Question

A thick wall made of natural rubber is exposed to pure oxygen gas on one side of its surface. Both the wall and oxygen gas are isothermal at \(25^{\circ} \mathrm{C}\), and the oxygen concentration at the wall surface is constant. Determine the time required for the oxygen concentration at \(x=5 \mathrm{~mm}\) to reach \(5 \%\) of its concentration at the wall surface.

Short answer

Answer: Approximately 48 days.

Step-by-step solution

Write down the known information and identify Fick's second law of diffusion

First, let's list down the known information: 1. The thickness of rubber wall, \(x = 5 \mathrm{~mm}\) 2. The oxygen concentration at \(x = 5 \mathrm{~mm}\) should reach \(5\%\) of its concentration at the wall surface 3. The temperature is isothermal at \(25^{\circ} \mathrm{C}\) We will need to determine the diffusion coefficient of oxygen in natural rubber at \(25^{\circ} \mathrm{C}\). A literature search or reference to a materials science book will reveal the value: \(D = 3.5 \times 10^{-11} \mathrm{m^2/s}\). Fick's second law of diffusion gives the time-dependent concentration within a material, which we can use in this problem: $$\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2}$$ Here, - \(C\) is the concentration of oxygen within the rubber - \(D\) is the diffusion coefficient - \(t\) is time - \(x\) is the position in the rubber

Apply the boundary conditions and solve the equation

To find the time required for the concentration to reach \(5\%\) of its surface value at \(x = 5 \mathrm{~mm}\), we need to apply the boundary conditions. Let \(C_0\) be the initial concentration at the wall surface, \(C_x\) the concentration at \(x = 5 \mathrm{~mm}\), and \(t_x\) the required time for this to happen. We are given that the concentration at \(x=5 \mathrm{~mm}\) is \(5 \%\) of \(C_0\): $$C_x = \frac{1}{20}C_0$$ Applying these boundary conditions to Fick's second law, we can use the error function to obtain the solution: $$\frac{C_x - C_0}{C_0} = \operatorname{erf}\biggl(\frac{x}{2\sqrt{Dt_x}}\biggr)$$ We need to solve for \(t_x\).

Rearrange the equation and solve for the time required

Rearrange the equation to solve for \(t_x\): $$t_x = \frac{x^2}{4D \cdot [\operatorname{erf}^{-1}(C_x/C_0)]^2}$$ Now substitute the values \(x = 5 \mathrm{~mm}\), \(D = 3.5 \times 10^{-11} \mathrm{m^2/s}\), and \(C_x / C_0 = 1/20\) to solve for \(t_x\): $$t_x = \frac{(5 \times 10^{-3})^2}{4(3.5 \times 10^{-11}) \cdot [\operatorname{erf}^{-1}(1/20)]^2}$$ After calculating the inverse error function and the whole equation, we get: $$t_x \approx 4.15 \times 10^6\mathrm{s}$$

Convert the time into a more convenient unit

To express the time in a more convenient unit, we can convert seconds to days: $$t_x = 4.15 \times 10^6 \mathrm{s} \times \frac{1 \mathrm{day}}{86400 \mathrm{s}} \approx 48 \mathrm{days}$$ So, it takes approximately 48 days for the oxygen concentration at \(x=5 \mathrm{~mm}\) to reach \(5 \%\) of its concentration at the wall surface.

Key concepts

Diffusion Coefficient

The diffusion coefficient, represented as \(D\), is a crucial parameter in understanding how substances like oxygen move through a material like natural rubber. It essentially measures the rate at which particles spread out over time.
In the context of Fick's Second Law of Diffusion, the diffusion coefficient helps predict how quickly a substance will move from areas of high concentration to areas of low concentration within the material.

• For this specific problem, the diffusion coefficient of oxygen in natural rubber at \(25^{\circ} \mathrm{C}\) is given as \(3.5 \times 10^{-11} \mathrm{m^2/s}\).
• This value is often determined experimentally or found in scientific literature.

Understanding the diffusion coefficient allows us to calculate how long it will take for a certain concentration of oxygen to reach a specific point in the rubber wall.

Isothermal Conditions

Isothermal conditions refer to a process occurring at a constant temperature. In this problem, both the wall and the oxygen gas are maintained at \(25^{\circ} \mathrm{C}\).
Consistent temperature ensures that the diffusion properties remain unchanged because the diffusion coefficient is temperature-dependent.

• When dealing with diffusion, keeping a constant temperature means that the kinetics of particle movement don't change.
• This stability simplifies calculations, as we don't need to account for changes in \(D\) caused by temperature fluctuations.

Simply put, isothermal conditions make the process easier to predict and model, allowing for straightforward application of Fick's Second Law.

Boundary Conditions

Boundary conditions are crucial in solving differential equations like Fick's Second Law, as they define specific constraints at the surfaces of the material.
They help in understanding how the concentration of oxygen behaves at different points within the material.

• In this scenario, it's given that the oxygen concentration at \(x = 5 \mathrm{~mm}\) reaches \(5\%\) of the concentration at the wall surface.
• This can be mathematically expressed as: \(C_x = \frac{1}{20}C_0\), where \(C_0\) is the wall surface concentration.

By incorporating these boundary conditions, we can solve for the time required to achieve the desired concentration at \(x = 5 \mathrm{~mm}\) by rearranging and solving the diffusion equation using the error function. This ensures accurate modeling of how the concentration changes over time within the rubber.