O2 saturation

A body of water that has absorbed oxygen and other gases from the environment until it has reached equilibrium is at 100% saturation. At this point the percentage of each gas in the water is equal to the percentage of each gas in the atmosphere, this is referred to as partial pressure. Thus the % saturation of water is equal to the ratio of the oxygen partial pressure in the water (P_O2) to the oxygen partial pressure in the air (P_O2air).

$//<![CDATA[ \begin{array}{l}Sat O_2 = \frac{P_{O2}}{P_{O_2air}}\end{array} //]]>$

The partial pressure of oxygen in the air is equivalent to the total pressure multiplied by the mole fraction of oxygen in the air. With the total pressure being the atmospheric pressure minus the vapour pressure.

$//<![CDATA[ \begin{array}{l}P_{O_2air} = x_{O_2} \cdot (P_{atm} - P_v)\end{array} //]]>$

The partial pressure of oxygen in the water is dependent on the concentration of oxygen (Cc),  the mole fraction of oxygen in the water at equilibrium (x_O2), the total pressure (P), the solubility of air in the water (Sol) and the moles in the gas determined by the ideal gas law.

$//<![CDATA[ \begin{array}{l}P_{O2} = Cc \cdot x_{O_2} \cdot \frac{(10.1325 - P_v)}{Sol} \cdot \exp{\bigg (\frac{V_m\cdot P}{R \cdot T_{abs}}}\bigg )\end{array} //]]>$

Solubility is calculated via Gordon and Garcia as:

$//<![CDATA[ \begin{array}{l}\begin{equation*} \begin{aligned} \text{Sol} = exp((&Ga_0 + Ga_1 \cdot Ts + Ga_2 \cdot Ts^2 + Ga_3 \cdot Ts^3 + Ga_4 \cdot Ts^4 + Ga_5 \cdot Ts^5) \\ + S \cdot (&Gb_0 + Gb_1 \cdot Ts + Gb_2 \cdot Ts^2 + Gb_3 \cdot Ts^3) + Gc_0 \cdot S^2) \end{aligned} \end{equation*}\end{array} //]]>$

where S is the salinity in PSU and Ts is defined as:

$//<![CDATA[ \begin{array}{l}Ts = ln(\dfrac{298.15 - T}{273.15 + T})\end{array} //]]>$

with T being the water temperature in °C and:

$//<![CDATA[ \begin{array}{l}\begin{equation*} \begin{aligned} Ga_0 & = 2.00856 \\ Ga_1 & = 3.22400 \\ Ga_2 & = 3.99063 \\ Ga_3 & = 4.80299 \\ Ga_4 & = 9.78188e-1 \\ Ga_5 & = 1.71069 \\ Gb_0 & = -6.24097e-3 \\ Gb_1 & = -6.93498e-3 \\ Gb_2 & = -6.90358e-3 \\ Gb_3 & = -4.29155e-3 \\ Gc_0 & = -3.11680e-7 \end{aligned} \end{equation*}\end{array} //]]>$

Air vapour pressure (in dbar) is calculated as:

$//<![CDATA[ \begin{array}{l}P_v = \dfrac{exp(52.57 - \dfrac{6690.9}{T + 273.15} - 4.6818 \cdot ln(T + 273.15))}{100}\end{array} //]]>$

Resulting with Saturation of O₂ defined as:

$//<![CDATA[ \begin{array}{l}Sat O_2 = \frac{x_{O_2} \cdot{(10.1325 - P_v)}\cdot Cc}{x_{O_2} \cdot (P_{atm} - P_v)\cdot {Sol}}\cdot \exp({\frac{V_m\cdot P}{R\cdot T_{abs}}})\end{array} //]]>$

Conversion of units:

$//<![CDATA[ \begin{array}{l}Sat O_2 = \frac{{(10.1325 - P_v)dbar}\cdot Cc (\mu mol / L)}{(P_{atm} - P_v)dbar\cdot{Sol (cm^3/dm^3)}}\cdot\frac{10^{-6}mol}{\mu mol}\cdot\frac{L}{dm^3}\cdot\frac{cm^3}{0.001L}\cdot22.414\bigg (\frac{L}{mol}\bigg )\cdot \exp\bigg ({\frac{V_m\cdot P}{R\cdot T_{abs}}}\bigg )\end{array} //]]>$

Final:

$//<![CDATA[ \begin{array}{l}Sat O_2 = \frac{{(10.1325 - P_v)}\cdot Cc}{(P_{atm} - P_v)\cdot{Sol}}\cdot 2.24457\end{array} //]]>$