Equimolar counter-diffusion is occurring at steady state in a tube 0.11 m long containing N2 and CO gases at a total pressure of 1.0 atm absolute. The partial pressure of N2 is 80 mm Hg at one end and 10 mm Hg at the other end. D_AB = 2.05 times 10^5 m^2/s.

a. Calculate the flux in kg mol/s. m^2 at 298 K for N2.
b. Repeat part a, at 473 K. Does the flux increase?
c. Repeat part a, at 298 K, but for a total pressure of 3.0 atm abs. The partial pressures of N2 remain 80 and 10 mm Hg. Does the flux change?
d. Calculate the CO flux for part c.

a.  Therefore, the flux in kg mol/s.m² at N₂ =

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 =

Explanation:

Given that :

Equimolar counter-diffusion  = 0.11 m

T = 298 K

For ideal gas equation:

PV = nRT

Making V the subject of the formula:

where the number of moles (n) =

∴ the V =

        V  =

        V  =

        V  =

        V  =

        V  =

        V  =

        V  =

        V  =

The volume of N₂ =  

Density of  

=

=

Now, The flux (J) =

J =

J =

J =

J =

J =

Therefore, the flux in kg mol/s.m² at N₂ =

b.  At T = 473 K

J =

J =

J =

J =

J =

Therefore; when temperature = 473k , the flux (J) decreases.

c. At P = 3 atm = 3×101325 Pa

T = 298 K

J =

J =

J =

To moles; we have:

J =

J =

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:

J =

J =

J =

J =

The CO flux for part C =