Calculation of pO₂ in Water Equilibrated with a Mixture of Room Air and Nitrogen

Abstract

When gas A is dry air and mixed with an oxygen-free dry gas or gas mixture B (e.g., N₂), then the ${p_{{{\rm{O}}_{\rm{2}}}}}$ in water equilibrated with this gas mixture is

$$ {p_{{{\rm{O}}_{\rm{2}}}}} = {(_{\rm{b}}}p - {p_{{{\rm{H}}_2}}}{\rm{o,}}{T_{\rm{s}}}) \times {\phi _{\rm{A}}} \times \phi _{{{\rm{o}}_{\rm{2}}}}^*, $$

((1))

where _bp is barometric pressure [kPa], ${p_{{{\rm{H}}_{\rm{2}}}}}{\rm{o,}}{T_{\rm{s}}}$ is the saturation water vapor pressure at the temperature of the water sample, Φ_A is the volume fraction of gas A in the gas mixture leaving the gas mixing pump,

$$ {\phi _{\rm{A}}} = {{{p_{\rm{A}}}} \over {_{\rm{b}}p}}{\rm{and1 - }}{\phi _{\rm{A}}} = {{{p_{\rm{B}}}} \over {_bp}}, $$

((2))

$\phi _{{{\rm{o}}_{\rm{2}}}}^*$ is the volume fraction of oxygen in dry air which can be taken as constant $\phi _{{{\rm{o}}_{\rm{2}}}}^* = 0.20946$.