Question

18.1-1. Diffusion of Methane Through Helium. A gas of CH4 and He is contained in a tube at 101.32 kPa pressure and 298 K. At one point, the partial pressure of methane is pA1 = 60.79 kPa, and at a point 0.02 m distance away, pA2 = 20.26 kPa. If the total pressure is constant throughout the tube, calculate the flux of CH4 (methane) at steady state for equimolar counterdiffusion.

Answer 1

Answer:

5.521 × 10⁻² mol/m².s

Explanation:

Given:

Pressure of the Methane and Helium gas = 101.32 kPa

Temperature of the Methane and Helium gas = 298 K

Partial pressure of Methane,  pA₁ = 60.79 kPa

Partial pressure of Methane at point 0.02 m away,  pA₂  = 20.26 kPa

Now,

Molar flux is given as:

$J_A^* = -D_{AB} \times\frac{pA_2 - pA_1}{RT(z_2 - z_1)}$

Here,

$D_{AB}$= 0.675 × 10⁻⁴ m²/s (for He-CH4 at 101.32 kPa and 298 K)

Z₂ - Z₁ = 0.02 m

R is the ideal gas constant = 8.314 J/mol.K

T is the temperature =  298 K

On substituting the respective values, we get

$J_A^* = -0.675\times10^{-4} \times\frac{20.26 - 60.79}{8.314\times298(0.02)}$

or

= 5.521 × 10⁻² mol/m².s