(a) Explanation Section 1: To determine: The expression for t in terms of γ . Answer: The expression for t in terms of γ is t = − 1 γ . Given Info: Pressure at the triple point is 4.80 × 10 4 Pa and at boiling point is 6.50 × 10 4 Pa . Formula to calculate pressure is, P = P o ( 1 + γ t ) I P is pressure at t ° C . P o is pressure at 0 ° C . γ is pressure coefficient. t is temperature. Above equation can be written for different temperatures and pressures. P 1 = P o ( 1 + γ t 1 ) II t 1 is initial temperature. P 1 is pressure at temperature t 1 . P 2 = P o ( 1 + γ t 2 ) III t 2 is final temperature. P 2 is pressure at temperature t 2 . Equate equation II and III to find γ . γ = P 2 − P 1 P 1 t 2 − P 2 t 1 IV Substitute 0 for P in equation I to find t 0 = P o ( 1 + γ t ) 0 = ( 1 + γ t ) γ t = − 1 t = − 1 γ Section 2: To determine: The temperature in Celsius at which the gas pressure becomes zero. Answer: The temperature in Celsius at which the gas pressure becomes zero is − 282 ° C (b) To determine The gas in the thermometer obeys expression T 1 T 2 = P 1 P 2 or not and also the pressure that the experimenter had measured at 0.01 ° C .

University Physics with Modern Physics Plus Mastering Physics with eText -- Access Card Package (14th Edition)

14th Edition

ISBN: 9780321982582

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

See similar textbooks

Concept explainers

Maxwell speed distribution

Maxwell-Boltzmann distribution or commonly known as Maxwell speed distribution, explains the division of the energy levels of the molecules of an ideal gas on the basis of the statistical theory. In 1859, Scottish physicist James Clerk Maxwell establishe…

Question

Chapter 17, Problem 17.9E

(a)

To determine

The expression for t in terms of γ and temperature in Celsius at which the gas pressure becomes zero.

(b)

To determine

The gas in the thermometer obeys expression T1T2=P1P2 or not and also the pressure that the experimenter had measured at 0.01°C .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 17, Problem 17.8E

Chapter 17, Problem 17.10E

Students have asked these similar questions

The pressure in a constant-volume gas thermometer is 0.700 atm at 1.00 x 102°C and 0.512 atm at 0°C. (a) What is the temperature when the pressure is 0.0400 atm? (b) What is the pressure at 450°C?

In a student experiment, a constant-volume gas thermometer is calibrated in liquid nitrogen (−196°C) and in boiling methyl alcohol (64.7°C). The separate pressures are 0.344 atm and 1.556 atm. Hint: Use the linear relationship P = A + BT, where A and B are constants. (a) What value of absolute zero does the calibration yield?? °C(b) What pressure would be found at the freezing point of water? ?atm(c) What pressure would be found at the boiling point of water? ?atm

At the end of a warm autumn day, you measure the air pressure in your car's tires and write down their pressure as 2.41 × 105 Pa (about 35 psi) when the outside temperature is 23°C. It cools down quite a bit overnight to -5°C. If you were to measure the air pressure in the tire on that cool morning, what pressure would you measure? Assume that you can ignore the change in the volume of the tire and that no air is released from the tire overnight.

Chapter 17 Solutions

University Physics with Modern Physics Plus Mastering Physics with eText -- Access Card Package (14th Edition)

Ch. 17.1 - You put a thermometer in a pot of hot water and...Ch. 17.2 - Which of the following types of thermometers have...Ch. 17.3 - Prob. 17.3TYU Ch. 17.4 - In the bimetallic strip shown in Fig. 17.3a, metal...Ch. 17.5 - Prob. 17.5TYU Ch. 17.6 - You take a block of ice at 0C and add heat to it...Ch. 17.7 - A room has one wall made of concrete, one wall...Ch. 17 - Explain why it would not make sense to use a...Ch. 17 - If you heat the air inside a rigid, scaled...Ch. 17 - Many automobile engines have cast-iron cylinders...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

The expression for t in terms of γ and temperature in Celsius at which the gas pressure becomes zero.

Solution Summary: The author explains the expression for t in terms of gamma and temperature in Celsius at which the gas pressure becomes zero.

Concept explainers

The expression for t in terms of γ and temperature in Celsius at which the gas pressure becomes zero.

The gas in the thermometer obeys expression T1T2=P1P2 or not and also the pressure that the experimenter had measured at 0.01°C .

Want to see the full answer?

Chapter 17 Solutions