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 xA + xB = 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}

Return to Exercise 5


Flux Equations for Gas Diffusion in Porous Media Copyright © 2021 by David B. McWhorter. All Rights Reserved.