Answer to Question #227810 in Chemical Engineering for ika

Question

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

/s, 

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.

Answer 1

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

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

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

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

$\dot{m} =MD^ 2 \frac{ π}{4} \frac{N_A}{ ˙ A} ​ ​$

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