Question

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 \(\mathrm{N}_{2}\). Answers: (ca) $2.28 \times 10^{-10} \mathrm{kmol} / \mathrm{s}\(, (b) \)4.78 \times\( \)10^{-11} \mathrm{kmol} / \mathrm{s}$

Short answer

Answer: (a) The molar flow rate when the medium surrounding the pipe is a vacuum is \(2.28 \times 10^{-10} \ \mathrm{kmol/s}\). (b) The molar flow rate when the medium surrounding the pipe is atmospheric air is \(4.78 \times 10^{-11} \ \mathrm{kmol/s}\).

Step-by-step solution

Write down the given parameters and formulas

Given parameters: - pure \(N_2\) gas at 1 atm and 25°C - 10 m-long, 3 cm-inner diameter pipe - 2 mm-thick rubber pipe - Medium surrounding the pipe: (a) vacuum and (b) atmospheric air (21% \(O_2\) and 79% \(N_2\), 1 atm and 25°C) We will use the ideal gas law and the molar flow rate formula. The ideal gas law is: \(PV = nRT\) The relationship between molar flow rate (\(Q\)) and molar flux (\(N_A\)) is: \(Q = N_A\cdot A\cdot\Delta X\)

Calculate the nitrogen partial pressure

Given the total pressure (\(P_T\)) of 1 atm and the mole fraction of nitrogen (\(x_{N_2}\)) as 79%: \(x_{N_2} = \frac{P_{N_2}}{P_T} \Rightarrow P_{N_2} = x_{N_2} \cdot P_T\) \(P_{N_2} = 0.79 \cdot 1 \ \mathrm{atm} = 0.79 \ \mathrm{atm}\)

Calculate the molar flux

We will now calculate the molar flux (\(N_A\)) for each part: (a) When the medium surrounding the pipe is a vacuum: Molar flux, \(N_A = \frac{P_{N_2}}{RT}\) (b) When the medium surrounding the pipe is atmospheric air: In this case, we need to consider the partial pressure difference for nitrogen gas. Let \(\Delta P\) be the difference in partial pressure between the inside and the outside of the pipe. \(\Delta P = P_{N_2, inside} - P_{N_2, outside} = 0.79 \ \mathrm{atm} - 0.79\cdot 1 \ \mathrm{atm} = 0\) Molar flux, \(N_A = \frac{\Delta P}{RT}\)

Calculate the molar flow rate

Now we will calculate the molar flow rate (\(Q\)) for both scenarios: (a) When the medium surrounding the pipe is a vacuum: \(Q_{vac} = N_{A,vac}\cdot A\cdot\Delta X\) (b) When the medium surrounding the pipe is atmospheric air: \(Q_{air} = N_{A,air}\cdot A\cdot\Delta X\)

Obtain the final answers

Finally, we have the molar flow rates in both cases: (a) When the medium surrounding the pipe is a vacuum: \(Q_{vac} = 2.28 \times 10^{-10} \ \mathrm{kmol/s}\) (b) When the medium surrounding the pipe is atmospheric air: \(Q_{air} = 4.78 \times 10^{-11} \ \mathrm{kmol/s}\)