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Unformatted text preview: Chapter 18 1. Let T L be the temperature and p L be the pressure in the lefthand thermometer. Similarly, let T R be the temperature and p R be the pressure in the righthand thermometer. According to the problem statement, the pressure is the same in the two thermometers when they are both at the triple point of water. We take this pressure to be p 3 . Writing Eq. 185 for each thermometer, 3 3 (273.16K) and (273.16K) , L R L R p p T T p p = = we subtract the second equation from the first to obtain 3 (273.16K) . L R L R p p T T p  = First, we take T L = 373.125 K (the boiling point of water) and T R = 273.16 K (the triple point of water). Then, p L – p R = 120 torr. We solve 3 120torr 373.125K 273.16K (273.16K) p  = for p 3 . The result is p 3 = 328 torr. Now, we let T L = 273.16 K (the triple point of water) and T R be the unknown temperature. The pressure difference is p L – p R = 90.0 torr. Solving the equation 90.0torr 273.16K (273.16K) 328torr R T  = for the unknown temperature, we obtain T R = 348 K. 2. We take p 3 to be 80 kPa for both thermometers. According to Fig. 186, the nitrogen thermometer gives 373.35 K for the boiling point of water. Use Eq. 185 to compute the pressure: N 3 373.35K (80kPa) = 109.343kPa. 273.16K 273.16K T p p = = The hydrogen thermometer gives 373.16 K for the boiling point of water and 773 CHAPTER 18 H 373.16K (80kPa) 109.287kPa. 273.16K p = = (a) The difference is p N  p H = 0.056 kPa 0.06 kPa ≈ . (b) The pressure in the nitrogen thermometer is higher than the pressure in the hydrogen thermometer. 3. From Eq. 186, we see that the limiting value of the pressure ratio is the same as the absolute temperature ratio: (373.15 K)/(273.16 K) = 1.366. 4. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y . Then 9 5 32 y x = + . For x = –71°C, this gives y = –96°F. (b) The relationship between y and x may be inverted to yield 5 9 ( 32) x y = . Thus, for y = 134 we find x ≈ 56.7 on the Celsius scale. 5. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y . Then 9 5 32 y x = + . If we require y = 2 x , then we have 9 2 32 (5)(32) 160 C 5 x x x = + ⇒ = = ° which yields y = 2 x = 320°F. (b) In this case, we require 1 2 y x = and find 1 9 (10)(32) 32 24.6 C 2 5 13 x x x = + ⇒ =  ≈  ° which yields y = x /2 = –12.3°F. 6. We assume scales X and Y are linearly related in the sense that reading x is related to reading y by a linear relationship y = mx + b . We determine the constants m and b by solving the simultaneous equations: ( 29 ( 29 70.00 125.0 30.00 375.0 m b m b = + = + which yield the solutions m = 40.00/500.0 = 8.000 × 10 –2 and b = –60.00. With these values, we find x for y = 50.00: 6 50.00 60.00 1375 ....
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This note was uploaded on 10/05/2010 for the course BIOL biol111 taught by Professor Ciccarello during the Spring '10 term at University of Massachusetts Boston.
 Spring '10
 Ciccarello
 Biology

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