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Chapter 14: Problem 9

An old-fashioned glass apothecary jar contains a patent medicine. The neck is closed with a rubber stopper that is \(20 \mathrm{~mm}\) tall, with a diameter of \(10 \mathrm{~mm}\) at the bottom end, widening to \(20 \mathrm{~mm}\) at the top end. The molar concentration of medicine vapor in the stopper is \(2 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) at the bottom surface and is negligible at the top surface. If the mass diffusivity of medicine vapor in rubber is \(0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\), find the rate \((\mathrm{kmol} / \mathrm{s})\) at which vapor exits through the stopper.

Short Answer

Expert verified
Using Fick's first law of diffusion, the rate of vapor exiting the stopper (\(R_{vapor}\)) is calculated as: \[R_{vapor} = J \times A_{average}\] where \(J = -D\frac{\mathrm{d}C}{\mathrm{d}x}\) is the diffusion rate, \(D = 0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\) is the diffusivity, and \(A_{average}\) is the average cross-sectional area of the stopper.

Step by step solution

01

1. Find the cross-sectional area at the bottom end

First, we need to find the cross-sectional area at the bottom end of the rubber stopper. Since it has a circular shape, we can calculate the area using the formula: \[A_{bottom} = \pi (r_{bottom})^2\] Where \(A_{bottom}\) is the cross-sectional area at the bottom end, \(r_{bottom}\) is the radius of the bottom end and \(\pi\) is a mathematical constant which is approximately equal to 3.14159. The diameter of the bottom end is given as 10 mm. So the radius is half of that, which is 5 mm or 0.005 m.
02

2. Find the cross-sectional area at the top end

Similar to step 1, we need to find the cross-sectional area at the top end of the rubber stopper. Since it has a circular shape, we can calculate the area using the formula: \[A_{top} = \pi (r_{top})^2\] Where \(A_{top}\) is the cross-sectional area at the top end, \(r_{top}\) is the radius of the top end. The diameter of the top end is given as 20 mm. So the radius is half of that, which is 10 mm or 0.010 m.
03

3. Calculate the average cross-sectional area

We can find the average cross-sectional area of the rubber stopper by taking the arithmetic mean of the bottom and top areas. \[A_{average} = \frac{A_{bottom} + A_{top}}{2}\]
04

4. Apply Fick's first law of diffusion

Fick's first law states that the rate of diffusion (\(J\)) is proportional to the concentration gradient and the mass diffusivity (\(D\)). Mathematically, it can be written as: \[J = -D\frac{\mathrm{d}C}{\mathrm{d}x}\] where \(x\) is the distance through the rubber stopper, \(D\) is the diffusivity of the vapor in the rubber stopper, and \(\frac{\mathrm{d}C}{\mathrm{d}x}\) is the concentration gradient. In our case, the concentration gradient is the difference in molar concentration divided by the height of the stopper: \[\frac{\mathrm{d}C}{\mathrm{d}x} = \frac{C_{bottom} - C_{top}}{h}\] where \(C_{bottom}\) is the molar concentration at the bottom end, \(C_{top}\) is the molar concentration at the top end (\(C_{top} \approx 0\)) and \(h = 20\) mm or 0.020 m is the height of the stopper.
05

5. Calculate the rate of vapor

Now that we have all the necessary variables to find the rate of diffusion, we can substitute them into the Fick's first law equation given in step 4: \[J = -D\frac{\mathrm{d}C}{\mathrm{d}x}\] where \(D = 0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\) is the diffusivity, and \(A_{average}\) is the average cross-sectional area calculated in step 3. Since the rate of vapor exiting the stopper (\(R_{vapor}\)) is equal to the diffusion rate (\(J\)) times the average cross-sectional area, we can now find the rate of vapor: \[R_{vapor} = J \times A_{average}\]

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Key Concepts

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

Fick's First Law
Fick's First Law is fundamental in understanding how substances move through various mediums. It helps us predict and calculate the rate at which a substance diffuses from one area to another.
The law, in simple terms, states that the diffusion process is driven by the difference in concentration between two points. This difference is what we term as the 'concentration gradient'. The greater the difference, the faster the substance moves.
Mathematically, Fick's First Law can be expressed as:
  • Rate of diffusion (\(J\)) is equal to the negative product of the mass diffusivity (\(D\)) and the concentration gradient (\(\frac{dC}{dx}\)).
  • The formula is given as \(J = -D\frac{dC}{dx}\).
This negative sign indicates that diffusion moves from high concentration to low concentration. It's like rolling downhill, moving naturally from an area of excess to one that has less. In our exercise, using Fick's First Law, we determined the rate at which vapor exits the stopper based solely on these principles.
Concentration Gradient
The concentration gradient is a crucial concept that acts as the driving force for diffusion. Simply put, it's the variation in concentration across a certain distance. Imagine you have a balloon filled with perfume. Once you pop it, the perfume spreads, moving from an area of high concentration inside the balloon to the lower concentration outside.
In terms of mathematics, the concentration gradient (\(\frac{dC}{dx}\)) is the change in concentration (\(C\)) over a change in position (\(x\)).
Key aspects to consider include:
  • It tells us how concentration changes over a certain distance.
  • The steeper the gradient, the faster diffusion happens.
  • It's the essence of why substances move from one place to another.
In the exercise context, the rubber stopper had a significant concentration difference at its bottom and top, acting as a force propelling the vapor to move and exit through the stopper.
Mass Diffusivity
Mass Diffusivity, often denoted as \(D\), is a property that indicates how quickly a substance spreads through another medium. It acts similarly to a speed limit on a highway. A higher diffusivity suggests that substances can spread quickly through the medium, while lower diffusivity implies slower movement.
  • Mass diffusivity depends on many factors, such as the nature of the substance and the medium through which it diffuses.
  • It's measured in square meters per second (\(m^2/s\)).
  • Use of mass diffusivity in our exercise allowed calculation of how quickly the medicine vapor diffused through the rubber stopper.
A crucial point to remember is that mass diffusivity is not constant; it may vary based on temperature, pressure, and the specific materials involved. In our exercise, knowing the mass diffusivity of the medicine vapor in rubber was essential for applying Fick's First Law to find the vapor's exit rate.

