I am seeing two different formulas for the mixing ratio, which differ by a factor 1000. To be more specific:

$w = 0.622 \cdot (e/p)$


$w = 1000 \cdot 0.622 \cdot (e/p)$

where $e$ and $p$ are the water vapour pressure and the atmospheric pressure, respectively.

I assume the factor 1000 has to do with the fact that the mixing ratio is theoretically a ratio of masses, so probably the numerator can be expressed in grams and has to be converted to kilograms, but I'm confused because I don't see masses in these formulas.

What's the difference between the two ?


The factor of 1000 may be related to the expected units. The mass 'units' can be found in the derivation.

The full derivation of mixing ratio goes like this $$w=\epsilon\frac{e}{P}$$ since, by the ideal gas law, $$e=\rho_vR_vT$$ and $$P_d=\rho_dR_dT$$ then the mixing ratio is expressed as $$w=\epsilon\frac{\rho_vR_vT}{\rho_dR_dT}=\epsilon\frac{\rho_vR_v}{\rho_dR_d}$$ Since $\epsilon=\frac{R_d}{R_v}=0.622$ (see above hyperlink), $$w=\frac{\rho_v}{\rho_d}$$

To return back to the question 'what is the difference between your two equations?' The units reported are $g/g$ and $g/kg$ for the first and second equations, respectively.

  • $\begingroup$ Thanks for clarifying ! Although to be coherent it'd probably help to refer to the units of the first equation as $kg/kg$, so that the factor 1000 converts them to $g/kg$ $\endgroup$ – duff18 Feb 4 at 8:55

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