I wonder if there is an effect of pressure on the solubility a liquid in another liquid. Generally such solubilities show no or negligible dependence on the pressure, rather these are usually increased with the rising temperature.

Generally hydrocarbons and especially long chain hydrocarbons and water are considered as immiscible based on the practical observations at wide ranges of temperature and pressure.   

To my best knowledge, n-octane, a short chain normal alkane, is not miscible in water at ordinary temperature and pressure has not sinificant effect on liquid state. However, could you, please, indicate the P and T values ? 

This can be thought of from two different points of view.

Using a somewhat loose argument, you can look at this from a molecular viewpoint, adding pressure forces molecules that attract more strongly together preferentially, essentially causing a squeezing out effect. This is a result of interaction energies, which increase in magnitude as the separation becomes smaller to a point. (A prototype is the Lennard-Jones model.). Since water-water attractions are stronger than water-octanol attractions, forcing molecules even slightly closer will increase the strength of water-water more than water-octanol.  

from a more thermodynamic point of view, the dependence of the solubility on the pressure for a poorly soluble solute goes as something similar to

d(ln S)/dP = - d Chi/dP - Vm/RT

where S is the octane solubility, Chi is the water octane interaction parameter (as in the Hildebrand-Scott models), Vm is the partial molar volume of the octane, R is the universal gas constant, and T is the temp in Kelvin. (Vm, R and T are positive)

For octane, the dependence of Chi on the pressure is probably small, so the -Vm/RT term is more important and results in a negative value for the right hand side.  The result is that increase the pressure decreases the solubility. 

(Take this model as an approximation for a poorly soluble solute... all systems can vary, but this gives the essential idea.)

Hope this helps!

Extremely thought provoking explanation Mr. Bellantone.

Whether the miscibility decreases or increases depends very much on the pressure and temperature. (Octane + water) belongs to phase diagram class III (nomenclature of Scott and van Konynenburg) or 1C1Z (rational nomenclature), and there to a subclass known as gas–gas equilibria of the 2nd kind (critical curve: E. Brunner, J. Chem. Thermodyn. 22 (1990) 335–353); you must have increasing solubility at high temperatures.

This kind of phase behaviour is typical for mixtures where the molecules differ much in size and where the interaction energy between unlike molecules is weaker than that between like molecules. For Lennard-Jones particles, one would look at 2 epsilon12 – epsilon11 – epsilon22.

At high densities (low temperature, high pressure) you tend to reduce the influence of mixing entropy (as Robert observed), and if then the energetic interaction is unfavourable, demixing becomes more likely.

This is another piece of very scientific explanation that is extremely helpful for beginners.

Dear colleagues, let me also share my understanding of this problem. As far as I understand, in the case of two practically immiscible liquids (such as octane and water), the solubility of octane in water (So/w) is its inversed activity coefficient "gamma" (when referred to the pure liquid state; So/w=1/gamma(o/w)). The dependence of the activity coefficient "gamma" of octane in water on pressure is known: the relevant derivative is defined by the excess molar volume of octane V(ex,o) in an aqueous solution:

d(ln(gamma(io/w)))/dP=V(ex,o)/RT (where temperature is considered constant). therefore,

dln(So/w)/dP=-V(ex,o)/RT where the excess molar volume of octane V(ex,o) is the difference between its partial molar volume (Vpartial) of octane in aqueous solution and molar volume of pure octane liquid (Vmolar); therefore,

dln(So/w)/dP=-V(partial)/RT+V(molar)/RT  (please, compare it with the equation provided by Mr. Robert A Bellantone' it is' very similar)

Would  the octane solution in water  be ideal, there would not be a difference between partial molar volume V(partial) in a solution and V(molar) in the pure liquid state. Usually, these quantities, V(partial)  and V(molar), are quite similar by magnitudes, and typically we neglect the differences between them. But, there is such a difference: the partial molar volume of octane in water is slightly higher than its pure liquid molar volume. The positive excess molar volume of octane in water leads to that the derivative is negative, and the increase of the pressure decreases octane solubility in water.

Why is that? Most probably, I would say, it is due to the same "hydrophobic effect". Octane dissolved in water results in a "freezing" of the aqueous shell around; "ice-like moieties" or "water icebergs" around octane are characterized by expanded volume, and that is the increasing contribution to the whole volume of all the solution, and therefore, to the partial molar volume of octane in water. And this physical phenomenon, i.e., decreasing octane solubility in water with increasing pressure, stands together with another one, i.e., the effect of the applied pressure on the interface tension between octane and water!

What is funny, is that I recalled now that once I have read in a some old paper with P. Huyskens (from 80ths), I guess) that the partial molar volume of water in alkanes is 24 cm3/mol, which is much higher that water molar volume (18 cm3/mol). It clearly shows how dissimilar liquids such as water and alkanes change their partial volumes in mutual solutions. However, I am not sure that my above explanation dealing with hydrophobic effect (of octane in water) will match this expanded partial volume of water in alkane. Maybe, yes, it is possible to connect.... 

I wonder what will happen at supercritical pressure?

@Mangalya: That depends on the temperature:

• At or above the critical temperature of water you will have complete miscibility of octane and water. But above 40 MPa a separation will occur.
• If you start from a 2-phase state slightly below the critical temperature, miscibility will improve with increasing pressure, then you will have total miscibility, and then demixing again.
• At lower temperatures you will get increasing, then decreasing miscibility with increasing pressure.
• At very low temperatures, you will always have decreasing miscibility.

@ Mikhail: The “iceberg effect” is probably one of the causes of the pressure effect. But it is relevant at low temperatures only.

Thanks Ulrich.