Answer to Question #163564 in Chemical Engineering for Joseph

Question

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

Answer 1

Considering O₂ to be non diffusing and with T = 273 K, P_t = 755 mm Hg, Z = 17.1 cm,

0.0208 cm³ of CCl₄ is evaporating in 10 hours, density of liquid CCl₄ = 1.59 g/cm³.

(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/m² 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 10⁵) ln{[(755/769 x 1.013 x 10⁵) - 0)]/[(755/769 x 1.013 x 10⁵) - (33/769 x 1.013 x 10⁵)]} = 6.355 x 10^-6 m²/s