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41E

Expert-verified
Found in: Page 405

### Modern Physics

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

# Show that the rms speed of a gas molecule, defined as ${{\mathbf{v}}}_{{\mathbf{rms}}}{\mathbf{\equiv }}\sqrt{{\mathbf{v}}^{\mathbf{2}}}$, is given by $\sqrt{\frac{\mathbf{3}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}{\mathbf{m}}}$.

The rms speed of a gas molecule is $\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}}$

See the step by step solution

## Step 1: Maxwell Probability Distribution

${{\mathbf{v}}}_{{\mathbf{rms}}}{\mathbf{=}}\sqrt{\overline{{\mathbf{v}}_{\mathbf{2}}}}$

${{\mathbf{v}}}_{{\mathbf{rms}}}{\mathbf{=}}{\left({\int }_{0}^{\infty }{v}^{2}P\left(v\right)\mathrm{dv}\right)}^{\frac{\mathbf{1}}{\mathbf{2}}}$ ……(1)

Where,

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

${\mathbf{P}}\left(v\right){\mathbf{=}}{\left(\frac{m}{2{\mathrm{\pi k}}_{B}T}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}{\mathbf{4}}{{\mathbf{\pi v}}}^{{\mathbf{2}}}{{\mathbf{e}}}^{\frac{\mathbf{-}{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}}$…..(2)

Where,

m is the mass of the particle.

v is velocity of particle.

kB is Boltzmann constant.

## Step 2: determine the speed of gas

${\mathrm{v}}_{\mathrm{rms}}={\left({\int }_{0}^{\infty }{\mathrm{v}}^{2}{\left(\frac{\mathrm{m}}{2{\mathrm{\pi k}}_{\mathrm{B}}\mathrm{T}}\right)}^{\frac{3}{2}}4{\mathrm{\pi v}}^{2}{\mathrm{e}}^{-\frac{{\mathrm{m}}^{2}}{2{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}\mathrm{dv}\right)}^{\frac{1}{2}}$

Let $\mathrm{b}=\frac{1}{2{\mathrm{a}}^{2}}=\frac{\mathrm{m}}{2{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{rms}}& =& {\left(4\mathrm{\pi }{\left(\frac{\mathrm{b}}{\mathrm{\pi }}\right)}^{\frac{3}{2}}{\int }_{0}^{\infty }{\mathrm{v}}^{4}{\mathrm{e}}^{-{\mathrm{bv}}^{2}}\mathrm{dv}\right)}^{\frac{1}{2}}\\ & =& {\left(4\mathrm{\pi }{\left(\frac{\mathrm{b}}{\mathrm{\pi }}\right)}^{\frac{3}{2}}\frac{3}{8}\sqrt{\frac{\mathrm{\pi }}{{\mathrm{b}}^{5}}}\right)}^{\frac{1}{2}}\\ & =& {\left(\frac{3}{2\mathrm{b}}\right)}^{\frac{1}{2}}\\ & =& \sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}}\end{array}$

Therefore, The rms speed of a gas molecule is $\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}}$.