ideally. 75. V, T, and P are all constant, so n must be constant. Because we have equal moles of gas in each container, gas B molecules must be heavier than gas A molecules. a. Both gas samples have the same number of molecules present (n is constant). b. Because T is constant, KEave must be the same for both gases (KEave = 3/2 RT). c. The lighter gas A molecules will have the faster average velocity. d. The heavier gas B molecules do collide more forcefully, but gas A molecules, with the faster average velocity, collide more frequently. The end result is that P is constant between the two containers. 76. Boyle's law: P 1/V at constant n and T In the kinetic molecular theory (KMT), P is proportional to the collision frequency which is proportional to 1/V. As the volume increases, there will be fewer collisions per unit area with the walls of the container, and pressure will decrease (Boyle's law). Charles's law: V T at constant n and P When a gas is heated to a higher temperature, the velocities of the gas molecules increase and thus hit the walls of the container more often and with more force. In order to keep the pressure constant, the volume of the container must increase (this increases surface area, which decreases the number of collisions per unit area, which decreases the pressure). Therefore, volume and temperature are directly related at constant n and P (Charles’s law). Avogadro’s law: V n at constant P and T As gas is added to a container (n increases), there will be an immediate increase in the number of gas particle collisions with the walls of the container. This results in an increase in pressure in the container. However, the container is such that it wants to keep the pressure constant. In order to keep pressure constant, the volume of the container increases in order to reduce the collision frequency, which reduces the pressure. V is directly related to n at constant P and