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

Answer:

$\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

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 $38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)$

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

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

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