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Chapter 5 OWL study questions

Chapter 5 OWL study questions - The following conversion...

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The following conversion factors may be helpful in solving the pressure conversion problems:  1 atm = 101.3 kPa = 1.013 bar = 760 torr = 760 mm Hg = 14.7 psi Liquid  chlorotribromomethane  has a density of  2.71  g/mL. A barometer is constructed using  chlorotribromomethane  instead of mercury.  If the atmospheric pressure  0.982  atm, the height of the  chlorotribromomethane  column in the barometer is mm. The density of liquid mercury is 13.6 g/mL. The pressure exerted by a column of a liquid can be expressed as the product of its density and the column height: P = D x h For two liquids, A and B, that exert the same pressure P = D A  x h A  = D B  x h B Since the conversion factor between mm Hg and atmospheres is known, first express the pressure of  0.982  atmospheres in mm Hg: 0.982  atm x (760 mm Hg / atm) =  746  mm Hg Substituting A for  chlorotribromomethane  and B for mercury gives h A  = (D Hg /D A ) x h Hg  = (13.6 g/mL /  2.71 g/mL) x  746  mm =  3.75E+3  mm  An automobile tire is inflated to a pressure of  28.1  psi. Express this pressure in atmospheres, kilopascals, bars, and millimeters Hg. Hint: 1 atm = 101.3 kPa = 1.013 bar = 760 torr = 760 mm Hg = 14.7 psi 1.91  atm 194  kPa 1.94  bar 1.45E+3  mm Hg Feedback: The following conversion factors may be used: 1 atm = 101.3 kPa = 1.013 bar = 760 mm Hg = 14.7 psi Convert 28.1 psi to atm: 28.1 psi x (1 atm / 14.7 psi) =  1.91  atm Convert 28.1 psi to kPa: 28.1  psi x (101.3 kPa / 14.7 psi) =  194  kPa
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Convert 28.1 psi to bar: 28.1  psi x (1.013 bar / 14.7 psi) =  1.94  bar Convert 28.1 psi to mm Hg: 28.1  psi x (760 mm Hg / 14.7 psi) =  1.45E+3  mm Hg A student reads a barometer in the laboratory and finds the prevailing atmospheric pressure to be  732  mm Hg. Express this pressure in  atmospheres, kilopascals, torrs, and bars. Hint: 1 atm = 101.3 kPa = 1.013 bar = 760 torr = 760 mm Hg = 14.7 psi 0.963  atm 97.6  kPa 732  torr 0.976  bar Feedback: The following conversion factors may be used: 1 atm = 101.3 kPa = 1.013 bar = 760 torr = 760 mm Hg Convert 732 mm Hg to atm: 732  mm Hg x (1 atm / 760 mm Hg) =  0.963  atm Convert 732 mm Hg to kPa: 732  mm Hg x (101.3 kPa / 760 mm Hg) =  97.6  kPa Convert 732 mm Hg to torr: 732  mm Hg x (760 torr / 760 mm Hg) =  732  torr Convert 732 mm Hg to bar: 732  mm Hg x (1.013 bar / 760 mm Hg) =  0.976   In the winter, a heated home in the Northeast might be maintained at a temperature of  80   o F . What is this temperature on the Celcius and  Kelvin scales? 27   o 300  K Feedback:
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GIVEN:   80   o WANTED:   o EQUATION:  T o F  - 32 = 1.8 T o C T o C  =  T o F  - 32 80   o F - 32  27   o C 1.8  1.8  GIVEN:   27   o WANTED:  K  EQUATION:  K =  o C + 273 K =  o C + 273 =  27   o C + 273 =  300  K 
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