Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration of high-pressure side is [tex]C_1 = 0.722 \ kg/m^3[/tex]
Explanation:
From the question we are told that
The thickness of the polyethylene is [tex]d = 1.5 \ mm = 0.0015 \ m[/tex]
The temperature is [tex]T = 600 \ K[/tex]
The flux is [tex]JA = 2.48 *10^{-5} \ kg/m^2\cdot s[/tex]
The concentration on the low-pressure side is [tex]C_2 = 0.5 \ kg/m^3[/tex]
The initial diffusivity is [tex]D_o = 6.2 *10^{-4} \ m^2 /s[/tex]
The activation energy for diffusion is [tex]Q_d = 41 \ kJ /mol = 41*10^3 J /mol[/tex]
Generally the diffusivity of the oxygen at 600 K can be mathematically evaluated as
[tex]D = D_o * e^{- \frac{Q_d}{R * T } }[/tex]
substituting values
[tex]D = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600 } }[/tex]
[tex]D = 1.671 *10^{-7} \ m^2 /s[/tex]
Generally the flux is mathematically represented as
[tex]JA = D * \frac{C_1 -C_2}{d}[/tex]
Where [tex]C_1[/tex] is the concentration of oxygen at the higher side
So
[tex]C_1 = d * \frac{JA}{D} + C_2[/tex]
substituting values
[tex]C_1 = 0.0015 * \frac{2.48*10^{-5}}{1.671*10^{-7}} + 0.5[/tex]
[tex]C_1 = 0.722 \ kg/m^3[/tex]
