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I would like to know the answer to the question "why do materials preventing heterogeneous nucleation of $CO_2$ aren't used for soda bottles and glasses?".

Two possible answers so far that I thought about: Either it is too hard to make such a material or polish it "near perfectly", or it would not be worth it. By "worth it" I mean the motivation below would be false.

The motivation to use such materials would be that bubbles of $CO_2$ would not form at all, because if I'm not wrong bubbles are forming on the walls of the bottle or glass thanks to heterogeneous nucleation due to microscopic cracks or "imperfections" (i.e. the crystal isn't plane. In a common glass the atoms aren't ordered like in a crystal and this favors heterogeneous nucleation) while homogeneous nucleation never occurs because the radius of the bubble required for it to occur is too big and so the probability that it's created spontaneously is almost nil. Therefore the bottle or glass would, I believe, only slowly lose $CO_2$ gas through the interface liquid/air which is due to diffusion and occurs because the chemical potential "$\mu$" of the soda is higher than the chemical potential of the air. So the process will end when there is no more $CO_2$ in the soda, if I assume that there's no $CO_2$ in the air which is a good approximation.

To sum up the motivation: no bubbles formed. The loss of $CO_2$ would be very slow and so we could drink soda with plenty of "gas" even if the bottle has been opened for a long while compared to what we're currently used to.

Now I would like to use some maths to show how much slower the rate of decrease of $CO_2$ would be if we were to use such bottle or glass, compared to a normal bottle or glass.

More precisely: let $c(t)$ be the concentration of $CO_2$ in function of time. To settle numbers I'll assume that when $c(t)=0.1\cdot c(0)$ then there is too few "gas" for the soda to be drank. I want to estimate by calculations via a model how much time it takes until this low concentration threshold is reached in both cases.

Let's assume the bottle or glass is a cylinder of 20 cm height, 3.5 cm radius. This means the area of the interface soda/air is $A_\text{interface soda/air} \approx 38.5 \text{ cm}^2$.

Model 1 (no heterogeneous nucleation occurs):

In this case we only have a diffusion equation for $c(t): \frac{dc(t)}{dt}=-rc(t)$ where $r$ is a positive constant proportional to $A_\text{interface soda/air}$, yielding a solution of the form $c(t)=c(0)e^{-rt}$. I can now solve for $t_c$ so that $c(t_c)=0.1\cdot c(0)$.

Model 2 (heterogeneous nucleation occurs):

In this case I have the same diffusion equation for c(t) except that I have a new term. I am unsure how to write it. I'll assume that there are 2 bubbles per $\text {cm} ^2$. The total area of the container is $A_\text{bottle}=2 \pi R \cdot 20 \text{ cm} + A_\text{interface soda/air} \approx 478 \text{ cm}^2$, so there are about 957 bubbles in total.

Again to simplify things I'll assume that a bubble is most of its existing time stuck on the glass rather than going up in the soda. So that nucleation only occurs to bubbles growing on the glass. I'll also assume that all bubbles start with a zero radius and all have the same maximum radius, say $\text{ 0.1 cm}$ before they detach from the wall and get replaced instantly by a $0 \text{ cm}$ radius bubble.

Now I believe the bubbles growth rate depend on the current value of $c(t)$, but I am not sure whether it is the rate of change of the volume or area or radius that depends linearly on $c(t)$. I'd appreciate a comment here. So that I can set up the differential equation for $c(t)$, solve it and compare it with the first model.

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    $\begingroup$ One might suggest that having a liquid supersaturated with CO2 in a glass in your hand while chatting up a pretty woman (or man) is not a desirable thing unless you like the look of a wet hand and pants... $\endgroup$ – Jon Custer Jan 15 '16 at 17:43
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    $\begingroup$ Hmm why would such a thing happen? It wouldn't be different than from opening a freshly opened soda bottle for the first time and drank right away. Am I wrong? $\endgroup$ – thermomagnetic condensed boson Jan 16 '16 at 23:04
  • $\begingroup$ I suspect any movement would cause some type of nucleation of the bubbles within the liquid (via cavitation or some other pressure induced nucleation mechanism) and all the polishing would be for naught. So I'd say it would not be worth it. $\endgroup$ – Vendetta Mar 23 '18 at 22:46

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