10
$\begingroup$

I am trying to understand the behavior of oxygen dissolved in seawater. I have always been under the impression that if a gas is in equilibrium between its gaseous and aqueous forms, that it will always have the same partial pressure in the air as in the water. For example, both air and air saturated seawater have a pO2 of 21.227 kPa (101.325 kPa x 20.95%). I recently, however, came across this tool from a company that creates dissolved O2 measuring devices which indicates that temperature has an effect on this relationship. It indicates that if atmospheric pressure is 101.3 kPa (pO2 in the atmosphere = 21.222 kPa) then the pO2 in seawater could be anything from 21.094 kPa at 0*C to 20.331 at 30*C.

If this is true, what causes this relationship? How can I calculate expected seawater pO2 given atmospheric pressure and temperature?

Bonus if you know of a function or package in R that could do this. Perhaps in the marelac package?

Improve this question
$\endgroup$
1
  • $\begingroup$ Turns out the issue I was dealing with has to do with the vapor pressure changing with temperature. $\endgroup$ – CephBirk Feb 11 '15 at 2:43
4
$\begingroup$

have always been under the impression that if a gas is in equilibrium between its gaseous and aqueous forms, that it will always have the same partial pressure in the air as in the water. For example, both air and air saturated seawater have a pO2 of 21.227 kPa (101.325 kPa x 20.95%).

That's a very bad impression, and your example is just not the case. Oxygen solubility in water at 0°C at one atmosphere of pressure is 14.6 mg/liter in fresh water, 11.2 mg/liter in salt water (35 g/kg salt). Those values correspond to 1036 pascals and 794 pascals, respectively.

Oxygen solubility in water is highly temperature and pressure dependent. Like most (but definitely not all) gases, oxygen solubility in water is greatest at 0°C.

Oxygen solubility comes very close to following Henry's law for typical pressures. Henry's law says that "at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." The constant of proportionality is not constant. It instead is a function of temperature and the nature of the liquid.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I agree with you that air saturated seawater holds 11.2 mg/l but could you explain how this corresponds to 794 pascals? $\endgroup$ – CephBirk Jan 27 '15 at 14:13
  • 1
    $\begingroup$ Actually, 795 pascals. $P=\rho R^\ast T = \rho \frac R m T$ Here's the calculation: wolframalpha.com/input/… $\endgroup$ – David Hammen Jan 27 '15 at 14:58
  • $\begingroup$ This is a bit of a paradigm shift for me. Do you understand, then, why the tool I reference above gives a different value from what you suggest? Unfortunately, the equations are locked so I cannot see the underlying logic, so I'm at a bit of a loss regarding how to reconcile these ideas... $\endgroup$ – CephBirk Jan 27 '15 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.