Question

Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3 kPa and 9 kPa, which are maintained constant. Assuming conditions of steady state, what is the diffusion flux, in [(cm3 STP)/cm2-s], at 298 K

Answer 1

`\boxed{0.000000266 \frac {cm^(3).STP}{cm^(2).s}}`

Step-by-step explanation:

Diffusion flux of a gas, J is given by

`J=P_m\frac {\triangle P}{\triangle x}` where
`P_m` is permeability coefficient,
`\triangle` P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is
`38* 10^(-13)(cm^(3). STP)(cm)/(cm^(2).s.Pa)`

Since the thickness of sheet is given as 1mm= 0.1 cm then

`J=38* 10^(-13)(cm^(3). STP)(cm)/(cm^(2).s.Pa)* \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^(3).STP}{cm^(2).s}`

Therefore, the diffusion flux is
`\boxed{0.000000266 \frac {cm^(3).STP}{cm^(2).s}}`