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# solut05 - CHAPTER FIVE GASES Pressure 21 4.75 cm 10 mm 1...

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74 CHAPTER FIVE GASES Pressure 21. 4.75 cm × cm mm 10 = 47.5 mm Hg or 47.5 torr; 47.5 torr × torr 760 atm 1 = 6.25 × 10 -2 atm 6.25 × 10 -2 atm × atm Pa 10 _ 1.013 5 = 6.33 × 10 3 Pa 22. If the levels of Hg 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 mm will be equal to the difference in pressure in mm Hg 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. P flask < P atm ; P flask = 760. - 118 = 642 torr 642 torr × torr 760 atm 1 = 0.845 atm 0.845 atm × atm Pa 10 _ 1.013 5 = 8.56 × 10 4 Pa b. P flask > P atm ; P flask = 760. torr + 215 torr = 975 torr 975 torr × torr 760 atm 1 = 1.28 atm 1.28 atm × atm Pa 10 _ 1.013 5 = 1.30 × 10 5 Pa

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75 c. P flask = 635 - 118 = 517 torr; P flask = 635 + 215 = 850. torr 23. Suppose we have a column of Hg 1.00 cm × 1.00 cm × 76.0 cm = V = 76.0 cm 3 : mass = 76.0 cm 3 × 13.59 g/cm 3 = 1.03 × 10 3 g × g 1000 kg 1 = 1.03 kg F = mg = 1.03 kg × 9.81 m/s 2 = 10.1 kg m/s 2 = 10.1 N ÷ ø ö ç è æ m cm 100 _ cm N 10.1 = area Force 2 2 = 1.01 × 10 5 m N 2 or 1.01 × 10 5 Pa (Note: 76.0 cm Hg = 1 atm = 1.01 × 10 5 Pa.) To exert the same pressure, a column of water will have to contain the same mass as the 76.0 cm column of Hg. Thus, the column of water will have to be 13.59 times taller or 76.0 cm × 13.59 = 1.03 × 10 3 cm = 10.3 m. 24. 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). volume mass = density = d ; The volume of silicon oil is the same as the volume of Hg in Exercise 5.22. V = d m ; V Hg = V oil ; d m = d m oil oil Hg Hg , m oil = d d m Hg oil Hg Since P is proportional to the mass of liquid: P oil = P Hg ÷ ÷ ø ö ç ç è æ d d Hg oil = P Hg ÷ ø ö ç è æ 13.6 1.30 = 0.0956 P Hg This conversion applies only to the column of silicon oil. a. P flask = 760. torr - (118 × 0.0956) torr = 760. - 11.3 = 749 torr
CHAPTER 5 GASES 76 749 torr × torr 760 atm 1 = 0.986 atm; 0.986 atm × atm Pa 10 _ 1.013 5 = 9.99 × 10 4 Pa b. P flask = 760. torr + (215 × 0.0956) torr = 760. + 20.6 = 781 torr 781 torr × torr 760 atm 1 = 1.03 atm; 1.03 atm × atm Pa 10 _ 1.013 5 = 1.04 × 10 5 Pa 25. 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.

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CHAPTER 5 GASES 77 Gas Laws 26. a. PV = nRT b. PV = nRT c. PV = nRT PV = Constant P = ÷ ø ö ç è æ V nR × T = Const × T T = ÷ ø ö ç è æ nR P × V = Const × V d. PV = nRT e. P = V Constant = V nRT f. PV = nRT PV = Constant P = Constant × V 1 T PV = nR = Constant
CHAPTER 5 GASES 78 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 y-