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Suppose I mix oil and water by stirring the liquid, then which state has the higher entropy, the mixed or the unmixed state?

Is this similar to:
When mixing coffee and milk, which state has the higher entropy, the mixed or the unmixed state?

Nat
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Rene Kail
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    If the fluids are miscible and chemically distinguishable, then the entropy in the mixed state is higher than the entropy in the unmixed state. – Chet Miller Aug 04 '18 at 12:14
  • is mixing reversible? – JEB Aug 04 '18 at 15:40
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    But what if the fluids are immiscible like oil and water? Consider the cycle: (1) starting with separated oil and water stir the mixture (stirrer or paddle work), (2) let stand and allow the mixture to spontaneously separate. Is the oil/water system now in its original state?. If so, wouldn't the change in entropy of the oil/water system be zero? But owing to the irreversible process (2) wouldn't the total entropy of the system plus surroundings increase? – Bob D Aug 04 '18 at 16:54
  • Possible duplicate: https://physics.stackexchange.com/q/421056/44126 and links therein. – rob Aug 04 '18 at 22:28

2 Answers2

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The entropy of a mole of a mixture is $$S=RT \sum_i x_i \log(x_i)$$ where $R$ is the universal gas constant, $T$ is the temperature and $x_i$ is the molar fraction of substance $i$; $\sum_i x_i =1$. So generally a mixture will have a higher entropy (and it will take energy to purify it into the component substances).

However, the change in free energy of a system $\Delta H = \Delta U-T\Delta S$ matters for whether a change can happen spontaneously: this happens if $\Delta H<0$ ($U$ is the internal energy). Normally a big increase in entropy ($\Delta S \gg 0$) is enough to cause this and can even overcome an increase in internal energy (as is the case of dissolving ammonium nitrate in water, which is endothermic and cools the mixture).

But if the increase in internal energy would be big enough then even a big entropy gain may not be enough to outweigh it. In the case of oil and water there is a surface energy along the interface between the substances that makes $\Delta U$ big and proportional to the interface surface area - hence converting oil and water into an even mixture is not favoured. Milk and water have far less surface energy so they mix easily. Add a surfactant to the oil and it can also be turned into an emulsion.

  • In this case are you discussing just the free energy of the water? If the system is isolated then there can be no change in internal energy (just energy transfer between the oil and the water). I just wanted to make sure I understand correctly. Also your definition of spontaneous only is true for a system at constant temperature and volume, but not all of the time, right? – BioPhysicist Aug 04 '18 at 18:58
  • The comparison would be between water+oil in separate layers and water+oil dissolved in each other. You can use the Gibbs free energy ($U+PV-TS$) when temperature, volume and pressure changes. – Anders Sandberg Aug 04 '18 at 19:26
  • But if you mixed the two together and let it separate naturally, the overall internal energy would be constant right? Since the entire system is isolated, spontaneous does mean entropy increase. This is why I am asking if you are just looking at the water, which is not an isolated system here. – BioPhysicist Aug 04 '18 at 20:13
  • @AaronStevens - If you mix the oil and water into an emulsion, you greatly increase the surface area and hence boost the internal energy. – Anders Sandberg Aug 04 '18 at 20:33
  • I'm talking about after you mix it and let it sit as an isolated system. The internal energy of the entire can't change after that. – BioPhysicist Aug 04 '18 at 20:41
  • @AaronStevens - Yes, the internal energy does change. Consider two oil droplets merging: the surface area decreases (from $2\times 4\pi r^2$ to $4 \pi 2^{2/3} r^2$ if they are equal sizes). – Anders Sandberg Aug 05 '18 at 09:06
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Generally, the entropy will go up when you mix materials. Each bit can now be in more places, so there are more possible states of the final product. In the example of the coffee and milk, that mixture is complete and irreversible, so the entropy has gone up.

But there are lots of special cases:

  • If the materials are the sane, I.e. water and water, entropy doesn’t change: the situation hasn’t changed. Unless you also stirred vigorously, in which case you added energy making more motion states available, so entropy went up.

  • if the materials react with themselves or each other, that can increase or decrease thermal energy, which can cause the resulting entropy to increase a lot (combustion, I.e. hypergolic rocket fuels) or almost not at all (oil and water, which self-separate).

Edit: Given the back-and-forth in the comments, here's an image to clarify the point:

enter image description here

From right to left are four initial states at the top, and final states at the bottom. Imagine there's initially a barrier between the blue and yellow liquids, which is then removed and they're allowed to mix.

In the left-most case, they mix entirely; that's an irreversible process, and clearly entropy goes up.

In the right-most case, which is meant to represent oil on top and water on the bottom, nothing changes. The barrier can be put back; the situation is completely reversible. There need be no increase of entropy in that case even when the materials are mixed. (The two states in the middle are intermediate, in that they're not obviously reversible but oil-water separate still greatly reduced the entropy gain over the left-hand cast; that kind of situation has a greatly reduced entropy rise over the complete-mixing case)

It's been argued in the comments that when you mechanically mix something, you must add energy, and that adds entropy. That can happen, but it need not: If the components are in a constant-temperature environment, for example, the entropy of the mixture (remember the original question is asking about the entropy of the liquids, not of the universe) is unchanged by adding mechanical energy, because the energy of the mixture is unchanged after that mechanical energy turns to heat and exits in the the environment.

