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Suppose you perpare a jar of salt water and another of sugar water and invert one on top of the other with a divider between them, and then carefully remove that divider so the liquids are in contact.

Will the concentrations of salt and sugar reach equilibrium, despite the fact that the salt or sugar in the bottom jar has to overcome gravity to rise into the upper jar? How would you calculate the relevant forces here?

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  • $\begingroup$ I am calling Gorilla vs. shark on the title. $\endgroup$
    – Wrzlprmft
    yesterday
  • $\begingroup$ @Wrzlprmft: I don't think so. Gorilla vs Shark is asking a subjective question, which ultimately comes down to personal taste. This question is about physical forces, which can in principle be calculated objectively. $\endgroup$ 15 hours ago
  • $\begingroup$ Well, another aspect of gorilla vs. shark is that it strongly depends on context, just like the titular case. But then it turns out that your question is about something slightly different anyway. (Hence me only calling gorilla vs. shark on the title, not the entire question.) $\endgroup$
    – Wrzlprmft
    14 hours ago

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Osmotic pressure is not really relevant here. In osmosis, only the solvent can move. In your scenario, both the solvent and the solutes can move. You are asking about ordinary mixing.

It will always be possible for equilibrium to be reached eventually, but in general we don't know how to calculate from first principles how long it will take. The solute particles in the bottom jar are not confined to the bottom jar; they have kinetic energy and will occasionally cross over into the top jar. The mixing will be more rapid at higher temperatures.

At equilibrium, the solute will have a very slightly higher concentration closer to the ground. The heavier the solute is, the more pronounced the gradient. If we treat the solution as an ideal solution, we can calculate the difference in concentrations explicitly using the Boltzmann distribution. For example, let's assume that each jar has a height of 10 cm. The mass of a sucrose molecule is $5.68 \times 10^{-25}$ kg. Let's assume an ambient temperature of 293 K. The ratio of probabilities for it to be found in the bottom jar rather than the top jar is $\exp(\frac{mgh}{kT})$ which comes out to about 1.000138. That means the sugar will be more concentrated in the bottom jar by a ratio of 1.000138; a very slight difference. You might be able to measure it using a very sensitive spectrophotometer.

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  • $\begingroup$ I'm not sure I follow the distinction between solvent and solutes moving. Isn't the solvent in both cases the water, and this is in both jars so it won't move much? But in this mixing situation, it sounds like the mixing force is stronger than gravity, or else the solute in the lower jar wouldn't be able to go up and mix. But gravity has some effect, causing the concentration gradient you mention. Is that a reasonable way to think about it? $\endgroup$ 15 hours ago
  • $\begingroup$ @JoshuaFrank I wanted to make sure you understand the correct usage of terminology even though in this case, if both jars are full, then it doesn't really matter anyway. If there is a semipermeable membrane, it is not gravity that prevents osmosis, but rather the incompressibility of water. If the top jar is not full and has lower osmotic pressure, osmosis can pull some water into the top jar against gravity. But you would need to add another question if you want to hear more about that. $\endgroup$
    – Brian Bi
    8 hours ago
  • $\begingroup$ @JoshuaFrank There is no mixing force per se, it's just entropy. There are accessible microstates with solute particles in the upper jar, so in order to maximize entropy, we should predict that the system may be found in such microstates. But a particle has slightly more microstates that involve being in the lower jar than ones that involve being in the upper jar, because when it moves to the lower jar, it picks up speed so it has more momentum phase space available. $\endgroup$
    – Brian Bi
    8 hours ago

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