Ammonia gas (A) and nitrogen gas (B) are diffusing in counterdiffusion through a

Question

Ammonia gas (A) and nitrogen gas (B) are diffusing in counterdiffusion through a straight glass tube 2.0 ft (0.610 m) long with an inside diameter of 0.080 ft (24.4 mm) at 298 K and 101.32 kPa. Both ends of the tube are connected to large mixed chambers at 101.32 kPa. The partial pressure of NH_3 is constant at 20.0 kPa in one chamber and 6.666 kPa in the other. The diffusivity at 298 K and 101.32 kPa is 2.30 times 10^-5 m^2/s. Calculate the diffusion of NH_3 in lb mol/h and kg mol/s. Calculate the diffusion of N_2. Calculate the partial pressures at a point 1.0 ft (0.305 m) in the tube and plot p_A, p_B and P versus distance z.

Explanation / Answer

We know than NA= xA*(NA+NB)+ JA ,xA= mole fraction of A

Where NA= flux due to bulk diffusion for NH3, JA= molar flux due to molecular diffusion

For equimolar counter diffusion NA=-NB

NA= JA= -DAB*dCA/dZ

JA*dZ= -DAB*dCA

When integrated, the expression becomes

JA*(Z2-Z1)= DAB*(CA1-CA2) =DAB/RT*(PA1-PA2)

Where PA1= 20Kpa =20*103 N/m2 and PA2= 6.6*103 N/m2, R = 8314 N.m/kmol

JA= 2.3*10-5*(20-6.6)*103/ (8314*298)*0.610 = 2.039*10-7 kmol/m2.s

If r= radius of tube= 24.4/2000

Rate of diffusion of NH3 = JA* Pi*r2= 2.039*10-7*(22/7)*(24.4/2000)2 =9.5*10-11 kmol/s

Lb= 0.4535 kg or 1 kg= 2.2 lb mole,rate of diffusio 9.5*10-11*2.2 lbmol/s = 2.1*10-10 lbmol/s

= 7.55*10-7 lbmol/hr

Rate of diffusion of N2= -9.5*10-11 kmol/s

At point 1 ft,   2.039*10-7 =(DAB/RT)*( 20*1000- PA2)/0.305 = 2.3*10-5/(8314*298) *(20*1000-PA2)/0.305

2.039*10-7 = 9.3*10-12*(20*1000-PA2)/0.305

PA2= 13312 Kpa =13.3*104 pa

Since JA= DAB*(PA1-PA2)/RTZ

JA= 2.039*10-7, Calculate PA2 by varying Z. The calculations are shown in the plot.

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