Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
13
Temperature and Kinetic Theory
Change chapter

13-1: Atomic Theory
13-2: Temperature and Thermometers
13-4: Thermal Expansion
13-5: Gas Laws; Absolute Temperature
13-6 and 13-7: Ideal Gas Law
13-8: Ideal Gas Law in Terms of Molecules; Avogadro's Number
13-9: Molecular Interpretation of Temperature
13-11: Real Gases; Phase Changes
13-12: Vapor Pressure and Humidity
13-13: Diffusion

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 46
Q

A gas is at 20C20 ^\circ \textrm{C}. To what temperature must it be raised to triple the rms speed of its molecules?

A
2365C2365^\circ\textrm{C}
Giancoli 7th Edition, Chapter 13, Problem 46 solution video poster
Padlock

In order to watch this solution you need to have a subscription.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The rms speed at the first temperature is going to be square root of 3 times Boltzmann's constant times the first temperature, T1, divided by the mass of a single molecule. And then the rms speed of the second temperature is going to be square root of 3 K T2 over m. And the m doesn't need a subscript because it's the same molecule, it's just a different temperature. So, we put a subscript on the T only and we know the V rms2, there's gonna be 3 times the rms speed of the first temperature. And so we substitute for V rms2 that's 3 K T2 over m square rooted. And that equals 3 times V rms1, 3 K T1 over m. And the square, well... Whole bunch of things cancel on both sides, you can multiply both sides by square root m over 3 K, and then that cancels away everything except for this square root 2 or T2 equals 3 times square root T1. And then square both sides. And you get T2 is 9 times T1. So, 9 times the first temperature written in kelvin, that's 20 degree Celsius plus 273.15. And that gives 2638.135 kelvin. And then we subtract away 273.15. And that gives 236.5 degrees Celsius.

Find us on:

Facebook iconTrustpilot icon
Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.