Question

5. Consider the absorption of hydrogen into molten aluminum at 750°C, according to the reaction; H2 = 2 H. in which hydrogen molecules are adsorbed and dissolve into liquid aluminum, in atomic form, according to Sieverts Law of solubility. Given that we know the solubility of hydrogen in molten aluminum at this temperature is lcc/100 gm of liquid aluminum for hydrogen at unit activity (PH2 = 1 atm), calculate this solubility in terms of ppm H (parts per million H on a wt% basis). If the actual concentration of hydrogen in aluminum is 1ppm (wt) for a particular commercial casting operation, estimate the partial pressure of hydrogen that would be manifested immediately above the molten aluminum surface.

Answer 1

The volume of hydrogen = 1 cc = 1 mL = 10^-3 L

According to the ideal gas equation: PV = nRT

i.e. 1 atm * 10^-3 L = n * 0.0821 L.atm/mol.K * (750+273) K

i.e. The no. of moles of hydrogen (n) = 1.19*10^-5 mol

i.e. The mass of hydrogen = 1.19*10^-5 mol * 2 g/mol = 2.38*10^-5 g

i.e. The mass of monoatomic hydrogen (H) = 2*2.38*10^-5 g = 4.76*10^-5 g

Now, the solubility of hydrogen = 4.76*10^-5 g H/100 g Al = 0.476 g H/10⁶ g Al = 0.476 ppm

If the solubility of hydrogen (monoatomic, i.e. H) is 1 ppm, then the partial pressure of hydrogen = (1/0.476)*1 atm = 2.1 atm