Respuesta :
Answer:
a. Therefore, the flux in kg mol/s.m² at N₂ = [tex]35.398 *10^{-29}kgmol/sm^2[/tex]
b. Therefore; when temperature = 473k , the flux (J) decreases.
c. Hence, when T = 298K, but the total pressure = 3.0 atm , the flux increases.
d. The CO flux for part C = [tex]212.6*10^{-29}kgmol/sm^2[/tex]
Explanation:
Given that :
Equimolar counter-diffusion [tex](d_x)[/tex] = 0.11 m
[tex]D_{AB} =[/tex] [tex]2.05 *10^{-5}m^2s^{-1}[/tex]
T = 298 K
For ideal gas equation:
PV = nRT
Making V the subject of the formula:
[tex]V = \frac{nRT}{P}[/tex]
where the number of moles (n) = [tex]\frac{mass}{molarmass}[/tex]
∴ the V = [tex]\frac{mass of N_2}{Molar mass of N_2} *\frac{RT}{P}[/tex]
V = [tex]\frac{14 gmol^{-1}}{28 gmol^{-1}} *\frac{8.314KPadm^3K^{-1}mol^{-1}}{101325Pa}[/tex]
V = [tex]\frac{0.5*2477.572KPadm^3mol^{-1}}{101325Pa}[/tex]
V = [tex]0.01223Kdm^{3}mol^{-1}[/tex]
V = [tex]12.23 dm^3mol^{-1}[/tex]
V = [tex]\frac{12.23dm^3mol^{-1}}{6.023*10^{23}mol^{-1}}[/tex]
V = [tex]2.031*10^{23}dm^3[/tex]
V = [tex]2.031*10^{-23}*10^{-3}m^3[/tex]
V = [tex]2.031*10^{-26}m^3[/tex]
The volume of N₂ = [tex]2.031*10^{-26}m^3[/tex]
Density of [tex]P_{N_2} = \frac{mass of N_2}{Volume}[/tex]
= [tex]\frac{\frac{14gmol^{-1}}{6.023*10^{23}mol^{-1}} }{2.031*10^{-26}m^3}[/tex]
[tex]P_{N_2} = \frac{2.324*10^{-23}}{2.031*10^{-26}m^3}[/tex]
= [tex]1.44*10^3gm^{-3}[/tex]
[tex]P_{N_2} = 1144 gm^{-3}[/tex]
Now, The flux (J) = [tex]D_{AB}*\frac{P_{N_2}}{d_x}[/tex]
J = [tex]\frac{2.05*10^{-5}m^2s^{-1}*1144gm^{-3}}{0.11m}[/tex]
J = [tex]21320*10^{-5}g/sm^2[/tex]
J = [tex]21320*10^{-8}kg/sm^2[/tex]
J = [tex]3539.8 *10^{-31}kgmol/sm^2[/tex]
J = [tex]35.398 *10^{-29}kgmol/sm^2[/tex]
Therefore, the flux in kg mol/s.m² at N₂ = [tex]35.398 *10^{-29}kgmol/sm^2[/tex]
b. At T = 473 K
[tex]V_{N_2}= \frac{0.5*8.314kPaK^{-1}mol^{-1}*473}{101325Pa}[/tex]
[tex]V_{N_2} =19.41 dm^3mol^{-1}[/tex]
[tex]V_{N_2} = 19.41 *10^{-3}m^3mol^{-1}[/tex]
[tex]P_{N_2}= \frac{14gmol^{-1}}{19.41*10^{-3}m^3mol^{-1}}[/tex]
[tex]P_{N_2}= 0.7123*10^3gm^{-3}[/tex]
J = [tex]\frac{2.05*10^{-5}m^2s^{-1}*0.7213*10^3gm^{-3}}{0.11m}[/tex]
J = [tex]13.44 *10^{-2} g/sm^2[/tex]
J = [tex]2.232*10^{-25}gmol/sm^2[/tex]
J = [tex]2.232*10^{-28}kgmol/sm^2[/tex]
J = [tex]22.32 *10^{-29} kgmol/sm^2[/tex]
Therefore; when temperature = 473k , the flux (J) decreases.
c. At P = 3 atm = 3×101325 Pa
T = 298 K
[tex]V_{N_2} = \frac{0.5 *8.314KPa K^{-1}mol^{-1}*298K}{3*101325Pa}[/tex]
[tex]V_{N_2} = 4.08dm^3mol^{-1}[/tex]
[tex]V_{N_2} = 4.08*10^{-3}m^3mol^{-1}[/tex]
[tex]P_{N_2} = \frac{14g/mol}{4.02*10^{-3}m^3mol^{-1}}[/tex]
[tex]P_{N_2} =3.43 *10^3gm^{-3}[/tex]
J = [tex]\frac{2.05*10^{-5}m^2s^{-1}*3.43*10^3gm^{-3}}{0.11m}[/tex]
J = [tex]63.92*10^{-2}g/sm^2[/tex]
J = [tex]63.92*10^{-5}kg/sm^2[/tex]
To moles; we have:
J = [tex]10.61*10^{-28}kgmol/sm^2[/tex]
J = [tex]106.1 *10^{-29}kgmol/sm^2[/tex]
Hence, when T = 298K, but the total pressure = 3.0 atm , the flux increases.
d. We can determine the CO flux for part c as follows:
[tex]P_{CO} = \frac{mass of CO}{Volume}[/tex]
[tex]P_{CO} =[/tex] [tex]\frac{28.01g/mol}{4.08*10^{-3}m^3mol^{-1}}[/tex]
[tex]P_{CO} =[/tex] [tex]6.87*10^3gm^{-3}[/tex]
J = [tex]\frac{2.05*10^{-5}m^2s^-{1}*6.87*10^3gm^{-3}} {0.11m}[/tex]
J = [tex]128.03*10^{-2}g/sm^2[/tex]
J = [tex]128.03*10^{-5}kg/sm^2[/tex]
J = [tex]212.6*10^{-29}kgmol/sm^2[/tex]
The CO flux for part C = [tex]212.6*10^{-29}kgmol/sm^2[/tex]
