A constant-volume gas thermometer registers an absolute pressure corresponding to 325 mm of mercury when in contact with water at the triple point. What pressure does it read when in contact with water at the normal boiling point?
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Solution 8E In triple point of water all the three phases of water can exist simultaneously in thermal equilibrium with each other. The required temperature is 273.16 K and the required pressure is 611.73 Pa. The gas thermometer is in contact with the water in triple point. Then it is again in contact with the normal water at it’s boiling point. The boiling point is 100 C = 373.15 K The mercury reading when the thermometer was in contact with the water at triple point was 325 mm. We know that, as the volume is constant, P1 T1 P2 = T2--------------------(1) 325= 273.16 P2 373.15 P = (325 × 373.15) / 273.16 = 443.96 = 444 mm (approximately). 2 Conclusion: It will read 444 mm pressure.
Textbook: University Physics
Author: Hugh D. Young, Roger A. Freedman
Since the solution to 8E from 17 chapter was answered, more than 430 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: Water, pressure, contact, point, corresponding. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 8E from chapter: 17 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460. The answer to “A constant-volume gas thermometer registers an absolute pressure corresponding to 325 mm of mercury when in contact with water at the triple point. What pressure does it read when in contact with water at the normal boiling point?” is broken down into a number of easy to follow steps, and 38 words.