contestada

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

Respuesta :

opudodennis

Answer:

[tex]\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}[/tex]

Explanation:

Diffusion flux of a gas, J is given by

[tex]J=P_m\frac {\triangle P}{\triangle x}[/tex] where [tex]P_m[/tex] is permeability coefficient, [tex]\triangle[/tex] 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 [tex]38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)[/tex]

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

[tex]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}[/tex]

Therefore, the diffusion flux is [tex]\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}[/tex]

Otras preguntas