# Factor 1000 in mixing ratio formula

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

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

and

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

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 ?

## 1 Answer

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.

• 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 $g/kg$ – duff18 Feb 4 at 8:55