104 CHAPTER 5 GASES Pressure 21. 4.75 cm × cmmm10 = 47.5 mm Hg or 47.5 torr; 47.5 torr × torr760atm1 = 6.25 × 102 atm 6.25 × 102 atm × atmPa10013.15 = 6.33 × 103 Pa 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 × torr760atm1 = 0.845 atm 0.845 atm × atmPa10013.15 = 8.56 × 104 Pa b. Pflask > Patm; Pflask = 760. torr + 215 torr = 975 torr 975 torr × torr760atm1 = 1.28 atm 1.28 atm × atmPa10013.15 = 1.30 × 105 Pa b. Pflask = 635  118 = 517 torr; Pflask = 635 + 215 = 850. torrCHAPTER 5 GASES 105 23. 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 × g1000kg1 = 1.03 kg F = mg = 1.03 kg × 9.81 m/s2 = 10.1 kg m/s2 = 10.1 N 22mcm100cmN1.10AreaForce= 1.01 × 105 2mN 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. 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 =volumemass; the volume of silicon oil is the same as the volume of mercury in Exercise 22. V = dm; VHg = Voil; oiloilHgHgdmdm, moil = HgoilHgddm Because P is proportional to the mass of liquid: Poil = PHgHgoildd = PHg6.1330.1= (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 × torr760atm1 = 0.986 atm; 0.986 atm × atmPa10013.15 = 9.99 × 104 Pa b. Pflask = 760. torr + (215 × 0.0956) torr = 760. + 20.6 = 781 torr 781 torr × torr760atm1 = 1.03 atm; 1.03 atm × atmPa10013.15 = 1.04 × 105 PaCHAPTER 5 GASES 106b. 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. a. 4.8 atm × atmHgmm760= 3.6 × 103 mm Hg; b. 3.6 × 103 mm Hg × Hgmmtorr1 = 3.6 × 103 torr c. 4.8 atm × atmPa10013.15 = 4.9 × 105 Pa; d. 4.8 atm × atmpsi7.14 = 71 psi Gas Laws 26. a. PV = nRT b. PV = nRT c. PV = nRT PV = constant P = VnR× T = const × T T = nRP× V = const × V d. PV = nRT e. P = VconstantVnR f. PV = nRT PV = constant P = constant × V1 TPV = nR = constant TVTPVPVPV1/VPPVTPCHAPTER 5 GASES 107 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. 27. 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. 28. 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 = 211VVP = X3X2P1 = 1P32 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 = 211VVP = X2XP1 = 2P1 The final pressure in both flasks will be one-half that of the initial pressure in the left flask. 29. Treat each gas separately, and use the relationship P1V1 = P2V2 (n and T are constant). For H2: P2 = 211VVP = 475 torr × L00.3L00.2 = 317 torr For N2: P2 = 0.200 atm × L00.3L00.1 = 0.0667 atm; 0.0667 atm × atmtorr760 = 50.7 torr Ptotal = 22NHPP  = 317 + 50.7 = 368 torr 30. For H2: P2 = 211VVP = 360. torr × L00.3L00.2 = 240. torr Ptotal = 2222HtotalNNHPPP,PP  = 320. torr  240. torr = 80. torr For N2: P1 = 122VVP = 80. torr × L00.1L00.3 = 240 torrCHAPTER 5 GASES 10831. PV = nRT, RVPnT= constant, 222111PTnPTn; moles × molar mass = mass 222111222111PTmassPTmass,PT)massmolar(nPT)massmolar(n mass2 = psi.2050K299psi.650K291g1000.1PTPTmass312211= 309 g 32. PV = nRT, n is constant. TPV= nR = constant, 12211222111TPTPVV,TVPTVP2 V2 = 1.00 L × K)23273(K)31273(torr.220torr.760 = 2.82 L; ΔV = 2.82  1.00 = 1.82 L 33. P = L00.4K.300molKatmL08206.0g01.44mol1g0.22VRTnP22COCO = 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 = airCOPP2 = 3.08 atm + torr760atm1torr.740= 3.08 + 0.974 = 4.05 atm 34. TPV = nR = constant, 222111TVPTVP P2 = 12211TVTVP = 710 torr × K.)30273(K)820273(mL25mL100.52 = 5.1 × 104