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A thought experiment:

A U-shaped tube with semi-permeable membrane at the base. The tube is completely thermally isolated from its surroundings. The liquid (solvent) is at some temperature $T$. When solute is added to one side of the tube the fluid level on that side rises until the increased pressure due to gravity ($\rho gh$) equals the osmotic pressure.

If I understand correctly, osmotic pressure could be used to do work (i.e. lift a weight as the solvent level in one side of U-shaped tube increases). Where does this energy come from? I guess it must be from the internal energy of the fluid.

At this point has the temperature of the fluid decreased?? If so, has it decreased by exactly the work done against gravity divided by the specific heat of the fluid? What does the quantitative treatment of this problem look like?


marked as duplicate by Aaron Stevens, GiorgioP, John Rennie thermodynamics Apr 6 at 9:51

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Yeah, I think you have already answered your own question.

In the general kinetic theory, when a pressure acts to do work, it is because the pressurized molecules have some kinetic energy due to $k_\text B T$ that then gets reduced as a result of impacting a surface that is moving away from them. Assuming the salt molecules re-thermalize very quickly, this does indeed take energy from the solution and ply it into doing work. That energy comes out of the temperature, or more precisely, what the temperature would have otherwise been.

Whether the temperature will have decreased depends on the interaction between the solvent and solute molecules, and whether anything else is going on in the solution. Some solvents like being near certain solutes more than they like being near their own atoms, so as they get more dilute they also release energy, warming up the solution. If you draw only the tiniest about of work out of such a system surely it must still warm up in the osmotic setup. In addition of course the solution could be actively or passively temperature-controlled by other factors, so if you had ice water as your solution of course the energy would ultimately come from the induced phase change.

  • $\begingroup$ Don’t forget the enthalpy of solution! The temperature does not necessarily go down overall... $\endgroup$ – Duncan Harris Apr 5 at 19:41
  • $\begingroup$ That's fair, I will edit it $\endgroup$ – CR Drost Apr 5 at 19:47
  • $\begingroup$ Thanks. Are you saying it's only the salt molecules that are impacting the surface and doing the work? Somehow the solvent molecules are contributing nothing because they're on both sides of the tube? Is there a way to make that clear? $\endgroup$ – Alex Apr 5 at 23:53
  • $\begingroup$ @Alex why isn't it clear? The whole point of osmosis is that the solvent can travel through the membrane; if the solvent comes to the same partial pressure on both sides by passing through it, then it will not contribute any pressure difference in equilibrium... $\endgroup$ – CR Drost Apr 6 at 13:59
  • $\begingroup$ Well there's a difference between the work done being $W=P_\mathrm{salt}dV$ or $W=(P_\mathrm{salt}+P_\mathrm{solvent})dV$. You seemed to imply the first, where I feel like it should be the second. I guess the resolution is that as the level goes down on the solvent-only side of the tube there's a negative amount of work done of $W=-P_\mathrm{solvent}dV$. So the overall work is just due the salt term (and making use of what you say that $P_\mathrm{solvent}$ is the same on both sides of the tube. Clearer now. $\endgroup$ – Alex Apr 7 at 15:36

For a semi-permeable membrane, the molecule that it is permeable to goes through the membrane in both directions. For the sake of the argument, assume a semi-permeable membrane that is permeable to water, but not permeable to sugar.

When distilled water is placed on one side of this membrane, and a concentrated sugar solution is placed on the other side of the membrane, the membrane sees an interesting effect at the molecular level. On the distilled water side of the membrane, a lot of water molecules are going across the membrane, with the rate of mass transfer being dependent on the temperature (water molecule velocity is a function of temperature) and the characteristics of the membrane. On the sugar side of the membrane, a lot of sugar molecules are trying to go through the membrane, but they can't. These molecules are hindering water molecules on that side of the membrane because they are getting in the way of the water molecules, so a lower rate of water molecules can diffuse from the sugar side of the membrane to the distilled water side of the membrane. This imbalance means that there is a net flow of water from the distilled water side (higher concentration of water) to the sugar side (lower concentration of water).

As noted, this effect continues, and the level of solution on the sugar side of the membrane continues rising, until the pressure on the sugar side of the membrane is equal to the osmotic pressure of the system. At that point, there are equal rates of water mass transfer across the membrane, and the water level stops rising on the sugar side of the membrane.

The energy for this process is due to the kinetic energy of the water molecules in the system, but it is also due to the fact that sugar disturbs the rate of mass transfer of water on that side of the membrane. Note that the temperature of the fluid on both sides of the membrane can be expected to remain constant as the above process takes place.

  • $\begingroup$ Why do we expect the temperature to remain constant? $\endgroup$ – Duncan Harris Apr 5 at 20:13
  • $\begingroup$ The process is slow enough to maintain thermal equilibrium with the environment, and there is no reason to expect that the temperature will drop just because more water molecules make it to one side of the membrane vs. the other. $\endgroup$ – David White Apr 5 at 20:28
  • $\begingroup$ So if it is in an insulated container the temperature does change, but if it is attached to a heat sink the temperature is roughly constant. Hmm. Doesn’t that describe all processes of any kind?? $\endgroup$ – Duncan Harris Apr 5 at 21:20
  • $\begingroup$ I feel like thermodynamically your answer amounts to: “the power involved in this process is so low that no one should care about the energy or where it comes from” $\endgroup$ – Duncan Harris Apr 5 at 21:26
  • $\begingroup$ I seriously doubt that osmosis in an insulated container would result in a temperature change, but I will admit that I have never seen this experiment done. Also, this process involves liquids, so any PV work will be VERY small. BTW, what kind of answer are you looking for? You obviously have something in mind. $\endgroup$ – David White Apr 5 at 22:12

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