Sublimation of a 20 cm long naphthalene cylinder of mass 10 g occurs in a large volume of air in a room at 45°C and 1 atm pressure. The vapor diffuses through a stagnant film of air of thickness of 3 mm surrounding the cylinder. The air beyond the film is well-mixed. 

Calculate the time required for the sublimation of half of the naphthalene. Loss of mass because of sublimation from the flat ends may be neglected. The diffusivity of naphthalene in air under the given conditions is 6.92 x 10-6 m2/s, its density is 1140 kg/m3, and its sublimation pressure at 45°C is 0.8654 mm Hg.

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We know that NA DAB PT RTZ PBM Where PAI 08654 ommity PA0 T 318 ... View the full answer