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• Vidhi Thakar

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The diffusivity of carbon tetrachloride, CCl4 through oxygen, O2 was determined in a steady state Arnold evaporating cell. The cell, having a cross sectional area of 0.82 cm2, was operated at 273 K and 755 mmHg pressure. The average length of the diffusion path was 17.1 cm. If 0.0208 cc of CCl4 was evaporated in 10 hours of steady state operation, what should be the value of the diffusivity of CCl4 through oxygen? Vapour pressure of CCl4 at 273 K = 33 mm Hg Density of liquid CCl4 = 1.59 g/cm3. Considering O2 to be non diffusing and with T = 273 K, Pt =755 mm Hg, Z = 17.1 cm 0.0208 cc of CCl4 is evaporating in 10 hours. (i.e.) 0.0208 × 1.59 = 2.147 × 10-5 gm mol /hr 154 × 10 Flux = NA = 2.147 ×10-5 ×10-3 = 7.27 × 10-8 kg mol /m2 s 3600 × 0.82 × 10-4 NA = DAB Pt ln [(Pt – pA2)/ (Pt – pA1)] ZRT DAB = ______NA × Z ×RT______ Pt ln [(Pt – pA2)/ (Pt – pA1)]

7.27 ×10-8 ×10- 2 × 8314 ×273 == -----------------------------------------------------------------------------------------------[(755/760)×1.013 ×105] ln{[(755/760×1.013 ×105) – 0]/ [(755/760×1.013 ×105)– (33/760 ×1.013 ×105)]}

DAB = 6.355 ×10- 6 m2/s The diffusivity of carbon tetrachloride, CCl4 through oxygen, O2 was determined in a steady state Arnold evaporating cell. The cell, having a cross sectional area of 0.82 cm2, was operated at 273 K and 755 mmHg pressure. The average length of the diffusion path was 17.1 cm. If 0.0208 cc of CCl4 was evaporated in 10 hours of steady state operation, what should be the value of the diffusivity of CCl4 through oxygen?

Vapour pressure of CCl4 at 273 K = 33 mm Hg Density of liquid CCl4 = 1.59 g/cm3. Considering O2 to be non diffusing and with T = 273 K, Pt =755 mm Hg, Z = 17.1 cm 0.0208 cc of CCl4 is evaporating in 10 hours. (i.e.) 0.0208 × 1.59 = 2.147 × 10-5 gm mol /hr 154 × 10 Flux = NA = 2.147 ×10-5 ×10-3 = 7.27 × 10-8 kg mol /m2 s 3600 × 0.82 × 10-4 NA = DAB Pt ln [(Pt – pA2)/ (Pt – pA1)] ZRT DAB = ______NA × Z ×RT______ Pt ln [(Pt – pA2)/ (Pt – pA1)]

7.27 ×10-8 ×10- 2 × 8314 ×273 == -----------------------------------------------------------------------------------------------[(755/760)×1.013 ×105] ln{[(755/760×1.013 ×105) – 0]/ [(755/760×1.013 ×105)– (33/760 ×1.013 ×105)]}

DAB = 6.355 ×10- 6 m2/s