Does Shaking a Soda Can Increase Pressure?

[Astronuc]

Yes, really!

For example,

Carbonated beverages are supersaturated solutions of CO2.

http://www.Newton.dep.anl.gov/askasci/gen01/gen01191.htm

Thermodynamic equilibrium, when internal processes of a system cause no overall change in temperature or pressure. Wikipedia - http://en.wikipedia.org/wiki/Equilibrium

I would prefer a more reliable source than Wikipedia, but that'll have to suffice for now.

A 'sealed' bottle of carbonated beverage - left to itself - is in equilibrium. Shack it (an external influence) can release some gas - pressure increase. Then over time the CO₂ will diffuse/dissolve back into the liquid and internal pressure decreases to a new equilibrium.

I did not quantify the degree of supersaturation.

 

[f91jsw]

Yes, really!

For example,
http://www.Newton.dep.anl.gov/askasci/gen01/gen01191.htm

The web page you quote implies that carbonated beverages are supersaturated solutions of CO2 at atmospheric pressure, which is correct. At the elevated pressure in the unopened can I would argue that it is not supersatured. I guess it depends on how you define "supersaturated." Obviously a supersaturated system is not in its lowest energy state. After a long time the unopened soda will be in its lowest state (more or less) and thus it cannot be supersaturated.

A 'sealed' bottle of carbonated beverage - left to itself - is in equilibrium. Shack it (an external influence) can release some gas - pressure increase. Then over time the CO₂ will diffuse/dissolve back into the liquid and internal pressure decreases to a new equilibrium.

This is the core question of this discussion - is the process you describe above physical or unphysical? Intuitively I would say it's unphysical. Can you refer to the theory which explains how the equilibrium would change by shaking?

/J

 

[Bystander]

Yes, really!

For example,
http://www.Newton.dep.anl.gov/askasci/gen01/gen01191.htm
(snip)

f91jsw has read your link correctly; an open container of carbonated beverage exposed to the atmosphere is "supersaturated" with respect to the equilibrium state of the same beverage with the atmosphere. That is, the concentration dissolved in the beverage, 0.1m to 0.2m, is greater than the 15-20 micromolality at equilibrium.

In a sealed container, closed system, the equilibrium state is for a carbon dioxide activity of around 150-200 kPa (I've never gauged it). It is NOT supersaturated. The definition of the system is that the container is sealed, closed, not open to atmosphere, and the OP's question has to do with the response of the system to mechanical agitation.

 

[f91jsw]

Entropy argument: let's assume we have a situation where we have a sealed container where one part of the volume is occupied by pure water and the remaining part is occupied by pure CO2 at a certain pressure. If we now wait for a sufficiently long time the system will reach an equilibrium state where some of the CO2 will be dissolved in the water. The entropy has increased and we have reached thermodynamic equilibrium.

Now, what happens if we shake the system? According to some posters here some of the CO2 will return to gaseous state and the pressure will increase. I.e., the entropy will decrease again. How could it be possible that a random process such as shaking would lower the entropy of the system?

/J

 

[beluluk]

Now, what happens if we shake the system? According to some posters here some of the CO2 will return to gaseous state and the pressure will increase. I.e., the entropy will decrease again. How could it be possible that a random process such as shaking would lower the entropy of the system?
/J

I don't understand entropy very much , but i believe it's similar to shaking a bottle with water and oil in it. The water and the oil will mix temporarily. After a short period, the water will go down and the oil will go up. I've read nothing about its explanation, but i suspect gravity responsible for this.

 

[dav2008]

Now, what happens if we shake the system? According to some posters here some of the CO2 will return to gaseous state and the pressure will increase. I.e., the entropy will decrease again. How could it be possible that a random process such as shaking would lower the entropy of the system?

/J

Why would the entropy decrease if the CO₂ is going into the gaseous state?

 

[f91jsw]

Why would the entropy decrease if the CO₂ is going into the gaseous state?

If you accept that we start in a state of thermodynamic equilibrium the entropy would decrease by definition.

/J

 

[dav2008]

If you accept that we start in a state of thermodynamic equilibrium the entropy would decrease by definition.

/J

Not necessarily. It's the free energy that has to be minimized (F=U-TS)--Helmholtz Free Energy. The soda can in question is at constant temp/volume.

While a gas has a higher entropy than a liquid, there is an energy barrier that has to be overcome for the liquid to evaporate into a gas. (Or I guess in this case for a dissolved gas to...evaporate...into a gas. The word for undissolving doesn't come to mind right now).

It's this trade-off between internal energy and entropy that leads to an equilibrium where each is at a level where the free energy is minimized.

 

[Bystander]

If you accept that we start in a state of thermodynamic equilibrium the entropy would decrease by definition.

/J

Container sits in a room for a month, it's pretty close to equilibrium. Pick it up and start shaking, it is not; the action of shaking establishes pressure gradients throughout the container, a non-equilibrium condition. There are zones of pressure greater than the "sitting" equilibrium pressure, and zones of lower pressure; in the lower pressure volumes, the solution will begin to approach an equilibrium with the lower pressure by exsolving gas.

 

[f91jsw]

Container sits in a room for a month, it's pretty close to equilibrium. Pick it up and start shaking, it is not; the action of shaking establishes pressure gradients throughout the container, a non-equilibrium condition. There are zones of pressure greater than the "sitting" equilibrium pressure, and zones of lower pressure; in the lower pressure volumes, the solution will begin to approach an equilibrium with the lower pressure by exsolving gas.

One could also argue the exact opposite: in the zones of greater pressure the gas will dissolve in the liquid. Which effect wins? Will there be a net change?

/J

 

[f91jsw]

Not necessarily. It's the free energy that has to be minimized (F=U-TS)--Helmholtz Free Energy. The soda can in question is at constant temp/volume.

You are right of course. I think my main argument was trying to bracket the equilibrium state between extremes. The extremes being pure water/pure CO2 on one end, totally dissolved CO2 on the other. Neither of these states are obviously stable. Somewhere between these two states there will be an equilibrium. Will that equilibrium change in either direction by applying a random motion to the system? If yes, why would increased outgassing be the preferred direction of change?

/J

 

[Bystander]

One could also argue the exact opposite: in the zones of greater pressure the gas will dissolve in the liquid. Which effect wins? Will there be a net change?

/J

Covered this already --- remember? Rate of dissolution is proportional to interfacial area (nearly constant), and rate of exsolution is proportional to volume that can be underpressured (a function of shake rate, acoustic energy, etc.).