$\gamma$ should not be number of particles, but molar amount of particles. Usually, $\gamma /V$ is replaced by amount concentration $c$. Additionally, for the case of electrolytes, there is the van't Hoff factor $i$ related to dissociation (>1) or association (<1) of particles.
So $\Pi = icRT$.
Osmotic pressures are at equilibrium equal only if hydrostatic pressures are equal. If a solution has osmotic pressure $\pu{1 atm}$, but has hydrostatic pressure greater by $\pu{1 atm}$ compared to pure water, osmosis does not occur.
If there is pure water and some solution on each side of semipermeable membrane, then ongoing osmosing causes growing difference of hydrostatic pressure that acts against the osmosis and finally balances it.
Response to feedback:
In biology, there is aside of osmolarity also tonicity. Osmolarity considers all solutes, tonicity just those not able to pass the membrane.
If osmolarity is equal, but tonicity is not, there is not equilibrium yet.
The same osmolarity but different tonicity leads to diffusion of solutes that takes with them solvent to rebalance osmolarity.
I read your updated response...I'm still not sure I understand why D is the correct answer to the U-tube problem presented in this post. Your responses so far would make me think that equality of osmolarity at equilibrium DOES NOT occur. I feel like there must be an assumption here about how the solution is sufficiently dilute and, therefore, no meaningful quantity of additional hydrostatic pressure is accumulated on the sucrose side.
If glucose concentrations are different, glucose would diffuse until they are the same. If summary glucose + sucrose concentrations are different, water would diffuse until they are the same.
Note that for $\pu{1 osmol/L}$, osmotic pressure is about $\pu{25 atm}$, or $\pu{250 m}$ of water column. The effect caused by hydrostatic pressure difference is therefore often negligible.
I think I must have a misunderstanding. Does your statement about the osmotic pressure mean that, at equilibrium conditions, for a U-tube with column radius $R$ meters, a net volume of $\pi R^2 \cdot 250 \pu{m3}$ of water flows from the right side (referring to this link ) of the U-tube to the left side??
No, it has nothing with volume nor flow. I see now it was rather a misleading, tanget info.
It means, that if you had "water||$\pu{1 osmol/L}$ of solute not passing membrane, and if the right part of U-tube had water level 250 m above the water level (or accordingly less based on density), then osmosis would not be ongoing, as counteracted by the hydrostatic pressure.
By other words, kinetic pressure of water molecules, decreased by solute presence, is counterbalanced by hydrostatic pressure. but there is low chance the membrane would survive such hydrostatic pressure.
If we take the meantioned U-tube, with passing glucose and not passing sucrose,
- with 2 M sucrose + 1 M glucose on one side
- with 1 M sucrose + 2 M glucose on the other side
- with the equal initial volume
then
- As the membrane is permeable for glucose, the final glucose concentration is 1.5 M, regardless of the final volumes on each side.
- As the osmolarity on both sides must be the same (otherwise water would flow to balance them), sucrose concentration must be 1.5 M on both sides as well.
- The same sucrose concentration is achieved by migration of water together with glucose, What increases glucose concentration and decreases sucrose concentration on the receiving side.
- The equilibrium osmolarity and tonicity at equilibrium is the same.
- The eventual effect of the optional external pressure is like $\Delta h \rho g = \Delta c RT$, but is mostly negligible.
The macroscopic conditions of osmotic equilibrium are:
- concentrations (therefore chemical potential) of solutes, for which the membrane is permeable, are the same on both sides.
- $\Delta h \rho g = \Delta c RT$, what means that
- At zero difference of hydrostatic pressure, osmolarity on both sides is the same.
- With the hydrostatic pressure difference, there is the difference in the equivalent osmolarity $\Delta c = \dfrac{\rho g}{RT} \Delta h$. This way is the solvent chemical potential of water(solvent) the same.
- Tonicity is the same, or follows the same pattern as osmolarity wrt hydrostatic pressure difference.
$\Pi = h \rho g$ is then relation of the height of solution column of osmolarity $\Pi$, that is needed to stop osmosis of pure water on the other membrane side.
For components passing the membrane:
$$\mu_{i,\text{L}} = \mu_{i,\text{R}}$$
$$c_{i,\text{L}} \approx c_{i,\text{R}} $$
For solvent water(w), solute(s) and sides left(L) and right(R):
$$\mu_\text{w,L} = \mu_\text{w,R}$$
If taken $a_i \approx x_i$:
$$\mu^\circ_\text{w} + V_\text{M,w}p_\text{L} + RT \ln{x_\text{w,L}}= \mu^\circ_\text{w} + V_\text{M,w}p_\text{R} + RT \ln{x_\text{w,R}}$$
$$V_\text{M,w}p_\text{L} + RT \ln{x_\text{w,L}}= V_\text{M,w}p_\text{R} + RT \ln{x_\text{w,R}}$$
$$V_\text{M,w}p_\text{L} + RT \ln{(1 - x_\text{s,L})}= V_\text{M,w}p_\text{R} + RT \ln{(1- x_\text{s,R})}$$
for $x_\text{s,R} \ll 1$ and $x_\text{s,R} \ll 1$:
$$\ln{(1- x_\text{s,R})} \approx = - x_\text{s,R}$$
$$\ln{(1- x_\text{s,L})} \approx = - x_\text{s,L}$$
$$V_\text{M,w}p_\text{L} - RT x_\text{s,L}= V_\text{M,w}p_\text{R} - RT x_\text{s,R}$$
if $x_i=ac_i$ then
$$V_\text{M,w}p_\text{L} - RTac_\text{s,L}= V_\text{M,w}p_\text{R} - RT ac_\text{s,R}$$
$$V_\text{M,w}(p_\text{L}-p_\text{R}) = RTa (c_\text{s,L} - c_\text{s,R})$$
