Answer:
The molar average velocity is 0.0588 cm/s
The N₂ diffusion velocity relative to the mole average velocity is -0.1428 cm/s
The molar diffusional flux of N₂ is -3.9x10⁻³
Explanation:
Given data:
T = temperature = 265 K
O₂ = 25%
N₂ = 60%
CO₂ = 15%
vO₂ = -0.084 cm/s
vN₂ = 0.12 cm/s
vCO₂ = 0.052 cm/s
The molar average velocity is equal:
[tex]v_{av} =(0.25*(-0.084))+(0.6*0.12)+(0.15*0.052)=0.0588cm/s[/tex]
The N₂ diffusion velocity relative to the molar average velocity is:
[tex]v_{i} -v_{av} =-0.084-0.0588=-0.1428cm/s[/tex]
The molar diffusional flux of N₂ is:
[tex]N_{N_{2} } =\frac{P}{RT} y_{A} (v_{i} -v_{av} )=-3.9x10^{-3}[/tex]
