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Insects do not have lungs as we do, nor do they breathe through their mouths. Instead, they have a system of tiny tubes, called tracheae, through which oxygen diffuses into their bodies. The tracheae begin at the surface of the insect's body and penetrate into the interior. Suppose that a trachea is \(1.9 \mathrm{~mm}\) long with a cross-sectional area of \(2.1 \times 10^{-9} \mathrm{~m}^{2}\). The concentration of oxygen in the air outside the insect is \(0.28 \mathrm{~kg} / \mathrm{m}^{3}\) and the diffusion constant is \(1.1 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). If the mass per second of oxygen diffusing through a trachea is \(1.7 \times 10^{-12} \mathrm{~kg} / \mathrm{s}\), find the oxygen concentration at the interior end of the tube.

Step by step solution

01

Understand Fick's Law of Diffusion

Fick's Law of Diffusion describes the rate at which a substance diffuses across a space. The formula is:\[ J = -D \frac{(C_2 - C_1)}{L} \cdot A \]Where:- \( J \) is the diffusion flux (mass per unit area per time, \( \text{kg/s} \)).- \( D \) is the diffusion constant (\( \text{m}^2/\text{s} \)).- \( C_1 \) and \( C_2 \) are the concentrations at two locations (\( \text{kg}/\text{m}^3 \)).- \( L \) is the length of the diffusion path (\( \text{m} \)).- \( A \) is the cross-sectional area (\( \text{m}^2 \)).

02

Rearrange the Equation to Solve for C2

We know from Fick's Law that the mass flux is given by:\[ J = \frac{1.7 \times 10^{-12} \text{ kg/s}}{2.1 \times 10^{-9} \text{ m}^2} \]This gives us:\[ J = 0.8095 \times 10^{-3} \text{ kg/s/m}^2 \]Using the equation \( J = D \frac{(C_2 - C_1)}{L} \), we solve for \( C_2 \):\[ C_2 = C_1 + \frac{J \cdot L}{D} \]

03

Substitute Known Values

Substitute the known values into the equation:- \( C_1 = 0.28 \text{ kg/m}^3 \) (outside concentration)- \( L = 1.9 \times 10^{-3} \text{ m} \)- \( D = 1.1 \times 10^{-5} \text{ m}^2/\text{s} \)- \( J = 0.8095 \times 10^{-3} \text{ kg/s/m}^2 \)\[ C_2 = 0.28 \text{ kg/m}^3 + \frac{0.8095 \times 10^{-3} \times 1.9 \times 10^{-3}}{1.1 \times 10^{-5}} \]

04

Calculate C2

Calculate the inner concentration \( C_2 \):\[ C_2 = 0.28 \text{ kg/m}^3 + \frac{1.53805 \times 10^{-6}}{1.1 \times 10^{-5}} \]\[ C_2 = 0.28 \text{ kg/m}^3 + 0.13982 \]\[ C_2 \approx 0.41982 \text{ kg/m}^3 \]

05

Conclusion

The concentration of oxygen at the interior end of the tube is approximately \(0.42 \text{ kg/m}^3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tracheae in Insects

Insects have a fascinating respiratory system that differs significantly from humans. Instead of lungs, they rely on a network of small tubes termed tracheae. These tracheae enable the distribution of oxygen throughout the insect's body.

Here's how they work:

• The tracheae start at tiny openings on the insect's exoskeleton called spiracles.
• From these spiracles, the tracheae branch into smaller tubes, delivering oxygen directly to muscle tissues and organs.
• This system allows for an efficient diffusion process, providing oxygen without the need for blood-based oxygen transport, like in humans.

As a result of this direct diffusion, insects can maintain efficient respiration despite their small size. However, it also limits the maximum size they can achieve, as diffusion suffices for smaller organisms rather than larger ones.

Oxygen Concentration

Oxygen concentration plays a vital role in the diffusion process within an insect's tracheae. It determines how much oxygen is available for an insect to utilize for survival and activity.

The exercise you've worked on illustrates this concept well. The outside oxygen concentration is given as 0.28 kg/m³, which serves as the originating point of diffusion into the insect.

• Initially, there's more oxygen outside than inside, driving the diffusion process inward.
• The concentration of oxygen at the interior end of the trachea, obtained through calculations, reveals how much oxygen reaches the insect’s tissues, which, in this example, is approximately 0.42 kg/m³.
• The difference between the initial and resulting concentrations highlights how diffusion alters oxygen levels as it moves through the tracheae.

Understanding the concentration differences helps identify the efficiency of an insect's respiratory system and how external oxygen conditions might affect them.

Diffusion Constant

A key component that influences the rate of diffusion is the diffusion constant (denoted as D). In the exercise, this value is crucial for calculating how quickly oxygen travels along the tracheae.
What exactly is a diffusion constant?

• It describes how easily a substance moves through a medium, in this case, oxygen through air within a trachea.
• The given diffusion constant, 1.1 x 10^-5 m²/s, is specific to the medium and substance involved.
• Higher values indicate faster diffusion, whereas lower values suggest a slower process.

In the example, this constant helps determine how effectively oxygen diffuses from the outside through the length of the trachea to the animal's interior.
It is an essential factor in calculating real-life scenarios, allowing predictions and understanding of how insects breathe based on environmental conditions and biological structure.