Problem 41 - Physics

Question

The diffusion constant of for the alcohol ethanol in water is \(12.4 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s} . \mathrm{A}\) cylinder has a cross- sectional area of \(4.00 \mathrm{~cm}^{2}\) and a length of \(2.00 \mathrm{~cm}\). A difference in ethanol concentration of \(1.50 \mathrm{~kg} / \mathrm{m}^{3}\) is maintained between the ends of the cylinder. In one hour, what mass of ethanol diffuses through the cylinder?

Step-by-Step Solution

Answer

In one hour, the mass of ethanol diffuses through the cylinder is \(1.34 \times 10^{-8} \) kg.

Step 1: Identify the key formula

The equation for the mass of substance diffusing through a cylinder is given by Fick's first law of diffusion: \[ J = -D \frac{\Delta C}{\Delta x} \] Where \( J \) is the diffusion flux, \( D \) is the diffusion constant, \( \Delta C \) is the concentration difference, and \( \Delta x \) is the length of the diffusion path.

Step 2: Calculate diffusion flux

Plug the given values into Fick's first law:\[ D = 12.4 \times 10^{-10} \text{ m}^2/ ext{s} \] \[ \Delta C = 1.50 \text{ kg/m}^3 \] \[ \Delta x = 0.0200 \text{ m} \] Calculating diffusion flux:\[ J = - (12.4 \times 10^{-10}) \left(\frac{1.50}{0.0200}\right) \]\[ J = - (12.4 \times 10^{-10}) (75) \]\[ J = - 9.30 \times 10^{-9} \text{ kg/m}^2/ ext{s} \] (The negative sign indicates direction but doesn't affect magnitude here.)

Step 3: Find total mass diffused

Use the formula to find the total mass diffused: \[ M = J \cdot A \cdot t \] Where \( A \) is the cross-sectional area and \( t \) is time.Given:- \( A = 4.00 \text{ cm}^2 = 4.00 \times 10^{-4} \text{ m}^2 \)- \( t = 1 \text{ hour} = 3600 \text{ seconds} \)Plug these into the formula:\[ M = (9.30 \times 10^{-9}) \cdot (4.00 \times 10^{-4}) \cdot 3600 \]\[ M = 1.34 \times 10^{-8} \text{ kg} \]

Step 4: Finalize and interpret the result

The mass of ethanol that diffuses through the cylinder in one hour is calculated to be \(1.34 \times 10^{-8} \) kg. This means only a very tiny amount of ethanol travels from one side to the other due to the small concentration gradient and diffusion constant.

Key Concepts

Fick's First Law of Diffusion

Diffusion is the process by which molecules spread from areas of high concentration to areas of low concentration. This movement continues until the molecules are evenly distributed. Fick’s first law of diffusion helps us understand how substances move. It states that the diffusion flux, \( J \), which is the amount of substance moving through a unit area per unit time, is proportional to the concentration gradient \( \Delta C/\Delta x \). The formula for Fick’s first law is:

• \( J = -D \frac{\Delta C}{\Delta x} \)

Where:

• \( D \) is the diffusion constant, a measure of how easily molecules move through a medium.
• \( \Delta C \) is the difference in concentration.
• \( \Delta x \) is the distance over which the concentration changes.

The negative sign indicates that diffusion occurs in the direction of decreasing concentration. It's important to remember that this law applies to steady-state diffusion, where conditions remain constant over time.

Diffusion Constant

The diffusion constant \( D \) is crucial when examining how quickly molecules can move through a material. For ethanol diffusing in water, \( D \) is given as \( 12.4 \times 10^{-10} \text{ m}^2/ ext{s} \). This value is specific to the ethanol-water pair and reflects how easily ethanol molecules navigate through water.
This constant can vary depending on factors like:

• The type of molecules involved. Larger or more complex molecules usually have lower diffusion constants.
• The medium through which diffusion takes place. Different mediums offer different levels of resistance to molecular movement.

A higher diffusion constant means molecules spread out more quickly, whereas a lower diffusion constant indicates slower movement. The diffusion constant integrates with Fick's law to predict how much substance diffuses over time.

Concentration Gradient

The concentration gradient, \( \Delta C \), is the difference in the amount of molecules over a specific distance, \( \Delta x \). In our example, this is \( 1.50 \text{ kg/m}^3 \) over \( 2 \text{ cm} \). The gradient provides a measure of how strong the driving force for diffusion is.
A steeper concentration gradient signifies a more significant difference in concentrations across a distance.

• This generally results in a faster rate of diffusion.
• The molecules move more quickly from the area of higher concentration to the area of lower concentration.

Conversely, a less steep concentration gradient will slow down the diffusion process as the driving force becomes weaker. Such conditions must be controlled by the experimenter depending on the desired diffusion rate.

Cylinder Geometry

Understanding the geometry of the system is essential when calculating diffusion rates. A cylinder, for instance, has a specific cross-sectional area and length which directly affect how substances move through it. In the given exercise:

• The cross-sectional area \( A \) is \( 4.00 \text{ cm}^2 = 4.00 \times 10^{-4} \text{ m}^2 \).
• The length \( \Delta x \) is \( 2.00 \text{ cm} \) or \( 0.0200 \text{ m} \).

The area impacts the total quantity of ethanol that can pass through at any time.
When applying Fick's law, the geometry ensures you're calculating how much ethanol diffuses through this specific spatial setup over time. Larger cross-sectional areas allow more molecules to traverse simultaneously compared to smaller ones, thus increasing the overall diffusion rate.

Ethanol in Water

When considering ethanol diffusing in water, it’s important to understand both substances' interactions. Ethanol is a small organic molecule, and water is a polar solvent.
The diffusion process in such systems can be influenced by factors like:

• Molecular size - Ethanol's small size facilitates its movement through water.
• Polarity - Both ethanol and water are polar, which can affect their interactions and diffusion.

In the discussed exercise, the diffusion of ethanol is quantitatively analyzed using Fick's first law to establish how much ethanol moves through the cylinder over a set time. By understanding the specific properties of ethanol and water, better predictions of diffusion behaviors can be made, which are crucial to processes in chemical and biological contexts.