Osmosis and membranes

Cut a wilted stalk of celery and stand it in a glass of plain water — within an hour it stiffens back to life. Sprinkle salt on a slice of eggplant and, minutes later, beads of water sit on its surface. Both are the same invisible traffic: water molecules slipping through a membrane too fine for anything else to cross, always drifting toward wherever the dissolved particles are more crowded. That one-way, solvent-only traffic is osmosis, and the thin barrier that permits it while blocking solute is a semipermeable membrane. This article stays inside the ISC Class 12 Solutions unit's actual demand — the mechanism, the hypotonic/isotonic/hypertonic vocabulary applied to cells, and reverse osmosis as a pressure-driven reversal of the natural flow — building on the $\Pi = MRT$ relation already fixed under Osmotic Pressure.

The core insight

Osmosis is the spontaneous, membrane-restricted flow of solvent alone from a region of lower osmotic pressure to one of higher osmotic pressure, and applying an external pressure greater than that osmotic pressure — $P_{applied} > \Pi = MRT$ — reverses the flow, which is reverse osmosis.

The core idea

A semipermeable membrane (SPM) has pores that pass small solvent molecules (almost always water) freely but exclude larger or differently-charged solute particles. Cellophane, parchment paper, an animal's bladder, and every living cell membrane behave this way for aqueous solutions.

Osmosis is not diffusion. Diffusion needs no membrane at all — any molecule or ion spreads from where it is concentrated to where it is dilute, in any fluid. Osmosis is far more restrictive: it requires a semipermeable membrane, and only the solvent moves, always from the side with lower solute concentration toward the side with higher solute concentration. The driving force is not the solute's own concentration gradient but the difference in osmotic pressure the two solutions exert across the membrane. That osmotic pressure is quantified by the relation already established for dilute solutions:

\Pi = MRT

where: $\Pi$ is the osmotic pressure, $M$ is the molar concentration of solute particles, $R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$ is the gas constant, and $T$ is the absolute temperature. Here $\Pi$ is read as the "pull" a solution exerts on solvent trying to cross into it — the higher a solution's $\Pi$ , the more strongly it draws water in.

Tonicity governs which way water goes. Compare a cell's internal osmotic pressure, $\Pi_{cell}$ , with that of its surroundings, $\Pi_{outside}$ : if $\Pi_{outside} < \Pi_{cell}$ the surrounding solution is hypotonic and water enters the cell (red blood cells swell and can burst — haemolysis; wilted plant cells regain turgidity); if $\Pi_{outside} > \Pi_{cell}$ the surroundings are hypertonic and water leaves the cell (red blood cells shrink — crenation; plant cells in salty soil lose turgor and the plant wilts); if $\Pi_{outside} = \Pi_{cell}$ the solutions are isotonic and there is no net flow, which is why intravenous fluids are formulated isotonic (about 0.9% saline) with blood plasma.

Figure

Two beakers separated by a membrane symbol showing water molecules crossing from the dilute side to the concentrated side while solute particles stay trapped on the concentrated side

Reverse osmosis (RO) exploits this direction on purpose, backwards. Left alone, solvent flows spontaneously from the dilute side to the concentrated side. If an external mechanical pressure greater than the solution's own osmotic pressure is applied on the concentrated side, that natural direction is overridden and pure solvent is forced from the concentrated solution through the membrane into the dilute side. Household and industrial RO purifiers use a synthetic semipermeable membrane, commonly cellulose acetate, and the same principle underlies seawater desalination — only the definitional principle is in scope; plant engineering is not.

Derivation AWhy reverse osmosis needs an applied pressure greater than the osmotic pressure

The goal is to show, starting only from the equilibrium condition for ordinary osmosis, exactly why reverse osmosis demands an applied pressure strictly exceeding the solution's osmotic pressure $\Pi$ , and then to turn that condition into a usable number using $\Pi = MRT$ , the same relation fixed under Osmotic Pressure.

Step 1. In ordinary osmosis with no external pressure, solvent flows spontaneously from the dilute side toward the concentrated side because the concentrated solution's osmotic pressure $\Pi$ represents its "pull" on solvent. Net inflow stops only when an opposing pressure equal to $\Pi$ has built up — this is precisely the equilibrium condition used to define and measure $\Pi$ in the first place.

Step 2. Now apply an external mechanical pressure $P_{applied}$ on the concentrated side deliberately, rather than letting it build up from osmosis alone. The net direction of solvent flow depends on the balance between $P_{applied}$ and $\Pi$ : if $P_{applied} < \Pi$ , the solution's own pull still wins and solvent keeps entering the concentrated side (normal osmosis, only slowed). If $P_{applied} = \Pi$ , the two effects exactly cancel and the system sits at equilibrium — no net flow either way.

Step 3. Only once $P_{applied} > \Pi$ does the applied pressure overpower the solution's pull, and pure solvent is squeezed out of the concentrated solution through the membrane — this is reverse osmosis, and $P_{applied} > \Pi$ is its necessary condition.

