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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 135
2.8 3.0 3.2 3.4 3.6
3.45
3.50
3.55
3.60
(103/T)/(K−1)
ln
S/
(g
(1
00
g
so
lv
)−
1 )
Figure 5.5
P5B.6 �e de�nition of the chemical potential of component A, µA, is
µA = ( ∂G
∂nA
)
nB ,p ,T
To use this de�nition an expression for G as a function of the nJ is required.
�e excess Gibbs energyGE is de�ned in [5B.5–156],GE = ∆mixG−∆mixG ideal,
therefore ∆mixG = GE + ∆mixG ideal. �e Gibbs energy of mixing is also ex-
pressed as ∆mixG = G − Gunmixed, where G is the Gibbs energy of the mixture
and Gunmixed is the Gibbs energy of the unmixed components. It follows that
G = ∆mixG +Gunmixed, and because ∆mixG = GE + ∆mixG ideal
G =
∆mixG³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
GE + ∆mixG ideal +Gunmixed
�ese quantities are all molar, so the Gibbs energy of a mixture of nA moles of
A and nB moles of B is
G = (nA + nB)GE + (nA + nB)∆mixG ideal + nAG∗m,A + nBG∗m,B
where G∗m,A is the molar Gibbs energy of pure A.
�e idealGibbs energy ofmixing (permole) is given by [5B.3–155], ∆mixG ideal =
RT(xA ln xA+xB ln xB), and the expression forGE is given in the problem; this
latter is rewritten gRTxAxB using xB = (1 − xA).�e �nal expression for G is
G = (nA+nB)gRTxAxB+(nA+nB)RT(xA ln xA+xB ln xB)+nAG∗m,A+nBG∗m,B