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In the figure given below:-

enter image description here

Since only solvent particles can move through semipermeable membrane, if we manage to keep pressure on both the containers equal say by keeping weights, will osmosis go on till pure solvent container reaches zero volume?

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    $\begingroup$ It depends on the geometry of the vessels. If tall enough, osmotic pressure will be balanced by hydrostatic pressure difference. $\endgroup$
    – Poutnik
    Sep 24, 2023 at 18:49
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    $\begingroup$ Osmosis phenomena produces a height difference which tends to a maximum ℎ h. If both solvent and solution are at atmospheric pressure, this maximum is such that 𝑔ℎ is equal to $1$ atm, for a $1$M solution. This height is also proportional the concentration of the solution. $\endgroup$
    – Maurice
    Sep 24, 2023 at 19:59
  • $\begingroup$ @Poutnik I actually meant that if I apply and keep changing pressure on the other (solvent) container as well such that pressure remains equal on both the containers, will the osmosis take place? $\endgroup$ Sep 25, 2023 at 12:16
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    $\begingroup$ If the membrane pressure difference (= external pressure difference + hydrostatic pressure difference) balances the osmotic pressure, the net osmosis is zero. $\endgroup$
    – Poutnik
    Sep 25, 2023 at 12:23
  • $\begingroup$ @Poutnik Thanks I understood that. Just wanted like confirmation that osmosis will keep going on if there difference in membrane pressure difference and osmotic pressure, right? $\endgroup$ Sep 25, 2023 at 15:14

1 Answer 1

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Let $H_l$ = Hydrostatic pressure in solution jar
Let $O_l$ = Osmotic pressure in solution jar
Let $P_l$ = Pressure from the weight applied to the solution jar

Let $H_s$ = Hydrostatic pressure in solvent jar
Let $O_s$ = Osmotic pressure in solvent jar
Let $P_s$ = Pressure from the weight applied to the solvent jar

Osmosis will take place until ... $H_l + O_s + P_l = H_s + O_l + P_s$

We know $O_s = 0$ and $P_l = P_s$. So the equation for when the osmosis will stop simplifies to ...

$H_l = O_l + H_s$

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