# Answer to Question #227811 in Chemical Engineering for Ika

Question #227811

A 3 cm diameter tube is used to measure the binary diffusion coefficient of water vapor in air at

20°C at an elevation of 1600 m where the atmospheric pressure is 83.5 kPa. The tube is partially

filled with water, and the distance from the water surface to the open end of the tube is 40 cm.

Dry air is blown over the open end of the tube so that water vapor rising to the top is removed

immediately and the concentration of vapor at the top of the tube is zero. In 15 days of

continuous operation at constant pressure and temperature, the amount of water that has

evaporated is measured to be 1.23 g. Determine the diffusion coefficient of water vapor in air at

20°C and 83.5 kPa. Given, the pressure of water at 20°C is 2.34 kPa.

1
2021-08-23T04:52:03-0400

"P_{A2} =vapor \\space pressure \\space for \\space water \\space at \\space the \\space top \\space of \\space vertical \\space tube =0 (flow \\space rate \\space of \\space Bis high)\\\\\nP_{A1} =v.p of \\space water \\space at \\space 200C = 2.34 KPa\\\\\nP= total \\space pressure= 83.5 KPa\\\\\nP_{B1}= P- p_{A1} = 83.5 - 2.34 = 81.16 KPa (air)\\\\\np_{B2} = P- p_{A2} = 83.5 Kpa\\\\\npB,lm = logarthimic \\space mean \\space partial \\space pressure \\space of \\space air =[ p_{B2} - p_{B1}] \/ln( \\frac{p{B2}}{p_{B1}})\\\\\np_{B,lm} = 2.34 \/ln(83.5\/81.16) =82.3244 KPa\\\\\nAt t=0 ; Z = Zo = 40 cm\\\\\nAt t=15 \\space days Zt ,\\\\\n1.23 gms \\space evaporated \\space so \\space volume \\space of \\space the \\space circular \\space tube \\space will \\space be,\\\\\n1.23 gms \/ 1 gm\/cm3 = 1.23 cm3 = 4\/3 \u03c0r^3\\\\\nradius \\space of \\space the \\space tube = r = 0.6646691 cm\\\\\narea \\space of \\space the \\space circular \\space tube = \u03c0r2 = 1.3879085 cm2\\\\\nVolume of the tube\/area of the tube = 1.23cm3\/ 1.3879cm2 = 0.886225 cm\\\\\nheight reduced after evaporation will be = 0.886225 cm\\\\\nZt = 40cm + 0.886225cm = 40.886225 cm at t' =15 days\\\\\nMA = 18 gms\/mol\\\\\nT= 20+273.15 = 293.15 K\\\\\nplug in all the values in equation below ,\\\\\nDAB = [\u03c1ARTpB,lm(Zt-Zo)2] \/ 2PMA(pA1-pA2) t'\\\\\n[\u03c1ARTpB,lm(Zt-Zo)2] = [\u03c1A8.314*103cm3KPa mol-1K-1*293.15K*82.3244 KPa*(40.886225-40)2cm2]\n = 157585584.6\u03c1A (\u03c1A =1 gm\/cm3)\\\\\n= 157585584.6 cm2 Kpa2 gm mol-1\\\\\n2PMA(pA1-pA2) t' = 2* 83.5 KPa* 18 g\/mol * (2.34-0)KPa *15 days\\\\\n= 105510.6 Kpa2 days gm mol-1\\\\\nD_{AB} = (157585584.6 cm2 Kpa2 gm mol-1) \/ (105510.6 Kpa2 days gm mol-1)\\\\\nD_{AB} = 1493.5521 cm2\/day =1493.5521 \/ 360*60*60 sec = 1.152432 *10-3 cm2\/sec\\\\\nD_{AB} \\space of \\space water \\space vapor \\space in \\space air \\space at \\space 20^oC and 83.5 KPa is 1.152432 *10-3 cm^2\/sec\\\\"

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