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I sort of know how carbonated beverages are carbonated: a lot of $\ce{CO2}$ gets pushed into the liquid, and the container is sealed. There are at least two things I don't know. First, how much carbon dioxide is actually dissolved in the liquid? Second, what is the resulitng partial pressure of $\ce{CO2}$ in the headspace and the total pressure in the headspace? I'm interested in cans, plastic bottles, and glass bottles. I know from experience that there is some variation among manufacturers even for the same beverage, so I will be happy with general numbers or a good estimate.

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up vote 7 down vote accepted

General estimates have placed a can of Coca-Cola to have 2.2 grams of $\ce{CO2} $ in a single can. As a can is around 12 fluid ounces, or 355 ml, the amount of $\ce{CO2}$ in a can is:

$$\text{2.2 g} \ \ce{CO2}* \frac{\text{1 mol} \ \ce{CO2}}{\text{44 g} \ \ce{CO2} } = 0.05 \ \text{mol}$$

$$ \text{355 mL} * \frac{\text{1 L}}{\text{1000 mL}} = 0.355 \ \text{L} $$

So here we can see we have about 0.05 mol/0.355 L or about 0.14 mol of carbon dioxide per liter of soda. Of course this value varies by manufacturer, type of drink, container, etc.

Looking at Wikipedia, inside Coca-Cola is:

Carbonated water, Sugar (sucrose or high-fructose corn syrup depending on country of origin), Caffeine, Phosphoric acid, Caramel color (E150d), Natural flavorings

A can of Coke (12 fl ounces/355 ml) has 39 grams of carbohydrates (all from sugar, approximately 10 teaspoons), 50 mg of sodium, 0 grams fat, 0 grams potassium, and 140 calories.

Thus, we can calculate the pressure of $CO_2$ gas using the Ideal Gas equation if we store our coke at, say, 20 Celsius:

$$ P = \frac{nRT}{V} $$ $$ P = \frac{\text{0.05 mol} * \text{0.08206} \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K} } * \text{293.15 K}}{\text{0.355 L}}$$ $$ P = 3.39 \ \text{atm} $$

According to this website,

On average, the 12 ounce soda cans sold in the US tend to have a pressure of roughly 120 kPa when canned at 4 °C, and 250 kPa when stored at 20 °C.

$$\text{3.39 atm} * \frac{\text{760 torr}}{\text{1 atm}} * \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 342.66 \ \text{kPa}$$

Water vapor exerts it's own partial pressure. Looking at standard tabulated values for water vapor pressure, water exerts a pressure of 17.5 torr at 20 Celsius.

$$\text{17.5 torr}* \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 2.3275 \ \text{kPa} $$

Knowing that our total pressure is the sum of all our pressures:

$$P_{total} = \text{342.66 kPa} + \text{2.3275 kPa} = 344.99 \ \text{kPa}$$

Here, we are roughly about 100 kPa off from the data provided by the website. This is just an approximation. A more accurate way would be to calculate the moles of each product inside the soda, and knowing the total pressure or partial pressure of one of the parts, we can calculate the pressures more accurately. However, that information is proprietary. It's their secret recipe!


We have to determine the volume of the headspace - again, I am not sure of exact data - which is between 1/2 inch to 1 and 1/2 inches depending on the container and what it holds. I will assume that the headpsace occupies 6% of the total volume of the can which is 21.3 mL.

At manufacturing and at storage, the can is at different temperatures. Taking the above data, we'll say it is manufactured at $4^\circ C$ and say, stored at $20^\circ C$

Furthermore, when carbon dioxide is solubilized in water, it forms carbonic acid. I will neglect that as the ionization constant is small.

Assuming our 2.2 grams of carbon dioxide is the maximum amount of carbon dioxide that can be placed inside, some of the carbon dioxide is soluble in water while the rest exerts pressure inside the headspace to force the carbon dioxide inside the liquid. The pressure is necessary inside this closed volume as, once you open the cap, the carbon dioxide tries to achieve equilibrium.

$$ \ce{CO_2 (solution) <=> CO2 (g)} $$

In general, the solubility of gases increases at lower temperatures and decreases at higher temperatures. A notable exception is the noble gases. In regards to pressure, Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.

The solubility of carbon dioxide in water shifts according to Le Chatlier's principle.

For the solution to this problem, we would need to know several things about the manufacturing of soda's, or rather, more quantitative data. At manufacturing, the pressure is high and temperature low, so as much carbon dioxide as possible is solubilized in the water.

Once shipped out and stored, the pressure inside the headspace increases as the pressure is decreased from manufacturing, the carbon dioxide then leaves solution and enters the headspace. Increasing temperature also decreases solubility.

