Exercise 5 Solution

Start with Equation 27 for species B.

$\displaystyle N_{B}^{D}=-\frac{DC\frac{dx_{B}}{dl}}{\left ( 1+\frac{D}{D_{B}^{k}} \right )-(1-M_{BA}^{0.5})x_{B}}$

Then, because x_A + x_B = 1 and $M_{BA}^{0.5}=M_{AB}^{-0.5}$ , the denominator is written as follows.

$\displaystyle \frac{D}{D_{B}^{K}}+1-(1-M_{BA}^{0.5})x_{B}$ $\displaystyle =\frac{D}{D_{B}^{K}}+\left\{1-(1-M_{AB}^{0.5})x_{A} \right\}M_{AB}^{-0.5}$

With $\frac{D}{D_{B}^{K}}M_{AB}^{0.5}=\frac{D}{D_{A}^{K}}$ from Equation 19, the flux of species B is shown below.

$\displaystyle N_{B}^{D}=\frac{M_{AB}^{0.5}DC\frac{dx_{A}}{dl}}{\frac{D}{D_{A}^{K}}+1-(1-M_{AB}^{0.5})x_{A}}$

This combines with Equation 27 to form Graham’s law.

$\displaystyle N_{B}^{D}=-(M_{AB})^{0.5}N_{A}^{D}$

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