Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 2 kPa and 11 kPa, which are maintained constant. Assuming conditions of steady state, what is the diffusion flux, in [(cm3 STP)/cm2-s], at 298 K?

Respuesta :

Answer:

1.52 x [tex]10^{-7}[/tex] x  (cm3 STP) / [tex]cm^{2}[/tex]-s

Explanation:

This is a permeability problem ithat asked to compute the diffusion flux of water vapor through a  2-mm  thick  sheet  of  polypropylene.  The permeability  coefficient  of  H2O  through  PP  is                            

38 x [tex]10^{-13}[/tex] ([tex]cm^{3}[/tex]STP)-cm / [tex]cm^{2}[/tex]-s-Pa.  

Thus, from Equation

J= [tex]P_{M}[/tex] x ΔP/Δx  = [tex]P_{M}[/tex] x [tex]P_{2} - P_{1}[/tex] / Δx

here

P1= 2 kPa  =2,000 Pa

P2= 11 kPa = 11,000 Pa

J = (38 X [tex]10^{13}[/tex] x (cm3 STP)(cm) / [tex]cm^{2}[/tex]-s-Pa) (10,000 Pa - 2,000 Pa  / 0.2 cm)

   = 1.52 x [tex]10^{-7}[/tex] x  (cm3 STP) / [tex]cm^{2}[/tex]-s

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