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Chapter 14: Problem 53

A 1-mm-thick square \((100 \mathrm{~mm} \times 100 \mathrm{~mm})\) sheet of polymer is suspended from a precision scale in a chamber characterized by a temperature and relative humidity of \(T=300 \mathrm{~K}\) and \(\phi=0\), respectively. Suddenly, at time \(t=0\), the chamber's relative humidity is raised to \(\phi=0.95\). The measured mass of the sheet increases by \(0.012 \mathrm{mg}\) over \(24 \mathrm{~h}\) and by \(0.016 \mathrm{mg}\) over \(48 \mathrm{~h}\). Determine the solubility and mass diffusivity of water vapor in the polymer. Preliminary experiments have indicated that the mass diffusivity is greater than \(7 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\).

Short Answer

Expert verified
The solubility \(S\) and mass diffusivity \(D\) of water vapor in the polymer can be determined by: \[S = \frac{\Delta C_2 - \Delta C_1}{\phi_1}\] \[D = \frac{\Delta C_1 \times (\Delta x)^2}{\Delta t_1 \times \Delta^2 C_1}\] Substitute the given values and solve for \(S\) and \(D\).

Step by step solution

01

Convert dimensions and mass increase

Given dimensions of the polymer sheet: - Thickness: \(1 \mathrm{~mm}\); convert to meters: \(1 \times 10^{-3} \mathrm{~m}\) - Side length: \(100 \mathrm{~mm}\); convert to meters: \(0.1 \mathrm{~m}\) Given mass increases: - Over 24h: \(0.012 \mathrm{~mg}\); convert to kg: \(1.2 \times 10^{-8} \mathrm{~kg}\) - Over 48h: \(0.016 \mathrm{~mg}\); convert to kg: \(1.6 \times 10^{-8} \mathrm{~kg}\) Step 2 鈥� Fick's second law of diffusion
02

Apply Fick's second law of diffusion

Fick's second law of diffusion describes the diffusion of a substance inside a medium: \[\frac{\partial C}{\partial t} = D \times \frac{\partial^2 C}{\partial x^2}\] Where: - \(C\) is the concentration (mass/volume) - \(t\) is time - \(D\) is the mass diffusivity - \(x\) is the spatial coordinate (distance inside the polymer) Step 3 鈥� Concentration profile
03

Obtain concentration profile

We need to obtain an expression for the concentration profile of water vapor in the polymer. Since the sheet is very thin and the mass diffusivity is small, we can assume a linear concentration profile with the concentration increasing linearly with depth into the sheet. This simplifies Fick's second law to: \[\frac{\Delta C}{\Delta t} = D \times \frac{\Delta^2 C}{(\Delta x)^2}\] Step 4 鈥� Solving for solubility and mass diffusivity
04

Determine solubility and mass diffusivity

The mass increase over time is proportional to the change in concentration. Using the given mass increases and the known sheet dimensions, we can determine the rate of mass increase per unit area: 1. Over 24h: \(\frac{1.2 \times 10^{-8} \mathrm{~kg}}{0.1 \mathrm{~m} \times 0.1 \mathrm{~m} \times 24 \mathrm{~h} \times 3600 \mathrm{~s/h}} = A_1\) 2. Over 48h: \(\frac{1.6 \times 10^{-8} \mathrm{~kg}}{0.1 \mathrm{~m} \times 0.1 \mathrm{~m} \times 48 \mathrm{~h} \times 3600 \mathrm{~s/h}} = A_2\) Let the solubility of water vapor in the polymer be \(S\). The initial concentration of water vapor in the chamber is \(\phi_0 \times S\), and the final concentration is \(\phi_1 \times S\), where \(\phi_0 = 0\) and \(\phi_1 = 0.95\). Then, the change in concentration over time is given by: 1. Over 24h: \(\Delta C_1 = A_1 \times \Delta t_1\) 2. Over 48h: \(\Delta C_2 = A_2 \times \Delta t_2\) Solving for the solubility \(S\) and mass diffusivity \(D\), we get: \[S = \frac{\Delta C_2 - \Delta C_1}{\phi_1}\] \[D = \frac{\Delta C_1 \times (\Delta x)^2}{\Delta t_1 \times \Delta^2 C_1}\] Now, we substitute the values obtained earlier for \(A_1\), \(A_2\), \(\Delta C_1\), and \(\Delta C_2\) and obtain the solubility and mass diffusivity of water vapor in the polymer.

