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.

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

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 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: $$t_x \approx 4.15 \times 10^6\mathrm{s}$$

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.