Osmosis in Desalination — Osmotic Pressure, Reverse Osmosis and the Real Numbers
Colligative properties look like a small, harmless chapter until you realise that one equation from it — the osmotic pressure equation — sets the operating pressure, the pump size and the electricity bill of every reverse osmosis desalination plant in the world. This article takes the same formula you learn for IIT-JAM, GATE and CSIR-NET and pushes it through the numbers a membrane engineer actually works with. Every value below is calculated in front of you, not quoted.
The equation everything rests on
- π — osmotic pressure (atm, or Pa if you use SI R)
- i — van't Hoff factor, the number of particles one formula unit gives on dissolving: NaCl → 2, CaCl₂ → 3, glucose → 1
- C — molar concentration of the solute, mol L⁻¹
- R — gas constant, 0.0821 L atm K⁻¹ mol⁻¹ (or 8.314 J K⁻¹ mol⁻¹)
- T — absolute temperature in kelvin, never °C
Osmosis is water moving through a semipermeable membrane from the dilute side to the concentrated side. π is the pressure you would have to apply to the concentrated side just to stop that flow. Reverse osmosis (RO) is what happens when you apply more than π: water is squeezed backwards out of the salt water, leaving the salt behind. So π is not a curiosity — it is the minimum pressure the plant must beat.
Worked example 1 — the osmotic pressure of seawater
Take seawater with about 35 g of dissolved salt per litre and treat it as sodium chloride, M(NaCl) = 22.990 + 35.45 = 58.44 g mol⁻¹.
C = 35 ÷ 58.44 = 0.599 mol L⁻¹
i = 2 (Na⁺ and Cl⁻)
T = 25 °C = 298 K
π = 2 × 0.599 × 0.0821 × 298
0.599 × 0.0821 = 0.04918
0.04918 × 298 = 14.66
14.66 × 2 = 29.3 atm
In SI units: 29.3 × 101325 = 2.97 × 10⁶ Pa ≈ 2.97 MPa ≈ 29.7 bar.
Honest caveat. That is the ideal-solution answer. Real seawater is concentrated enough that ion–ion interactions matter, activity coefficients fall below 1, and the effective van't Hoff factor comes out a little under 2 — so measured osmotic pressures are somewhat lower than this estimate. Use π = iCRT for the order of magnitude and for exam problems; a plant designer uses activity-based models. Stating the assumption is part of the chemistry, not a weakness in it.
Worked example 2 — brackish water is a completely different job
Inland brackish groundwater carrying 2.0 g L⁻¹ of dissolved salt:
C = 2.0 ÷ 58.44 = 0.0342 mol L⁻¹
π = 2 × 0.0342 × 0.0821 × 298
0.0342 × 0.0821 = 0.002810
0.002810 × 298 = 0.8373
0.8373 × 2 = 1.67 atm
Seawater needs roughly seventeen times the pressure of this brackish water for the same separation. That single ratio explains why inland brackish-water plants are far cheaper to build and to run than coastal seawater plants, and why the two use different membranes and very different pumps.
Worked example 3 — why the pressure keeps rising along the module
An RO module does not treat one fixed solution. As pure water leaves, the salt left behind becomes more concentrated, so π climbs as you move down the pressure vessel.
Suppose the plant recovers 40% of the feed as fresh water (recovery ratio r = 0.40) and the membrane rejects essentially all the salt. All the salt now sits in the remaining 60% of the volume:
C(brine) = C(feed) ÷ (1 − 0.40) = C(feed) ÷ 0.60
= 1.67 × C(feed)
π(brine) = 29.3 ÷ 0.60 = 48.8 atm
The pump must therefore beat about 49 atm at the tail of the array, not the 29 atm you calculated for the feed. Push recovery higher and π rises faster still — which is why recovery is capped in practice, and why the concentrated brine leaving the plant is a design problem in its own right rather than a waste stream you can ignore.
Worked example 4 — the thermodynamic minimum energy
Work done against a pressure difference is W = pV. For the very first drop of water removed from seawater — before the feed has concentrated at all — the minimum work per cubic metre of product is:
W = 2.97 × 10⁶ Pa × 1 m³ = 2.97 × 10⁶ J
= 2.97 MJ per m³
In kilowatt-hours: 2.97 × 10⁶ ÷ 3.6 × 10⁶
= 0.825 kWh per m³
That figure is a floor set by thermodynamics; no membrane, pump or clever piping can go below it. A working plant always uses more — because it runs at finite recovery (Example 3), because pumps and energy-recovery devices are not perfect, and because pre-treatment, cleaning and post-treatment all consume energy too. The chemistry sets the minimum; engineering decides how close you get to it.
Where the research work actually happens
Because π is fixed by nature, the whole research field organises itself around the gap between that minimum and reality. Broadly, and without claiming any particular result:
- Membrane materials — raising water permeability without losing salt rejection, since those two normally trade off against each other.
- Concentration polarisation — salt piles up in a thin layer right at the membrane face, so the local π is higher than the bulk value calculated above. Module geometry is designed to disrupt that layer.
- Fouling and scaling — sparingly soluble salts reach their solubility product in the concentrating brine and deposit on the membrane. That is a Ksp problem, managed with antiscalants and recovery limits.
- Energy recovery — the high-pressure brine still carries most of the pump work, and devices exist to hand that pressure back to the incoming feed.
- Brine management — deciding what to do with a stream that is, by Example 3, considerably saltier than the sea it came from.
Every one of those is a physical-chemistry problem wearing an engineering coat: solution thermodynamics, transport, solubility equilibria and interfacial chemistry.
Mistakes that cost marks — and would cost money
- Forgetting i. Using C instead of iC halves the answer for NaCl and cuts it to a third for CaCl₂. For a mixture, what matters is the total particle concentration, not the molarity of any single salt.
- Using °C. T must be in kelvin. Substituting 25 for 298 makes the answer nonsense, and it is the most common slip in colligative-property questions.
- Mismatched R. 0.0821 goes with litres, atmospheres and moles; 8.314 goes with cubic metres, pascals and moles. Mixing them silently changes the answer by orders of magnitude.
- Assuming the operating pressure equals the feed π. It must exceed the π of the most concentrated brine at the membrane surface, which is higher on two counts: recovery and polarisation.
- Getting the direction wrong. In natural osmosis water flows into the salty side; RO reverses that only because the applied pressure exceeds π.
- Treating π as ideal at high salinity. Acceptable in an exam, wrong in design. State the assumption instead of hiding it.
Where this appears in your exam
| Exam | How it is asked |
|---|---|
| IIT-JAM | Osmotic pressure from concentration; molar mass of a solute from a measured π; van't Hoff factor and degree of dissociation |
| GATE | Colligative properties, membrane separation, recovery and mass balance across a separation unit |
| CSIR-NET | Non-ideal solutions, activity coefficients, deviations from van't Hoff behaviour |
| CUET-PG | Direct substitution into π = iCRT and comparison of i across electrolytes |
Run these numbers yourself. The suite has no dedicated osmotic-pressure tool, so there is nothing to send you straight to. Open the calculator and use the Ideal Gas Law tool: πV = nRT is the same algebra as PV = nRT, so entering your concentration and temperature there gives the osmotic pressure directly — just remember to multiply by the van’t Hoff factor i yourself for an electrolyte.
Open the Ideal Gas Law Calculator →Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and online for students across India — details at abcchemistry.in.