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Key Concepts

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

Fick's Second Law of Diffusion
In the realm of material science, particularly when examining polymers, one of the pivotal concepts is Fick's second law of diffusion. This law gives us a mathematical expression to describe the time-dependent concentration changes within a material. Imagine dropping a single drop of food coloring into a glass of water; as time passes, the color slowly spreads out. Fick's second law gives us the equations to predict this process quantitatively.

For the second law, we use the formula: \[\frac{\partial C}{\partial t} = D \times \frac{\partial^2 C}{\partial x^2}\]
Here, \(C\) represents the concentration of the diffusing species (such as water vapor in our case) at a specific location and time, \(D\) is the diffusivity or how easily molecules spread out, and \(\frac{\partial^2 C}{\partial x^2}\) denotes the change in concentration with distance, \(x\). What we understand from this law is that the spread of particles is not just random; it follows a pattern that can be modeled mathematically, which is fundamental when predicting how a polymer will behave when exposed to substances like water vapor.
Solubility
Solubility is a measure of the capacity of a material to dissolve in another. In the case of polymer science, solubility refers to how much of a particular substance can be absorbed by the polymer. Considering the exercise at hand, the substance is water vapor and the material is a polymer sheet. The solubility is crucial for predicting not only how much water can be absorbed but also influences subsequent material properties such as its flexibility, durability, or even its electrical conductivity.

The exercise aims to determine the solubility of water vapor in the polymer, which has practical implications. For instance, if we are looking at packaging materials, knowing how much moisture can be absorbed can be critical to preserving the quality of the goods being protected. Moreover, the interaction between water molecules and polymer chains can lead to swelling, which is an important aspect in the field of biomedical applications, such as drug delivery systems where controlled swelling is often desirable.
Concentration Profile
The concentration profile within a material describes how the concentration of a substance varies across the material. In other words, it's a snapshot of where particles of the diffused substance are within the host material at a given time. For our exercise, we examine water vapor's concentration profile within a polymer sheet as the humidity in the surrounding environment changes.

In many cases, especially when the diffusing substance has low mobility, or the diffusion occurs over short times or through thin materials, we can simplify the profile as being linear. This means we can expect a steady increase or decrease in concentration from one side of the material to the other. For thicker materials or over longer periods, the profile may take on more complex shapes, reflecting various rates of diffusion at different depths. Understanding these profiles helps engineers and scientists predict the behavior of materials in real-world applications such as in moisture barriers or in situations where gradient-driven diffusion plays a key role.
Water Vapor Diffusion
Water vapor diffusion is particularly interesting and relevant in polymer science because many polymers are used in contexts where they are exposed to moisture. Water vapor diffusion refers to the movement of water vapor molecules through a polymer matrix. This can affect a polymer's physical properties, such as causing swelling or even hydrolyzing bonds within the polymer chain, resulting in degradation.

Understanding how water vapor diffuses through polymers is key for designing materials with the right properties for their intended use. Whether it's creating water-resistant coatings or designing films that allow for controlled release of water vapor, being able to predict and control the rate of diffusion is crucial. This understanding can lead to innovations such as breathable materials in clothing, efficient drying processes in manufacturing, or protective packaging that prevents moisture damage to sensitive electronic components.

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Most popular questions from this chapter

Consider a spherical organism of radius \(r_{o}\) 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-order, homogeneous chemical reaction. (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{o}\right)=C_{\mathrm{A}, o}\) is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen, \(C_{\mathrm{A}}(r)\), within the organism. From your solution, can you discern any limits on applicability of the result? (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an organism of radius \(r_{o}=0.10 \mathrm{~mm}\) and a diffusion coefficient for oxygen transfer of \(D_{\mathrm{AB}}=10^{-8} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{A}, o}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\)and \(k_{0}=1.2 \times 10^{-4} \mathrm{kmol} / \mathrm{s}^{*} \mathrm{~m}^{3}\), what is the molar concentration of \(\mathrm{O}_{2}\) at the center of the organism?

