The purification of hydrogen gas is to be achieved by diffusion through a palladium sheet. Calculate the number of kilograms of hydrogen that passes per hour through an 8-mm thick sheet of palladium having an area of 0.2 m2 at 520°C. Assume a linear concentration profile with a diffusion coefficient of 1.10x10-8 m2/s. The high and low-pressure sides of the plate are 2 and 0.5 kg/m3, respectively. Include a drawing with your answer. Please give your answer with 4 significant figures.

We are tasked with calculating the number of kilograms of hydrogen gas that passes per hour through a palladium sheet using Fick's first law of diffusion. The formula for the diffusion rate through a planar sheet is given by:

\[
J = \frac{D (C_2 - C_1)}{L}
\]

Where:
- \( J \) is the diffusion flux in \( \text{kg/m}^2 \cdot \text{s} \)
- \( D \) is the diffusion coefficient in \( \text{m}^2/\text{s} \)
- \( C_2 \) and \( C_1 \) are the concentrations of hydrogen on the high-pressure and low-pressure sides, respectively, in \( \text{kg/m}^3 \)
- \( L \) is the thickness of the sheet in meters

Once we calculate \( J \), we can find the total mass of hydrogen passing per second by multiplying the diffusion flux \( J \) by the area \( A \) of the sheet:

\[
\dot{m} = J \times A
\]

The mass per hour can then be found by multiplying the mass per second by the number of seconds in an hour (3600 seconds).

Let's break it down and calculate the value. I'll now calculate this.# Given data
D = 1.10e-8  # diffusion coefficient in m^2/s
C2 = 2.0     # high pressure side concentration in kg/m^3
C1 = 0.5     # low pressure side concentration in kg/m^3
L = 8e-3     # thickness of the palladium sheet in meters
A = 0.2      # area of the palladium sheet in m^2
seconds_per_hour = 3600  # number of seconds in one hour

# Calculating the diffusion flux J (kg/m^2/s)
J = D * (C2 - C1) / L

# Calculating the total mass per second (kg/s)
mass_per_second = J * A

# Calculating the total mass per hour (kg/hour)
mass_per_hour = mass_per_second * seconds_per_hour

# Provide the result with 4 significant figures
mass_per_hour_kg = round(mass_per_hour, 4)
mass_per_hour_kgThe number of kilograms of hydrogen gas that passes through the palladium sheet per hour is \( 0.0015 \, \text{kg} \).

### Drawing:
Let's illustrate the system:

- The palladium sheet has a thickness \( L = 8 \, \text{mm} \).
- The surface area through which diffusion occurs is \( 0.2 \, \text{m}^2 \).
- On one side of the sheet, the hydrogen concentration is \( 2 \, \text{kg/m}^3 \) (high-pressure side).
- On the other side, the concentration is \( 0.5 \, \text{kg/m}^3 \) (low-pressure side).

This creates a concentration gradient across the sheet, driving the diffusion of hydrogen gas.

### Diagram:

Here’s a conceptual sketch:

```
High-Pressure Side (C2 = 2 kg/m³)
    |---------------------------|
    |                           |    Palladium Sheet (Thickness = 8 mm)
    |---------------------------|
Low-Pressure Side (C1 = 0.5 kg/m³)
```

Would you like further clarification on any aspect, or do you have any questions?

### Related Questions:
1. How would the diffusion rate change if the thickness of the sheet were halved?
2. How does temperature affect the diffusion coefficient in this scenario?
3. What if the surface area were doubled—how would the hydrogen flow rate change?
4. How would the concentration gradient change if the pressures on both sides were equal?
5. How does the type of metal used for the sheet affect the diffusion rate of hydrogen?

### Tip:
Remember that diffusion is highly dependent on the temperature and material properties. Increasing temperature often increases the diffusion coefficient significantly!

Learn how to calculate the diffusion of hydrogen gas through a palladium sheet using Fick's first law of diffusion. Given specific conditions like diffusion coefficient, concentration gradient, sheet thickness, and temperature, find out the mass of hydrogen passing through per hour. Includes step-by-step calculations and a clear explanation. Ideal for academic studies and practical applications.

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