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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffusion Coefficients
Diffusion coefficients are key to understanding how substances spread or move through different media. They are symbolized as \(D_{AB}\) in Fick's law of diffusion. This coefficient essentially measures how easily molecules of a species \(A\) move in a medium comprising species \(B\).

A few factors affect diffusion coefficients:
  • Temperature: Higher temperatures increase diffusion rates, as molecules move more rapidly.
  • Medium: The nature of the medium, whether it's air, liquid, or solid, also influences how freely molecules can move.
  • Size of Molecules: Larger molecules tend to diffuse more slowly than smaller ones.
In the given exercise, we see diffusion coefficients in both mass and mole basis expressions. Importantly, these coefficients remain the same in both formulations. This is because \(D_{AB}\) remains constant when shifting between mass and mole fractions, ensuring the consistency of diffusion measurement across different contexts.
Mass Fraction
Mass fraction plays a crucial role in expressing the concentration of a particular component in a mixture. It is often denoted as \(w_A\), where \(A\) represents the species of interest. The mass fraction is the ratio of the mass of a specific component in relation to the total mass of all components in the mixture.

Here's how to think about mass fraction:
  • The mass fraction is unitless, making it easy to apply across different systems and scales.
  • It ranges from 0 to 1, where 0 indicates that the component is not present, and 1 means the mixture is purely that particular component.
  • Since it involves mass, it links closely with real, tangible quantities in systems like industrial processes and chemical reactions.
In the context of Fick's law, mass fraction helps in representing concentrations from a mass perspective. This can be especially helpful in engineering applications where mass conservation is crucial.
Mole Fraction
The mole fraction, denoted as \(y_A\), represents another way to express the concentration of a species within a mixture. Instead of focusing on mass, the mole fraction considers the number of moles present. It provides a way to express concentration from a chemical standpoint.

Important aspects of mole fraction include:
  • Similar to the mass fraction, the mole fraction is also a unitless quantity.
  • It offers accurate concentration descriptions, particularly in chemical contexts, as reactions often depend on the number of participating molecules.
  • Mole fractions add up to 1 across all components in a mixture, facilitating easy comparison and calculation.
In Fick's law of diffusion, using mole fractions emphasizes the role of chemical stoichiometry and reaction mole balances, offering a versatile way to engage with chemical processes.

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Most popular questions from this chapter

Carbon at \(1273 \mathrm{~K}\) is contained in a \(7-\mathrm{cm}\)-innerdiameter cylinder made of iron whose thickness is \(1.2 \mathrm{~mm}\). The concentration of carbon in the iron at the inner surface is \(0.5 \mathrm{~kg} / \mathrm{m}^{3}\) and the concentration of carbon in the iron at the outer surface is negligible. The diffusion coefficient of carbon through iron is \(3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). The mass flow rate of carbon by diffusion through the cylinder shell per unit length of the cylinder is (a) \(2.8 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (b) \(5.4 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (c) \(8.8 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(1.6 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\) (e) \(5.2 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\) 14-185 The surface of an iron component is to be hardened by carbon. The diffusion coefficient of carbon in iron at \(1000^{\circ} \mathrm{C}\) is given to be \(3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). If the penetration depth of carbon in iron is desired to be \(1.0 \mathrm{~mm}\), the hardening process must take at least (a) \(1.10 \mathrm{~h}\) (b) \(1.47 \mathrm{~h}\) (c) \(1.86 \mathrm{~h}\) (d) \(2.50 \mathrm{~h}\) (e) \(2.95 \mathrm{~h}\)

Consider a brick house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is \(85 \mathrm{kPa}\). The walls of the house are made of 20 -cm-thick brick whose permeance is \(23 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\). Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a \(3-\mathrm{m} \times 5-\mathrm{m}\) section of a wall during a 24-h period.

A heated piece of steel, with a uniform initial carbon concentration of \(0.20 \%\) by mass, was exposed to a carburizing atmosphere for an hour. Throughout the entire process, the carbon concentration on the surface was \(0.70 \%\). If the mass diffusivity of carbon in steel in this process was uniform at \(1 \times\) \(10^{-11} \mathrm{~m}^{2} / \mathrm{s}\), determine the percentage of mass concentration of carbon at \(0.2 \mathrm{~mm}\) and \(0.4 \mathrm{~mm}\) below the surface after the process.

Liquid methanol is accidentally spilt on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) laboratory bench and covered the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) air flow parallel over the bench surface. The air is maintained at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), and the concentration of methanol in the free stream is negligible. If the methanol vapor at the air-methanol interface has a pressure of \(4000 \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\), determine the evaporation rate of methanol in molar basis.

The basic equation describing the diffusion of one medium through another stationary medium is (a) \(j_{A}=-C D_{A B} \frac{d\left(C_{A} / C\right)}{d x}\) (b) \(j_{A}=-D_{A B} \frac{d\left(C_{A} / C\right)}{d x}\) (c) \(j_{A}=-k \frac{d\left(C_{A} / C\right)}{d x}\) (d) \(j_{A}=-k \frac{d T}{d x}\) (e) none of them
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