Chemistry by Whitten, Atwood, Morrison Chapter 12 solutions

Chemistry by Whitten, Atwood, Morrison Chapter 12 solutions...

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107 Gases and the Kinetic-Molecular Theory 12 12-1 Pressure is a measure of a force applied per square unit of surface area. Pressure is the force per unit area of a gas (or other fluid) confined to a container. 12-3 A mercury barometer consists of a glass tube, sealed at one end, filled with mercury, and then inverted into a dish of mercury. The mercury in the tube falls until the pressure due to its height matches the pressure due to the air (the atmosphere) on the surface of the mercury in the dish. The height of the mercury column is a measure of the external air pressure. 12-5 (a) ? mm Hg = 675 torr x 1 mm Hg 1 torr = 675 mm Hg (b) ? atm = 675 torr x 1 atm 760. torr = 0.888 atm (c) ? Pa = 675 torr x 1 atm 760. torr x 1.013 x 10 5 Pa 1 atm = 9.00 x 10 4 Pa (d) ? kPa = 9.00 x 10 4 Pa x 1 kPa 1000 Pa = 90.0 kPa 12-7 We use conversion factors based on the equivalences 1 atm = 760 torr = 101.3 kPa. atm torr Pa kPa Standard atmosphere 1 760 1.013 x 10 5 101.3 Partial pressure of nitrogen in the atmosphere 0.780 593 7.90 x 10 4 79.0 A tank of compressed hydrogen 1.59 1.21 x 10 3 1.61 x 10 5 1.61 x 10 2 Atmospheric pressure at the summit of Mt. Everest 0.333 253 3.37 x 10 4 33.7 12-9 The height of liquid is inversely proportional to its density; the lower the density of the liquid, the taller the column would be. height of oil column = 760. mm Hg x 13.5 g/mL 0.92 g/mL x 1 m 1000 mm = 11 m 12-11 31 psi x 1 atm 14.70 psi = 2.1 atm guage pressure total pressure = 1.0 atm + 2.1 atm = 3.1 atm
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12-13 Yes, because for a given substance, the number of moles is directly proportional to the mass of substance. 12-15 V = 2.50 L x 1.00 atm 2.50 atm = 1.00 L 12-17 P 1 = 0.500 atm V 1 = 75.0 mL P 2 is given in each part V 2 = ? For each part, we use Boyle’s Law, P 1 V 1 = P 2 V 2 or V 2 = P 1 V 1 P 2 (a) V 2 = 0.500 atm x 75.0 mL 5.00 atm = 7.50 mL (b) V 2 = 0.500 atm x 75.0 mL 0.0500 atm = 7.50 x 10 2 mL (c) V 2 = 0.500 atm x 75.0 mL 555 torr x 1 atm 760 torr = 51.4 mL (d) V 2 = 0.500 atm x 75.0 mL 5.00 torr x 1 atm 760 torr = 5.70 x 10 3 mL (e) V 2 = 0.500 atm x 75.0 mL 5.5 x 10 -2 torr x 1 atm 760 torr = 5.2 x 10 5 mL 12-19 (a) Yes. From graphs analogous to A, B, or C in Figure 12-5, but with temperature expressed in o F, we could find the value on the Fahrenheit scale to which the lines extrapolate. This would provide the constant that we could use to convert from the Fahrenheit to the “absolute Fahrenheit” scale. (b) Yes. By the reasoning in part (a), it would not matter what size the degree was; the same approach would still give an absolute scale. 12-21 K o C o F Normal boiling point of water 373 100. 212. Reference for thermodynamic data 298.15 25.00 77.00 Dry ice becomes a gas at atmospheric pressure 194.6 –78.5 –109.3 The center of the sun (estimated) 1.53 x 10 7 1.53 x 10 7 2.75 x 10 7 12-23 At approximately 25 o C or 298 K, the volume is 20 mL. At approximately 325
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This note was uploaded on 03/17/2009 for the course CHEM 1212 taught by Professor Suggs during the Spring '08 term at University of Georgia Athens.

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Chemistry by Whitten, Atwood, Morrison Chapter 12 solutions...

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