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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 169
I5.10 In Section 5B.2(e) on page 162 the derivation of the expression for the osmostic
pressure starts by equating the chemical potential of A as a pure liquid subject
to pressure pwith that of A in a solution ofmole fraction xA containing solute B
and subject to pressure p+Π: µ∗A(p) = µA(xA , p+Π).�e chemical potential of
A in the solution is then expressed as µA(xA , p + Π) = µ∗A(p + Π) + RT ln xA,
which assumes ideality. If the solution is not ideal, then the mole fraction is
replaced by the activity aA to give
µ∗A(p) = µ∗A(p + Π) + RT ln aA
�e derivation thenproceeds as before yielding the intermediate result−RT ln aA =
VmΠ for the non-ideal solution.
�e osmotic coe�cient ϕ is de�ned as ϕ = −(xA/xB) ln aA, hence ln aA =
−ϕxB/xA. �is expression for ln aA is substituted into −RT ln aA = VmΠ to
give RTϕxB/xA = VmΠ. �e �nal steps assume that the solution is dilute so
that xA ≈ 1 and xB = nB/(nA + nB) ≈ nB/nA
RTϕ
xB
xA
= VmΠ
hence RTϕ
nB
nA
= VmΠ
RTϕ
nB
nAVm
= Π
RTϕ[B] = Π
To go to the last line V = nAVm and [B] = nB/V are used.