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 5-mm-thick sheet of palladium having an area of 0.20 m2 at 500C. Assume a diffusion coefficient of 1.0 108 m2 /s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained

This question is incomplete, the complete question is;

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 5-mm-thick sheet of palladium having an area of 0.20 m² at 500C.

Assume a diffusion coefficient of 1.0 × 10⁻⁸ m²/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained

the number of kilograms of hydrogen that pass per hour through is 2.592 × 10⁻³ kg

Explanation:

Given the data in the question;

thickness of sheet t = 5 mm

Area A = 0.20 m²

Temperature T = 500°C

diffusion coefficient ∝ = 1.0 × 10⁻⁸ m²/s

concentration high pressure side C = 2.4

concentration low pressure side C = 0.6 kg

from the question, we calculate the concentration gradient

dc/dx = (C -  C)/dt

so we substitute

dc/dx = (2.4 - 0.6) / ( - 5 × 10⁻³ )

dc/dx = -360

now, mass of hydrogen per hour that diffuse through a pd sheet

M = -∝AT(dc/dx)

where time t is 1 hour ( 3600 sec )

we substitute

M = -(1.0 × 10⁻⁸) × 0.20 × 3600 (-360)

M = 0.002592

M = 2.592 × 10⁻³ kg per one hour

Therefore, the number of kilograms of hydrogen that pass per hour through is 2.592 × 10⁻³ kg

neutrons are in the middle of the atom and the electrons, which are much smaller.

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