Pure \(\mathrm{N}_{2}\) gas at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) is flowing through a 10-m-long, 3-cm-inner diameter pipe made of 2 -mm-thick rubber. Determine the rate at which \(\mathrm{N}_{2}\) leaks out of the pipe if the medium surrounding the pipe is \((a)\) a vacuum and \((b)\) atmospheric air at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) with 21 percent \(\mathrm{O}_{2}\) and 79 percent \(\mathrm{N}_{2}\).

First, need to find the area of the pipe wall to determine where diffusion takes place. Calculate the area using the following formula: \(A = 2 \pi L(D+d)\), where \(L\) is the length of the pipe, and \(D\) and \(d\) are the inner and outer diameters, respectively. In this case, the inner diameter is 3 cm, and since the rubber is 2 mm thick, the outer diameter is 3.2 cm. $$A = 2 \pi (10~m)(0.032~m) = 2.011 ~\text{m}^2$$

Before using Fick's law, we need to find the concentration of \(\mathrm{N}_{2}\) gas inside and outside the pipe. We know that the pressure inside the pipe is \(1~atm\) and the temperature is \(25^{\circ}C\). Using the ideal gas law, \(PV = nRT\), we can calculate the concentration of \(\mathrm{N}_{2}\) inside the pipe, \(C_{inside}\): $$C_{inside} = \frac{n}{V} = \frac{P}{RT} = \frac{1~atm}{0.0821 \cdot (273.15+25)} = 40.66 ~\text{mol/m}^{3}$$ For part \((b)\), we also need the concentration of \(\mathrm{N}_{2}\) outside the pipe, \(C_{outside}\): $$C_{outside} = 0.79 \cdot \frac{1~atm}{0.0821 \cdot (273.15+25)} = 32.12 ~\text{mol/m}^{3}$$

Now, we use Fick's law for diffusion to calculate the mass flow rate (\(\dot{m}\)). Fick's law is given by: $$\dot{m} = -D_{AB}A\frac{\Delta C}{\delta}$$ where \(D_{AB}\) is the diffusion coefficient of the gas, \(\Delta C\) is the concentration difference between inside and outside, and \(\delta\) is the thickness of the rubber pipe wall. For part \((a)\), we have a vacuum outside, so the concentration difference is simply the concentration inside the pipe: $$\Delta C_a = C_{inside}$$ For part \((b)\), we have atmospheric air outside, so the concentration difference is the difference between the concentrations inside and outside: $$\Delta C_b = C_{inside} - C_{outside}$$ Using the given diffusion coefficient of \(\mathrm{N}_{2}\) through rubber (\(D_{AB} = 1.6 \times 10^{-9} \mathrm{~m^2/s}\)), we can now calculate the mass flow rate for both scenarios: $$\dot{m}_a = -(1.6 \times 10^{-9}~\text{m}^{2}/\text{s})(2.011 ~\text{m}^2)(40.66 ~\text{mol/m}^{3})/(2 \times 10^{-3}~\text{m})$$ $$\dot{m}_a = -6.55 \times 10^{-8} ~\text{mol/s}$$ $$\dot{m}_b = -(1.6 \times 10^{-9}~\text{m}^{2}/\text{s})(2.011 ~\text{m}^2)(8.54 ~\text{mol/m}^{3})/(2 \times 10^{-3}~\text{m})$$ $$\dot{m}_b = -1.37 \times 10^{-8} ~\text{mol/s}$$

The mass flow rate of \(\mathrm{N}_{2}\) leaking out of the pipe is: \((a)\) When surrounded by vacuum: \(\dot{m}_a = -6.55 \times 10^{-8} ~\text{mol/s}\). The negative sign indicates that the flow is from inside to outside the pipe. \((b)\) When surrounded by atmospheric air: \(\dot{m}_b = -1.37 \times 10^{-8} ~\text{mol/s}\). The negative sign also indicates that the flow is from inside to outside the pipe. Note that the mass flow rate is faster when surrounded by vacuum, as expected.