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 48

Q

A

$43.9^\circ\textrm{C}$

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We're going to raise the rms speed of a molecule by 4% and the first rms speed that it'll have is going to be square root of 3 times Boltzmann's constant times the first temperature divided by mass. And the second rms speed is going to be square root of 3 *K T2*, second temperature divided by the same mass cause it's the same molecule, the only thing we're changing is the temperature. So, it increases speed by 4% means that the difference in speeds in the second case versus the first case divided by the first speed is going to be 4% or 4 divided by 100. And we can multiply both sides by *V rms1* and that means *V rms2* minus *V rms1* is 4% times *V rms1*. 4.0 dividede by 100 is 0.040. And then take this term to the right hand side. And you get *V rms2* equals *V rms1* plus 0.04 *V rms1* which is 1.04 *V rms1*. So, let's substitute into this line, on the next line. *V rms2* is square root 3 *K T2* over *m*. And that equals 1.044 times square root of 3 *K T1* over *m*. The... You can multiply both sides by square root *m* over 3 *K* to cancel a bunch of things. And that leaves us with square root *T2* equals 1.04 times square root *T1*. So, the temperature in the second case to increase the rms speed by 4% is going to be 1.04 squared times *T1* 1.04 squared times the temperature in kelvin, 20 degrees Celsius plus 273.15, which gives 317 kelvin. And we take away 273.15 to find the temperature in degrees Celsius of 43.9.

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