Concise Physical Chemistry

c12 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come

OSMOTIC PRESURE 191

freezing point depression brought about by dissolution of a known (small) weight of

solute to produce a dilute solution, we can calculate the number of moles of solute

from
Tf, hence its molar mass in units of kg mol−^1.

12.8 OSMOTIC PRESURE

Many membranes have the property that they pass small molecules but not large

ones. They aresemipermeable. The human kidney contains semipermeable mem-

branes that separate salt and urea from the bloodstream for excretion but do not pass

blood protein. If a semipermeable membrane separates two arms of a U tube and a

solution of a protein is placed in one arm of the tube with the solvent in the other

arm, solvent molecules will pass through the membrane into the protein solution

because of the spontaneous tendency of the protein solution to evolve toward a more

dilute state of higher entropy. The process is calledosmosis. The liquid level in the

protein arm of the U tube will rise until the downward pressure it exerts on the

membrane due to gravity is just sufficient to balance the entropy-drivenosmotic flow

across the membrane. Thisosmotic pressure,π, follows a law similar to the ideal

gas law:

πV=nRT

whereVis the volume of a solution containingnmoles of themacromolecular

solute.

We have already made the point that liquids, contrary to gases, are subject to strong

intermolecular interactions. The internal structure of the liquid state is distinct and

different from that of the gaseous state, so why should a property of solutions follow

what appears to be an ideal gas law? The derivation of the osmotic law to arrive at

this remarkable and somewhat counterintuitive conclusion is quite distinct from ideal

gas law theory, and it reveals some of the power of the condition that, at equilibrium,

the Gibbs chemical potential must be the same over different macroscopic segments

of the same thermodynamic system.

To apply this idea to a real system, we find the osmotic pressure on the solution

in the right-hand chamber of Fig. 12.6 by a two-step process. First we shall calculate

the total pressure, which is the ambient pressurepplus an arbitrary pressureπon

pure solvent, through our knowledge of the variation of the Gibbs chemical potential

with variation in pressure. Second we shall find the variation of the Gibbs chemical

potential brought about by addition of a small amount of solute. These two steps are

then summed to find the total pressure on the solution of solvent plus solute.

Step 1

Start with pure solvent and apply some pressureπat constant temperature. The

Gibbs free energy of the system is the Gibbs free energy of the pure solvent at