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

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error