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) \(3.58 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

Step by step solution

01

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: \(D=1.5\times10^{-10}\,\mathrm{m}^2/\mathrm{s}\)

02

Calculate the concentration gradient

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

03

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: \(J= -(1.5\times10^{-10}\,\mathrm{m}^2/\mathrm{s}) \left(\frac{0.058\,\mathrm{kg/m}^3}{0.04\,\mathrm{m}}\right)\) \(J= -2.175\times10^{-9}\,\mathrm{kg/m}^2\,\mathrm{s}\)

04

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: \(A=6(2\,\mathrm{m})^2=24\,\mathrm{m}^2\)

05

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

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