Question

Calculate the diffusion coefficient of N₂ at 100 Pa, 100 kPa an 20 MPa (temperature T = 20 degrees celsius).

Answer 1

(a)

= kT / (2^0.5)p)

=> = [(1.381 x 10^-23 J K^-1) (293 K)] / [(2^0.5) (0.36 x 10^-18m^2) (100 Pa)]

[We know J/K = Kg m2 s^-1    and      Pa = Kg m^-1 s^-2

= 7.94 x 10^-5 m

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Average speed = (8RT / M)^0.5

= (8 x 8.314 J K^-1 mol^-1 x 293 K / 28 x 10^-3 Kg mol)^0.5

[We know J L^-1 mol^-1 = kg m^2 s^-1]

= (221543.86 m/s)^0.5

= 470.68 m/s

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Diffusion D = 1/3 ( C)

= 1/3 ( 7.94 x 10^-5 m) (470.68 m /s)

= 0.0125 m^2/s

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b) at 100 k Pa = 100 x 10^3 Pa

= kT / (2^0.5)p)

=> = [(1.381 x 10^-23 J K^-1) (293 K)] / [(2^0.5) (0.36 x 10^-18m^2) (100 x 10^3 Pa)]

[We know J/K = Kg m2 s^-1    and      Pa = Kg m^-1 s^-2

= 7.94 x 10^-8 m

D= (1/3)(7.94 x 10^-8 m) (470.68 m/s)

= 1.245 x 10^-5 m^2/s

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c) at 20 MPa = 20 x 10^6 Pa

= kT / (2^0.5)p)

=> = [(1.381 x 10^-23 J K^-1) (293 K)] / [(2^0.5) (0.36 x 10^-18m^2) (20 x 10^6 Pa)]

[We know J/K = Kg m2 s^-1    and      Pa = Kg m^-1 s^-2

= 3.974 x 10^-10 m

D= (1/3)(3.974 x 10^-10 m) (470.68 m/s)

= 6.23 x 10^-8 m^2/s