I mean, if I had the option, I would display the specific humidity for my nerdy atmospheric science friends. Conversion is easy. Copied here for brevity. I may submit a PR for this if I have time this weekend.

What we want: specific humidity, $q=\frac{w}{w+1}$
What we have: percent relative humidity, $\%RH=100\cdot \frac{e_v}{e_s}$

$e_s$ can be calculated from the Clausius-Clapeyron equation. The equation below has the latent heat of vaporization and gas constant of water substituted in (from Atmospheric Thermodynamics: Elementary Physics and Chemistry, North & Erukhimova (2009), page 110).

$$\begin{align} e_s(T)=2.497\times10^9e^{-5417/T} (hPa) \end{align}$$

So, given $T$ in Kelvin, we can get $e_s$ and then find $e_v=\%RH\cdot e_s \cdot 10^{-2} (hPa)$. Once we have that, and the current pressure reading $p$, $w$ can be calculated from this equation

$$w=\frac{e_vR_d}{R_v(p-e_v)}=\frac{287\cdot e_v}{461(p-e_v)}.$$

At this point, we can calculate $q$.

The only potential pitfalls are to ensure that $p$ and $e_v$ have the same units and that we calculate using Kelvin ($K=^\circ C+273.15$). 100 pascals = 1 hectopascal

$e_s$=saturation vapor pressure $hPa$
$e_v$=vapor pressure of water $hPa$
$m_v$=mass of water vapor
$m_d$=mass of dry air
$q$=specific humidity
$R_d$=gas constant for dry air = $287.0 J K^{-1} kg^{-1}$
$R_v$=gas constant for water vapor= $461.0J K^{-1} kg^{-1}$
$T$=temperature in $K$
$w$=mixing ratio

To display pressure in millibars, conveniently, 1 millibar = 1 hectopascal.

Add specific humidity option #46

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