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

A
$2365^\circ\textrm{C}$

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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.

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