GATE MT

This is mass transfer case, which can be solved by using FICK’s 1^st law.

            Mass flux rate (mol/sec.m²) = D*(C/L)

            Hence, rate of oxidation (mol/””min””) = area*(D/L)*C*””60″”

            Given D/L = 0.03 m/s

            C = n/V = P/RT = 0.4/1173*8.205*.00001 = 0.4156 mol/m³ (these are moles of oxygen in air)

            Area = 4*pie*(0.025)²

           Rate of oxidation (mol/min) = 4*pie*(0.025)²*0.03*0.4156*60 = 0.05875 mols of oxygen/min

            Now, this we have calculated in terms of moles of oxygen, but requirement is in terms of moles of Nickel.

            2Ni + O₂ = 2NiO

           1 mol of O₂ = 2 mol of Ni

           Hence, rate of oxidation in terms of moles of nickel per min = 2*0.05875 = 0.11 mol/min