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A flask contains a mixture of neon (Ne), krypton (Kr), and

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 30E Chapter 18

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 30E

A flask contains a mixture of neon (Ne), krypton (Kr), and radon (Rn) gases. Compare (a) the average kinetic energies of the three types of atoms and (b) the root-mean-square speeds. (?Hint?: Appendix D shows the molar mass (in g/mol) of each element under the chemical symbol for that element.)

Step-by-Step Solution:

Solution to 30E Step 1 M(Ne)=20.18g/mol M(Kr)=83.80 g/mol M(Rn)=222g/mol Average kinetic energy of molecules is 3 KE = kT 2 Thus average kinetic energy depends upon the temperature of the gas. (a) Since Neon, Krypton, and Radon are in a container with a temperature of T kelvin, the average kinetic energies of all three gases will be same. KE(Ne)=KE(Kr)=KE(Rn) Step 2 0 Let us assume T=20 C=298K RMS speed of molecules V = 3RT/M rms 3 Vrmse)= 3 × 8.314 × 298/20.18 × 10 Vrmse)=606.9m/s 3 Vrmsr)= 3 × 8.314 × 298/83.80 × 10 Vrmsr)=297.82m/s V (Rn) = 3 × 8.314 × 298/222 × 103 rms Vrmsn)=182.98m/s

Step 3 of 3

Chapter 18, Problem 30E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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