Question

A gas of CH 4 and He is contained in a tube at 101.32 kPa pressure and 298 K. At one point the partial pressure of methane is p A1 = 60.79 kPa, and at the point 0.02 m distance away, p A2 = 20.26 kPa. If the total pressure is constant throughout the tube, calculate the flux of CH 4 (methane) at steady state for equimolar counterdiffusion

Answer 1

Let CH₄ be denoted by "A" .
Let He be denoted by "B".

Let "1" and "2" be two distinct points on the tube separated by distance 0.02m .

Then, p_A1 = 60.79 kPa , p_B1 = 40.53 kPa

p_A2 = 20.26 kPa , p_B2 = 81.06 kPa

Considering steady state equimolar counter-diffusion of Ideal Gases , we use the Fick's Law (Ideal Gas):

Molar Flux JA^* = -D_AB x dC_A/ dZ ,

where dC_A/ dZ can be expressed for ideal gases as , dP/RTdZ

Here, J_A^* = -D_AB x (p_A2 - p_A1) / {RT(z₂ - z₁)}

At 298K and 1 atm pressure ( conditions stated in the problem) , D_AB= 0.675 x 10^-4 m²/sec (for He-CH4) .

• p_A2 - p_A1 = -40.53 kPa
• z2 - z1 = 0.02m
• R = 8.314 J/mol.K
• T = 298K

Substituing the values in the above equation,

J_A^* = 5.52108 x 10^-2 mol/m².sec