# 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?

## 1 Answer

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 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.