Question

In the early 1990s, fusion involving hydrogen dissolved in palladium at room temperature, or cold fusion, was proposed as a new source of energy. This process relies on the diffusion of H2 into palladium. The diffusion of hydrogen gas through a 0.005-cm-thick piece of palladium foil with a cross section of 0.810 cm2 is measured. On one side of the foil, a volume of gas maintained at 298 K and 1 atm is applied, while a vacuum is applied to the other side of the foil. After 24 h, the volume of hydrogen has decreased by 16.0 cm3 . What is the diffusion coefficient of hydrogen gas in palladium?

Answer 1

Fick's first law: J = -D (dC/dx)

where, J = "diffusion flux," amount of substance that will flow per unit area per unit time

D = diffusion coefficient

dC = change in concentration

dx = change in position

Here,

Amount of the gas that flows

= PV/RT = 16cm³ * 1 atm / 0.082 lit.atm.mol^-1.K^-1 * 298 K

=0.016 lit * 1 atm / 0.082 lit.atm.mol^-1.K^-1 * 298 K

= 0.000655 moles

J = Amount(mol) of the gas that flows/area*time

= 0.000655 moles / 0.810 cm² * 24 h

= 0.000655 moles *100² / 0.810 m² * 24 * 3600 s

= 9.356 *10^-5 moles.m^-2.s^-1

From ideal gas law, PV = nRT

or, density = n/V = P/RT

Initial concentration of gas

= P/RT

=1 atm/ 0.082 lit.atm.mol^-1.K^-1 * 298 K

= 0.041 lit^-1.mol

Concentration initially on the vacuum side = 0

So, dC = 0 - 0.041 lit^-1.mol = - 0.041 lit^-1.mol

dx = 0.005 cm = 0.00005 m

So,

J = -D (dC/dx)

or, (9.356 *10^-5 moles.m^-2.s^-1)= -D [(- 0.041 lit^-1.mol)/0.00005m]

or, (9.356 *10^-5 moles.m^-2.s^-1)= -D [(- 0.000041 mol.m^-3 )/0.00005m]

or, D = (9.356 *10^-5 moles.m^-2.s^-1)/ [(0.000041 mol.m^-3 )/0.00005m]

= 1.141*10^-4 m².s