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Modern Physics
Found in: Page 405
Modern Physics

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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Short Answer

Show that the rms speed of a gas molecule, defined as vrmsv2, is given by 3kBTm.

The rms speed of a gas molecule is 3kBTm

See the step by step solution

Step by Step Solution

Step 1: Maxwell Probability Distribution


vrms=(0v2Pvdv)12 ……(1)


P(v) is the Maxwell probability distribution, which is given by



m is the mass of the particle.

v is velocity of particle.

kB is Boltzmann constant.

Step 2: determine the speed of gas


Let b=12a2=m2kBT


Therefore, The rms speed of a gas molecule is 3kBTm.

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