Question

Pheromones are compounds secreted by females of many insect species to attract mates. Typically, \(1.0 \times 10^{-8} \mathrm{~g}\) of a pheromone is sufficient to reach all targeted males within a radius of 0.50 mi. Calculate the density of the pheromone (in grams per liter) in a cylindrical air space having a radius of \(0.50 \mathrm{mi}\) and a height of \(40 \mathrm{ft}\).

Short answer

The density of the pheromone in the given cylindrical airspace is \( \frac{1.0 \times 10^{-8} \, \text{g}}{\pi \times (0.5)^{2} \times \frac{40}{5280} \times 4.16818183 \times 10^{9}}\) g/L.

Step-by-step solution

Convert Units of Cylinder Dimensions to Common Units

The dimensions of the cylinder are given in different units, so the first step is to convert them to a common unit. Convert height from feet to miles considering that one mile is approximately 5280 feet. So, \( 40 \, \text{ft} = \frac{40}{5280} \, \text{miles} \).

Calculate the Volume of the Cylinder in Miles

The volume of a cylinder is given by the formula \( \pi \times r^{2} \times h \), where \( r \) is the radius and \( h \) is the height. Substituting the given radius and the converted height into the formula, the volume in miles would be \( \pi \times (0.5)^{2} \times \frac{40}{5280} \) miles\(^{3}\)

Convert Volume in Miles to Liters

To get the density in grams per liter, convert the volume from miles \(\,^{3}\) to liters. The exact conversion is complex and not a round number, but for practical purposes we can say that 1 mile\(^{3}\) is approximately equal to \(4.16818183 \times 10^{9}\) liters. So, the volume in liters would be \( \pi \times (0.5)^{2} \times \frac{40}{5280} \times 4.16818183 \times 10^{9}\) liters.

Calculate the Density

Finally, with the mass and volume in the appropriate units, the density can be calculated using the formula density = mass/volume. So, the density of the pheromones in this cylinder is \( \frac{1.0 \times 10^{-8} \, \text{g}}{\pi \times (0.5)^{2} \times \frac{40}{5280} \times 4.16818183 \times 10^{9}}\) g/L.

Key concepts

Understanding Cylinder Volume

To calculate the volume of a cylinder, which is a fundamental concept in geometry, we use the formula:

• Volume = \( \pi \times r^{2} \times h \)

Here, \( \pi \) is a constant approximately equal to 3.14159, \( r \) is the radius, and \( h \) is the height.
In our problem, the pheromone is dispersed in a cylinder-shaped area with a given radius and height.
Given:

• The radius \( r = 0.50 \text{ mi} \)
• The height needs conversion from feet to miles (we'll handle this in the unit conversion section).

Understanding the volume allows us to comprehend how much space the pheromone occupies.
Once we have the volume in a consistent unit, like miles cubed, we can move on to more accurate measurements like liters, which we'll discuss next.

The Necessity of Unit Conversion

In many scientific calculations, different measurements are often given in varying units. To simplify our calculations, it is essential to convert all units to a standard form.
In our scenario, the dimensions of the cylindrical air space are given in miles (radius) and feet (height).
To calculate the volume correctly:

• Convert height: \( 40 \text{ feet} = \frac{40}{5280} \text{ miles} \).

After standardizing these measurements in miles, we need to convert the volume from miles cubed to liters for the density calculation.
1 mile cubed is roughly equal to \( 4.16818183 \times 10^{9} \) liters.
This conversion is crucial because density is commonly expressed in grams per liter (g/L), making it easier to understand the concentration of substances like pheromones.

Pheromone Concentration in Density Terms

Pheromone concentration is crucial in understanding how a tiny amount of these compounds can be effective over large distances.
The concentration of pheromones is often described in terms of density, which is the mass of a substance per unit of volume.

• Density formula: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \).
• In our exercise, the mass of the pheromone is \( 1.0 \times 10^{-8} \text{ g} \).

By substituting the mass and our calculated volume from previous sections into this formula, we get the pheromone density:
The density helps us understand how little mass of the pheromone is needed to fill a large volume of air.
This understanding is critical in studying pheromones' role in communication among insects in nature. It also emphasizes the precision needed in the chemical analysis of natural compounds.