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    Derived from the Greek word for push or impulsion, osmosis denotes the phenomenon in which a solvent separated from a solution by a semi-permeable barrier flows across the barrier to dilute the solution. It seems almost magical, but in fact it gives us an excellent example of the workings of chemical thermodynamics.

    As an example consider seawater separated from fresh water by a semi-permeable membrane. On average every kilogram of seawater contains 35 grams of sodium chloride and smaller amounts of other salts. If our osmotic system contains seawater at that level of salinity, then fresh water will flow across the membrane with a measured osmotic pressure of 390 pounds per square inch (267.69 newtons per square centimeter), the overpressure that must be imposed upon the solution to stop osmosis from happening. To explain osmosis we need to account for that pressure.

    We can make an analogy between the sodium and chlorine ions suspended in water and the particles in an ideal gas. In this analogy the water plays the role of the container holding the gas. To see whether thatís a good analogy we calculate the pressure that the sodium/chlorine pseudo-gas exerts upon the walls of its container.

    We use the ideal gas equation (p=nRT), in which we multiply the density of the gas particles by the ideal gas constant and the absolute temperature of the gas. This particular use of the ideal gas equation to describe a liquid solution was first described by J. H. vanít Hoff in 1885. In this case we have seawater consisting of 469 moles of sodium ions and 546 moles of chloride ions dissolved in 53,600 moles of water (one tonne of seawater). We take the absolute temperature of the system to be standard room temperature, T=293 degrees Kelvin (20 degrees Celsius). With those units set, we have R=8.314 joules per mole per degree Kelvin. We thus multiply 1015 moles of sodium and chloride ions per cubic meter of water by 2436.002 joules per mole and thereby calculate an osmotic pressure of 247.25 newtons per square centimeter, which compares favorably with the measured osmotic pressure of the seawater/fresh water system. If we add in the partial pressures due to sulfate, magnesium, calcium, potassium, and bicarbonate ions in the seawater, we calculate the osmotic pressure as 272.6 nt/cm2.

    Our calculation gives us a correct value for the osmotic pressure, so we can feel confident that our analogy with an ideal gas comes close to the truth of the matter. Itís good enough for engineering, certainly: the vanít Hoff equation gives us a correct calculation of the osmotic pressure. As physicists, though, we want to know more; we want to understand whatís causing the solvent to flow across the membrane. So given that a solution behaves like a gas in a container, we ask How does it work? How does this model produce the phenomenon of osmosis?

    Having analogized the solute ions to the particles in an ideal gas, letís look at an actual ideal gas (or a reasonable facsimile of one). Imagine a chamber filled with pure hydrogen gas at room temperature and a pressure of P(H2). Take a rubber balloon and fill it with pure nitrogen at a pressure of P(N2) and put it into the chamber. In a short time the balloon will expand due to hydrogen passing through the rubber skin and into the balloon. The expansion will stop when the partial pressure of the hydrogen inside the balloon equals the pressure of the hydrogen outside the balloon. The total pressure inside the balloon thus equals P(H2) plus the pressure caused by the stretched rubber, that latter pressure being equal to the partial pressure of the nitrogen P(N2).

    We are accustomed to the concept of different pressures equalizing when their containers are connected. If we have compressed air in a cylindrical tank (of the kind used by scuba divers perhaps) and we open the valve, we know that some air in the tank will come out and that the air will continue to come out of the tank until the pressure inside the tank equals the pressure of the surrounding atmosphere. But the idea that partial pressures equalize separately involves a bit more subtlety. We need to look at our theory of gases on the molecular level.

    Of course we understand in our present example that hydrogen molecules pass through submicroscopic pores in the rubber skin and that the nitrogen molecules are too big to pass through those pores. We also know that not all of the hydrogen molecules that approach the rubber pass through it; most hit the rubber and bounce off it, contributing to the pressure that the gas exerts upon the rubber. Now we remember that the rate at which the hydrogen molecules pass through the pores stands, in part, proportional to the density of the hydrogen. In the case of our balloon hydrogen will seep into the balloon faster than it seeps out, so there will be a net flow of hydrogen into the balloon until the hydrogen density on both sides of the rubber skin equalizes or until some applied pressure (as, for example, caused by the stretching of the rubber) stops the process.

    If the gas both inside and outside the balloon has the same temperature, then equalization of the hydrogen density corresponds to an equalization of the pressure. If the balloon had started out empty, then the seepage of hydrogen into it would stop as soon as the rubber went taut. Any stretching of the rubber would produce a pressure that would increase the density of the hydrogen and cause a net outward seepage of the gas until the additional pressure went away.

    Another assumption that we incorporate into the theory of ideal gases, which assumption conforms very closely to the nature of real gases, tells us that the molecules interact with each other only rarely. So, to a good approximation, the hydrogen molecules and the nitrogen molecules inside the balloon donít have any direct effect on one another. The nitrogen only has an indirect effect on the hydrogen by diluting it and that dilution causes the osmosis.

