Question

The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the following equation:
PM=PM0exp(−Qp/RT)

where PM0 and Qp are constants for a given gas-polymer pair. Consider the diffusion of water through a polystyrene sheet 30 mm thick. The water vapor pressures at the two faces are 20 kPa and 1 kPa, which are maintained constant. Compute the diffusion flux [in(cm^3 STP)/cm^2⋅s] at 350 K? For this diffusion system,

PM0=9.0×10^−5(cm^3 STP)(cm)/cm^2. s. Pa
Qp=42,300J/mol

Assume a condition of steady-state diffusion.

Answer 1

Answer:

Answer for the question is explained in the attachment.

Explanation:

Diffusion flux at 350K for water in polystyrene is equal to

2.8 * 10^-7 .

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Answer 2

Answer:

The diffusion flux is 2.8x10⁻⁷cm³STP/cm²s

Explanation:

The permeability coefficient of water is

$P_{M} =P_{0} exp(-\frac{Q}{RT} )$

where

P0 = permeability coefficient of water at 273 K = 9x10⁻⁵

Q = 42300 J/mol

$P_{M} =9x10^{-5} exp(\frac{-42300}{8.314*350} )=4.4x10^{-11} \frac{cm^{3}STP*cm }{cm^{2}*s*Pa }$

The diffusion flux at 350 K is

$J=P_{M} \frac{P_{2} -P_{1} }{deltax} =4.4x10^{-11} (\frac{20000-1000}{3} )=2.8x10^{-7} \frac{cm^{3}STP }{cm^{2}*s }$