<blockquote>
  <blockquote>
    <p>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%).</p>
  </blockquote>
</blockquote>

<p>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.</p>

<p>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.</p>

<p>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.</p>

CC BY-SA 3.0
2015-01-27T14:13:22.613Z
I agree with you that air saturated seawater holds 11.2 mg/l but could you explain how this corresponds to 794 pascals?
CephBirk
CC BY-SA 3.0
2015-01-27T14:58:42.780Z
Actually, 795 pascals. $P=\rho R^\ast T = \rho \frac R m T$ Here's the calculation: http://www.wolframalpha.com/input/?i=convert+%2811.2+mg%2Fliter%29%28ideal+gas+constant%29%2F%2831.998+grams%2Fmole%29%28273.15+K%29+to+pascal
David Hammen
CC BY-SA 3.0
2015-01-27T23:46:35.430Z
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...
CephBirk
CC BY-SA 3.0
2015-01-27T12:55:36.020Z
2015-01-27T13:01:16.070Z
2015-01-27T13:01:16.070Z
David Hammen
<p>This is pretty old, but I wanted to correct the record since it is still a top result for this question:</p>
<p>Short answer: CephBirk is correct in both the initial approximation and the source of the minor deviations from that assumption. David Hammen's answer is way off track.</p>
<p>Long answer:
David Hammen's answer confuses solubility (which relates partial pressure and concentration) with the thermal effect of partial pressure equilibrium that CephBirk asked about. Moreover, even if that were the question, the ideal gas law is not appropriate for aqueous solutions, which is why his pO2 calculations are off by a factor of more than 20.</p>
<p>CephBirk's initial understanding is approximately right--pO2 (or pN2, pAr etc) is pretty close to constant with temperature when in equilibrium with a constant gas phase composition. And his comment is right on the money--the differences are caused by the vapor pressure of water, which increases with temperature and comprises a larger portion of the (wet) total pressure, leaving a smaller partial pressure for oxygen. This is one reason aquatic chemists often correct their readings for water vapor pressure and report the dry partial pressures (which are the same as the overlying gas, just as CephBirk expected). Extending this, pCO2 is a bit of a special case, because the speciation between dissolved inorganic carbon chemical species is also temperature dependent, such that the pCO2 of seawater (in equilibrium with a constant atmospheric partial pressure) actually increases slightly with temperature in net.</p>
<p>Why you should believe me: I'm an ocean chemist specializing in dissolved gases, with ~20 years experience in this field, and I regularly do these sorts of calculations.</p>

CC BY-SA 4.0
2024-11-21T01:36:51.510Z
2024-11-21T01:40:23.573Z
2024-11-21T01:40:23.573Z
Usmivka
<p>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 <a href="http://www.presens.de/fileadmin/user_upload/downloads/Tools_Utilities/Oxygen_Unit_Calculation_dv1_1.xls">this</a> 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.</p>

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

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

CC BY-SA 3.0
2015-02-11T02:43:07.370Z
Turns out the issue I was dealing with has to do with the vapor pressure changing with temperature.
CephBirk
<atmosphere><ocean><barometric-pressure>
Does temperature affect partial pressure of gas dissolved in water?
CephBirk
# Does temperature affect partial pressure of gas dissolved in water?

(Asked by: username_0 on 2015-01-26)

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](http://www.presens.de/fileadmin/user_upload/downloads/Tools_Utilities/Oxygen_Unit_Calculation_dv1_1.xls) 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?

**Tags:** atmosphere, ocean, barometric-pressure

> **Comments:**
> **username_0**: Turns out the issue I was dealing with has to do with the vapor pressure changing with temperature.  

---

## 2 answers

### Answer by username_1 on 2015-01-27. Score: 4

> > 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.

> **Comments:**
> **username_0**: I agree with you that air saturated seawater holds 11.2 mg/l but could you explain how this corresponds to 794 pascals?  
> **username_1**: Actually, 795 pascals. $P=\rho R^\ast T = \rho \frac R m T$ Here's the calculation: http://www.wolframalpha.com/input/?i=convert+%2811.2+mg%2Fliter%29%28ideal+gas+constant%29%2F%2831.998+grams%2Fmole%29%28273.15+K%29+to+pascal  
> **username_0**: 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...  

---

### Answer by username_2 on 2024-11-21. Score: 1

This is pretty old, but I wanted to correct the record since it is still a top result for this question:

Short answer: username_0 is correct in both the initial approximation and the source of the minor deviations from that assumption. username_1 username_1's answer is way off track.

Long answer:
username_1 username_1's answer confuses solubility (which relates partial pressure and concentration) with the thermal effect of partial pressure equilibrium that username_0 asked about. Moreover, even if that were the question, the ideal gas law is not appropriate for aqueous solutions, which is why his pO2 calculations are off by a factor of more than 20.

username_0's initial understanding is approximately right--pO2 (or pN2, pAr etc) is pretty close to constant with temperature when in equilibrium with a constant gas phase composition. And his comment is right on the money--the differences are caused by the vapor pressure of water, which increases with temperature and comprises a larger portion of the (wet) total pressure, leaving a smaller partial pressure for oxygen. This is one reason aquatic chemists often correct their readings for water vapor pressure and report the dry partial pressures (which are the same as the overlying gas, just as username_0 expected). Extending this, pCO2 is a bit of a special case, because the speciation between dissolved inorganic carbon chemical species is also temperature dependent, such that the pCO2 of seawater (in equilibrium with a constant atmospheric partial pressure) actually increases slightly with temperature in net.

Why you should believe me: I'm an ocean chemist specializing in dissolved gases, with ~20 years experience in this field, and I regularly do these sorts of calculations.


---

