Question #163564

diffusivity of gas pair oxygen carbon tetrachloride is determined by observing steady state evaporation of carbon tetrachloride.The distance between carbon tetrachloride liquid level and the top of the tube is 17.1cm.the total pressure of the system 755mmHg.The vapour pressure of carbon tetrachloride at this temperature is 33mmHg.The cross sectional area of the two is 0.82cm^2.If it is found that 0.0208cm^3 of CCL4 evaporate in 10hours period.What is the diffusivity of gas pass carbon tetrachloride

Expert's answer

Considering O_{2} to be non diffusing and with T = 273 K, P_{t} = 755 mm Hg, Z = 17.1 cm,

0.0208 cm^{3} of CCl_{4} is evaporating in 10 hours, density of liquid CCl_{4} = 1.59 g/cm^{3}.

(0.0208 x 1.59) / (154 x 10) = 2.147 x 10^{-5} gm mol/hr

Flux N_{A} = (2.147 x 10^{-5} x 10^{-3}) / (3600 x 0.82 x 10^{-4}) = 7.27 x 10^{-8} kg mol/m^{2} s

N_{A} = D_{AB} P_{t} ln [(P_{t} - P_{A2})/(P_{t} - P_{A1})] / ZRT

D_{AB} = N_{A} x Z x RT / P_{t} ln [(P_{t} - P_{A2})/(P_{t} - P_{A1})] = 7.27 x 10^{-8} x 17.1 x 10^{-2} x 8314 x 273 / (755/769 x 1.013 x 10^{5}) ln{[(755/769 x 1.013 x 10^{5}) - 0)]/[(755/769 x 1.013 x 10^{5}) - (33/769 x 1.013 x 10^{5})]} = 6.355 x 10^{-6} m^{2}/s

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