Bottom line: Most mixing greatly increases entropy because it's irreversible. The more irreversible, the more the entropy increase.

Bob Jacobsen
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  • Your characterization of the oil-water mixture is clearly wrong: given that oil and water self-separate, the mixture must have a lower entropy than the separated layers. If it didn't, then the self-separation would be a violation of the second law of thermodynamics. – Emilio Pisanty Aug 04 '18 at 14:54
  • @EmilioPisanty reference please? Put oil and water in the usual “two boxes horizontally adjacent with a barrier that’s then removed”. They merge, a spontaneous irreversible process. But they remain separated; entropy hasn’t increased as much as total mixing. If it had been a vertical stack, removing the barrier wouldn’t have chafed anything. – Bob Jacobsen Aug 04 '18 at 14:59
  • This is being discussed right now here: https://physics.stackexchange.com/questions/421056/if-entropy-is-a-measure-of-disorder-how-come-mixing-water-and-oil-finishes-in-a/421065?noredirect=1#comment943629_421065 – BioPhysicist Aug 04 '18 at 15:12
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    If you consider a system with oil and water molecules initially interspersed (somehow), you'll find london dispersion forces and non-negligible interaction between an induced dipole and the polar molecules. It's just that the state with both parts separated is preferable since the bonds between 2 polar molecules is stronger. –  Aug 04 '18 at 15:13
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    @EmilioPisanty I agree with your specific response that the oil-water separation has higher entropy, but I disagree with your general statement "the mixture must have a lower entropy than the separated layers. If it didn't, then the self-separation would be a violation of the second law of thermodynamics" It sounds like you are saying entropy can never decrease, which is not the case when your system is not isolated. Not every spontaneous process has to raise entropy in all parts of that process. – BioPhysicist Aug 04 '18 at 15:21
  • @AaronStevens Entropy can never decrease in a spontaneous process that happens in an isolated system that's not receiving work from the outside (or acting as a conduit between two external heat reservoirs at different temperatures). All of those conditions are satisfied for the separation of mixed oil and water, so the conclusion also holds. Or perhaps you're saying that the system is not isolated, and you have an interesting argument to share regarding why that's the case? – Emilio Pisanty Aug 04 '18 at 17:59
  • @EmilioPisanty You are exactly right, and I never said you were wrong. You need to read my comment again more carefully. I said you are right about the oil-water system. I am saying that your reasoning was not specific enough. You said since they are separating the entropy must be increasing. This argument is not true in general, which is what I was talking about. You brought up the sufficient assumptions in your new comment, so it's all good now :) – BioPhysicist Aug 04 '18 at 18:46
  • @AaronStevens I would ask you to read my initial comment with the same care - the logical structure and coverage are exactly the same. – Emilio Pisanty Aug 04 '18 at 19:43
  • @EmilioPisanty You do not make the specifications you made in your second comment in your first comment. You first comment says, without any specifications, (paraphrasing) "since the process happens, the entropy must be higher after it takes place." This is not always true without additional assumptions, of which you correctly specify in the second comment. I just didn't want others to be mislead. – BioPhysicist Aug 04 '18 at 20:11
  • @EmilioPisanty the mixture is not the initial state. The initial state is before you mix them. And, depending on how that mixing is done, it can be completely reversible. Hence no entropy change. – Bob Jacobsen Aug 04 '18 at 20:39
  • @BobJacobsen This is not true. When you mix the system you perform work and decrease the entropy, since the water molecules now have fewer configurations with the hydrogen bonds they can make. Then if you let the system sit as an isolated system the entropy increases. – BioPhysicist Aug 04 '18 at 20:49
  • @AaronStevens again, it depends on how. See earlier thought experiment. Yes, if you add energy you generally increase entropy. But how does that make the answer wrong? – Bob Jacobsen Aug 04 '18 at 21:08
  • @AaronStevens I don't see how the phase "you perform work and decrease the entropy" can be consistent with the 2nd law of thermodynamics. I can get that work (from e.g. an electrical store) without any entropy increase. So the process you're describing is decreasing the entropy of the Universe. I must be misunderstanding your point somehow, but I don't see it. – Bob Jacobsen Aug 04 '18 at 22:17
  • @BobJacobsen The second law deals with spontaneous entropy increases in isolated systems. One way to fight this locally in a system that is not isolated is to do work. So this does not violate anything. The second law does not say "entropy increases all the time no matter what" – BioPhysicist Aug 04 '18 at 22:46
  • @AaronStevens I disagree with that statement of the 2nd law, but no matter. A way to use work alone to reduce entropy in an isolated system would allow better-than-Carnot efficiency, so that must not be what you're making. Maybe I'm misunderstanding your argument about "fewer configurations with the hydrogen bonds they can make". Isn't that less than the $N R ln{V_o/V_i}$ increase in entropy due to number of physical states? – Bob Jacobsen Aug 04 '18 at 22:54
  • First, if you are doing work on the system then it is not isolated. Second, this does not violate any Carnot efficiency. This is how a refrigerator operates. It uses work to decrease entropy locally. Violating Carnot efficiency would mean converting all energy transferred in a heat flow completely into work. – BioPhysicist Aug 04 '18 at 23:34