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.