Does Shaking a Soda Can Increase Pressure?

[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.).