/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A thin plastic membrane is used ... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin plastic membrane is used to separate helium from a gas stream. Under steady-stare conditions the concentration of helium in the membrane is known to be \(0.02\) and \(0.005 \mathrm{kmol} / \mathrm{m}^{3}\) at the inner and outer surfaces, respectively. If the membrane is \(1 \mathrm{~mm}\) thick and the binary diffusion coefficient of helium with respect to the plastic is \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), what is the diffusive flux?

Short Answer

Expert verified
The diffusive flux is \( 1.5 \times 10^{-4} \; \mathrm{kmol/m^2/s} \).

Step by step solution

01

Understand the Variables

Before you start solving the problem, identify all the given values and what they represent.- Concentration of helium at the inner surface: \( C_1 = 0.02 \; \mathrm{kmol/m^3} \)- Concentration of helium at the outer surface: \( C_2 = 0.005 \; \mathrm{kmol/m^3} \)- Thickness of the membrane: \( \,L = 1 \; \mathrm{mm} = 0.001 \; \mathrm{m} \)- Binary diffusion coefficient: \( D = 10^{-5} \; \mathrm{m^2/s} \)These will be used in calculating the diffusive flux using Fick's Law.
02

Apply Fick's First Law of Diffusion

Fick's First Law of Diffusion states that the diffusive flux \( J \) is proportional to the concentration gradient across the membrane.The formula is: \[ J = -D \cdot \frac{C_2 - C_1}{L} \]Substitute the known values into this formula.
03

Calculate the Concentration Gradient

Calculate the concentration gradient \( \frac{C_2 - C_1}{L} \):\[ C_2 - C_1 = 0.005 \kmol/m^3 - 0.02 \kmol/m^3 = -0.015 \kmol/m^3 \]\[ \frac{C_2 - C_1}{L} = \frac{-0.015}{0.001} = -15 \; \mathrm{kmol/m^4} \]
04

Compute the Diffusive Flux using Fick's Law

Now, calculate the diffusive flux \( J \) by substituting the concentration gradient into Fick's Law:\[ J = -10^{-5} \cdot (-15) = 1.5 \times 10^{-4} \; \mathrm{kmol/m^2/s} \]
05

Interpret the Result

The positive value of the flux (\( 1.5 \times 10^{-4} \; \mathrm{kmol/m^2/s} \)) indicates the net flow of helium diffuses through the membrane from the inner surface to the outer surface due to the concentration difference.

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Key Concepts

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

Fick's Law
Fick's Law of Diffusion is a fundamental principle in understanding how substances spread. It all started with Adolf Fick, who developed this law in 1855. Fick's Law states that the diffusive flux, or the flow of particles across a unit area, is proportional to the concentration gradient. This is like saying water flows faster where the slope is steeper. The law is usually written as follows:
\[ J = -D \frac{dC}{dx} \].
Here, \( J \) represents the diffusive flux, \( D \) is the diffusion coefficient, which measures how easily the substance spreads, and \( \frac{dC}{dx} \) is the concentration gradient, which we will cover shortly. This mathematical relation shows that the flux goes from high to low concentration, getting its negative sign from this direction of flow.
- Diffusion coefficient, \( D \), is a material-specific property and varies with temperature.- The minus sign in the formula indicates directionality from higher to lower concentration.- Fick’s Law can be applied in various fields like chemistry, biology, and environmental science.By applying Fick's Law, scientists and engineers can predict how substances will move over time. This principle helps design better materials and processes in industries such as pharmacology and environmental science.
Concentration Gradient
The concentration gradient is a term that describes a change in the concentration of a substance over a distance. Imagine it like a hill that molecules "roll" downwards from higher to lower concentration areas. In simpler terms, it's just the rate at which concentration changes from one point to another and is key in the diffusion process.
The gradient can be calculated by subtracting the concentration at the lower surface from the concentration at the upper surface, and then dividing by the distance between these points:
\[ \frac{C_2 - C_1}{L} \].
- \( C_1 \) and \( C_2 \) represent the concentrations at two different locations.- \( L \) is the distance over which the concentration change occurs.A steeper concentration gradient typically means a faster rate of diffusion because there's a "larger hill" causing the substance to spread out more quickly.
Understanding concentration gradients is crucial in predicting how molecules move, whether in the atmosphere, in living organisms, or even between the layers of a membrane in industrial applications. This helps us control the speed and efficiency of diffusion-related processes.
Diffusive Flux
Diffusive flux is essentially the speed and amount at which molecules spread from one area to another. Think of it like how fast and intensely one perfume would spread through the air in a room. In scientific terms, the diffusive flux \( J \) is the amount of substance that passes through a unit area per unit time due to diffusion.
The formula for diffusive flux according to Fick's First Law is:
\[ J = -D \frac{(C_2 - C_1)}{L} \].The main components in this formula are:- \( J \): Diffusive flux (measured in \( \mathrm{kmol/m^2/s} \)).- \( D \): Diffusion coefficient determining how easily particles spread.- \( C_2 - C_1 \): Difference in concentration between two areas.- \( L \): Distance between the areas, or thickness of the layer.In our exercise, the calculation of this flux allowed us to determine how fast helium travels through a plastic membrane, an important fact for applications in gas separation technologies.
Understanding diffusive flux is vital in predicting and controlling how substances move in various environments. This has practical implications in many fields, including medicine, manufacturing, and environmental science.

