Question

Nitrogen gas at high pressure and \(298 \mathrm{~K}\) is contained in a \(2-\mathrm{m} \times 2-\mathrm{m} \times 2-\mathrm{m}\) cubical container made of natural rubber whose walls are \(4 \mathrm{~cm}\) thick. The concentration of nitrogen in the rubber at the inner and outer surfaces are \(0.067 \mathrm{~kg} / \mathrm{m}^{3}\) and \(0.009 \mathrm{~kg} / \mathrm{m}^{3}\), respectively. The diffusion coefficient of nitrogen through rubber is \(1.5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). The mass flow rate of nitrogen by diffusion through the cubical container is (a) \(8.24 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (b) \(1.35 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (c) \(5.22 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(9.71 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (e) \(3.58 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

Short answer

Answer: (c) \(5.22\times10^{-9}\,\mathrm{kg/s}\)

Step-by-step solution

Gather given information

: We have: - The concentration of nitrogen at the inner surface: \(C_1=0.067\,\mathrm{kg/m}^3\) - The concentration of nitrogen at the outer surface: \(C_2=0.009\,\mathrm{kg/m}^3\) - The thickness of the rubber walls: \(x_2-x_1 =0.04\,\mathrm{m}\) - The diffusion coefficient of nitrogen through rubber: \(D=1.5\times10^{-10}\,\mathrm{m}^2/\mathrm{s}\)

Calculate the concentration gradient

: The concentration gradient \(\Delta C = C_1 - C_2\). This gives us: \(\Delta C = 0.067\,\mathrm{kg/m}^3 - 0.009\,\mathrm{kg/m}^3 = 0.058\,\mathrm{kg/m}^3\)

Apply Fick's first law of diffusion

: We can calculate the mass flow rate per unit area \(J\) using Fick's first law of diffusion: \(J= -D \frac{\Delta C}{\Delta x}\) Plugging in the given values: \(J= -(1.5\times10^{-10}\,\mathrm{m}^2/\mathrm{s}) \left(\frac{0.058\,\mathrm{kg/m}^3}{0.04\,\mathrm{m}}\right)\) \(J= -2.175\times10^{-9}\,\mathrm{kg/m}^2\,\mathrm{s}\)

Calculate the surface area of the container

: The surface area of a cube is given by \(A=6L^2\), where \(L\) is the side length. In this case, \(L=2\,\mathrm{m}\). So: \(A=6(2\,\mathrm{m})^2=24\,\mathrm{m}^2\)

Calculate the total mass flow rate

: To find the total mass flow rate of nitrogen through the container, we need to multiply the mass flow rate per unit area by the surface area. The total mass flow rate is given by: \(\text{Mass flow rate}= J \times A\) Plugging in the values from Steps 3 and 4: \(\text{Mass flow rate}= (-2.175\times10^{-9}\,\mathrm{kg/m}^2\,\mathrm{s})(24\,\mathrm{m}^2)= -5.22\times10^{-8}\,\mathrm{kg/s}\) The mass flow rate is negative, indicating that nitrogen is diffusing out of the container. To report the rate of diffusion, we can take the absolute value of the mass flow rate: \(|\text{Mass flow rate}|= 5.22\times10^{-8}\,\mathrm{kg/s}\) Comparing this result to the given options, we find that the correct answer is: (c) \(5.22\times10^{-9}\,\mathrm{kg/s}\)

Key concepts

Fick's first law of diffusion

Fick's First Law of Diffusion is a fundamental concept in physics and chemistry that describes how molecules move from areas of high concentration to areas of low concentration. The law is often represented by the formula:
\[J = -D \frac{\Delta C}{\Delta x}\]where:

• \(J\) is the diffusion flux, representing the mass flow rate per unit area, measured in \(\text{kg/m}^2\text{s}\).
• \(D\) is the diffusion coefficient, a constant that measures how easily a substance diffuses through another, typically given in \(\text{m}^2/\text{s}\).
• \(\Delta C\) is the change in concentration across a distance, given in \(\text{kg/m}^3\).
• \(\Delta x\) is the thickness of the material the substance is diffusing through, measured in meters.

In essence, Fick's First Law tells us that the flow of particles between two points is directly proportional to the difference in concentration between those points and inversely proportional to the distance between them. This principle is crucial in many fields such as materials science, biology, and environmental science, where understanding the movement of particles is essential.

concentration gradient

The concentration gradient is a key concept in the study of diffusion. It is the driving force behind the movement of particles from one region to another. In mathematical terms, the concentration gradient is expressed as the difference in concentration of a substance between two points, divided by the distance separating those points. For example:
\[\Delta C = C_1 - C_2\]where:

• \(C_1\) is the concentration at the starting point.
• \(C_2\) is the concentration at the ending point.

A large concentration gradient means a steep slope, indicating a rapid change in concentration over a short distance, leading to a faster diffusion rate. Conversely, a smaller gradient indicates a slower rate of diffusion. The concept is critical in understanding how substances move across different materials, such as gases diffusing through a membrane or solutes moving in liquid solutions.

mass flow rate calculation

Calculating the mass flow rate of a diffusing substance involves a few simple steps, often leveraging Fick's First Law. In the context of the exercise provided, the mass flow rate calculation steps were as follows:
1. **Identify the diffusion flux**: Calculate the mass flow rate per unit area \(J\) using the diffusion coefficient, concentration gradient, and thickness.2. **Determine the surface area**: Ascertain the total surface area through which diffusion occurs. For a cube, this is calculated using the formula \(A = 6L^2\) where \(L\) is the length of one side of the cube.3. **Calculate the total mass flow rate**: Multiply the diffusion flux \(J\) by the surface area \(A\) to find the total mass of the substance that diffuses per unit time. This value is expressed in terms of mass per unit time \(\text{kg/s}\).
This process shows how linked these concepts are, as working through each step requires an understanding of diffusion, concentration gradients, and dimensions of the containment material.