chapter5solutions

# chapter5solutions - CHAPTER 5 GASES Pressure 21 4.75 cm 10...

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96 CHAPTER 5 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 10 013 . 1 5 " = 6.33 × 10 3 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. P flask < P atm ; P flask = 760. ! 118 = 642 torr 642 torr × torr 760 atm 1 = 0.845 atm 0.845 atm × atm Pa 10 013 . 1 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 10 013 . 1 5 " = 1.30 × 10 5 Pa c. P flask = 635 ! 118 = 517 torr; P flask = 635 + 215 = 850. torr

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CHAPTER 5 GASES 97 23. Suppose we have a column of mercury 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 2 2 m cm 100 cm N 1 . 10 Area Force # \$ % & ( " ) = 1.01 × 10 5 2 m N 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 mercury. 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. 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 ; V Hg = V oil ; oil oil Hg Hg d m d m ) , m oil = Hg oil Hg d d m Because P is proportional to the mass of liquid: P oil = P Hg # # \$ % & & ( Hg oil d d = P Hg # \$ % & ( 6 . 13 30 . 1 = (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 749 torr × torr 760 atm 1 = 0.986 atm; 0.986 atm × atm Pa 10 013 . 1 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 10 013 . 1 5 " = 1.04 × 10 5 Pa
CHAPTER 5 GASES 98 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. a. 4.8 atm × atm Hg mm 760 = 3.6 × 10 3 mm Hg; b. 3.6 × 10 3 mm Hg × Hg mm torr 1 = 3.6 × 10 3 torr c. 4.8 atm × atm Pa 10 013 . 1 5 " = 4.9 × 10 5 Pa; d. 4.8 atm × atm psi 7 . 14 = 71 psi Gas Laws 26. a. PV = nRT b. PV = nRT c. PV = nRT PV = constant P = # \$ % & ( V nR × T = const × T T = # \$ % & ( nR P × V = const × V d. PV = nRT e. P = V constant V nR ) f. PV = nRT PV = constant P = constant × V 1 T PV = nR = constant T V T P V PV P V 1/V P PV T P

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CHAPTER 5 GASES 99 Note : The equation for a straight line is y = mx + b where y is the y axis and x is the x axis.
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