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 $v_{rms} \equiv \sqrt{v^{2}}$ , is given by $\sqrt{\frac{3 k_{B} T}{m}}$ .

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Maxwell Probability Distribution

$v_{rms} = \sqrt{\bar{v_{2}}}$

$v_{rms} = {(\int_{0}^{\infty} v^{2} P (v) dv)}^{\frac{1}{2}}$ ……(1)

Where,

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

$P (v) = {(\frac{m}{2 {πk}_{B} T})}^{\frac{3}{2}} 4 {πv}^{2} e^{\frac{- {mv}^{2}}{2 k_{B} T}}$ …..(2)

Where,

m is the mass of the particle.

v is velocity of particle.

k_B is Boltzmann constant.

Step 2: determine the speed of gas

$v_{rms} = {(\int_{0}^{\infty} v^{2} {(\frac{m}{2 {πk}_{B} T})}^{\frac{3}{2}} 4 {πv}^{2} e^{- \frac{m^{2}}{2 k_{B} T}} dv)}^{\frac{1}{2}}$

Let $b = \frac{1}{2 a^{2}} = \frac{m}{2 k_{B} T}$

$\begin{array}{rcl} v_{rms} & = & {(4 π {(\frac{b}{π})}^{\frac{3}{2}} \int_{0}^{\infty} v^{4} e^{- {bv}^{2}} dv)}^{\frac{1}{2}} \\ = & {(4 π {(\frac{b}{π})}^{\frac{3}{2}} \frac{3}{8} \sqrt{\frac{π}{b^{5}}})}^{\frac{1}{2}} \\ = & {(\frac{3}{2 b})}^{\frac{1}{2}} \\ = & \sqrt{\frac{3 k_{B} T}{m}} \end{array}$

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

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Modern Physics

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Short Answer

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

Step by Step Solution

Step 1: Maxwell Probability Distribution

Step 2: determine the speed of gas

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