The purification of hydrogen gas by diffusion through a palladium sheet was discussed in section 5.3. compute the number of kilograms of hydrogen that pass per hour through a 6-mm thick sheet of palladium having an area of 0.25 m2at 600°c. assume a diffusion coefficient of 1.7 × 10–8m2/s, that the respective concentrations at the high- and low-pressure sides of the plate are 2.0 and 0.4 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

NA = 6.8 E-12 Kg H2(g) / hour

Explanation:

steady-state diffusion of A through non-diffuser B:

∴ (A): H2(g)    

∴ (B): Pd

∴ DAB = 1.7 E-8 m²/s

∴ p*A1 = 2.0 Kg H2 / m³ Pd

∴ p*A2 = 0.4 Kg H2 / m³ Pd

∴ z = 6 mm = 6 E-3 m

∴ T = 600°C ≅ 873 K

∴ R = 8.314 J/mol.K = 8.314 N.m/mol.K

⇒ NA = ((1.7 E-8)/(8.314)(873)(6 E-3))(2.0 - 0.4)

⇒ NA = 6.246 E-10 mol/s.m³

for A = 0.25 m²

⇒ volume (v) = A×z = (0.25)(6 E-3) = 1.5 E-3 m³

∴ Mw H2(g) = 2.016 g/mol

⇒ NA = (6.246 E-10 mol/s.m³)(1.5 E-3 m³)(2.016 g/mol)(Kg/1000 g)(3600 s/h)

⇒ NA = 6.8 E-12 Kg H2(g)/h

answer; i believe the correct answer is (solid gold);