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

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:

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

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:

$N_{N_{2} } =\frac{P}{RT} y_{A} (v_{i} -v_{av} )=-3.9x10^{-3}$