Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

Using the Maxwell speed distribution, determine the most probable speed of a particle of mass m in a gas at temperature T
How does this compare with v_rms ? Explain.

The most probable speed is $\sqrt{\frac{2 k_{B} T}{m}}$ .
The most probable speed is $\sqrt{\frac{2}{3}}$ times $v_{rms}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Maxwell Probability Distribution.

$P (v) = {(\frac{m}{2 {πk}_{B} T})}^{\frac{3}{2}} 4 {πv}^{2} e^{- \frac{{mv}^{2}}{2 k_{B} T}}$ …..(1)

Where,

m is the mass of the particle.

v is velocity of particle.

T is temperature.

k_B is Boltzmann constant.

$\begin{array}{rcl} \frac{d P}{d v} & = & \frac{d}{dv} [\sqrt{\frac{2}{π}} {(\frac{m}{k_{B} T})}^{\frac{3}{2}} v^{2} e^{- \frac{{mv}^{2}}{2 k_{B} T}}] \\ \frac{d P}{d v} & = & \sqrt{\frac{2}{π}} {(\frac{m}{k_{B} T})}^{\frac{3}{2}} \frac{d}{dv} [v^{2} e^{- \frac{{mv}^{2}}{2 k_{B} T}}] \\ \frac{d P}{d v} & = & \sqrt{\frac{2}{π}} {(\frac{m}{k_{B} T})}^{\frac{3}{2}} [2 {ve}^{- \frac{{mv}^{2}}{2 k_{B} T}} - (2 v) (\frac{m}{k_{B} T}) v^{2} e^{- \frac{{mv}^{2}}{2 k_{B} T}}] \\ \frac{d P}{d v} & = & 2 \sqrt{\frac{2}{π}} {(\frac{m}{k_{B} T})}^{\frac{3}{2}} [1 - \frac{{mv}^{2}}{2 k_{B} T}] {ve}^{- \frac{{mv}^{2}}{2 k_{B} T}} \\ 0 & = & 2 \sqrt{\frac{2}{π}} {(\frac{m}{k_{B} T})}^{\frac{3}{2}} [1 - \frac{{mv}^{2}}{2 k_{B} T}] {ve}^{- \frac{{mv}^{2}}{2 k_{B} T}} \\ 0 & = & 1 - \frac{{mv}^{2}}{2 k_{B} T} \\ \frac{{mv}^{2}}{2 k_{B} T} & = & 1 \\ v^{2} & = & \frac{2 k_{B} T}{m} \\ v & = & \sqrt{\frac{2 k_{B} T}{m}} \end{array}$

Therefore, the most probable speed is $\sqrt{\frac{2 k_{B} T}{m}}$ .

Step 2: Mathematical Expression of rms Speed.

$v_{rms} = \sqrt{\frac{3 k_{B} T}{m}}$

Rearrange for $\sqrt{\frac{k_{B} T}{m}}$ ,

$\sqrt{\frac{k_{B} T}{m}} = \frac{v_{rms}}{\sqrt{3}}$

The mathematical expression for the most probable speed is,

$v = \sqrt{\frac{2 k_{B} T}{m}}$

Substitute $\frac{v_{rms}}{\sqrt{3}}$ for $\sqrt{\frac{k_{B} T}{m}}$ ,

$\begin{array}{rcl} v & = & \sqrt{2} \frac{v_{rms}}{\sqrt{3}} \\ = & \sqrt{\frac{2}{3}} v_{rms} \end{array}$

Therefore, the most probable speed is $\sqrt{\frac{2}{3}}$ times v_rms.

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Modern Physics

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Short Answer

Using the Maxwell speed distribution, determine the most probable speed of a particle of mass m in a gas at temperature T
How does this compare with v_rms ? Explain.

Step by Step Solution

Step 1: Maxwell Probability Distribution.

Step 2: Mathematical Expression of rms Speed.

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Modern Physics

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Short Answer

Using the Maxwell speed distribution, determine the most probable speed of a particle of mass m in a gas at temperature THow does this compare with vrms ? Explain.

Step by Step Solution

Step 1: Maxwell Probability Distribution.

Step 2: Mathematical Expression of rms Speed.

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Using the Maxwell speed distribution, determine the most probable speed of a particle of mass m in a gas at temperature T
How does this compare with v_rms ? Explain.