Concise Physical Chemistry

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

OSMOTIC PRESURE 193

Step 2

In the second step, add a small amount of solute. This brings about a change in the

Gibbs free energy function^1

μ◦ 1 (p+π,X 2 )=μ◦ 1 (p+π)+RTlnX 1

We already have an expression for the first term on the right, so

μ◦ 1 (p+π,X 2 )=μ◦ 1 (p)+πVm, 1 +RTlnX 1

At equilibrium, the pressure on the solution is the same as the pressure on the pure

solvent:

μ◦ 1 (p)=μ◦ 1 (p+π,X 2 )

Substitution forμ◦ 1 (p+π,X 2 )on the right gives

μ◦ 1 (p)=μ◦ 1 (p)+πVm, 1 +RTlnX 1

which means that

πVm, 1 +RTlnX 1 = 0

or

πVm, 1 =−RTlnX 1

In a dilute binary solution, we have

lnX 1 ∼=−X 2

so

πVm, 1 =RTX 2

whereVm, 1 is the molar volume of the solvent. This can be expressed in terms of the

total volume asVm, 1 =V/n 1 ,so

πVm, 1 =π

V

n 1

=RTX 2

(^1) In a binary solution,μ◦ 1 can be expressed either as a function ofX 1 orX 2.