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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
The diffusion coefficients in both the mass basis and mole basis relations of Fick's law of diffusion are the same.

Step by step solution

01

Analyze mass basis equation

The mass basis equation of Fick's law of diffusion is given as: \(\dot{m}_{\mathrm{diff}, A}=-\rho A D_{A B}\left(d w_{A} / d x\right)\) Here, \(\dot{m}_{\mathrm{diff}, A}\) represents the rate of mass transfer of component A by diffusion (kg/s), \(\rho\) is the density of the mixture (kg/m³), \(A\) is the cross-sectional area (m²), \(D_{AB}\) is the diffusion coefficient on a mass basis (m²/s), \(w_A\) is the mass fraction of component A, and \(x\) denotes the spatial direction (m).
02

Analyze mole basis equation

The mole basis equation of Fick's law of diffusion is given as: \(\dot{N}_{\mathrm{diff}, A}=\) \(-C A D_{A B}\left(d y_{A} / d x\right)\) Here, \(\dot{N}_{\mathrm{diff}, A}\) represents the rate of molar transfer of component A by diffusion (mol/s), \(C\) is the molar concentration of the mixture (mol/m³), \(A\) is the cross-sectional area (m²), \(D_{AB}\) is the diffusion coefficient on a mole basis (m²/s), \(y_A\) is the mole fraction of component A, and \(x\) denotes the spatial direction (m).
03

Compare the diffusion coefficients

Let's compare the two diffusion coefficients by analyzing their units: - Diffusion coefficient on a mass basis (\(D_{AB}\) in the mass basis equation): m²/s - Diffusion coefficient on a mole basis (\(D_{AB}\) in the mole basis equation): m²/s Both diffusion coefficients have the same units, which indicates that they represent the same quantity.
04

Conclusion

The diffusion coefficients \(D_{AB}\) in both the mass basis and mole basis relations of Fick's law of diffusion are the same.

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Most popular questions from this chapter

Liquid methanol is accidentally spilled on a $1-\mathrm{m} \times 1-\mathrm{m}$ laboratory bench and covers the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) airflow 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 on a molar basis.

What is the relation \((f / 2) \operatorname{Re}=\mathrm{Nu}=\) Sh known as? Under what conditions is it valid? What is the practical importance of it?

Air flows in a 4-cm-diameter wet pipe at \(20^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}\( with an average velocity of \)4 \mathrm{~m} / \mathrm{s}$ in order to dry the surface. The Nusselt number in this case can be determined from \(\mathrm{Nu}=0.023 \operatorname{Re}^{0.8} \mathrm{Pr}^{0.4}\) where \(\operatorname{Re}=10,550\) and \(\mathrm{Pr}=\) \(0.731\). Also, the diffusion coefficient of water vapor in air is $2.42 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$. Using the analogy between heat and mass transfer, the mass transfer coefficient inside the pipe for fully developed flow becomes (a) \(0.0918 \mathrm{~m} / \mathrm{s}\) (b) \(0.0408 \mathrm{~m} / \mathrm{s}\) (c) \(0.0366 \mathrm{~m} / \mathrm{s}\) (d) \(0.0203 \mathrm{~m} / \mathrm{s}\) (e) \(0.0022 \mathrm{~m} / \mathrm{s}\)

A 1-in-diameter Stefan tube is used to measure the binary diffusion coefficient of water vapor in air at \(80^{\circ} \mathrm{F}\) and \(13.8\) psia. The tube is partially filled with water with a distance from the water surface to the open end of the tube of \(10 \mathrm{in}\). Dry air is blown over the open end of the tube so that water vapor rising to the top is removed immediately and the concentration of vapor at the top of the tube is zero. During 10 days of continuous operation at constant pressure and temperature, the amount of water that has evaporated is measured to be $0.0025 \mathrm{lbm}$. Determine the diffusion coefficient of water vapor in air at \(80^{\circ} \mathrm{F}\) and \(13.8\) psia.

Reconsider Prob. 14-82. In order to reduce the migration of water vapor through the wall, it is proposed to use a \(0.051-\mathrm{mm}\)-thick polyethylene film with a permeance of $9.1 \times 10^{-12} \mathrm{~kg} / \mathrm{s}^{2} \mathrm{~m}^{2}$.Pa. Determine the amount of water vapor that will diffuse through the wall in this case during a \(24-\mathrm{h}\) period. Answer: \(26.4 \mathrm{~g}\)

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