Step 4. Since $\Pi = MRT$ , the threshold pressure for any given solution can be found directly from its concentration and temperature. Take seawater with $\Pi = 27\ \text{atm}$ at $T = 300\ \text{K}$ : its effective solute molarity is $M = \dfrac{\Pi}{RT} = \dfrac{27}{0.0821 \times 300} = \dfrac{27}{24.63} = 1.10\ \text{mol/L}$ . Any $P_{applied}$ above $27\ \text{atm}$ will reverse the flow for this seawater; below it, normal osmosis still dominates. Verified: $M = \Pi/(RT) = 27/24.63 = 1.10\ \text{mol/L}$ , consistent with $\Pi = MRT = 1.10 \times 0.0821 \times 300 = 27.1\ \text{atm} \approx 27\ \text{atm}$ .

Worked example

A seawater RO plant starts with feed water at $\Pi = 27\ \text{atm}$ , $T = 300\ \text{K}$ (as above). As the plant runs, pure water is withdrawn and the remaining seawater becomes concentrated to $1.5$ times its original molarity. Find the new osmotic pressure and the new minimum applied pressure needed to continue reverse osmosis, and comment on why real RO plants run at 55–80 atm rather than at the bare minimum.

Step 1. From the derivation above, the original seawater has effective molarity $M = 1.10\ \text{mol/L}$ at $T = 300\ \text{K}$ , giving $\Pi = MRT = 27\ \text{atm}$ — the starting point for this problem.

Step 2. As water is withdrawn, the same dissolved solute is left in less solvent, so the molarity rises by the stated factor: $M' = 1.5 \times 1.10 = 1.65\ \text{mol/L}$ .

Step 3. The new osmotic pressure follows directly from $\Pi = MRT$ at the same temperature: $\Pi' = M'RT = 1.65 \times 0.0821 \times 300 = 1.65 \times 24.63 = 40.6\ \text{atm}$ .

Step 4. By the condition derived above, the minimum applied pressure to keep reverse osmosis going must now exceed this new, higher $\Pi'$ , not the original $27\ \text{atm}$ . The osmotic pressure rises to about 40.6 atm, so the minimum pressure needed also rises to about 40.6 atm — and this is exactly why commercial plants deliberately run at 55–80 atm: as freshwater is continuously extracted, the remaining brine keeps concentrating and its $\Pi$ keeps climbing, so a pressure fixed at the original $27\ \text{atm}$ minimum would stall partway through the run. A generous margin above the highest $\Pi$ expected, plus extra margin for a practical flow rate, is why the operating pressure sits well above the textbook minimum.

Common traps

Treating diffusion and osmosis as the same process — diffusion needs no membrane and moves any species; osmosis specifically requires a semipermeable membrane and moves only solvent.
Believing solute particles cross a semipermeable membrane during osmosis — by definition only solvent passes; solute stays blocked on its original side.
Assuming any applied pressure reverses osmosis — reversal needs $P_{applied}$ strictly greater than $\Pi$ ; at $P_{applied} = \Pi$ the system is merely at equilibrium, and below it normal osmosis still wins.
Losing track of which way water moves for hypotonic/hypertonic pairs — always trace back to "water moves toward the side of higher osmotic pressure," whether the context is a blood cell, a plant cell, or an RO membrane.

Ask the tutor: want to work through more tonicity scenarios — like why a grape swells in plain water but shrivels in syrup — using this same osmotic-pressure comparison?

Why does a plant wilt in over-fertilised or salty soil even though there is plenty of water around

Over-fertilising or salting the soil raises the osmotic pressure of the soil water until it exceeds the osmotic pressure inside the root cells, making the soil solution hypertonic relative to the roots. Water then flows the wrong way — out of the root cells into the soil — even though liquid water is abundant nearby. The plant is not short of water in the ordinary sense; it is short of water that is free to move toward it, because the concentration gradient across the root-cell membranes now favours outflow.

Is reverse osmosis really the opposite process of osmosis, or something completely different

It is the same physical machinery — the same semipermeable membrane, the same $\Pi = MRT$ relation, the same solvent-only passage — run against its natural direction. Ordinary osmosis is what happens when nothing opposes the solution's pull; reverse osmosis is what happens when an external pump supplies a pressure exceeding that pull. Nothing about the membrane or the underlying physics changes; only which pressure wins the tug-of-war changes, so it is genuinely the same phenomenon reversed by force rather than a distinct one.

Why do RO plants use pressures far above the calculated minimum instead of just the exact osmotic pressure

Two separate reasons. First, running exactly at $P_{applied} = \Pi$ gives zero net flow — you need pressure strictly above $\Pi$ to get any flow at all, and the closer you sit to $\Pi$ the slower that flow is, so a comfortable margin is needed just for a usable production rate. Second, as fresh water is continuously drawn off, the remaining seawater keeps concentrating and its $\Pi$ keeps rising through the run, so the pressure must stay above the worst-case, most-concentrated $\Pi$ reached near the end, not just the starting value. Both effects push real operating pressures to roughly double or triple the initial minimum.

Osmosis and membranes

The core idea

Worked example

Common traps

Before this

Where this leads