The diffusion constant for the amino acid glycine in water is 1.06 × 10⁹ m²/s. In a 2.0-cm-long tube with a cross-sectional area of 1.2 x 10-4 m², the mass rate of diffusion is m/t = 4.2 x 10-14 kg/s, because the glycine concentration is maintained at a value of 9.0 × 10-³ kg/m³ at one end of the tube and at a lower value at the other end. What is the lower concentration? C₁ =

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**Diffusion Constant and Concentration Calculation**

The diffusion constant for the amino acid glycine in water is \(1.06 \times 10^{-9} \, \text{m}^2/\text{s}\). In a 2.0-cm-long tube with a cross-sectional area of \(1.2 \times 10^{-4} \, \text{m}^2\), the mass rate of diffusion is given as \(\frac{m}{t} = 4.2 \times 10^{-14} \, \text{kg/s}\). The glycine concentration is maintained at a value of \(9.0 \times 10^{-3} \, \text{kg/m}^3\) at one end of the tube and at a lower value at the other end. The problem is to find the lower concentration.

**Equation to consider:**

Fick's First Law of Diffusion is often used in such problems, which can be given by:

\[ J = -D \frac{\Delta C}{\Delta x} \]

Where:
- \( J \) is the diffusion flux (\(\text{kg/m}^2\text{s}\)),
- \( D \) is the diffusion constant (\(\text{m}^2/\text{s}\)),
- \(\Delta C\) is the concentration difference (\(\text{kg/m}^3\)),
- \(\Delta x\) is the distance (\(\text{m}\)).

By using these values, the lower concentration at the other end of the tube can be calculated.

(Note: The diagram in the image is a simple entry box labeled "C₁ =" for entering the final calculated value.)
Transcribed Image Text:**Diffusion Constant and Concentration Calculation** The diffusion constant for the amino acid glycine in water is \(1.06 \times 10^{-9} \, \text{m}^2/\text{s}\). In a 2.0-cm-long tube with a cross-sectional area of \(1.2 \times 10^{-4} \, \text{m}^2\), the mass rate of diffusion is given as \(\frac{m}{t} = 4.2 \times 10^{-14} \, \text{kg/s}\). The glycine concentration is maintained at a value of \(9.0 \times 10^{-3} \, \text{kg/m}^3\) at one end of the tube and at a lower value at the other end. The problem is to find the lower concentration. **Equation to consider:** Fick's First Law of Diffusion is often used in such problems, which can be given by: \[ J = -D \frac{\Delta C}{\Delta x} \] Where: - \( J \) is the diffusion flux (\(\text{kg/m}^2\text{s}\)), - \( D \) is the diffusion constant (\(\text{m}^2/\text{s}\)), - \(\Delta C\) is the concentration difference (\(\text{kg/m}^3\)), - \(\Delta x\) is the distance (\(\text{m}\)). By using these values, the lower concentration at the other end of the tube can be calculated. (Note: The diagram in the image is a simple entry box labeled "C₁ =" for entering the final calculated value.)
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