Question

The evaporation rate of chloroform (CHCl3) in air has been experimentally measured in an Arnold diffusion cell at 25°C and 1.00 atm. Over a period of 10 hr, the chloroform dropped 0.44 cm from the initial surface 7.40 cm from the top of the tube. The density of liquid chloroform is 1.485 g/cm3 and its vapor pressure at 25°C is 200 mm Hg. 1. Assuming air is an ideal gas, determine the molar concentration of air in mol/cm3 . 2. Determine the mole fraction of chloroform in air at the liquid-air interface. 3. Determine the flux of chloroform through the air column in mol/(cm2 s) 4. Determine the diffusion coefficient of chloroform in air in cm2 /s.

Answer 1

Answer:

1. C = 0.041 mol/L, 2. Mole Fraction = 0.2632, 3.  N(A) = 9.322 x 10⁻⁵ mol/cm².s, 4. D(AB) = 5.688 x 10⁻⁵ cm²/s

Explanation:

Use the ideal gas equation

PV = nRT

Can be re4arranged as

P = nRT/V, where n/V = C

Hence

P = CRT

1. Concentration of air

C = P/RT

C = 101325/(8.314 x 298)

C = 40.89mol/m³ = 0.041 mol/L

2. Mole Fraction of chloroform in air at the liquid- air interface

Mole fraction = vapor pressure(chloroform)/vapor pressure(total)

Mole Fraction = 200/760 = 0.2632

3. Flux of chloroform

N(A) = (D(AB)//Z) . (P(T)/RT) .(P(A1) – P(A2))/P(BM)

P(BM) = (760 – 560)/ln(760/560) = 655 mm Hg

D(AB) = RTP(BM) (Z²₁ – Z²₀)/{2PM(A)(P(A1) – P(A2))}

D(AB) = 8.314 x 298 x 655 (0.0758² – 0.074²)/(2 x 101325 x 0.1195 x 200 x 36000)

D(AB) = 5.688 x 10⁴ = 5.688 x 10⁻⁵cm²/s (multiplying by 10⁴ to convert into cm2)

N(A) = (5.688 x 10⁻⁵)/(7.62) x (101325/8 .314 x 298) x (200/655)

N(A) = 9.322 x 10⁻⁵ mol/cm².s

4. D(AB) has already been calculated in the solution of 3

D(AB) = 5.688 x 10⁻⁵ cm²/s