chemical engineering Equimolar counter-diffusion is occurring at steady state in

Question

chemical engineering

Equimolar counter-diffusion is occurring at steady state in a tube 0.11 m long containing N_2 and CO gases at a total pressure of 1.0 atm absolute. The partial pressure of N_2 is 80 mm Hg at one end and 10 mm Hg at the other end. D_AB = 2.05 times 10^5 m^2/s. Calculate the flux in kg mol/s.m^2 at 298 K for N_2. Repeat part a, at 473 K. Does the flux increase? Repeat part a, at 298 K, but for a total pressure of 3.0 atm abs. The partial pressures of N_2 remain 80 and 10 mm Hg. Does the flux change? Calculate the CO flux for pan c.

Explanation / Answer

for equimolar counter diffusion

Na= Ja= -DAB* dCA/dz

JA= DAB*(CA1-CA2)/ thickness

CA1= PA1/RT = (80/760)/ (0.0821L.atm.mole.K*298=0.004302 moles/L =0.004302*103 moles/M3

CA2= (10/760)/ (0.0821*298) =0.000538 moles/L =0.000538*103 moles/m3

Flux = 2.05*10-5*(0.004302-0.000538)*1000/0.11=0.0007015 moles/m2.s

b) Difusivity of gase varies with temperatrue as DAB is proportional to (T)3/2, DAb at 473= 2.05*10-5*(473/298)1.5=4.10*10-5

accordingly flux = =4.10*10-5*(0.004302-0.000538)*1000/0.11= 0.0014-29moles/m2.s

c) diffusivity varies with pressure as DAB is DAb at 3 atm = (1/3)* 2.05*10-5 =0.68333*10-5 m2/s

accordingly flux at 298K= =0.6833*10-5*(0.004302-0.000538)*1000/0.11=0.0002338 mol/m2.s

d) flux of CO= -flux of N2 for equi-molar counter diffusion diffusion . flux of CO= -0.0002338 mol/m2.s

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