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Chapter 14: Problem 9

Fick's law of diffusion is expressed on the mass and mole basis as \(\dot{m}_{\mathrm{diff}, A}=-\rho A D_{A B}\left(d w_{A} / d x\right)\) and \(\dot{N}_{\mathrm{diff}, A}=\) \(-C A D_{A B}\left(d y_{A} / d x\right)\), respectively. Are the diffusion coefficients \(D_{A B}\) in the two relations the same or different?

Short Answer

Expert verified
Answer: The diffusion coefficients in the mass-based and molar-based expressions of Fick's law of diffusion are the same.

Step by step solution

01

Rewrite the mass-based expression in terms of mole fraction #

To express the mass flux \(\dot{m}_{\mathrm{diff}, A}\) in terms of mole fraction, we can use the relationship between mass concentration and molar concentration: \(\rho = C*M\), where \(\rho\) is the mass concentration, \(C\) is the molar concentration, and \(M\) is the molar mass. Since \(w_A = \frac{m_A}{m_A + m_B}\), we can express the derivative of \(w_A\) as \(d w_A / d x = \frac{1}{(M_A + M_B)} * d C_A / d x\). Substituting this into the mass-based expression, we get: \(\dot{m}_{\mathrm{diff}, A}=-\rho A D_{A B} \left(\frac{1}{M_A + M_B} \frac{d C_A}{d x}\right)\). Now, let us rewrite it in terms of mole fraction: \(y_A = \frac{C_A}{C_A + C_B}\).
02

Rewrite the molar-based expression in terms of mass fraction #

We can express the derivative of \(y_A\) as \(dy_A /dx =\frac{1}{C_{total}} \frac{d C_A}{d x} - \frac{(C_A + C_B)}{C_{total}^2} \frac{d C_{total}}{d x}\). Substituting this into the molar-based expression, we have: \(\dot{N}_{\mathrm{diff}, A}=-C A D_{A B}\left(\frac{1}{C_{total}} \frac{d C_A}{d x} - \frac{(C_A + C_B)}{C_{total}^2} \frac{d C_{total}}{d x}\right)\). Now we have both expressions in terms of mole fractions.
03

Compare the diffusion coefficients #

Comparing the modified mass-based expression and the modified molar-based expression, we can see that the term involving the diffusion coefficients (\(D_{AB}\)) is the same in both expressions: \(-\rho A D_{A B} \left(\frac{1}{M_A + M_B} \frac{d C_A}{d x}\right) = - C A D_{A B}\left(\frac{1}{C_{total}} \frac{d C_A}{d x} - \frac{(C_A + C_B)}{C_{total}^2} \frac{d C_{total}}{d x}\right)\). Since we can derive one expression from the other, we can conclude that the diffusion coefficients \(D_{AB}\) are the same in both expressions.

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Most popular questions from this chapter

A recent attempt to circumnavigate the world in a balloon used a helium-filled balloon whose volume was \(7240 \mathrm{~m}^{3}\) and surface area was \(1800 \mathrm{~m}^{2}\). The skin of this balloon is \(2 \mathrm{~mm}\) thick and is made of a material whose helium diffusion coefficient is \(1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). The molar concentration of the helium at the inner surface of the balloon skin is \(0.2 \mathrm{kmol} / \mathrm{m}^{3}\) and the molar concentration at the outer surface is extremely small. The rate at which helium is lost from this balloon is (a) \(0.26 \mathrm{~kg} / \mathrm{h}\) (b) \(1.5 \mathrm{~kg} / \mathrm{h}\) (c) \(2.6 \mathrm{~kg} / \mathrm{h}\) (d) \(3.8 \mathrm{~kg} / \mathrm{h}\) (e) \(5.2 \mathrm{~kg} / \mathrm{h}\)

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