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Show that the rms speed of a gas molecule, defined as vrms≡v2, is given by3kBTm.

Short Answer

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The rms speed of a gas molecule is 3kBTm

Step by step solution

01

Maxwell Probability Distribution

vrms=v2¯

vrms=(∫0∞v2Pvdv)12…â¶Ä¦(1)

Where,

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

P(v)=(m2Ï€°ìBT)324Ï€±¹2e-mv22kBT…..(2)

Where,

mis the mass of the particle.

vis velocity of particle.

kBis Boltzmann constant.

02

determine the speed of gas

vrms=∫0∞v2m2Ï€°ìBT324Ï€±¹2e-m22kBTdv12

Letb=12a2=m2kBT

vrms=4πbπ32∫0∞v4e-bv2dv12=4πbπ3238πb512=32b12=3kBTm

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

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