Entropy change mixing oil and water question

[rockinrack]

My question is: when oil and water from separate containers are poured into the same container and given time to equilibrate, is the entropy change positive, negative, or zero?

I've been told that there is zero change in entropy because the water and oil do not mix at all. However, I suspect that there is a negative entropy change due to oil-water interactions at the oil-water interface. There will be interface tension (surface tension) in both the oil and water since the molecules at the interface because both species experience much greater cohesive forces than adhesive forces, but isn't it inevitable that there will be interactions between some small fraction of the billions of water and oil molecules at the interface? If there was some interaction, wouldn't this cause a negative change in entropy since water forms clathrate structures around hydrophobic regions of molecules?

So just to be clear, I'm asking:
1. What is the change in entropy when mixing oil and water after letting the system equilibrate?
2. Are there interactions between oil and water molecules at the oil-water interface?
3. If there are interactions, do they affect the entropy of the system?

Also, I'm new to physicsforums.com and I'm not totally sure of how to follow the template for posting an assignment that is totally theoretical and does not involve equations. I think I followed it for the most part, but I'm open to criticism!
Thanks!

 

[Chestermiller]

Some small amount of oil will dissolve in the water phase, and some small amount of water will dissolve in the oil phase. So, based on this, what is your conclusion?

 

[rockinrack]

Thank you for the reply! My conclusion is that there is a small decrease in entropy. When you say "dissolve," I take it to mean that some molecules will leave their own phase and become fully solvated by molecules of the other phase. It follows that at equilibrium, these dissolved molecules can be located anywhere in the solution. This seems totally plausible as an explanation for why there is a decrease in entropy when combining immiscible liquids, but please correct me if I am wrong about the nature of this dissolution.

I could also imagine that some molecules at the interface partially solvate molecules of the other phase while still remaining as part of the surface layer in their own phases. Does this occur and would it contribute to the decrease in entropy upon combining oil and water in the same container?

The basic question that I am very curious about is how the level of freedom of motion for molecules at the surface layer of there phase is affected by the identity of the substance at the interface. To make my question clear: Ignoring the additional freedom of motion that comes from the ability to vaporize, is the motion of water molecules at the water-air interface less restricted than the motion of water molecules at the water-oil interface?

Anyway, I completely understand if you want to move on from this topic, but any insight would be much appreciated.

 

[Chestermiller]

Thank you for the reply! My conclusion is that there is a small decrease in entropy. When you say "dissolve," I take it to mean that some molecules will leave their own phase and become fully solvated by molecules of the other phase. It follows that at equilibrium, these dissolved molecules can be located anywhere in the solution. This seems totally plausible as an explanation for why there is a decrease in entropy when combining immiscible liquids, but please correct me if I am wrong about the nature of this dissolution.

This is all correct.

I could also imagine that some molecules at the interface partially solvate molecules of the other phase while still remaining as part of the surface layer in their own phases. Does this occur and would it contribute to the decrease in entropy upon combining oil and water in the same container?

The basic question that I am very curious about is how the level of freedom of motion for molecules at the surface layer of there phase is affected by the identity of the substance at the interface. To make my question clear: Ignoring the additional freedom of motion that comes from the ability to vaporize, is the motion of water molecules at the water-air interface less restricted than the motion of water molecules at the water-oil interface?

What's happening at the interface can't significantly count in the entropy change because, for the same amounts of oil and water, you can make the interface area as small as you please, and it won't affect the amount of oil that dissolves in the water and the amount of water that dissolves in the oil.

Regarding the ability of water to mix in air compared to water dissolving in the oil, the water vapor can get into the air with very little difficulty, but, in the case of the oil, the oil molecules tend to attract each other and prevent all but a minimum amount of water from entering.

 

[rockinrack]

I just read that when oil is mixed with water, the systems equilibrates so that the surface area of the oil-water interface is minimized. This occurs because the conformation with the lowest surface area is the most energetically favorable, and it is energetically favorable because it introduces the least possible amount of order into the system. Therefore, increased surface area is related to increased order in the system. It seems to me that the interactions between oil and water at the interface must be causing increased order and therefore, increased entropy.

The only other possible explanation for the system equilibrating by minimizing the surface area of the oil-water interface is that somehow, minimizing the interface releases heat. It seems much more likely to me that the interactions between oil and water molecules at the interface introduce order.

Any thoughts on this? I can see why you think that the interface doesn't have anything to do with entropy because it's true that the area doesn't affect the solubility of each molecule in the other. But the fact that the equilibrium state has a minimized surface area really seems to indicate an impact of the interface.

 

[Chestermiller]

I just read that when oil is mixed with water, the systems equilibrates so that the surface area of the oil-water interface is minimized. This occurs because the conformation with the lowest surface area is the most energetically favorable, and it is energetically favorable because it introduces the least possible amount of order into the system. Therefore, increased surface area is related to increased order in the system. It seems to me that the interactions between oil and water at the interface must be causing increased order and therefore, increased entropy.

The only other possible explanation for the system equilibrating by minimizing the surface area of the oil-water interface is that somehow, minimizing the interface releases heat. It seems much more likely to me that the interactions between oil and water molecules at the interface introduce order.

Any thoughts on this? I can see why you think that the interface doesn't have anything to do with entropy because it's true that the area doesn't affect the solubility of each molecule in the other. But the fact that the equilibrium state has a minimized surface area really seems to indicate an impact of the interface.

Well, the surface must have some small effect on entropy (smaller than the dissolution effect) because, if you shake a container of oil and water, you generate lots of surface, but the globules will eventually coalesce into a minimum surface area configuration, ultimately with water surrounding oil or oil surrounding water. So a spontaneous process occurs, which entails increase in entropy. What is involved here is surface tension and viscous deformation. If you wanted to figure out what the change in entropy would be for the spontaneous reduction in surface area, you would have to be able to either devise a reversible path between the initial state and the final state (in such a way that you control manually what is happening so it occurs quasistatically) and determine the integral of dq/T for that path, or you would have to solve the fluid mechanics (transport) equations to calculate the local rates of entropy generation by viscous dissipation, and integrate that over the volume of the system. Either way, that's not typically considered the significant part of the thermodynamic changes, since, even in distillation towers where there is substantial bubbling and in liquid-liquid extraction operations where the are lots of globules, the systems are typically modeled assuming phase equilibrium on each contact stage (with no allowance for surface effects).

Chet

 

[rockinrack]

Interesting. I was thinking that the formation of oil globules suggests a significant difference in entropy compared to the oil spread evenly sitting on top of the water in a flat layer because the less dense oil must push itself down into the water in order to establish the globule with the lowest interface area, displacing some of the the denser water. But the amount of work required to form the globule must be very small. Thanks for all your help, Chet!

 

[Chestermiller]

Interesting. I was thinking that the formation of oil globules suggests a significant difference in entropy compared to the oil spread evenly sitting on top of the water in a flat layer because the less dense oil must push itself down into the water in order to establish the globule with the lowest interface area, displacing some of the the denser water. But the amount of work required to form the globule must be very small. Thanks for all your help, Chet!

The work reducing the surface is mostly a reversible effect (in my judgement). It's the viscous deformation of the liquids into their new configurations that generates most of the entropy change (again, in my judgement).

Also, in terms of considering densities (and gravitational effects), the consolidation of surface area would occur even in the absence of gravitational effects.