Volume Molar Parcial em Misturas

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124 5 SIMPLEMIXTURES
0 2 4 6 8
0
200
400
600
nA/(mol)
V
/(
cm
3 )
Figure 5.2
�e data fall on a good straight line which implies that the partial molar vol-
ume (the slope) is nearly constant with composition. However, for there to be
anything to do in the rest of the problem it is necessary to have some variation
in VA, so the data are �tted to a quadratic to give
V/(cm3) = 0.02524 (nA/mol)2 + 80.34 (nA/mol) + 74.03
From this an expression for VA is found by di�erentiation (recall that nB has
been held constant)
VA/(cm3mol
−1) = ( ∂V
∂nA
)
nB
= 0.05048 (nA/mol) + 80.34
Now that an expression for VA is available the integral in eqn 5.2 can be evalu-
ated.�e upper limit of the integral is (xA = 0.5, xB = 0.5) because the partial
molar volume at xA = 0.5 is required.�e integral will be the area under a plot
of xA/(1 − xA) against VA over this range from xA = 0 to xA = 0.5.
Equation 5.3 gives nA as a function of xA, so the partial molar volume can be
expressed as
VA/(cm3mol
−1) = 0.05048 ( xA
1 − xA
) + 80.34
�is expression is used to draw up a table of values of VA in the required range
and then to make the plot of xA/(1 − xA) against VA, shown in Fig. 5.3.
VA/(cm3mol
−1) xA xA/(1 − xA)
80.3406 0.000 0.000
80.3462 0.100 0.111
80.3533 0.200 0.250
80.3623 0.300 0.429
80.3743 0.400 0.667
80.3911 0.500 1.000