Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Suppose that we have two fluids $A$ and $B$ in a container $\Omega$, and we notice that $A,B$ do not mix.

  • Can you pleas explain to me what is the cause of this property?
  • What properties of the two fluids create the superficial tension between them?
  • Is the superficial tension between the two fluids independent of the 'sizes' of the two considered fluids? I mean, if we have a big container, with $100l$ of $A$ and $120l$ of $B$ and a small container with $40l$ of $A$ and $30l$ of $B$, the superficial tensions are the same?
  • Do walls of the container $\Omega$ influence the final equilibrium position of the considered fluids?

I know that I wrote many questions above. If you can, please give me some good references to understand better the given phenomena. If you can, please give references that do not contain too advanced physics, since I'm kind of a beginner in physics. (I'm a math student) Thank you.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

To see if a process will take place you need to calculate it's Gibbs free energy $\Delta G$. This is defined as:

$$\Delta G = \Delta H - T\Delta S$$

The quantity $\Delta H$ is the Helmholtz free energy and for liquids and solids is roughly the amount of energy released ($\Delta H$ is negative if energy is released and positive if energy is absorbed). $\Delta S$ is the entropy change for the process. For precise definitions of these quantities have a look on Wikipedia.

The equilibrium constant for the process is the exponent of the Gibbs free energy. If $\Delta G$ is large and negative the process goes to approximately 100% completion while if $\Delta G$ is large and positive the process hardly takes place at all.

Anyhow, for the mixing of two liquids the entropy change is always large and positive so it tends to make $\Delta G$ large and negative and this favours mixing. For two liquids to be immiscible the molecules of the two liquids must repel each other so strongly that the enthalpy of mixing overcomes the entropy.

As an example take water and mineral oil. Water hydrogen bonds very strongly to other water molecules but hardly at all to oil molecules. To mix oil and water you have to put energy in to break all those hydrogen bonds but you get no energy back when the water and oil mix. That makes $\Delta H$ large and positive and this disfavours mixing.

The surface tension arises from the same physics as the immiscibility. Suppose you want to increase the area of the interface. To do this you need to move water molecules from the bulk to the oil/water boundary. But this costs energy because in moving a water molecule to the bounday you break water-water bonds and replace them with water-oil bonds. Since it takes energy to increase the area of the boundary this means there is a force, and this is the surface tension.

The surface tension is a force per unit length (because it's an energy per unit area) and it is independant of the area of the interface. It depends only on the two liquids at the interface.

The walls of the container can have an effect because the liquids will generally interact differently with the container, but normally the surface tension is dominant and it would only be in very thin tubes that you'd see an effect from the walls.

You asked about articles on this subject, and as usual Wikipedia has some excellent articles. have a look at http://en.wikipedia.org/wiki/Entropy_of_mixing and http://en.wikipedia.org/wiki/Surface_tension

share|improve this answer
    
Great answer to read, but I struggled with this: "moving a water molecule to the bounday you break water-water bonds and replace them with oil-oil bonds", should that be "replace them with oil-water bonds"? It seems odd to talk about a bubble of water, and that distorting it (away from sphericity) would result in more oil-oil bonds. I would think the opposite would be the case. It would result in fewer oil-oil bonds because more oil molecules are now busy with water partners. –  AlanSE Feb 6 '12 at 17:51
    
Oops, quite correct, that was a a typo. I've corrected my answer and credited you in the "reason for editing". –  John Rennie Feb 6 '12 at 17:54
    
Thank you very much for your answer. Can you please explain to me in a few words why the entropy of mixing is always large and positive? –  Beni Bogosel Feb 7 '12 at 10:08
    
You can think of entropy as a way of measuring the disorder in a system. The more disordered a system the higher it's entropy. Two separate liquids are more ordered than a mixture of the two, so the entropy rises on mixing. The Wikipedia "Entropy of mixing" shows how to calculate the entropy of mixing for an ideal gas mixture, though if you don't have a maths background you may find their argument a bit hard to follow. –  John Rennie Feb 7 '12 at 10:29
    
@JohnRennie: Is there an book, which explains all what you have said in your answer with even more details? If yes, can you please advise it to me. Thank you. –  Beni Bogosel Feb 19 '12 at 19:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.