Answer:
The value is [tex]J = 1.71 *10^{-6} \ kmol/m^2\cdot s [/tex]
Explanation:
From the question we are told that
The length the tube is [tex]l = 0.20 \ m[/tex]
The diameter of the tube is [tex]d = 0.01 \ m[/tex]
The total pressure inside the tube is [tex]P = 101.32\ kPa = 101.32 *10^{3} \ Pa[/tex]
The partial pressure of CO2 at the first end is [tex]P_1 = 456 \ mmHg = 456 *133.322 = 60794.832 \ Pa[/tex]
The partial pressure of CO2 at the other end is [tex]P_2 = 76 \ mmHg = 76 *133 = 10132.472 \ Pa[/tex]
The temperature is T = 298 K
The diffusion coefficient is [tex]D_{1,2} = 1.67 * 10^{-5} \ m^2 /s[/tex]
Generally the molar flux of CO2 is mathematically represented as
[tex]J = \frac{D_{1,2} * [P_1 -P_2]}{R* T[l]}[/tex]
Here R is the gas constant with value [tex]R = 8.314 \ J/k mol[/tex]
So
[tex]J = \frac{ 1.67 * 10^{-5} * [60794.832 -10132.472 ]}{8.314 * 298 [0.20]}[/tex]
[tex]J = 1.71 *10^{-3} \ mol/m^2\cdot s [/tex]
Converting to kmol
[tex]J = \frac{1.71 *10^{-3}}{1000} [/tex]
[tex]J = 1.71 *10^{-6} \ kmol/m^2\cdot s [/tex]
