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Starting from the classical osmosis experiment, a U-shaped tube with a semi-permeable membrane, I would like to consider the case when the solute added to one of the compartments (labelled A) is composed of two species—or rather, the same species, e.g. dextran, but two very different molecular weights : the high-molecular weight dextran cannot cross the membrane, but the low-molecular weight can. Assume that low-molecular one is in large excess.

What will be the final state:

  • equal concentration of low-weight dextran in A and B with a column height difference so that hydrostatic and osmotic pressures equilibrate,
  • or, equal total concentration in A and B, that is, low-weight dextran has a higher concentration in B than in A so that the osmotic pressure, to which both low- and high-weight solute contribute, is the same on both sides. Column heights remain equal.

Equilibrating chemical potential across the membrane leads me to the second final state. Is that correct? Another question with only a comment as an answer seems to point to the first answer, and seems to make sense when considering the diffusion of the low-molecular weight solute.

I am interested also to read about interesting additional effects if the solutes differ in other ways in addition to their different permeation of the membrane.

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You can treat the problem using thermodynamics. You have three species: the solvent $S$, high molecular weight dextran $D_h$ and low molecular weight dextran $D_l$. Your membrane is semipermeable to $D_l$ and $S$ (the latter was omitted but I’ll assume it to be the case). I’ll also assume the membrane to be diathermal so that the temperature is the same on either side.

Equilibrium is given by balancing the chemical potential of $S$: $\mu_s$ and of $D_l$: $\mu_l$. There is no reason to balance the pressure on either side since the membrane is immovable (at least that’s what I understood from context). You therefore expect to develop an osmotic pressure resulting in a height difference.

Quantitatively, I’ll consider an ideal solution giving the chemical potential (from mixing entropy): $$ \mu=\mu^0(T,p)+RT\ln x $$ with $x$ the molar fraction of the species.

Using the fact that $S$ is a solvent so $x_s\to 1$, you get the familiar equations: $$ \begin{align} \mu_s &=\mu_s^0-RT(x_l+x_h) \\ \mu_l &=\mu_l^0+RT\ln x_l \end{align} $$ I will write the quantities with a superscript $A,B$ to refer to the relevant quantities. You have to view the previous equation as a system of equations determining the differences $\Delta p=p^A-p^B$ and $\Delta x_l=x_l^A-x_l^B$.

In absence of $D_h$, the solution is $\Delta x_l=\Delta p=0$ with the common values $p=p^A=p^B$ and $x_l=x_l^A=x_l^B$. A slight positive $x_h^A$ upsets the balance. You can solve this perturbatively. Taylor expanding $\mu^0$: $$ \mu^0(p,T)=\hat \mu^0(T)+pv(T) $$ your system is linear and solvable: $$ \begin{align} 0 &=v_s\Delta p-RT(\Delta x_l+x_h) \\ 0 &=v_l\Delta p+RT\frac{\Delta x_l}{x_l} \end{align} $$ giving: $$ \begin{align} \Delta p &= \frac{RT x_h}{v_s+x_lv_l} \\ \Delta x_l &= -\frac{x_hx_lv_l}{v_s+x_lv_l} \end{align} $$

Note that when $x_l=0$, you recover the usual expression of osmotic pressure. Qualitatively, the case $x_l >0$ is similar. You have an excess of pressure in $A$ so its column is taller than that if $B$. The novelty now is that there is slight inhomogeneity in $D_l$. It will be more concentrated in $B$.

Thus neither of your proposed solutions are correct. They are both incompatible with the equality $\mu_l$ on either side. For the first one, the inconsistency is: if there is an osmotic pressure, then $x_l$ cannot be equal on either sides. For the second attempt, you had the correct intuition that there will be more low weight dextran in $B$. However, you predict equal pressure on either side (columns of equal height), but in $\mu_l$, $D_h$ cannot compensate the difference in $x_l$.

Essentially, you were only considering the equation for $\mu_s$, which is why both alternatives looked plausible. The correct method is to include the second equation and solve for both at the same time which is not easy to guess using handwaving arguments.

Hope this helps.

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  • $\begingroup$ Many thanks. Before I try to analyse in depth, one quick question: why the minus sign in $\mu_l =\mu_l^0-RT\ln x_l$? $\endgroup$
    – Joce
    Commented Sep 2, 2023 at 14:32
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    $\begingroup$ That was a mistake, fixed it. Now the denominator has a tamer look. $\endgroup$
    – LPZ
    Commented Sep 2, 2023 at 15:45
  • $\begingroup$ Great, now the results completely make sense - more of $D_l$ moving into $B$ which results in a lower pressure difference than for usual osmotic pressure, and all those effects only appearing once $v_l x_l$ is sufficiently large to be comparable to $v_s$. $\endgroup$
    – Joce
    Commented Sep 2, 2023 at 18:30
  • $\begingroup$ Can you confirm that the formal assumption leading to the expression of the chemical potential is that we have an ideal solution? $\endgroup$
    – Joce
    Commented Sep 2, 2023 at 18:35
  • $\begingroup$ Yes exactly, added it in the answer $\endgroup$
    – LPZ
    Commented Sep 2, 2023 at 19:45

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