-1
$\begingroup$

Suppose we have a source of O$_2$ bubble formation in water at specific nucleation sites, how can we estimate the water vapour pressure inside the O$_2$ bubbles?

I know that:

(1): $p_\mathrm{O_2}+p_\mathrm{H_2O} = p_\mathrm{tot}$

(2): $c_\mathrm{O_2} = K_\mathrm{O_2}\cdot p_\mathrm{O_2}$ (Henry's law)

where $c_\mathrm{O_2}$ is the concentration of $O_2$ in the liquid phase, and $p_x$ are partial pressures. How can I get $p_\mathrm{H_2O}$ probably as a function of $O_2$ gas pressure?

|improve this question
$\endgroup$
  • 1
    $\begingroup$ If I understand your question then you're looking in the wrong place for a solution. If the bubbles are 100% oxygen then the question has nothing to do with partial pressure in the gas phase and should instead be about the surface tension of the solvent and the pressure will vary in inverse proportion to the size of the bubble. You could try editing your question so it can be answered here. Or ask on physics.se . $\endgroup$ – We are Monica. Dec 31 '18 at 21:22
  • 1
    $\begingroup$ There is still a partial pressure, because there is also a vapor pressure of water right? So it is not 100 % oxygen $\endgroup$ – Guiste Jan 1 at 3:47
  • $\begingroup$ I will also post it on physics.se maybe someone can help me there: physics.stackexchange.com/questions/451473/… $\endgroup$ – Guiste Jan 1 at 3:47
  • $\begingroup$ Quite right about water partial pressure, but unless you're working at 99 celsius or in a partial vacuum then compared to surface tension it'll be tiny in a tiny bubble. $\endgroup$ – We are Monica. Jan 1 at 10:35
0
$\begingroup$

The oxygen is acting as essentially an inert gas holding water. The equilibrium vapour pressure of water will be less than its saturation vapour pressure due to the concave surface. The more concave (ie the smaller the bubble) then the lower will be equilibrium vapour pressure. This is a surface tension effect and is called (by some) as the Kelvin equation

RTln(Pr/Po) = 2g Vm/ r

Where R is the gas constant, T the absolute temperature, Pr is the vapour pressure inside the bubble, Po the vapour pressure on a plane surface, g the surface tension, Vm the molar volume of the liquid, r the radius of the bubble.

Note that the sign of r is negative for bubbles and positive for drops, which show an increase in vapour pressure to that of a planar surface

Reference D J Shaw introduction to colloid and surface chemistry 2nd edition page 57, ISBN 0 408 70021 1

Sorry for the book reference as the web seems to be a bit more complicated and has omissions.

|improve this answer
$\endgroup$
  • $\begingroup$ I agree with this. This gives however the pressure relative to a planar surface. Is there a way that the oxygen pressure could also change the water vapour pressure for a planar interface? $\endgroup$ – Guiste Jan 8 at 14:18

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.