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

Answer:

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[tex]_A[/tex] = 2.4

concentration low pressure side C[tex]_B[/tex] = 0.6 kg

from the question, we calculate the concentration gradient

dc/dx = (C[tex]_A[/tex] -  C[tex]_B[/tex])/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

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 500°C is 2.592 × 10⁻³ kg.

Purification: It refers to the process of removing impurities or contaminants from a substance in order to make it cleaner, purer, or more refined. It also involves the elimination or reduction of unwanted substances or components that may be present in the original material.

According to the question, given data is:

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 [tex]C_A[/tex] = 2.4

Concentration low pressure side[tex]C_B[/tex]= 0.6 kg

from the question, we calculate the concentration gradient,

[tex]dc/dx = (C _A- C_B)/dt[/tex]

so, we substitute the values,

[tex]dc/dx[/tex] = (2.4 - 0.6)/ ( - 5 × 10⁻³)

[tex]dc/dx = -360[/tex]

Now, mass of hydrogen per hour that diffuse through a Palladium sheet,

M = -∝[tex]AT(dc/dx)[/tex]

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 hour.

Therefore, the number of kilograms of hydrogen that pass per hour through a palladium sheet is approximately 2.592 × 10⁻³ kg.

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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.

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