Derivation of mean speed from Maxwell–Boltzmann distribution

I found that if a velocity of a gas follows the Maxwell–Boltzmann distribution, the mean velocity is given by


$$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}}$$


where $R$ is the gas constant, $T$ is temperature, and $M$ is molar mass. How would one obtain this result from the Maxwell–Boltzmann distribution?




If one has a probability density function $P(x)$, then the expectation value of a quantity $f(x)$ is given by


$$\langle f \rangle = \int f(x)P(x)\,\mathrm{d}x$$


evaluated over the limits of the probability density function, i.e. if your PDF runs from $-\infty$ to $\infty$, then those are your limits of integration.


In our case, the PDF is the Maxwell–Boltzmann distribution (denoted $P(v)$ here) and the quantity we want to find the expectation value of is simply the velocity, $v$. Since velocity can only be positive,\* the limits of integration are from $0$ to $\infty$.


Therefore, the mean velocity $\langle v \rangle$ is given by the integral


$$\langle v \rangle = \int\_0^\infty v P(v)\,\mathrm{d}v$$


where $P(v)$ is the Maxwell-Boltzmann distribution


$$P(v) = \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi v^2 \exp{\left(-\frac{mv^2}{2kT}\right)}$$


So:


$$\begin{align}
\langle v \rangle &= \int\_0^\infty v P(v)\,\mathrm{d}v \\
&= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \int\_0^\infty v^3 \exp{\left(-\frac{m}{2kT}v^2\right)}\,\mathrm{d}v
\end{align}$$


The integral can be evaluated using integration by parts repeatedly. The process is not interesting and you could consult [a table of standard integrals](http://www.colby.edu/chemistry/PChem/notes/Integral.pdf) to find the result:


$$\int\_0^\infty v^3 \exp{(-\alpha v^2)}\,\mathrm{d}v = \frac{1}{2\alpha^2}$$


Setting $\alpha = m/2kT$,


$$\begin{align}
\langle v \rangle &= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \frac{4k^2T^2}{2m^2} \\
&= \sqrt{\frac{8kT}{\pi m}} \\
\end{align}$$


$m$ here refers to the mass of one molecule, whereas $M$ in your question refers to the molar mass of the compound. They are related by $M = N\_\mathrm{A}m$, where $N\_\mathrm{A}$ is the Avogadro constant. Since $R = N\_\mathrm{A}k$, you can multiply top and bottom by $N\_\mathrm{A}$ to obtain the desired result


$$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}}$$




---


**Footnote**


\* If the assertion that velocity is non-negative is confusing, it's worth noting that the $v$ in the *Maxwell–Boltzmann distribution* $P(v)$ is not the vector quantity $\vec{v} = (v\_x, v\_y, v\_z)$, but rather the *magnitude* of the velocity $v = |\vec{v}| = \sqrt{v\_x^2 + v\_y^2 + v\_z^2}$. Part of the derivation of the Maxwell–Boltzmann distribution involves converting from the vector quantity to its magnitude: that's where the $3/2$ exponent comes from (it's actually $1/2$ per dimension in three dimensions), and also where the factor of $4\pi v^2$ comes from ($4\pi v^2$ is the formula for the surface area of a sphere: loosely speaking, this factor represents the "collection" of all points $(v\_x, v\_y, v\_z)$ which have the same magnitude $v$). A fuller explanation may be found in any typical physical chemistry textbook.


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