Respuesta :
Answer:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by
[tex]N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })[/tex]
Where
[tex]D_{AB}[/tex] = Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
[tex]P_{A1}[/tex] = Pressure at first boundary
[tex]P_{A2}[/tex] = Pressure at the destination boundary
T = System temperature
[tex]P_{T}[/tex] = System pressure
Where [tex]P_{T}[/tex] = 101.3 kPa [tex]P_{A2} =0[/tex], [tex]P_{A1} =y_{A}[/tex], [tex]P_{T} =[/tex] 0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R = [tex]\frac{kJ}{(kmol)(K)} ,[/tex] T = 298 K and [tex]D_{AB}[/tex] = 1.18 [tex]\frac{cm^{2} }{s}[/tex] = 1.8×10⁻⁵[tex]\frac{m^{2} }{s}[/tex]
[tex]N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65})[/tex] = 5.153×10⁻⁴[tex]\frac{kmol}{m^{2}s }[/tex]
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
