Question

The probability density function for the component of the$\mathit{x}$ velocity of a molecule of an ideal gas is proportional torole="math" localid="1665039876656" ${\mathit{e}}^{\frac{\mathbf{-}\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\left(2kT\right)}}$ where$\mathit{v}$ is the$\mathit{x}$ component of the velocity,$\mathit{m}$ is the mass of the molecule, $\mathit{T}$is the temperature of the gas androle="math" localid="1665040106251" $\mathit{k}$ is the Boltzmann constant. By comparing this with (8.1), find the mean and standard deviation of $\mathit{v}$, and write the probability density function $\mathit{f}\left(v\right)$.

Short answer

The probability density function is $\frac{e^{{\displaystyle \frac{-m{v}^2}{\left(2kT\right)}}}}{\sqrt{{\displaystyle \frac{2\mathrm{\pi kT}}{m}}}}$.

Step-by-step solution

Given information

The velocity of a molecule of an ideal gas is proportional to$e^{\frac{-m{v}^2}{\left(2kT\right)}}$.

Definition of variance

A statistical measurement of the dispersion between values in a data collection is known as variance.

Find the average of the function

The function is $f\left(v\right)\infty e^{\frac{-m{v}^2}{\left(2kT\right)}}$…….(1)

role="math" localid="1665042441825" $$\begin{gathered} f\left(v\right)=c{e}^{\frac{-m{v}^2}{\left(2kT\right)}}f\left(x\right) \\ =\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{\left(2{\sigma }^2\right)}} \end{gathered}$$……………..(2)

Compare (1) and (2)

$\begin{array}{rcl}-\frac{1}{2{\sigma }^2}& =& -\frac{m}{2kT}\\ \sigma & =& \sqrt{\frac{kT}{m}}\\ & & \end{array}$

The value of average is $\begin{array}{rcl}\mu & =& 0\\ & & \end{array}$.

Substitute in (1).

$\begin{array}{rcl}f\left(v\right)& =& \frac{e^{{\displaystyle \frac{-m{v}^2}{\left(2kT\right)}}}}{\sqrt{{\displaystyle \frac{2\mathrm{\pi kT}}{m}}}}\\ & & \end{array}$

The variance is $\begin{array}{rcl}\sigma & =& \sqrt{\frac{kT}{m}}\\ & & \end{array}$.

The average is $\begin{array}{rcl}\mu & =& 0\\ & & \end{array}$.

The probability density function is$\begin{array}{rcl}f\left(v\right)& =& \frac{e^{{\displaystyle \frac{-m{v}^2}{\left(2kT\right)}}}}{\sqrt{{\displaystyle \frac{2\pi kT}{m}}}}\\ & & \end{array}$ .