Question

What is the pressure inside an alveolus having a radius of 2.50 × 10⁻⁴ , {m}if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as that created by a spherical bubble.

A. P = 2.00 , {kPa}
B. P = 4.00 , {kPa}
C. P = 6.00 , {kPa}
D. P = 8.00 , {kPa}

Answer 1

The pressure inside the alveolus with a radius of 2.50 × 10⁻⁴ m, assuming the surface tension of the fluid-lined wall is the same as for soapy water, is P = 4.00 kPa. Therefore, The correct Option is B because, according to Laplace's law, which governs the pressure inside a spherical bubble or alveolus, the pressure (P) is inversely proportional to the radius (r). As the radius decreases, the pressure increases.

Step-by-step explanation:

The pressure inside a spherical bubble or alveolus can be determined using the Laplace's law, which relates pressure (P), surface tension (T), and radius (r). The formula is given by:

`\[ P = (4T)/(r) \]`

In this case, the radius (r) is 2.50 × 10⁻⁴ m, and we are told the surface tension (T) is the same as for soapy water. Plugging in these values into the formula:

`\[ P = \frac{4 * T_{\text{soapy water}}}{2.50 * 10^(-4)} \]`

Now, the surface tension of soapy water is typically around 0.03 N/m. Substituting this value into the equation:

`\[ P = (4 * 0.03)/(2.50 * 10^(-4)) \]`

Calculating this gives us the pressure, which is approximately 4.00 kPa. Therefore, the final answer is option B.

In conclusion, the pressure inside the alveolus is directly influenced by the surface tension and inversely proportional to the radius. As the radius decreases, the pressure increases, following Laplace's law. Understanding this relationship is crucial in explaining physiological phenomena such as pulmonary surfactant's role in maintaining alveolar stability and preventing collapse. The calculation reinforces the importance of surface tension in determining pressure within small structures like alveoli.