Chapter18 - 1. Let TL be the temperature and pL be the...

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1. 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, 33 (273.16K) and (273.16K) , LR pp TT p p §· == ¨¸ ©¹ we subtract the second equation from the first to obtain 3 (273.16K) . p p 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.
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H 373.16K (80kPa) 109.287kPa. 273.16K p §· == ¨¸ ©¹ (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. 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: N3 373.35K (80kPa) = 109.343kPa. 273.16K T pp The hydrogen thermometer gives 373.16 K for the boiling point of water and
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3. 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.
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4. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y . Then 9 5 32 yx =+ . For x = –71°C, this gives y = –96°F. (b) The relationship between y and x may be inverted to yield 5 9 ( 32) xy =− . Thus, for y = 134 we find x 56.7 on the Celsius scale.
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5. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y . Then 9 5 32 yx =+ . If we require y = 2 x , then we have 9 2 32 (5)(32) 160 C 5 xx x ¡ == ° which yields y = 2 x = 320°F. (b) In this case, we require 1 2 y x = and find 19 ( 1 0 ) ( 3 2 ) 32 24.6 C 25 1 3 x ¡ =− ≈− ° which yields y = x /2 = –12.3°F.
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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: () 70.00 125.0 30.00 375.0 mb −= + + 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: 50.00 60.00 1375 . 0.08000 yb x X m −+ == = °
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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 ( 170) mb =− + + which yield the solutions m = 100/(170 – 53.5) = 0.858 and b = 419. With these values, we find x for y = 340: 340 419 92.1 . 0.858 yb x X m −− == = °
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8. The change in length for the aluminum pole is 6 01 (33m)(23 10 / C )(15 C)=0.011m.
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter18 - 1. Let TL be the temperature and pL be the...

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