Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 19 - The Kinetic Theory of Gases - Problems - Page 578: 18

Answer

$v_{rms}=9.53\times10^6\frac{m}{s}$

Work Step by Step

The root mean square speed $v_{rms}$ of a molecule can be determined as follows $v_{rms}=\sqrt\frac{3RT}{M}$....eq(1) To use this equation, we first have to find molar mass of the electron: $M=m_e \times N_a$ where $m_e=9.11\times10^{-31Kg}$ and Avogadro's Number $N_a=6.023\times10^{23}$ Molar mass of the electron:$M=9.11\times10^{-31}\times6.023\times10^{23}=5.49\times10^{-7}Kg$ Now substituting these values in eq(1) and solving: $v_{rms}=\sqrt\frac{3(8.314)(2\times10^6)}{5.49\times10^{-7}}$ $v_{rms}=9.53\times10^6\frac{m}{s}$
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