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 ]}

22.1

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