Giancoli's Physics: Principles with Applications, 7th Edition

13

Temperature and Kinetic Theory

Change chapter13-1: 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

$2365^\circ\textrm{C}$

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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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