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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?

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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 (as far as our model is concerned), 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

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

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  • 1
    $\begingroup$ A superb answer! $\endgroup$ – rainman Mar 25 '18 at 7:27

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