Question #227810

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

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 119.38Â

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

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

/s,Â

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/cmÂ³.

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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