Question

CHEMICAL ENGINEERING Two bulbs are connected by a straight tube, 0.001 m in diameter and 0.15 m in length. Initially the bulb at end 1 contains N2 and the bulb at end 2 contains H2. The pressure and temperature are maintained constant at 25°C and 1 atm. At a certain time after allowing diffusion to occur between the two bulbs, the nitrogen content of the gas at end 1 of the tube is 80 mol% and at end 2 is 25 mol%. If the binary diffusion coefficient is 0.784 cm2 /s, determine: (a) The rates and directions of mass transfer of hydrogen and nitrogen in mol/s (b) The species velocities relative to stationary coordinates, in cm/s

Answer 1

Let A = N₂, B= H₂

T = 273 + 25 = 298 K

P = 1 atm = 101325 Pa

Since P, T are kept constant we can assume equimolar counterdiffusion of A & B

P_A¹ = 101325 * 80 /100 = 81060 Pa, P_A² = 101325 * 25 /100 = 25331.25 Pa

P_B¹ = 101325 * 20 /100 = 20265 Pa, P_B² = 101325 * 75 /100 = 75993.75 Pa

Na = - (0.784 x 10^-4 m²/s) (P_A² - P_A¹ Pa) / (8.314 J/mol-K * 298 K * 0.15 m)

= 0.012 mol/m²-s

Nb = - Na = -0.012 mol/m²-s

A = pi*d²/4 = 3.14 * 0.001² / 4 = 7.85 x 10^-7 m2

Ra = Na. A = 0.012 * 7.85 x 10^-7 = 9.23 x 10^-9 mol/s

Rb = -9.23 x 10^-9 mol/s (B diffuses in opposite direction)

Mole density = P / RT = n/V (mol/m³)

Mole density (Rho) = 101325 / (8.314 * 298) = 40.9 mol/m³

Volumetric Flow = Ra/Rho = 9.23 x 10^-9 / 40.9 = 2.26 x 10^-10 m³/s

Species velocity of A = Vol. Flow / A = 2.26 x 10^-10 / 7.85 x10^-7 = 2.9 x 10^-4 m/s

Species velocity of B = - 2.9 x 10^-4 m/s