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
