**Unformatted text preview: **d be
dV = V 1 dn 1 + V 2 dn 2 ,which is simply another application of the chain rule in partial
differentiation.
However, Eq. (5.33) can indeed be justified on physical grounds as follows.
Consider a large volume of solution containing ethanol (E) and water (W). We
now add a small amount of water, say, ∆nW moles of water, to this solution. We
would want to express the new volume of the solution as
V new = V old + Dn W V ,m ,
W
where V ,m is the molar volume of pure water. However, this will give us the final
W
volume only in the case of an ideal solution. In the ethanol-water solution, the
effective molar volumes of both substances are different from their molar volumes
in the absence of the other substance. Designating the actual molar volume of
water in the presence of ethanol as VW,m, the change of volume of the solution is
DV = Dn W V W,m
Therefore, we get
V W,m = DV .
Dn W
The partial molar volume of water, VW, is defined as the value of the fraction on
the right hand side i...

View
Full
Document