Question

Example 2 : Diffusion of Water Through Stagnant, Nondiffusing Air Water in the bottom of a narrow metal tube is held at a constant temperature of 20°C. The total pressure of air (assumed dry) is 1.0 atm and the temperature is 20°C. Water evaporates and diffuses through the air in the tube, and the diffusion path 22-Z is 0.5 ft long. Calculate the rate of evaporation at steady state in lb mol/hr.ft. The diffusivity of water vapor at 20°C and 1 atm pressure is 0.969 ft2/hr. Assume that the system is isothermal (Ans: 1.176 x 10-4 lb mol/hr.ft2)

I need you to show the table

Answer 1

I hope you understand well.. vapour pressure of water vapour at 293k is 17.51 mmHg..

Given that t g Aris (6) 2 PAG=O At 293 K , DAG=0.969 f? hr. Zzzi DA = 0.969 ? In Iswates (A) PAI Zz Zi=0.5t Total Pressure (P+) = 1 atun At za3ki PAI=17.54 mmig X lotus 760mming you can find from stean tahke .=0.023 latus. IR = 108 IR=1.8 PA2 =O (Pure air only) atn.ft3 R= 0.73024 lb mol. R * Rate of evapolation at steady statet DAB PE (PAI-PA2) NAF RT (22-21) PB, LM Å 18=1-882983 =527.4 T=527.uk where PAI-PAL PBILM ² un P-PAZ P-P PAI (6.0231 - 0) PBilm = 50.988 atu 1-0 1-0.0231

0.969 NA= e x latach latyh (0.0231-0) atan 0.73024 anys. As *527.08X0.54x 0.988 atyn lb mol of hr NA= 1.1768104 tr lomo) Ab mol NA = 1.176 810-4 hr. tr. Rate of evaporation at steady state