In an Air-NH_3 gas mixture at 2.5 std atm, 25 degree C, the concentrations of am

Question

In an Air-NH_3 gas mixture at 2.5 std atm, 25 degree C, the concentrations of ammonia at two planes 15 mm apart are 65 and 10 vol%, respectively. Assume these gases are ideal gases. Calculate the diffusive flux of ammonia and report it in SI and English units for the case where: The air is non-diffusing. There is equimolar counter diffusion of two gases. Experiments shows that diffusivity of Butanol-Ammonia at 25 degree C and 2.5 atm is 0.081 times 10^-4 m^2/s, explain why the diffusivity for this system is lower than that of Air-Ammonia at the same condition. D_Air-NH3 at 25 degree C = 0.185 times 10^-4 m^2/s

Explanation / Answer

DAIR-NH3 at 25 °C = 0.185 x 10-4 m2/s

Distance between planes, Z2 – Z1 = 15 mm = 15 x 10-3 m

Concentration of ammonia at plane 1, C1 = 65 %

Concentration of ammonia at plane 2, C2 = 10 %

Total pressure, P = 2.5 * 1.013 x 105 = 2.53 x 105 Pa

Temperature, T = 25 + 273.15 = 298.15 K

[(P – PA2)/(P – PA1)] = (100 – 10) / (100 – 65) = 2.57

a)Air is non diffusing

NA = P DAIR-NH3 * ln [(P – PA2)/(P – PA1)] / [(Z2 – Z1) RT]

= 2.53 x 105 * 0.185 x 10-4 * ln (2.57) / (15 x 10-3 * 8.314 * 298.15)

= 0.119 mol/m2-s

b)Equimolar counter diffusion of two gases

Mole fraction at plane 1, Y1 = 65 / 100 = 0.65

Mole fraction at plane 2, Y2 = 10 / 100 = 0.1

NA = P DAIR-NH3 * (Y1 – Y2) / [(Z2 – Z1) RT]

= 2.53 x 105 * 0.185 x 10-4 * (0.65 – 0.1) / (15 x 10-3 * 8.314 * 298.15)

= 0.069 mol/m2-s

c)

Lower the molecular weight of components, higher the combined diffusivity coefficient.

Molecular weight of air (~29 g/mol) is less than molecular weight of butanol (~74 g/mol). Therefore Air-NH3 system has a higher diffusivity coefficient (DAIR-NH3) at same conditions of pressure and temperature.

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