Chapter 6: Q. 6.35 (page 246)

Verify from Maxwell speed distribution that the most likely speed of a molecule is $\sqrt{\frac{2 k T}{m}}$ .

Short Answer

Expert verified

The most likely speed of a molecule is $v_{mp} = \sqrt{\frac{2 k T}{m}}$ .

Step by step solution

Step 1. Given information

Maxwell speed distribution function is given by

$D (v) = {(\frac{m}{2 π k T})}^{\frac{3}{2}} 4 π v^{2} e^{- \frac{m v^{2}}{2 k T}} . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Step 2. Calculation

Differentiate both sides of equation (1) with respect to $v$ .

role="math" localid="1646971259346" $\begin{array}{rcl} \frac{d D (v)}{d v} & = & \frac{d}{d v} [{(\frac{m}{2 π k T})}^{\frac{3}{2}} 4 π v^{2} e^{- \frac{m v^{2}}{2 k T}}] \\ = & {(\frac{m}{2 π k T})}^{\frac{3}{2}} 4 π [2 v e^{- \frac{m v^{2}}{2 k T}} - \frac{2 m v^{3}}{2 k T} e^{- \frac{m v^{2}}{2 k T}}] \\ = & 8 π {(\frac{m}{2 π k T})}^{\frac{3}{2}} v e^{- \frac{m v^{2}}{2 k T}} [1 - \frac{m v^{2}}{2 k T}] . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) \end{array}$

At most likely speed of the gas molecules, the derivative of the Maxwell speed distribution function will be zero.

Substitute $v_{mp}$ for $v$ and $0$ for $\begin{array}{rcl} \frac{d D (v)}{d v} \end{array}$ into equation (2) and simplify to obtain the most likely speed.

$\begin{array}{rcl} 0 & = & 8 π {(\frac{m}{2 π k T})}^{\frac{3}{2}} v e^{- \frac{m v^{2}}{2 k T}} [1 - \frac{m v^{2}}{2 k T}] \\ [1 - \frac{m {v_{mp}}^{2}}{2 k T}] & = & 0 [As v \neq 0] \\ v_{mp} & = & \sqrt{\frac{2 k T}{m}} \end{array}$

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Verify from Maxwell speed distribution that the most likely speed of a molecule is $\sqrt{\frac{2 k T}{m}}$ .

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

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Verify from Maxwell speed distribution that the most likely speed of a molecule is2kTm.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Verify from Maxwell speed distribution that the most likely speed of a molecule is $\sqrt{\frac{2 k T}{m}}$ .