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

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