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 \(3 \mathrm{~cm}\) 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) $8.1 \times 10^{-10} \mathrm{~kg} / \mathrm{s}$ (b) \(3.2 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (c) \(3.8 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(7.0 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (e) \(1.60 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

Short answer

Answer: (e) \(1.60 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

Step-by-step solution

Write Fick's First Law of Diffusion

Fick's first law of diffusion states that the mass flow rate through a specific area (J) is proportional to the concentration gradient. It is given by: $$J = -D \frac{\Delta C}{\Delta x}$$ where \(J\) is the mass flow rate per unit area (in \(\frac{kg}{m^2 \cdot s}\)), \(D\) is the diffusion coefficient (in \(\frac{m^2}{s}\)), \(\Delta C\) is the change in concentration (in \(\frac{kg}{m^3}\)), and \(\Delta x\) is the thickness of the material (in meters).

Calculate the Concentration Gradient

The concentration gradient is the difference in concentrations between two points, divided by the distance between those points. To calculate the concentration gradient, we can subtract the concentration at the outer surface from the concentration at the inner surface, and divide the result by the thickness of the wall. $$\frac{\Delta C}{\Delta x} = \frac{(0.067 - 0.009) kg/m^3}{(0.03)m} = \frac{0.058 kg/m^3}{0.03m}$$

Calculate the Mass Flow Rate per Unit Area

Now, we can plug the concentration gradient and the diffusion coefficient (\(1.5\times10^{-10} \frac{m^2}{s}\)) into Fick's first law of diffusion to find the mass flow rate per unit area. $$J = -D \frac{\Delta C}{\Delta x} = -(-1.5\times10^{-10} \frac{m^2}{s})(\frac{0.058 kg/m^3}{0.03m}) = 2.9\times10^{-9} \frac{kg}{m^2 \cdot s}$$

Calculate the Mass Flow Rate for One Face of the Cube

To find the mass flow rate for one face of the cube, we multiply the mass flow rate per unit area by the area of one face, which is \(2m \times 2m = 4 m^2.\) $$\textrm{Mass flow rate for one face} = J \times \textrm{area} = (2.9\times10^{-9} \frac{kg}{m^2 \cdot s})(4m^2) = 1.16 \times 10^{-8} \frac{kg}{s}$$

Calculate the Total Mass Flow Rate

Since the mass flow rate is the same for all six faces of the cube, to find the mass flow rate through the entire container, we simply multiply the mass flow rate for one face by six. $$\textrm{Total mass flow rate} = 6 \times \textrm{Mass flow rate for one face} = 6 \times 1.16\times 10^{-8} \frac{kg}{s} = 6.96 \times 10^{-8} \frac{kg}{s}$$ The closest answer from the given choices is (e) \(1.60 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\).