Referring to Problem 14.34, a more representative model of respiration in a spherical organism is one for which oxygen consumption is governed by a firstorder reaction of the form \(\dot{N}_{\mathrm{A}}=-k_{1} C_{\mathrm{A}}\). (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{o}\right)=C_{\mathrm{A}, o}\) is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen, \(C_{\mathrm{A}}(r)\), within the organism. Hint: To simplify solution of the species diffusion equation, invoke the transformation \(y \equiv r C_{\mathrm{A}}\). (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an organism of radius \(r_{o}=0.10 \mathrm{~mm}\) and a diffusion coefficient of \(D_{\mathrm{AB}}=10^{-8} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{A}, o}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\) and \(k_{1}=20 \mathrm{~s}^{-1}\), estimate the corresponding value of the molar concentration at the center of the organism. What is the rate of oxygen consumption by the organism?

A solar pond operates on the principle that heat losses from a shallow layer of water, which acts as a solar absorber, may be minimized by establishing a stable vertical salinity gradient in the water. In practice such a condition may be achieved by applying a layer of pure salt to the bottom and adding an overlying layer of pure water. The salt enters into solution at the bottom and is transferred through the water layer by diffusion, thereby establishing salt-stratified conditions. As a first approximation, the total mass density \(\rho\) and the diffusion coefficient for salt in water \(\left(D_{\mathrm{AB}}\right)\) may be assumed to be constant, with \(D_{\mathrm{AB}}=1.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). (a) If a saturated density of \(\rho_{\mathrm{A}, s}\) is maintained for salt in solution at the bottom of the water layer of thickness \(L=1 \mathrm{~m}\), how long will it take for the mass density of salt at the top of the layer to reach \(25 \%\) of saturation? (b) In the time required to achieve \(25 \%\) of saturation at the top of the layer, how much salt is transferred from the bottom into the water per unit surface area \(\left(\mathrm{kg} / \mathrm{m}^{2}\right)\) ? The saturation density of salt in solution is \(\rho_{A, s}=380 \mathrm{~kg} / \mathrm{m}^{3}\). (c) If the bottom is depleted of salt at the time that the salt density reaches \(25 \%\) of saturation at the top, what is the final (steady-state) density of the salt at the bottom? What is the final density of the salt at the top?

A 100-mm-long, hollow iron cylinder is exposed to a \(1000^{\circ} \mathrm{C}\) carburizing gas (a mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) ) at its inner and outer surfaces of radii \(4.30\) and \(5.70 \mathrm{~mm}\), respectively. Consider steady-state conditions for which carbon diffuses from the inner surface of the iron wall to the outer surface and the total transport amounts to \(3.6 \times 10^{-3} \mathrm{~kg}\) of carbon over \(100 \mathrm{~h}\). The variation of the carbon composition (weight \(\%\) carbon) with radius is tabulated for selected radii. $$ \begin{array}{lllllllll} r(\mathrm{~mm}) & 4.49 & 4.66 & 4.79 & 4.91 & 5.16 & 5.27 & 5.40 & 5.53 \\ \text { Wt.C }(\%) & 1.42 & 1.32 & 1.20 & 1.09 & 0.82 & 0.65 & 0.46 & 0.28 \end{array} $$ (a) Beginning with Fick's law and the assumption of a constant diffusion coefficient, \(D_{\mathrm{C}-\mathrm{Fe}}\), show that \(d \rho_{\mathrm{C}} / d(\ln r)\) is a constant. Sketch the carbon mass density, \(\rho_{\mathrm{C}}(r)\), as a function of \(\ln r\) for such a diffusion process. (b) The foregoing table corresponds to measured distributions of the carbon mass density. Is \(D_{\mathrm{C}-\mathrm{Fe}}\) constant for this diffusion process? If not, does \(D_{\mathrm{C}-\mathrm{Fe}}\) increase or decrease with an increasing carbon concentration? (c) Using the experimental data, calculate and tabulate \(D_{\mathrm{C}-\mathrm{Fe}}\) for selected carbon compositions.

Beginning with a differential control volume, derive the diffusion equation, on a molar basis, for species A in a three-dimensional (Cartesian coordinates), stationary medium, considering species generation with constant properties. Compare your result with Equation 14.48b.
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