aqueous solution - Diffusion of CO2 and O2 in Water/Atmosphere System

Consider the following experiment:

• One reservoir of pure water is interfacing the atmosphere at standard condition.


• Both are at rest and at the same temperature.



• Both have no internal gradients (ie: constant concentrations, temperature, ...)


• The interface between them is a perfect flat surface (for sake of simplicity).


• The dissolved $\ce{CO2}$ in the water has a concentration somewhere between 0-50ppm.


• The dissolved $\ce{O2}$ in the water has a concentration somewhere between 0-11ppm.

Questions
in order of importance

1. How to mathematically model the $\ce{CO2}$ and $\ce{O2}$ rate of change per unit area (flux) in the water and in the atmosphere considering only diffusion?


2. Suppose the atmosphere is at constant velocity relative to the water reservoir. How this would change the mathematical model above?

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Notes: Analytical methods would be appreciated instead of 'heavy' computer simulations.

Answer

Such questions are relatively standard in discussing transport phenomena, so I'll suggest Bird, Stewart, and Lightfoot's Transport Phenomena as a good reference.

We assume a flat plane of contact infinite in extent. We assume that the bulk atmosphere has a constant concentration $c_A$ of some species of interest, and that the bulk liquid has a constant concentration $c_B$ of that species. The species is only allowed to mix between the two bulk phases for times $t > 0$. Then the problem reduces to the one-dimensional diffusion equation (Fick's first and second laws) $$\partial_t c(z,t) = D\,\partial_z^2 c(z,t)$$ subject to the initial condition $$c(z,0) = \frac{c_A+c_B}{2} + \frac{c_B-c_A}{2}\text{sgn}(z),$$ and the boundary conditions $$c(-\infty, t) = c_A, \quad c(+\infty, t) = c_B,$$ where $D$ is the species' diffusion constant and $\text{sgn}(z)$ is the signum function.

For ease of solution, make the substitutions $$\bar{c}(z,t) := c(z,t) - \frac{c_A+c_B}{2}, \quad \Delta c := \frac{c_B-c_A}{2}, \quad \phi(z,t) := \frac{\bar{c}(z,t)}{\Delta c},$$ which yields $$\partial_t\phi(z,t) = D\,\partial_z^2 \phi(z,t), \quad \phi(z,0) = \text{sgn}(z), \quad \phi(\pm\infty, t) = \pm1.$$ By definition $\phi$ is a dimensionless quantity, so it must depend on the variables relevant in the problem in a dimensionless form, for otherwise $\phi$ would have units. The relevant variables in our problem are the diffusion constant $D$, the length $z$, and the time $t$.

The only dimensionless variable that arises from this set of quantities is $\eta := z/\sqrt{Dt}$, so we know $\phi(z,t) = \phi(\eta)$. Substituting this result into our diffusion equation and applying the chain rule yields the ordinary differential equation $$-\frac{\eta}{2}\phi'(\eta) = \phi''(\eta).$$ Integrating twice yields the result $$\phi(\eta) = \alpha + \beta\int_0^\eta\text{d}\eta'\,\exp\left(-\frac{\eta'^2}{4}\right),$$ and it remains to determine the constants $\alpha$ and $\beta$ from our boundary conditions. Actually, for standardization, we will first take $\eta' \to 2\eta'$ and $2\eta \to \eta$, so $$\phi(\eta) = \alpha + 2\beta\int_0^\eta\text{d}\eta'\,\exp(-\eta'^2).$$ By symmetry $\alpha = 0$, and by explicit evaluation $\beta = 1/\sqrt{\pi}$, and hence we have $$\phi(\eta) = \text{erf}(\eta), \quad \text{or} \quad \phi(z,t) = \text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ Going back and untangling all our substitutions, we end up with $$c(z,t) = \frac{c_A+c_B}{2}+\frac{c_B-c_A}{2}\text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ The flux then follows as $$J(z,t) = -D\,\partial_zc(z,t) = -D\,\frac{c_B-c_A}{\sqrt{4\pi Dt}}\exp\left(-\frac{z^2}{4Dt}\right),$$ and we are done.

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Regarding your second question, I don't believe the movement of one phase relative to the other will have any effect. As we've seen above, the xy-plane essentially drops out of the picture, and you won't be introducing any further gradients in chemical potential by translating one phase relative to the other, since the concentration of each phase is a function of $z$ only. Of course, if you had a finite area of contact, then things would be different. Perhaps see BSL 18.5, in this case.

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In addition, keep in mind that there's important chemistry that's being neglected in this simplistic model. For example, the dissolution of carbon dioxide in water follows not just from diffusion but also from reaction with water, forming bicarbonate ion and related species in solution. This model doesn't account for that.