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

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


$$\begin{align} \langle v \rangle &= \int_0^\infty v P(v)\,\mathrm{d}v \\ &= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \int_0^\infty v^3 \exp{\left(-\frac{m}{2kT}v^2\right)}\,\mathrm{d}v \end{align}$$

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

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


* 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.