Question

Determine the binary diffusion coefficient of \(\mathrm{CO}_{2}\) in air at (a) \(200 \mathrm{~K}\) and \(1 \mathrm{~atm},(b) 400 \mathrm{~K}\) and $0.5 \mathrm{~atm}\(, and \)(c) 600 \mathrm{~K}\( and \)5 \mathrm{~atm}$.

Short answer

Question: Calculate the binary diffusion coefficients of CO₂ in air under the following conditions: (a) at 200 K and 1 atm; (b) at 400 K and 0.5 atm; (c) at 600 K and 5 atm. Answer: The binary diffusion coefficients of CO₂ in air under the given conditions are: (a) at 200 K and 1 atm, \(D_{(a)} = 1.46\times10^{-5}\mathrm{~m^{2} s^{-1}}\); (b) at 400 K and 0.5 atm, \(D_{(b)} = 3.23\times10^{-5}\mathrm{~m^{2} s^{-1}}\); (c) at 600 K and 5 atm, \(D_{(c)} = 5.00\times10^{-5}\mathrm{~m^{2} s^{-1}}\).

Step-by-step solution

Constants and data

First, collect all necessary constants and data for the problem. The molar masses of CO₂\((m_A)\) and air \((m_B)\) are: \(m_A = 44.01 \mathrm{~g/mol} = 44.01 \times10^{-3} \mathrm{~kg/mol}\); \(m_B = 29 \mathrm{~g/mol} = 29 \times10^{-3}\mathrm{~kg/mol}\) The collision diameter of CO₂-air \((\sigma_{AB})\) is: \(\sigma_{AB} = 3.95 \times 10^{-10}\mathrm{~m}\) Boltzmann's constant \((k)\) is: \(k = 1.380649\times10^{-23}\mathrm{JK^{-1}mol^{-1}}\) Calculate the binary diffusion coefficient for each case using the Chapman-Enskog equation and the given temperature and pressure values.

Case (a) at 200 K and 1 atm

For this case, T = 200 K, and the diffusion collision integral \((\Omega_{D,AB})\) is 0.899 (from a table or chart). Plug the known values into the equation and solve for \(D_{(a)}\): \(D_{(a)}= \frac{3}{16} \times \frac{[(2 \pi (44.01\times10^{-3} + 29\times10^{-3})(1.380649\times10^{-23}\times200))/((44.01 \times10^{-3} )(29\times10^{-3}))]^{1/2}}{\pi (3.95\times 10^{-10})^{2}(0.899)}\) After solving this equation, you obtain the binary diffusion coefficient for case (a): \(D_{(a)} = 1.46\times10^{-5}\mathrm{~m^{2} s^{-1}}\)

Case (b) at 400 K and 0.5 atm

For this case, T = 400 K, and the diffusion collision integral \((\Omega_{D,AB})\) is 0.772. Plug the known values into the equation and solve for \(D_{(b)}\): \(D_{(b)}= \frac{3}{16} \times \frac{[(2 \pi (44.01\times10^{-3} + 29\times10^{-3})(1.380649\times10^{-23}\times400))/((44.01\times10^{-3})(29\times10^{-3}))]^{1/2}}{\pi (3.95\times 10^{-10})^{2}(0.772)}\) After solving this equation, you obtain the binary diffusion coefficient for case (b): \(D_{(b)} = 3.23\times10^{-5}\mathrm{~m^{2} s^{-1}}\)

Case (c) at 600 K and 5 atm

For this case, T = 600 K, and the diffusion collision integral \((\Omega_{D,AB})\) is 0.709. Plug the known values into the equation and solve for \(D_{(c)}\): \(D_{(c)}= \frac{3}{16} \times \frac{[(2 \pi (44.01\times10^{-3} + 29\times10^{-3})(1.380649\times10^{-23}\times600))/((44.01\times10^{-3})(29\times10^{-3}))]^{1/2}}{\pi (3.95\times 10^{-10})^{2}(0.709)}\) After solving this equation, you obtain the binary diffusion coefficient for case (c): \(D_{(c)} = 5.00\times10^{-5}\mathrm{~m^{2} s^{-1}}\)

Summary of Results

For the given conditions, the binary diffusion coefficients of CO₂ in air are: (a) At 200 K and 1 atm: \(D_{(a)} = 1.46\times10^{-5}\mathrm{~m^{2} s^{-1}}\) (b) At 400 K and 0.5 atm: \(D_{(b)} = 3.23\times10^{-5}\mathrm{~m^{2} s^{-1}}\) (c) At 600 K and 5 atm: \(D_{(c)} = 5.00\times10^{-5}\mathrm{~m^{2} s^{-1}}\)