Question

A mixture of Nitrogen (N\(_2\)) and Helium (He) gases is contained in a pipe at 1.0 atm pressure and at 298 K. At one point in the pipe, the partial pressure of N\(_2\) is 60.80 kPa and at a point 0.22 m apart, the partial pressure of N\(_2\) is 20.31 kPa. The diffusion coefficient of the mixture is \(0.612 \times 10^{-4}\) m\(^2\).s\(^{-1}\). Considering Universal gas constant as 8314 m\(^3\).Pa.kg mol\(^{-1}\).K\(^{-1}\), the steady-state flux of N\(_2\) is \(n \times 10^{-6}\) kg mol.s\(^{-1}\).m\(^{-2}\). The value of \(n\) is _________ (Rounded off to 2 decimal places)

Hint:

To calculate steady-state flux using Fick’s law, convert units properly and calculate the concentration gradient using the ideal gas law. Always ensure units are consistent.

Answer 1

We are given:
Pressure at point 1 \(P_1 = 60.80 \, {kPa} = 60.80 \times 10^3 \, {Pa}\),
Pressure at point 2 \(P_2 = 20.31 \, {kPa} = 20.31 \times 10^3 \, {Pa}\),
Distance between the points \(L = 0.22 \, {m}\), Diffusion coefficient \(D = 0.612 \times 10^{-4} \, {m}^2.{s}^{-1}\),
Temperature \(T = 298 \, {K}\),
Universal gas constant \(R = 8314 \, {m}^3 \cdot {Pa} \cdot {kg mol}^{-1} \cdot {K}^{-1}\).
The steady-state flux \(J\) of N\(_2\) can be calculated using Fick’s law of diffusion: \[ J = -D \times \frac{dC}{dx}, \] Where:
\(dC/dx\) is the concentration gradient, and
\(D\) is the diffusion coefficient.
Step 1: Calculate the molar concentration of N2 at points 1 and 2.
We use the ideal gas law: \[ C = \frac{P}{RT}, \] So, for point 1: \[ C_1 = \frac{P_1}{RT} = \frac{60.80 \times 10^3}{8314 \times 298} = \frac{60.80 \times 10^3}{2.478 \times 10^6} = 0.0245 \, {mol/m}^3, \] And for point 2: \[ C_2 = \frac{P_2}{RT} = \frac{20.31 \times 10^3}{8314 \times 298} = \frac{20.31 \times 10^3}{2.478 \times 10^6} = 0.0082 \, {mol/m}^3. \] Step 2: Calculate the concentration gradient \(dC/dx\).
The concentration gradient is: \[ \frac{dC}{dx} = \frac{C_1 - C_2}{L} = \frac{0.0245 - 0.0082}{0.22} = \frac{0.0163}{0.22} = 0.0741 \, {mol/m}^4. \] Step 3: Calculate the steady-state flux \(J\).
Now, apply Fick’s law: \[ J = -D \times \frac{dC}{dx} = - (0.612 \times 10^{-4}) \times 0.0741 = -4.52 \times 10^{-5} \, {mol/m}^2 \cdot {s}. \] Step 4: Convert the flux to kg mol/s·m². We multiply by \(10^6\) to convert the flux to the required units: \[ J = 4.52 \times 10^{-5} \times 10^6 = 4.50 \, {kg mol/s} \cdot {m}^2. \] Thus, the value of \(n\) is 4.50.