If you can provide me with more information, I will be more than happy to help. :)

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Why are you adding water vapor partial pressure to the $\ce{CO2}$ pressure inside the soda ? – mannaia Mar 9 '14 at 21:57
@mannaia A mixture of gases that do not react with one another behaves like a single pure gas. The total pressure results from the sum of the pressure of carbon dioxide and water vapor. If I wanted to find just the pressure exerted by carbon dioxide, I would find the difference of the total pressure and water vapor. – anosdsd Mar 9 '14 at 22:06
That's fine, but water vapor is above (headspace) while the $\ce{CO2}$ you take (2.2 g) is mainly below, dissolved in the soda. Shouldn't you calculate the $\ce{CO2}$ vapor pressure rising from $\ce{CO2}$ dissolved in the soda and add it to the water vapor pressure ? – mannaia Mar 9 '14 at 22:14
@mannaia Ah, I didn't know what headspace was. Apparently, the pressure inside is maximized in the headspace in order to force more $\ce{CO_2}$ in the drink. I'll need to look up some numbers for that however. – anosdsd Mar 9 '14 at 22:19

It depends from drink and time, according to S. Teerasong, Analytica Chimica Acta 668 (2010) 47–53 for a normal Cola is about $3.1 \frac{g_{CO_2}}{L}$ (take a look even to Glevitzky, Chem. Bull. "POLITEHNICA" Univ. (Timişoara) Volume 50 (64), 1-2,2005 but his esteem is very high).

Regarding the pressure in the headspace this is very difficult to esteem theoretically because you have to take in account not only the temperature but the whole system. :you can shake the bottle and increase the pressure dramatically.

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@G M Shaking the bottle does not increase the pressure.… – anosdsd Mar 9 '14 at 22:20
@Jun-GooKwak interesting do you have the link for the next month? Sincerely I'm still skeptic I think the author doesn't know the time the the equilibrium needs! Thanks a lot! – G M Mar 9 '14 at 22:34
@G M Hi, here is the link… – anosdsd Mar 9 '14 at 22:44
@Jun-GooKwak Thanks a lot for correcting me! I will edit my answer I can't vote more then one time for your answer but If I could I will do it gladly! – G M Mar 9 '14 at 23:00
@G M Haha! Thanks, no problem! I initially thought the same until I saw that article. Right now my answer is still incomplete as I have to refer to the volume of the headspace and the pressure therin. In doing so, I have to make quite a few assumptions! – anosdsd Mar 9 '14 at 23:02

Through an interesting approach, authors estimate $\ce{CO2}$ pressure inside carbonated beverages, by measuring the freezing point ($fp$) depression caused by $\ce{CO2}$. In other words, when $\ce{CO2}$ is dissolved in water, a solution is formed and the freezing point is lowered.

The molality of $m_{\ce{CO2}}$ in water can be obtained, according to:

$$ m_{\ce{CO2}}=\Delta{T_{fp}}/k_{fp} $$

with $k_{fp}$ for water $=-1.86\,^{o}C \, kg \, mol^{-1}$. For a particular brand of sparkling water, a value of $\Delta{T_{fp}}$ = $-0.42^oC$ was measured, thus yielding a value for $m_{\ce{CO2}}$ equal to $0.23$ $mol \, kg^{-1}$. This first result allows to estimate the mass of ${\ce{CO2}}$ dissolved in one liter of that kind of beverage.

Said mass of ${\ce{CO2}}$ is approximately equal to $10\,g$ per liter (or $3.6\,g$ per can content volume).

Applying Henry's law,

$$ P_{\ce{CO2}}=m_{\ce{CO2}}/k_{H} $$

with $k_{H}$ being Henry's law constant ($~0.077\,mol\,kg^{-1}\,atm^{-1}$ at $0^{o}C$), one finally gets the pressure $P_{\ce{CO2}}$ exerted by the gas over the liquid, at $0^{o}C$:

$$ P_{\ce{CO2}} = 3.0\,\text{atm} $$

that corresponds to a mass of about $0.1\,g$ of ${\ce{CO2}}$ inside the headspace volume of a can (about $15\,ml$), mass calculated via $PV=nRT$.

EDIT: for most carbonated beverages, $\Delta{T_{fp}}$ should fall at around $-0.2^oC$ (as reported by Brooker): that would give a $P_{\ce{CO2}} = 1.5\,\text{atm}$ at $0^{o}C$, that rises at about $3\,\text{atm}$ at $25^{o}C$.

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protected by jonsca Nov 5 '14 at 19:24

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