Mixing of two different fluids is associated with an increase of entropy. Conversely, separation of two gases must be associated with a decrease of the entropy of the two fluids.

Is there a minimum requirement imposed by this thermodynamic constraint on the minimum energy needed to separate two fluids?

  • $\begingroup$ I am curious about the answers. The only meaningful nonzero result could be $\delta E = T\,\delta S$ where $T$ is the absolute temperature and $\delta S$ is the change of the entropy from mixing. However, there are many inherent subtleties. First of all, you don't mean a decrease of energy - the total energy is conserved during mixing as well as unmixing - you mean some work done by the other systems. But it's subtle what you count and what you don't. Moreover, if the molecules of 2 colors are just a bit different, I think you may always separate them for free. $\endgroup$ May 3, 2011 at 17:39
  • $\begingroup$ Thanks for the remarks, indeed I mean work done by the other systems. My intuition tells me that separation for free seems wrong, but intuition is not always to correct. $\endgroup$
    – Whelp
    May 9, 2011 at 21:51
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    $\begingroup$ Oil and vinegar separate "for free". The energy comes from attractions between the molecules themselves, no energy needs to be provided externally by the chemist. In this case, the entropy lowering from demixing is more than compensated by the entropy-raising effect of heat released to the outside when the molecules attracted to each other are allowed to come together and lower their enthalpy. $\endgroup$ Jun 26, 2011 at 18:09
  • $\begingroup$ I'm curious if this has something to do with the concept of a separative work unit (SWU) in Uranium enrichment. I think a major problem, however, is that different physical methods could reduce the required effort drastically. SWU, if I understand correctly, applies to using the mass difference in a mechanical system to separate them. If you use electrochemical mechanisms, it would be hard to apply a consistent minimum required work concept. $\endgroup$ Jul 14, 2011 at 14:40
  • $\begingroup$ @Steve B ""Oil and vinegar separate "for free". "" Because Oil and vinegar do not mix in the first place, this is not relevant. $\endgroup$
    – Georg
    Jul 14, 2011 at 15:39

2 Answers 2


This question can be answered, but only with the addition of some extra assumptions. What we know for sure is that the entropy of the universe as a whole cannot decrease, and hence fluids can't be unmixed without increasing the entropy of some other system by an amount equal to $\Delta S_\text{mixing}$. This doesn't immediately tell us anything about work.

However, let us now assume we have to hand a source of mechanical work, and a large heat reservoir at temperature $T$. I'll assume that we'll use the work (somehow) to unmix the fluids, and that any excess heat that gets generated will be dumped in the heat reservoir.

In general the mixed and unmixed states of the fluid system will have different energies. I'll call the difference $\Delta H$, with positive $\Delta H$ meaning that the unmixed state has a higher energy than the mixed one. (I'm guessing that for real miscible fluids this is usually a small positive value, but it needn't necessarily be.)

We'll assume it takes us an amount $W$ of work to unmix the fluids. $W$ has been lost from the work source, and $\Delta H$ has been gained from the fluid system, so the first law says that the energy of the heat bath must increase by $W-\Delta H$.

We can now calculate the total change in entropy. The fluids' entropy has decreased by $\Delta S_\text{mixing}$, and the heat bath's entropy has increased by $(W - \Delta H)/T$, so the total is

$\Delta S_\text{total} = \frac{W - \Delta H}{T} - \Delta S_\text{mixing} \ge 0,$


$W \ge T\Delta S_\text{mixing} + \Delta H.$

Changing this to an equality gives you the minimum amount of work required to unmix the fluids. But note that the expression involves $T$, which is the temperature of the heat bath that I assumed to exist. Without the heat bath there would be nowhere for the energy from the work source to go once it's used up.

Also note that $T$ is not necessarily the temperature of the mixed fluids, which could be different from that of the heat bath. I didn't need to make any assumptions about the fluids' temperature in order to work out the above. In practice you'd usually assume the fluids to be in contact with the heat bath, so that the two temperatures are in fact equal. In this case (and only in this case), the minimum work required is equal to the difference in Helmholtz free energy between the mixed and unmixed states.

It's possible for this minimum bound to be negative, meaning that you can actually get work out of the system by unmixing the fluids - but in this case the fluids would be immiscible, so the mixed state would be the unstable one.

Finally, let's make the additional assumption that the fluids are gases and that we have two very special semi-permeable pistons to hand, one at either end of the container. One is permeable to the first gas but completely blocks the second, and the other is permeable to the second gas but blocks the first. We can now very slowly push the two pistons in opposite directions until they meet in the middle, while holding the system in contact with a heat bath. If we do it slowly enough we will have reversibly unmixed the two gases by something similar to reverse osmosis, using an amount of work equal to the minimum $W$ calculated above. (I got that last idea from this brilliant paper on the entropy of mixing by Edwin Jaynes: http://bayes.wustl.edu/etj/articles/gibbs.paradox.pdf)

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    $\begingroup$ A couple of extra points that came to my mind when reading this back again. Firstly, I've assumed the mixed and unmixed states have the same volume. This is reasonable for gases but often not for liquids. If you instead assume that the pressure is kept constant then the expression becomes $W \ge T\Delta S_\text{mixing} + \Delta H - p\Delta V$. If you assume that the fluids' temperature is the same as the heat bath temperature, this is the difference in Gibbs free energy rather than Helmholtz free energy. $\endgroup$
    – N. Virgo
    Jan 14, 2012 at 13:17
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    $\begingroup$ Secondly, it's worth mentioning that for ideal gases, $\Delta H$ and $\Delta V$ would both be 0, so the minimum amount of work needed to unmix them is simply $T\Delta S_\text{mixing}$. $\endgroup$
    – N. Virgo
    Jan 14, 2012 at 13:20
  • $\begingroup$ I have some further questions regarding this topic and your answer here: physics.stackexchange.com/questions/137990/… $\endgroup$
    – mart
    Oct 2, 2014 at 19:30

If you are interested in minimal mechanical work required to move the system from one state to another it can be roughly arbitrary.

Indeed, during the process the system should change its state (and so its entropy and energy), however this change can be due to mechanical work or heat.

You can just cool the mixture so the constituents will separate (like cooling air). Maybe it is also valid for liquids --- one of the constituent will turn solid earlier. No mechanical work is required.

All you can say for sure is the change of the entropy of the outside world --- it is no less then the difference of entropies of mixed and unmixed states (2nd law). And of course you are sure about the energy change of the outside world --- it is exactly the energy difference of mixed and unmixed states (1st law).

You can't say anything specific about work or heat required, they are not functions of state. Functions of state are entropy and energy, they are specified.


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