Consider steady-state equimolar counterdiffusion through the converging-diverging tube. Given: r1=1cm; r2=2cm; NA=10^-6 kmol/m^2.s at the vessel 1 end of the tube. The rate of transport of B at the vessel 2 end of the tube is
(a) 2*10^-6 kmol/s, towards vessel 1
(b) 2.5*^-7 kmol/s, towards vessel 1
(c) 5*10^-6 kmol/s, towards vessel 2
(d) 9*10^-7 kmol/s, towards vessel 2
The question was asked by my school teacher while I was bunking the class.
This interesting question is from Diffusion through Variable Area in chapter Diffusion in Solids and Molecular Diffusion in Fluids of Mass Transfer