Chapter 10 - CHAPTER 10 Quick Quizzes 1. (b). The glass...

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C HAPTER 10 Quick Quizzes 1. (b). The glass surrounding the mercury expands before the mercury does, causing the level of the mercury to drop slightly. The mercury rises after it begins to get warmer and approach the temperature of the hot water, because its coefficient of expansion is greater than that for glass. 2. (c). Gasoline has the highest coefficient of expansion. 3. (c). A cavity in a material expands in exactly the same way as if the cavity were filled with material. Thus, both spheres will expand by the same amount. 4. (a) On a cold day, the trapped air in the bubbles would be reduced in pressure, according to the ideal gas law. Thus, the volume of the bubbles may be smaller than on a hot day, and the material would not be as effective in cushioning the package contents. 5. (a). It expands. Imagine the balloon rising into air at uniform temperature. The air cannot be uniform in pressure because the lower layers support the weight of all the air above them. The rubber in a typical balloon stretches or contracts until interior and exterior pressures are nearly equal. So as the balloon rises, it expands; this can be considered as constant temperature expansion with V increasing as P decreases by the same factor in P Vn R T = . If the rubber wall is strong enough, the buoyant force will eventually match the total weight of the balloon and helium so the balloon will stop rising. It is more likely that the rubber will stretch and rupture, releasing the helium, which in turn will escape from the Earth’s atmosphere. 6. (b). Helium, because it has the lower molar mass. 321
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CHAPTER 10 Problem Solutions 10.1 (a) () FC 99 32 273.15 55 TT =+ = + = -460 F ° (b) CF 98.6 =− = = 37.0 C ° (c) K 9 173.15 5 T = − + = + = 280 F ° 10.2 136 = 57.8 C ° and 127 = = 88.3 C ° 10.3 (a) Converting from Celsius to Fahrenheit, 252.87 = + = 423 F ° , and converting to Kelvin, C 273.15 252.87 273.15 = += −+= 20.28 K (b) 20 = + = 68 F ° , and C 273.15 20 273.15 = 293 K 10.4 134 = 56.7 C ° and C 5 79.8 9 T =− − = 62.1 C ° 10.5 Start with and convert to Celsius. F 40 F T ° 40 = = 40 C ° Since Celsius and Fahrenheit degrees of temperature change are different sizes, this is the only temperature with the same numeric value on both scales. 322
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CHAPTER 10 10.6 Since we have a linear graph, we know that the pressure is related to the temperature as , where A and B are constants. To find A and B , we use the given data: =+ C PAB T ( ) ° 0.900 atm -80.0 C AB (1) and ( ) ° 1.635 atm 78.0 C (2) Solving equations (1) and (2) simultaneously, we find: , and 1.27 atm A = 3 4.65 10 atm C B = ×° . Therefore, ( ) 3 C 1.27 atm 4.65 10 C P T × ° . (a) At absolute zero the gas exerts zero pressure ( ) 0 P = , so C 3 1.27 atm 4.65 10 atm C T == 273 C ° (b) At the freezing point of water, C 0 T = and 1.27 atm 0 P = 1.27 atm At the boiling point of water, C 100 C T = ° , so ( ) () 3 1.27 atm 4.65 10 C 100 C P ° = 1.74 atm 10.7 Apply FC 9 32 5 TT to two different Celsius temperatures, , to obtain C1 C2 and F1 C1 9 5 , (1) and F2 C2 9 5 . (2) Subtracting equation (1) from (2) yields F1 C2 C1 9 5 −= , or 95 ∆= 323
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CHAPTER 10 10.8 (a) Using the result of problem 7, gives () FC 9 95 450 5 TT ∆= = 810 F ° (b) Since the only difference in the Kelvin and Celsius temperature scales is the location
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Chapter 10 - CHAPTER 10 Quick Quizzes 1. (b). The glass...

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