Question 22.1: Consider a binary mixture of components 1 and 2, where B(T, ...
Consider a binary mixture of components 1 and 2, where B \left( T , P , x _{2}\right) is known. Calculate \overline{ B }_{1}\left( T , P , x _{2}\right) \text { and } \overline{ B }_{2}\left( T , P , x _{2}\right) and provide a geometrical interpretation of your results.
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Clearly, x _{1}=1- x _{2} was eliminated from the set of the two mole fractions. In order to compute \overline{ B }_{1}, we use Eq. (22.21) for a mole fraction that was eliminated (x _{1} in this case). Specifically,
\bar{B}_{1}= B -\sum_{ j \neq 1} x _{ j }\left(\frac{\partial B }{\partial x _{ j }}\right)_{ T , P , x [ j , i ]} (22.24)
or, because j = 2,
\overline{ B }_{1}= B – x _{2}\left(\frac{\partial B }{\partial x _{2}}\right)_{ T , P }=\overline{ B }_{1}\left( T , P , x _{2}\right) (22.25)
To compute \overline{ B }_{2}, we use Eq. (22.23), where k = 2 is in the set of mole fractions on which B depends. Specifically, for k = 2, i = 1, and j = 2, we obtain:
\overline{ B }_{2}= B +\left(\frac{\partial B }{\partial x _{2}}\right)_{ T , P }-\sum_{2 \neq 1}^{ n } x _{2}\left(\frac{\partial B }{\partial x _{2}}\right)_{ T , P } (22.26)
where in Eq. (22.26), the sum is redundant. In other words:
\overline{ B }_{2}= B +\left(\frac{\partial B }{\partial x _{2}}\right)_{ T , P }- x _{2}\left(\frac{\partial B }{\partial x _{2}}\right)_{ T , P } (22.27)
or
\overline{ B }_{2}= B +\left(1- x _{2}\right)\left(\frac{\partial B }{\partial x _{2}}\right)_{ T , P }=\overline{ B }_{2}\left( T , P , x _{2}\right) (22.28)
The expressions for \overline{ B }_{1} \text { and } \overline{ B }_{2} in Eq. (22.25) and Eq. (22.28), respectively, have a very nice geometrical interpretation that can be used to graphically compute \overline{ B }_{1} \text { and } \overline{ B }_{2} . Suppose that we are given B as a function of x2, at constant T and P, in graphical format (see Fig. 22.1). In Fig. 22.1, as well as in the derivation below, we have replaced x by X to enhance visualization.
Let us consider triangles I and II in Fig. 22.1:
Triangle I: Using Trigonometry
\frac{ B \left( T , P , X _{2}^{*}\right)- d _{1}}{ X _{2}^{*}-0}=\left(\frac{\partial B }{\partial X _{2}}\right)_{ T , P \mid X _{2}^{*}} (22.29)
where \mid X _{2}^{*} indicates that the partial derivative is evaluated at X _{2}= X _{2}^{*}.
Equation (22.29) can be rearranged as follows (see Fig. 22.1):
d _{1}= B \left( T , P , X _{2}^{*}\right)- X _{2}^{*}\left(\frac{\partial B }{\partial X _{2}}\right)_{ T , P \mid X _{2}^{*}} (22.30)
Triangle II: Using trigonometry
\frac{ d _{2}- B \left( T , P , X _{2}^{*}\right)}{1- X _{2}^{*}}=\left(\frac{\partial B }{\partial X _{2}}\right)_{ T , P \mid X _{2}^{*}} (22.31)
Rearranging Eq. (22.31) yields (see Fig. 22.1):
d _{2}= B \left( T , P , X _{2}^{*}\right)+\left(1- X _{2}\right)\left(\frac{\partial B }{\partial X _{2}}\right)_{ T , P \mid X _{2}^{*}} (22.32)
A comparison of Eq. (22.30) for d _{1} with Eq. (22.25) for \overline{ B }_{1} reveals that:
d _{1}=\overline{ B }_{1}\left( T , P , X _{2}^{*}\right) ( 22.33)
A comparison of Eq. (22.32) for d _{2} with Eq. (22.28) for \overline{ B }_{2} reveals that:
d _{2}=\overline{ B }_{2}\left( T , P , X _{2}^{*}\right) (22.34)
Accordingly, for any 0< X _{2}^{*}<1, a tangent to the \text { B vs. } X_{2} curve at X _{2}^{*} intercepts the X _{2}=0 axis at \overline{ B }_{1}\left( T , P , X _{2}^{*}\right) and the X _{2}=1 axis at \overline{ B }_{2}\left( T , P , X _{2}^{*}\right) (see Fig. 22.1). This provides a very nice geometrical interpretation. As the number of mole fractions increases, the geometrical depiction of our results becomes increasingly challenging, because it requires a multidimensional space, where tangent lines become tangent hyper planes in n dimensional space.
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Eq. (22.21) : \overline{ B }_{ i }= B -\sum_{ j \neq i } x _{ j }\left(\frac{\partial B }{\partial x _{ j }}\right)_{ T , P , x [ j , i ]}
Eq. (22.23) : \overline{ B }_{ k }= B +\left(\frac{\partial B }{\partial x _{ k }}\right)_{ T , P , x [ k , i ]}-\sum_{ j \neq i } x _{ j }\left(\frac{\partial B }{\partial x _{ j }}\right)_{ T , P , x [ j , i ]}