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Most popular questions from this chapter

A platinum catalytic reactor in an automobile is used to convert carbon monoxide to carbon dioxide in an oxidation reaction of the form \(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}\). Species transfer between the catalytic surface and the exhaust gases may be assumed to occur by diffusion in a film of thickness \(L=10 \mathrm{~mm}\). Consider an exhaust gas that has a pressure of \(1.2\) bars, a temperature of \(500^{\circ} \mathrm{C}\), and a \(\mathrm{CO}\) mole fraction of \(0.0012\). If the reaction rate constant of the catalyst is \(k_{1}^{\prime \prime}=0.005 \mathrm{~m} / \mathrm{s}\) and the diffusion coefficient of \(\mathrm{CO}\) in the mixture is \(10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), what is the molar concentration of \(\mathrm{CO}\) at the catalytic surface? What is the rate of removal of \(\mathrm{CO}\) per unit area of the catalyst? What is the removal rate if \(k_{1}^{\prime \prime}\) is adjusted to render the process diffusion limited?

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.

Consider the DVD of Problem 14.49, except now the reacting polymer is blended uniformly with the polycarbonate to reduce manufacturing costs. Assume that a first-order homogeneous chemical reaction takes place between the polymer and oxygen; the reaction rate is proportional to the oxygen molar concentration. (a) Write the governing equation, boundary conditions, and initial condition for the oxygen molar concentration after the DVD is removed from the oxygen- proof pouch, for a DVD of thickness \(2 L\). Do not solve. (b) The DVD will gradually become more opaque over time as the reaction proceeds. The ability to read the DVD will depend on how well the laser light can penetrate through the thickness of the DVD. Therefore, it is important to know the volume-averaged molar concentration of product, \(\bar{C}_{\text {prod }}\), as a function of time. Write an expression for \(\bar{C}_{\text {prod }}\) in terms of the oxygen molar concentration, assuming that every mole of oxygen that reacts with the polymer results in \(p\) moles of product.

As an employee of the Los Angeles Air Quality Commission, you have been asked to develop a model for computing the distribution of \(\mathrm{NO}_{2}\) in the atmosphere. The molar flux of \(\mathrm{NO}_{2}\) at ground level, \(N_{\mathrm{A}, 0}^{N}\), is presumed known. This flux is attributed to automobile and smoke stack emissions. It is also known that the concentration of \(\mathrm{NO}_{2}\) at a distance well above ground level is zero and that \(\mathrm{NO}_{2}\) reacts chemically in the atmosphere. In particular, \(\mathrm{NO}_{2}\) reacts with unburned hydrocarbons (in a process that is activated by sunlight) to produce PAN (peroxyacetylnitrate), the final product of photochemical smog. The reaction is first order, and the local rate at which it occurs may be expressed as \(\dot{N}_{\mathrm{A}}=-k_{1} C_{\mathrm{A}}\). (a) Assuming steady-state conditions and a stagnant atmosphere, obtain an expression for the vertical distribution \(C_{\mathrm{A}}(x)\) of the molar concentration of \(\mathrm{NO}_{2}\) in the atmosphere. (b) If an \(\mathrm{NO}_{2}\) partial pressure of \(p_{\mathrm{A}}=2 \times 10^{-6}\) bar is sufficient to cause pulmonary damage, what is the value of the ground level molar flux for which you would issue a smog alert? You may assume an isothermal atmosphere at \(T=300 \mathrm{~K}\), a reaction coefficient of \(k_{1}=0.03 \mathrm{~s}^{-1}\), and an \(\mathrm{NO}_{2}\)-air diffusion coefficient of \(D_{\mathrm{AB}}=0.15 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\).

Consider an ideal gas mixture of \(n\) species. (a) Derive an equation for determining the mass fraction of species \(i\) from knowledge of the mole fraction and the molecular weight of each of the \(n\) species. Derive an equation for determining the mole fraction of species \(i\) from knowledge of the mass fraction and the molecular weight of each of the \(n\) species. (b) In a mixture containing equal mole fractions of \(\mathrm{O}_{2}\), \(\mathrm{N}_{2}\), and \(\mathrm{CO}_{2}\), what is the mass fraction of each species? In a mixture containing equal mass fractions of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{CO}_{2}\), what is the mole fraction of each species?
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