To calculate this you need to know the solubility of methane in water at the temperature of the water. You can get this from a website called "The Engineers Toolbox".
Alan's answer is quite good. It is worth noting that salinity, and atmosphere pressure are effected on the solubility. I had done for oxygen.
You need to first convert your molar concentrations to a partial pressure so that you can ratio this to the methane partial pressure in air to get the saturation value.
To do this you need to divide your concentration by the methane solubility at your sample temperature and salinity. You will need to obtain methane solubility data in molar units per atmosphere . Weisenberg and Guinasso (1979) give Bunsen solubilities, and equations for calculating these at any given temperature and salinity. These are volumetric units so you must convert to molar units. You can do this reasonably easily if you assume (i) methane is an ideal gas and (ii) that at standard temperature and pressure 1mol of an ideal gas occupies 22.4 litres.
Next adjust the ”molar volume” of 22.4 liters to the equivalent at your sample temperature using the general gas law: V/T = constant. Divide your calculated Bunsen solubility by this molar volume value to give the equivalent solubility in units of mols per liter per atmosphere.
As concentration = solubility x partial pressure, divide your sample concentrations by the solubility you just calculated taking into account to adjust for the fact you have nmols per liter (convert these to mols) and you should end up with a partial pressure. If you want to you can adjust for non-ideal behaviour of methane but you will need to look that up! All you need do now is compare your in situ partial pressure to the partial pressure of methane in air to get the percent saturation.
Dear Robert,
Thank you so much for the detailed reply. This question is raised on behalf of a PhD student working within our group and your answer was very helpful. Thank you once again.
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