Question

Pure ethanol is contained at the bottom of a long vertical tube( cross sectional area is 0.9 cm²). Above the liquid is a quaescent layer of air.The liquid at the bottom of the tube is carefully adpuruted so that the distance from the air liquid interface to the open top of the tube is constant at 15 cm . No liquid is withdrawn from the tube. At the open end of the tube is perpendicular to the vertical tube so that the concentration of ethanol at the top of the tube is eventially zero. The enture apparatus is kept at 0 degree celcius and P_int= 0.98 atm. The tube is carefully arranged so that there is no convection iin the tube. Over the course of several days we find that the average evaporation rate is 0.9190*10^-3 cm³/. What is the value of the diffusion coefficient of ethanol in air at 0 degree celcius? (mass density is 789 kg*m³ and its molar massis 46.079 g/mol )

Answer 1

Data provided in the problem statement

C/S area of tube = 0.9 cm2 = 0.9 * 10-4 m2

Distance between air-liquid interface and top of tube(Z ) = ( L ) = 15 cm = 0.15 m

Concentration of ethanol at top of tube ( CL ) = 0

Temperature of system = 0 C = 273 K

The pressure of system = 0.98 atm = 744.8 mmHg

Average evaporation rate = 0.9190*10-3 cm3 / hr * ( 1 h / 3600 s ) * ( 1 m3 / 106 cm3) = 2.552 * 10-13 m3 /s

Mass density = 789 kg/m3

Molar mass = 46.079 gm /mol = 46.079 * 10-3 kg / mol

So, molar density of ethanol = Mass density / Molar mass = 789 kg/m3 / 46.079 * 10-3 kg / mol = 17122 mol / m3

Now

Molecular diffusion coefficient can be determined by,