Humidity Measures
In humidity: Calculate Water Vapor Measures from Temperature and Dew Point

Saturation Vapor Pressure $e_s$ {-}

Saturation vapor pressure $e_s$ is calculated from a given temperature $T$ (in $K$) by using the Clausius-Clapeyron relation. \begin{equation} e_s(T) = e_s(T_0)\times \exp \left(\frac{L}{R_w}\left(\frac{1}{T_0} - \frac{1}{T}\right)\right) (#eq:1) \end{equation} where $e_s(T_0) = 6.11 hPa$ is the saturation vapor pressure at a reference temperature $T_0 = 273.15 K$, $L = 2.5 \times 10^6 J/kg$ is the latent heat of evaporation for water, and $R_w = \frac{1000R}{M_w} = 461.52 J/(kg K)$ is the specific gas constant for water vapor (where $R = 8.3144621 J / (mol K)$ is the molar gas constant and $M_w = 18.01528 g/mol$ is the molar mass of water vapor). More details refer to @Shaman-Kohn:2009.

An alternative way to calculate saturation vapor pressure $e_s$ is per the equation proposed by @Murray:1967. \begin{equation} e_s = 6.1078\exp{\left[\frac{a(T - 273.16)}{T - b}\right]} \end{equation} where $\begin{cases} a = 21.8745584 \ b = 7.66 \end{cases}$ over ice; $\begin{cases} a = 17.2693882 \ b = 35.86 \end{cases}$ over water.

The resulting $e_s$ is in hectopascal ($hPa$) or millibar ($mb$).

Vapor Pressure $e$ {-}

When given dew point $T_d$ (in $K$), the actual vapor pressure $e$ can be computed by plugging $T_d$ in place of $T$ into equation \@ref(eq:1). The resulting $e$ is in millibar ($mb$).

Relative Humidity $\psi$ {-}

Relative humidity $\psi$ is defined as the ratio of the partial water vapor pressure $e$ to the saturation vapor pressure $e_s$ at a given temperature $T$, which is usually expressed in $\%$ as follows \begin{equation} \psi = \frac{e}{e_s}\times 100 (#eq:2) \end{equation}

Therefore, when given the saturation vapor pressure $e_s$ and relative humidity $\psi$, the partial water vapor pressure $e$ can also be easily calculated per equation \@ref(eq:2). $$ e = \psi e_s $$ The resulting $e$ is in $Pa$.

Absolute Humidity $\rho_w$ {-}

Absolute humidity $\rho_w$ is the total amount of water vapor $m_w$ present in a given volume of air $V$. The definition of absolute humidity can be described as follows $$ \rho_w = \frac{m_w}{V} $$

Water vapor can be regarded as ideal gas in the normal atmospheric temperature and atmospheric pressure. Its equation of state is \begin{equation} e = \rho_w R_w T (#eq:3) \end{equation}

Absolute humidity $\rho_w$ is derived by solving equation \@ref(eq:3). $$ \rho_w = \frac{e}{R_w T} $$ The resulting $\rho_w$ is in $kg/m^3$.

Mixing Ratio $\omega$ {-}

Mixing ratio $\omega$ is the ratio of water vapor mass $m_w$ to dry air mass $m_d$, expressed in equation as follows $$ \omega = \frac{m_w}{m_d} $$

The resulting $\omega$ is in $kg/kg$.

Specific Humidity $q$ {-}

Specific humidity $q$ is the ratio of water vapor mass $m_w$ to the total (i.e., including dry) air mass $m$ (namely, $m = m_w + m_d$). The definition is described as $$ q = \frac{m_w}{m} = \frac{m_w}{m_w + m_d} = \frac{\omega}{\omega + 1} $$

Specific humidity can also be expressed in following way. $$ \begin{equation} q = \frac{\frac{M_w}{M_d}e}{p - (1 - \frac{M_w}{M_d})e} (#eq:4) \end{equation} $$ where $M_d = 28.9634 g/mol$ is the molar mass of dry air; $p$ represents atmospheric pressure and the standard atmospheric pressure is equal to $101,325 Pa$. The details of formula derivation refer to Wikipedia.

Substitute $\frac{M_w}{M_d} \approx 0.622$ into equation \@ref(eq:4) and simplify the formula. $$ q \approx \frac{0.622e}{p - 0.378e} (#eq:5) $$ The resulting $q$ is in $kg/kg$.

Hence, by solving equation \@ref(eq:5) we can obtain the equation for calculating the partial water vapor pressure $e$ given the specific humidity $q$ and atmospheric pressure $p$.

$$ e \approx \frac{qp}{0.622 + 0.378q} (#eq:6) $$ Substituting equations \@ref(eq:1) and \@ref(eq:6) into equation \@ref(eq:2), we can get the equation for converting specific humidity $q$ into relative humidity $\psi$ at a given temperature $T$ and under atmospheric pressure $p$.