Problem 56 - Physics for Scientists and Engineers with Modern Physics

Question

One liter \(\left(1000 \mathrm{~cm}^{3}\right)\) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an oil slick just one molecule thick, with adjacent molecules just touching. estimate the diameter of the oil slick. Assume the oil molecules have a diameter of \(2 \times 10^{-10} \mathrm{~m}\)

Step-by-Step Solution

Answer

The estimated diameter of the oil slick is approximately 3.1 meters.

Step 1: Determine the Volume of a Single Molecule

Given that the oil forms a single layer of molecules, and each molecule is a sphere with a diameter of \(2 \times 10^{-10} \text{ m}\), we first find the radius \( r \) of one molecule. \[ r = \frac{2 \times 10^{-10}}{2} = 1 \times 10^{-10} \text{ m} \]Using the radius, compute the volume \( V_m \) of one molecule using the formula for the volume of a sphere:\[ V_m = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1 \times 10^{-10})^3 \]\[ V_m \approx 4.19 \times 10^{-30} \text{ m}^3 \]

Step 2: Calculate the Number of Molecules

Given the total volume of oil is \( 1000 \text{ cm}^3 = 1000 \times 10^{-6} \text{ m}^3 \), find the total number of molecules \( N \) by dividing the total oil volume by the volume of one molecule:\[ N = \frac{1000 \times 10^{-6}}{4.19 \times 10^{-30}} \approx 2.39 \times 10^{23} \]

Step 3: Relate Number of Molecules to Area

Each molecule is touching side to side in a single layer to form the oil slick, which implies that the area \( A \) of the slick is formed by \( N \) molecules each contributing an area equal to the area of a circle of diameter \( 2 \times 10^{-10} \text{ m} \). Hence, the area \( A \) of the slick is:\[ A = N \pi r^2 = 2.39 \times 10^{23} \times \pi \times (1 \times 10^{-10})^2 \]\[ A \approx 7.52 \text{ m}^2 \]

Step 4: Convert Area to Diameter

The area of the slick is a circle, and the area of a circle is given by \( A = \pi \left( \frac{d}{2} \right)^2 \). Solving for \( d \), the diameter of the circle:\[ A = \pi \left( \frac{d}{2} \right)^2 \Rightarrow 7.52 = \pi \left( \frac{d}{2} \right)^2 \]\[ \frac{d}{2} = \sqrt{\frac{7.52}{\pi}} \]\[ d = 2 \times \sqrt{\frac{7.52}{3.14159}} \approx 3.10 \text{ m} \]

Step 5: Final Calculation and Assumption Check

The calculated diameter \( d \approx 3.10 \text{ m} \) is based on the assumptions made regarding molecule size and the spherical spread of oil. Verify that this assumption holds, ensuring all calculations used consistent dimensions and conversion factors.

Key Concepts

Molecular Dimensions

When discussing geometric calculations, molecular dimensions are crucial. Molecules, often imagined as tiny spheres, have diameters measured in nanometers or picometers. In this exercise, each oil molecule is assumed to be a sphere with a diameter of \(2 \times 10^{-10} \text{ m}\). This small size is typical for molecular dimensions.
To work with such minuscule sizes, scientists frequently use scientific notation, as seen here, to express these dimensions conveniently. The sphere's radius, half of its diameter, is essential for further calculations. Here, dividing the molecule’s diameter by 2 gives us a radius of \(1 \times 10^{-10} \text{ m}\).
Understanding these basic molecular dimensions helps us visualize how these tiny particles create larger observable patterns, such as an oil slick.

Volume Calculation

Volume calculation is a fundamental aspect of geometry, especially when dealing with spherical objects like molecules. A sphere's volume is calculated using the formula:

• \( V = \frac{4}{3} \pi r^3 \)

Plugging in the radius \( r = 1 \times 10^{-10} \text{ m} \) gives us the volume of a single oil molecule.
After inserting the radius into the formula, one needs to be mindful of the calculation precision due to the small size of the molecule. This exercise simplifies our understanding of micro-scale quantities by converting them into a single comprehensible number. For a tiny oil molecule, this volume is approximately \(4.19 \times 10^{-30} \text{ m}^3\).
By tallying the volume of numerous molecules, we bridge the gap between micro and macro scales in geometry.

Area Calculation

Calculating the area of a region covered by molecules is essential in understanding how they arrange themselves to form larger patterns. Given a fixed amount of oil, we determine the number of molecules and the total area they cover.
We calculate the number of molecules by dividing the total volume of the oil by the volume of a single molecule:

• Total number of molecules \( N \approx 2.39 \times 10^{23} \)

Each molecule optimally arranges itself such that they just touch each other. With the number of molecules known, the area covered by these molecules can be calculated based on the area occupied by each molecule, estimated as a circle with radius \(1 \times 10^{-10} \text{ m}\). This results in a total area of approximately \(7.52 \text{ m}^2\).
Such calculations underscore the link between microscopic molecule size and macroscopic area.

Spherical Geometry

Spherical geometry comes into play when visualizing how liquid spreads over a surface—like oil on water. With the molecules forming a monolayer on the lake's surface, they make up a spherical section. The oil slick is essentially a circle on the surface.
We use the calculated area \( A \approx 7.52 \text{ m}^2 \) to estimate more substantial geometric properties like diameter. Using the area of a circle formula, \( A = \pi \left( \frac{d}{2} \right)^2 \), we deduce the diameter by rearranging:

• \(d = 2 \times \sqrt{\frac{7.52}{\pi}} \approx 3.10 \text{ m} \)

This calculation showcases spherical geometry’s role in connecting the dots from simple molecular arrangements to larger scale observations, like determining how far spilled oil might spread.