PHYSICAL CHEMISTRY IN BRIEF

CHAP. 7: PHASE EQUILIBRIA [CONTENTS] 225

7.11 Osmotic equilibrium

Let us have a solution (this phase will be denoted′) separated from a pure solvent (this phase
will be denoted′′) by a semipermeable membrane, i.e. a membrane through which only the
molecules of the solvent can permeate. In order for the equilibrium condition (7.2) for solvent
to be satisfied, both phases separated by the membrane must be under different pressures.
Their difference is calledosmotic pressure

π=p′−p′′. (7.72)

For osmotic pressure we can write

π=−

φ 1 RTlnx 1

V 1 •

, (7.73)

whereφ 1 is the osmotic coefficient [see (6.110)],x 1 is the molar fraction of the solvent, andV 1 •
is the molar volume of a pure solvent at mean pressure (p′+p′′)/2.
For very diluted, ideal and incompressible solutions, relation (7.73) simplifies to

π=RT

∑n

i=2

ci, (7.74)

wherecistands for the amount-of-substance concentrations of the particles (molecules or ions)
of the dissolved substances. This equation is called thevan’t Hoff equation. According to
this equation, osmotic pressure does not depend on the kind of solvent used.
Reverse osmosisallows for obtaining a pure liquid from a solution by way of increasing
pressure over the membrane.

Example

Calculate the osmotic pressure over the solution of sucrose in water, separated from pure water

by a semipermeable membrane, below which neither dilution nor reverse osmosis occurs at 60◦C.

The amount-of-substance concentration of the solution is 0.1923 mol dm−^3.

Solution

From the specification it follows thatc= 192. 3 mol m−^3. According to (7.74) it applies that

π=RT c 2 = 8. 314 × 333. 15 × 192 .3 = 532. 6 kPa.