Question

A gaseous mixture at 265K and 1.0 atm contains 25 O2 60 N2 and 15 CO2 mole basis The velocities of the components are 0.084 cm s O2 0.120 cm s N2 and 0.052 cm s CO2 Find the N2 diffusion velocity relative to the mole average velocity and the molar diffusional flux of N2

Answer 1

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

Step-by-step 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:

`v_(av) =(0.25*(-0.084))+(0.6*0.12)+(0.15*0.052)=0.0588cm/s`

The N₂ diffusion velocity relative to the molar average velocity is:

`v_(i) -v_(av) =-0.084-0.0588=-0.1428cm/s`

The molar diffusional flux of N₂ is:

`N_{N_(2) } =(P)/(RT) y_(A) (v_(i) -v_(av) )=-3.9x10^(-3)`