Question

Hydrogen gas is maintained at 3 bars and 1 bar on opposite sides of a plastic membrane which is .3 mm thick. The temperature is 25C, and the diffusion coefficient of hydrogen in the plastic at 25C is 9x10-8 m2/s. The solubility of hydrogen in the plastic membrane is 1.5x10-3 kmol/m3. What is the mass diffusive flux (in kg/m2.s) of hydrogen on the membrane

Answer 1

Answer:

$N_a=1.8*10^{-6}kg/sm^2$

Explanation:

From the question we are told that:

Thickness of plastic membrane $L_t=0.3mm$

The temperature of hydrogen $T_h=25 \textdegree C$

Diffusion coefficient of hydrogen in the plastic $25 \textdegree C \mu=9*10-8m2/s$

The solubility of hydrogen in the plastic membrane $\+x=1.5*10{-3} kmol/m3$

Generally the equation for molar conc is mathematically given by

$CA_1=x*bar$

3bars

$CA_1=1.5*10^{-3}*3$

$CA_1=4.5*10^{-3}kmol/m^3$

1bar

$CA_1=1.5*10^{-3}*1$

$CA_1=1.5*10^{-3}kmol/m^3$

Generally the equation for molar diffusion flux of Hydrogen N_a is mathematically given by

$N_a=\frac{D{AB}}{\L}(CA_1-CA_2)$

$N_a=\frac{9*10^{-8}}{0.3*10^-^3} (4.5*10^{-3}-1.5*10^{-3})kmoi/m^-3$

$N_a=9.*10^{-7}kmol/sm^2*2kh/kmole$

$N_a=1.8*10^{-6}kg/sm^2$