## 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. $\begin{equation} e_s(T) = e_s(T_0)\times \exp \left(\frac{L}{R_w}\left(\frac{1}{T_0} - \frac{1}{T}\right)\right) \tag{1} \end{equation}$ 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 and 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 (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 $\begin{equation} \psi = \frac{e}{e_s}\times 100 \tag{2} \end{equation}$

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 (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 $\begin{equation} e = \rho_w R_w T \tag{3} \end{equation}$

Absolute humidity $$\rho_w$$ is derived by solving equation (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. $\begin{equation} q = \frac{\frac{M_w}{M_d}e}{p - (1 - \frac{M_w}{M_d})e} \tag{4} \end{equation}$ 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 (4) and simplify the formula. $q \approx \frac{0.622e}{p - 0.378e} \tag{5}$ The resulting $$q$$ is in $$kg/kg$$.

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

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