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Consider the classical U-tube, semi-permeable membrane experiment. Assume I define osmotic pressure as $\Pi=\frac{RT}{V}\gamma$, where $\gamma$ denotes the number of total solute particles (across all species) residing in a solvent of volume $V$.

I have seen people say something to the effect of, "At equilibrium conditions, the osmotic pressure $\Pi$ of the left compartment $C_L$ is equal to the osmotic pressure of the right compartment $C_R$"...equivalently: $\Pi_{C_L}=\Pi_{C_R}$ (at equilibrium).

Defining osmolarity $\Omega$ as $\Omega = \frac{\gamma}{V}$, simple arithmetic then tells us that:

$$\Pi_{C_L}=\Pi_{C_R} \implies \frac{\gamma_{C_L}}{V_{C_L}}=\frac{\gamma_{C_R}}{V_{C_R}} \implies \Omega_{C_L}=\Omega_{C_R}$$

i.e. if osmotic pressure of the two compartments is equal, then osmolarity of the two compartments is equal.

This result immediately caught my eye because then I thought of the following edge case:

Suppose I have a U-tube filled with pure solvent and I add some amount of solute (have your pick) to the left side. Suppose that the semi-permeable membrane allows solvent diffusion but prohibits solute diffusion. What happens?

Because $\gamma_{C_R}=0$, I would need an infinite amount of water to begin with in order for the non-zero $\frac{\gamma_{C_L}}{V_{C_L}}$ to approach $0$.

So what's wrong with my derivation? More specifically, when is my derivation appropriate? Do you always need non-zero solute in both compartment and a sufficiently large reservoir of solvent in order for equilibrium to be achieved? What exactly are the implicit assumptions of the procedure for the U-tube, semipermeable experiment? Alternatively, is something wrong with my initial formulation of osmotic pressure?

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If the semi-permeable membrane prevents diffusion of the solute, then equilibrium isn't determined by balancing the osmotic pressures of both sides of the container: the membrane itself exerts whatever pressure is required to contain the solution (in practice, the membrane will tend to 'expand' into the pure solvent chamber until the surface tension balances the osmotic pressure.)

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  • $\begingroup$ Oh, this is very different than what I previously understood. Could you please clarify what exactly is 'in balance' for the two compartments (left and right) when the U-tube with a semi-permeable membrane (to solvent only) 'reaches equilibrium'. I understand that the rate of flow in both directions through the semi-permeable membrane perfectly cancel out (i.e. no more water level change in either compartment)...but is anything else 'in balance'? Or maybe I am misunderstanding how your parentheses statement is related to your paragraph's the prior sentence. $\endgroup$
    – S.C.
    Commented Nov 16 at 2:50
  • $\begingroup$ Is your response specific to if you have a pure solvent in one of the compartments? i.e. am i correct when i write, "If both compartments have non-zero solute, and there is a sufficiently large reservoir of water, then, at equilibrium, the osmotic pressures of each compartment are equal" $\endgroup$
    – S.C.
    Commented Nov 16 at 2:55
  • $\begingroup$ If the semi-permeable membrane blocks the solute but not the solvent, the solvent will equilibrate across the membrane (modulo solvation energy) but the solute will not. In the simple case where the solute is sufficiently dilute that interactions between it and the solvent can be ignored, the situation is analogous to an ideal gas trapped by a piston/wall on one side of a container (and the solvent would be analogous to the "vacuum" or free space.) In the case of a solution, you would match chemical potentials of the solvent but not the solute. $\endgroup$
    – TLDR
    Commented Nov 16 at 17:59

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