This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 18 1. 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 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. 2. 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. 3. 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 785 CHAPTER 18 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 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 = + . 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. 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 = + . 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. 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: 18 50.00 60.00 1375 ....
View
Full
Document
This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics

Click to edit the document details