We derived the expression for the Maxwell-Boltzmann speed distribution in a previous post, which was
\begin{equation}
f(v)\,dv = 4\pi v^2\left(\frac{m}{2\pi k_B T}\right)^{3/2}\exp\left[\frac{-mv^2}{2k_B T}\right]\,dv
\end{equation}This expression can be further simplified to get the characteristic speeds for the ensemble of the gas, i.e., the mean speed, the most probable speed, and the root-mean-square speed of the gas. In this post, we will derive the expression for the mean speed of the gases under the Maxwell-Boltzmann distribution.
Table of Contents
Deriving the mean speed expression
Starting from equation $(1)$ we take the expected value of the speed distribution which can be determined using the formula below.
\langle x\rangle = \int\limits_{0}^{\infty} x f(x)\,dxSo, for the speed distribution, we get,
\begin{equation}
\begin{aligned}
\langle v\rangle &= \int\limits_{0}^{\infty} v f(v)\,dv\\
\langle v\rangle &= 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2}\int\limits_{0}^{\infty} v\cdot v^2\exp\left[\frac{-mv^2}{2k_B T}\right]\,dv
\end{aligned}
\end{equation}Assuming $$b = \frac{m}{2k_B T}$$ we can simplify the expression as,
\begin{equation}
\begin{aligned}
\langle v\rangle &= 4\pi \left(\frac{b}{\pi}\right)^{3/2}\int\limits_{0}^{\infty}v^3\exp\left[-bv^2\right]\,dv\\
\langle v\rangle &= 4\pi\left(\frac{b}{\pi}\right)^{3/2}A(v)
\end{aligned}
\end{equation}We can now solve for $A(v)$ and get the value,
\begin{equation}
A(v) = \int\limits_{0}^{\infty}v\cdot v^2\exp\left[-bv^2\right]\,dv
\end{equation}Substituting $u = v^2$, we can reform the expression as,
\begin{equation*}
\begin{aligned}
u &= v^2\\ du &= 2v\,dv \\& \frac{du}{2v}=dv
\end{aligned}
\end{equation*}Rewriting equation $(4)$ with the substitutions,
\begin{equation*}
\begin{aligned}
A(v) &= \int\limits_{0}^{\infty} v\cdot u\exp\left[-bu\right]\,\frac{du}{2v}\\
A(v) &= \frac{1}{2}\left\{\int\limits_{0}^{\infty}u\exp\left[-bu\right]\,du\right\}
\end{aligned}
\end{equation*}We can use integration by parts to solve the above expression,
\begin{equation*}
\begin{aligned}
A(v) &= \frac{1}{2}\left\{u\int\limits_{0}^{\infty}\exp\left[-bu\right]\,du - \int\limits_0^\infty\left(\int\exp\left[-bu\right]\,du\right)\frac{d}{du}u\,du\right\}\\
A(v) &= \frac{1}{2}\left\{ u\left.\frac{\exp\left[-bu\right]}{-b}\right|_0^\infty - \int\limits_0^\infty -\frac{\exp\left[-bu\right]}{b}\,du\right\}\\
A(v) &= \frac{1}{2}\left\{\left.\frac{-u\exp\left[-bu\right]}{b}\right|_0^\infty + \frac{1}{b} \int\limits_0^\infty\exp\left[-bu\right]\,du\right\}\\
A(v) &= \frac{1}{2}\left\{\left.\frac{-u\exp\left[-bu\right]}{b}\right|_0^\infty + \frac{1}{b}\left.\frac{\exp\left[-bu\right]}{-b}\right|_0^\infty\right\}\\
A(v) &= -\frac{1}{2}\left\{\left.\frac{u\exp\left[-bu\right]}{b}\right|_0^\infty + \frac{1}{b^2}\left.\frac{\exp\left[-bu\right]}{1}\right|_0^\infty\right\}
\end{aligned}
\end{equation*}Working out the limits, we get
\begin{equation}
\begin{aligned}
A(v) &= -\frac{1}{2}\left\{0 + \frac{1}{b^2}\left(0 - 1\right)\right\}\\
A(v) &= \frac{1}{2b^2}
\end{aligned}
\end{equation}Substituting equation $(5)$ back in equation $(3)$, we get
\begin{equation*}
\begin{aligned}
\langle v\rangle &= 4\pi \left(\frac{b}{\pi}\right)^{3/2} \frac{1}{2b^2}\\
\langle v\rangle &= 2\frac{\pi}{\pi^{3/2}} \frac{b^{3/2}}{b^2}\\
\langle v \rangle &= \sqrt{\frac{4}{\pi b}}
\end{aligned}
\end{equation*}Substituting the value of $b$ back, we obtain the following expression,
\begin{equation}
\begin{aligned}
\langle v\rangle &= \sqrt{\frac{4}{\frac{m\pi}{2k_B T}}}\\
\langle v\rangle &= \sqrt{\frac{8k_B T}{m\pi}}
\end{aligned}
\end{equation}
great work. it would be very helpful for students.
Thank you for your kind response.