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ch18 - Chapter 18 1 From Eq 18-6 we see that the limiting...

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763 Chapter 18 1. From Eq. 18-6, 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. 2. We take p 3 to be 80 kPa for both thermometers. According to Fig. 18-6, the nitrogen thermometer gives 373.35 K for the boiling point of water. Use Eq. 18-5 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 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. Let T L be the temperature and p L be the pressure in the left-hand thermometer. Similarly, let T R be the temperature and p R be the pressure in the right-hand 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. 18-5 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
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CHAPTER 18 764 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. 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: which yield the solutions m = 40.00/500.0 = 8.247 10 –2 and b = –59.69. With these values, we find x for y = 50.00: 70.00 125.0 30.00 360.0 m b m b
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765 50.00 59.69 1330 . 0.08247 y b x X m 7. We assume scale X is a linear scale in the sense that if its reading is x then it is related to a reading y on the Kelvin scale by a linear relationship y = mx + b . We determine the constants m and b by solving the simultaneous equations: 373.15 ( 53.5) 273.15 ( 175) m b m b
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