<p>The factor of 1000 may be related to the expected units. The mass 'units' can be found in the derivation.</p>
<p><a href="https://www.e-education.psu.edu/meteo300/node/519#:%7E:text=The%20water%20vapor%20mixing%20ratio,to%20within%20a%20few%20percent." rel="noreferrer">The full derivation</a> of mixing ratio goes like this
<span class="math-container">$$w=\epsilon\frac{e}{P}$$</span>
since, by the ideal gas law,
<span class="math-container">$$e=\rho_vR_vT$$</span>
and
<span class="math-container">$$P_d=\rho_dR_dT$$</span>
then the mixing ratio is expressed as
<span class="math-container">$$w=\epsilon\frac{\rho_vR_vT}{\rho_dR_dT}=\epsilon\frac{\rho_vR_v}{\rho_dR_d}$$</span>
Since <span class="math-container">$\epsilon=\frac{R_d}{R_v}=0.622$</span> (see above hyperlink),
<span class="math-container">$$w=\frac{\rho_v}{\rho_d}$$</span></p>
<p>To return back to the question 'what is the difference between your two equations?' The units reported are <span class="math-container">$g/g$</span> and <span class="math-container">$g/kg$</span> for the first and second equations, respectively.</p>

CC BY-SA 4.0
2021-02-04T08:55:48.350Z
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
CC BY-SA 4.0
2021-02-03T18:54:13.330Z
2021-02-03T18:54:13.330Z
BarocliniCplusplus
<p>I am seeing two different formulas for the mixing ratio, which differ by a factor 1000. To be more specific:</p>
<p><span class="math-container">$w = 0.622 \cdot (e/p)$</span></p>
<p>and</p>
<p><span class="math-container">$w = 1000 \cdot 0.622 \cdot (e/p)$</span></p>
<p>where <span class="math-container">$e$</span> and <span class="math-container">$p$</span> are the water vapour pressure and the atmospheric pressure, respectively.</p>
<p>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.</p>
<p>What's the difference between the two ?</p>

<atmosphere><geophysics>
Factor 1000 in mixing ratio formula
duff18
# Factor 1000 in mixing ratio formula

(Asked by: username_0 on 2021-02-03)

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 ?

**Tags:** atmosphere, geophysics


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## 1 answers

### Answer by username_1 on 2021-02-03. Score: 5

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

[The full derivation](https://www.e-education.psu.edu/meteo300/node/519#:%7E:text=The%20water%20vapor%20mixing%20ratio,to%20within%20a%20few%20percent.) 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.

> **Comments:**
> **username_0**: 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 $g/kg$  

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