    As long as the balloon is limp, the pressure on the inside, the sum of the partial pressures of the nitrogen and the hydrogen, must equal the pressure of the hydrogen atmosphere outside the balloon. That fact necessitates that the partial pressure of the hydrogen and, therefore, the density of the hydrogen, be less than the pressure and density of the hydrogen outside the balloon. Thus hydrogen will seep into the balloon faster than it seeps out. The balloon will fill with hydrogen until the rubber goes taut and then fill further until the rubber stretches enough to create a pressure equal to the partial pressure of the nitrogen. When the system has come to equilibrium, the stretched rubber produces a pressure that balances the partial pressure of the nitrogen and the pressure of the outside atmosphere balances the partial pressure of the hydrogen.

    In a liquid solution one of our ideal-gas assumptions no longer operates. In a liquid the molecules interact with each other frequently and strongly. To gain an impression of how strongly we need only return to the pressure calculation that we carried out for sodium chloride in water. We had 1015 moles of sodium and chloride ions in one cubic meter of water producing 247.25 newtons per square centimeter (24.4 atmospheres) of osmotic pressure. That cubic meter contains 53,600 moles of water and if we apply the vanít Hoff equation to that fluid, we calculate a pressure of 13,057 newtons per square centimeter (1289 atmospheres). Of course, the only pressure that a cubic meter of water exerts comes from its weight: the mutual attraction of the water molecules cancels the 1289 atmospheres of gas pressure, thereby making water a liquid.

    Imagine two open containers connected to each other through a semipermeable membrane, one through which water molecules will pass but salt ions wonít. Fill both containers to the same level, one with fresh water and the other with salt water (35 grams of sodium chloride per kilogram of water). The mechanical pressure on the membrane will be almost zero, the pressure on the salt side being slightly higher than the pressure on the fresh side due to the extra weight of the dissolved salt. But we also expect to measure 24.4 atmospheres of osmotic pressure difference across the membrane.

    One cubic meter of seawater ponders 1025 kilograms and contains 35 kilograms of salt, so the water alone ponders 990 kilograms, ten kilograms less than we find in one cubic meter of fresh water. Even though seawater is heavier than an equal volume of fresh water, its water content is less dense than that of fresh water and therein lies our explanation of osmosis. In spite of the force drawing water molecules toward each other, the molecules are only loosely connected (so that they slide freely past one another, instead of being rigidly connected as in a solid), so we can describe their motions much as we describe the motions of particles in a gas. But unlike the case of a gas, whose particles move in more-or-less empty space (in accordance with our assumption of rare interactions), the molecules in a liquid rub up against each other: that fact makes the liquid effectively incompressible. Held together by electrostatic force, the molecules also repel each other if they come too close. If something, such as a piston, presses on the liquid, the force exerted on the molecules touching the piston will be exerted through those molecules onto their neighbors, which pass the force on to their neighbors, and so on. Because the molecules slide freely over one another, like greased ball bearings, the pressure exerted at any point in the fluid gets transmitted uniformly in all directions.

    At the membrane the rate at which water molecules pass through the pores stands in direct proportion to the particle density of the water, so water seeps more rapidly from fresh to salt than it does from salt to fresh, thereby producing a net flow of fresh water into the salt water. That net flow will make the free surface on the salt side rise and the free surface of the water on the fresh side fall. In consequence a mechanical pressure will grow on the membrane until the difference in the water levels produces a mechanical pressure that exactly counters the osmotic pressure and brings the osmosis to an end. Raising the level of the salt water necessitates that something do work upon the water in order to satisfy the conservation of energy law. Most of that work comes from the fall of the fresh water and the remainder comes from the kinetic energies of the water molecules, the heat of the water.

    By analogy with what happens with our balloon, we can attribute partial osmotic pressures to the components of the salt water. Those partial pressures must add up to a pressure equal to the osmotic pressure on the fresh water side of the membrane; otherwise, the difference in the total pressures would come manifest as a mechanical pressure. So on the salt side we have the partial pressure of the salt as 24.4 atmospheres and the partial pressure of the water as 1289-24.4 = 1264.6 atmospheres. As the hydrogen does in the balloon case, the water moves to equalize the two water partial pressures. Only two factors will stop the net flow of water from fresh to salt.

    The first factor is dilution. If enough fresh water flows into the salt water, it will dilute the salt and thereby lower its osmotic pressure. If the salt concentration is diluted effectively to zero, then osmosis will stop: the osmotic pressure of the water on both sides of the membrane has equalized.

    Application of a mechanical pressure to the salt side, as by pressing a piston down on the free surface, will also bring osmosis to a halt. The applied pressure increases the rate at which water seeps from the salt side of the membrane to the fresh side: it thus augments the seepage due to the osmotic pressure in the salt water. The amount of applied pressure that will bring the seepage from salt to fresh to equality with the seepage from fresh to salt equals the amount by which the osmotic pressure in the water on the salt side differs from the osmotic pressure in the fresh water, which difference equals the osmotic pressure that we attribute to the salt.

    If we increase the applied pressure, we will further enhance the rate of seepage from salt to fresh and thereby generate reverse osmosis. We will cause the water molecules on the salt side to push into the pores harder than the molecules on the fresh side do, so there will be net seepage from salt to fresh as the more forceful molecules push through.

    Thus we have a simple, straightforward description and explanation of osmosis.


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