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Most popular questions from this chapter

Steel is carburized in a high-temperature process that depends on the transfer of carbon by diffusion. The value of the diffusion coefficient is strongly temperature dependent and may be approximated as \(D_{c-3}\) \(\left(\mathrm{m}^{2} / \mathrm{s}\right)=2 \times 10^{-5} \operatorname{cxp}[-17,000 / T(\mathrm{~K})]\). If the process is effected at \(1000^{\circ} \mathrm{C}\) and a carbon mole fraction of \(0.02\) is maintained at the surface of the steel. how much time is required to elevate the carbon content of the steel from an initial value of \(0.1 \%\) to a value of \(1.0 \%\) at a depth of \(1 \mathrm{~mm}\) ?

Consider a spherical organism of radius \(r_{k}\) within which respiration occurs at a uniform volumetric rate of \(\dot{N}_{\mathrm{A}}=-k_{0}\). That is, oxygen (species A) consumption is governed by a zero-arder, homogeneous chemical reaction. (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{\mathrm{e}}\right)=C_{\mathrm{A}}\), is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen. \(C_{A}(r)\). within the organism. From your solution. can you discem any limits on applicability of the result? (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an crganism of radius \(r_{c}=0.10 \mathrm{~mm}\) and a diffusicn coefficient for oxygen transfer of \(D_{\mathrm{AB}}=10^{-1} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{An}}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\) and \(k_{0}=1.2 \times 10^{-4} \mathrm{kmol} / \mathrm{s} \cdot \mathrm{m}^{3}\), what is the molar concentration of \(\mathrm{O}_{2}\) at the center of the organism?

Ultra-pure hydrogen is required in applications ranging from the manufacturing of semiconductors to powcring fuel cells. The crystalline structure of palladium allows only the transfer of atomic hydrogen (H) through its thickness, and therefore palladium membranes are used to filter hydrogen from contaminated streams containing mixtures of hydrogen and other gases. Hydrogen molecules \(\left(\mathrm{H}_{2}\right)\) are first adserbed onto the palladium's surface and ate then dissociated into atoms (H), which subsequently diffuse through the metal. The H atoms recombine on the opposite side of the membrane, forming pure \(\mathrm{H}_{2}\). The surface concentration of H takes the form \(C_{\mathrm{H}}=K_{S} p_{H_{1},}^{09}\), where \(K_{,}=\) \(1.4 \mathrm{kmol}^{3} \mathrm{~m}^{3} \cdot\) bar \(^{\mathrm{ns}}\) is known as Sievert's constant. Consider an industrial hydrogen purifier consisting of an array of palladium tubes with one tube cnd connected to a collector plenum and the other end closed. The tube bank is inserted into a shell. Impure \(\mathrm{H}_{2}\) at \(T=600 \mathrm{~K}, p=15\) bars, \(x_{\mathrm{H}_{2}}=0.85\) is introduced into the shell while pure \(\mathrm{H}_{2}\) at \(p=6\) bars, \(T=600 \mathrm{~K}\) is extracted through the tubes. Determine the production rate of pure hydrogen \((\mathrm{kg} / \mathrm{h})\) for \(N=100\) tubes which are of inside diameler \(D_{0}=1.6 \mathrm{~mm}\), wall thickness \(t=\) \(75 \mu \mathrm{m}\), and length \(L=80 \mathrm{~mm}\). The mass diffusivity of hydrogen (H) in palladium at \(600 \mathrm{~K}\) is approximately \(D_{N A}=7 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Insulation degrades (experiences an increase in thermal conductivity) if it is subjected to water vapor condensation. The problem may occur in home insulation during cold periods, when vapor in a humidified room diffuses through the dry wall (plaster board) and condenses in the adjoining insulation. Estimate the mass diffusion rate for a \(3 \mathrm{~m}\) by \(5 \mathrm{~m}\) wall, under conditions for which the vapor pressure is \(0.03\) bar in the room air and \(0.0\) bar in the insulation. The dry wall is \(10 \mathrm{~mm}\) thick, and the solubility of water vapor in the wall material is approximately \(5 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\). bar. The binary diffusion coefficient for water vapor in the dry wall is approximately \(10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Consider air in a closed, cylindrical container with its axis vertical and with opposite ends maintained at different temperatures. Assume that the total pressure of the air is uniform throughout the container. (a) If the bottom surface is colder than the top surface, what is the nature of conditions within the container? For example, will there be vertical gradients of the species \(\left(\mathrm{O}_{2}\right.\) and \(\left.\mathrm{N}_{2}\right)\) concentrations? Is there any motion of the air? Does mass transfer occur? (b) What is the nature of conditions within the container if it is inverted (i.e., the warm surface is now at the bottom)?

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