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I found two versions of the Diffusion coefficient, first:

$$D=\frac{\pi \lambda }{8}\overline{c}$$

Where $ \overline{c}$ ist the particles mean thermal velocity and $\lambda$ the particles mean free path. (Found in W. C. Hinds, Aerosol Technology. Wiley Interscience (1999). S.156)

and second the version from my lecture (also found on wikipedia):

$$D=\frac{ \lambda}{3} \langle v\rangle$$

Where $\langle v\rangle$ is the particles mean thermal verlocity.

Hinds gives two more equations (S.154):

$$c_{\text{rms}}=\sqrt{\frac{3kT}{m}}$$

$$\overline{c}=\sqrt{\frac{8kT}{\pi m}}$$

With $c_{\text{rms}}$ being the root mean square velocity of a particle, $k$ being the Boltzmann-constant, $T$ being the temperature and $m$ the particles mass. My first thought was that maybe there is a typo in the book or the lecture notes so I calculated

$$\frac{c_{\text{rms}}}{\overline{c}}=\sqrt{\frac{3\pi}{8}}$$

which led to

$$\overline{c}=\sqrt{\frac{8}{3\pi}}c_{\text{rms}}$$

so I replaced $\overline{c}$ in

$$D=\frac{\pi \lambda }{8}\overline{c}$$

just to come to

$$D=\frac{\pi \lambda }{8}\sqrt{\frac{8}{3\pi}}c_{\text{rms}}$$

Which (set $\langle v\rangle=c_{\text{rms}}$, due to the potential typo) because of the square root isn't

$$D=\frac{ \lambda}{3} \langle v\rangle$$

Where did I make a mistake?

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So to start out, you're looking at the RMS and mean of the Maxwell-Boltzmann distribution; you can look them up on Wikipedia, where they match Hinds' expressions.

In the two expressions you have, you see they'd cancel out were you to get rid of the square root. So, I believe $\lambda$ to be defined differently in your source material: In the one in Hinds, $\lambda = \bar{c}\tau$, whereas on the page on Wikipedia you found, I take it $\lambda = c_\text{rms}\tau$.

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