Answer
$J = 2.71 \times 10^{-6} \frac{cm^{3}STP}{cm^{2}-s}$
Work Step by Step
Given:
Consider the diffusion of oxygen through a low density polyethylene (LDPE) sheet 15 mm thick. The pressures of oxygen at the two faces are 2000 kPa ($2.0 \times 10^{6} Pa$) and 150 kPa (150,000 Pa), which are maintained constant.
Required:
The diffusion flux (in $(cm^{3} STP/cm^{2})$ at 298 K, assuming steady state conditions.
Solution:
Using Equation 14.9 and using the permeability coefficient of oxygen through LDPE in Table 14.6, it follows:
$2.2 \times 10^{-13} (cm^{3} STP)-cm/cm^{2}-s-Pa$,
$J = P_{M} \frac{ΔP}{Δx} = P_{M} \frac{P_{2}-P_{1}}{Δx} = (\frac {2.2 \times 10^{-13} (cm^{3} STP)-cm}{cm^{2}-s-Pa}) \frac{(2.0 \times 10^{6} Pa) -150,000 Pa}{0.15 cm} \\= 2.71 \times 10^{-6} \frac{cm^{3}STP}{cm^{2}-s}$
