Answer to Question #227810 in Chemical Engineering for ika

Question #227810

A lab technician, working in a lab has poured liquid chloroform into a test tubes with inner 

diameter of 25 mm. The test tube holds an amount of 29.46 cm3

liquid chloroform at a level of

1 cm from the top (Figure 2). The chloroform molecular weight and vapor pressure is 119.38 

kg/kmol and 21 kPa respectively. The evaporation of chloroform to the surrounding air is 

occurring at 25 C and 1 atm. If the diffusion coefficient of chloroform in air is DAB is 0.104 cm2


determine the rate of evaporation and the time taken for half of the chloroform to be evaporated

at the stated condition. Given the density of the chloroform is 1.49 g/cm³.

Expert's answer

"y_{A,0}=\\frac{P_{A,0}}{P} = \\frac{101.325kPa}{21kPa} =0.20725"

The total molar density throughout the tube remains constant because of the constant temperature and pressure conditions and is determined to be

"C=\\frac{P}{R_uT} =\\frac{101.325kPa }{(8.314 kPa\u22c5 m^ 3\/kmol\u22c5K)(273+25) K} =0.0409 kmol\/m ^3"

Then, the diffusion rate per unit interface area for one test is

"\\frac{N_A}{A}= \\frac{CD_{AB}}{L} \\ln(\\frac{1-y_{A,L}}{1-y_{A,O}})"

"\\frac{N_A}{A}= \\frac{0.049*0.104*1*10^{-4}}{0.01} \\ln(\\frac{1-0}{1-0.20725})"

"\\frac{N_A}{A}=9.8789*10^{-6} kmol\/s*m^2"

The evaporation rate of chloroform from one test tube is

"\\dot{m}\n =MD^ \n2\n \\frac{\n\u03c0}{4}\n\\frac{N_A}{\n\u02d9\n \nA}\n\u200b\n \n\u200b"

"\\dot{m}=119.38*0.025^2 \\frac{\\pi}{4}*(9.8789*10^{-6}\\\\\n\\dot{m}=5.789*10^{-7} kg\/s= 34.734 mg\/min"

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