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Chapter 1: Problem 57

You can obtain a rough estimate of the size of a molecule by the following simple experiment: Let a droplet of oil spread out on a smooth surface of water. The resulting oil slick will be approximately one molecule thick. Given an oil droplet of mass \(9.00 \times 10^{-7} \mathrm{~kg}\) and density \(918 \mathrm{~kg} / \mathrm{m}^{3}\) that spreads out into a circle of radius \(41.8 \mathrm{~cm}\) on the water surface, what is the order of magnitude of the diameter of an oil molecule?

Short Answer

Expert verified
The order of magnitude of the diameter of an oil molecule, assuming that the oil droplet spreads into a one-molecule-thick layer, can be found by computing the volume of droplet from its mass and density, computing the area it covers, determining the thickness of that area (which would be the diameter of the molecule) and converting that value into an order of magnitude expression. This would entail calculating the base 10 logarithm of the diameter and disregarding any multiplying constants.

Step by step solution

01

Calculate the volume of the oil droplet

The volume of the droplet can be calculated from its mass and density. The volume \(V\) is calculated with the formula \(V = \frac{m}{\rho}\), where \(m\) is the mass and \(\rho\) is the density. Thus: \(V = \frac{9.00 \times 10^{-7} \mathrm{kg}}{918 \mathrm{kg/m^{3}}}\).
02

Calculate the area of the oil slick

The area \(A\) the oil droplet covers can be calculated with the formula \(A = \pi r^{2}\), where \(r\) is the radius of the circle. Given that the radius is 41.8 cm, we can convert it to meters as 0.418m and substitute it into the formula, thus: \(A = \pi (0.418^{2})\).
03

Determine thickness of oil layer

The thickness \(h\) can be calculated by dividing the volume of the droplet \(V\) by the area the oil is spread over \(A\). Thus: \(h = \frac{V}{A}\).
04

Calculate the order of magnitude

The thickness of the oil layer is equivalent to the diameter \(D\) of an oil molecule (since it was spread out to be one molecule thick). Therefore, the order of magnitude for the diameter of an oil molecule would be same as the computed thickness. But the order magnitude simply involves how many significant figures are in a number and is usually expressed in powers of ten for easier comparison and computation, given the large variation in size of things in the universe. Hence, convert the computed diameter \(D\) into a number in power of ten and ignore any constants multiplying the ten. Use the logarithmic definition of the order of magnitude: it is the exponent in the power-of-ten representation. So, the diameter converted into the order of magnitude is calculated by taking the base 10 logarithm of the calculated diameter, \(D\).

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Key Concepts

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

Understanding the Oil Drop Experiment
The oil drop experiment is a historically significant method for estimating molecular dimensions, which provides a visual, practical demonstration of the molecular scale of substances. In this experiment, a small droplet of oil is allowed to spread across a water surface, forming a thin film one molecule thick.

This exercise involves using the mass and density of the oil droplet to estimate the diameter of an individual oil molecule. It is a powerful example of how macroscopic measurements can provide insights into the microscopic world. The fact that the oil slick forms a layer one molecule thick is essential, as it allows for the direct calculation of a molecule's size from the volume and area covered by the droplet.
Volume and Density Relationship
Volume and density have an inverse relationship when it comes to calculating the size of a substance. Knowing the mass and density of a material allows us to calculate its volume using the formula
\[\begin{equation}\V = \frac{m}{\rho}\end{equation}\]\
where \(m\) is mass and \(\rho\) is density. In this case, understanding how to manipulate these measurements is key to solving for the dimensions of an oil molecule. As the density remains constant for the oil, any change in mass directly affects the volume it occupies.

This understanding allows us to approximate the volume of the oil droplet, which, after being spread into a thin film, will have a volume equal to the volume of the droplet before spreading.
Calculating the Area of a Circle
Calculating the area of a circle is fundamental for various scientific computations, including our oil drop experiment. The area \(A\) is determined by the formula
\[\begin{equation}\A = \pi r^{2}\end{equation}\]\
where \(r\) is the radius of the circle. To ensure accuracy, it's important to convert all measurements to the same units before performing calculations鈥� in this case, converting the radius from centimeters to meters before squaring it.

Once we have the area, it can be used in conjunction with the volume to find the thickness of the oil film, which is assumed to be equal to the diameter of a single oil molecule. Understanding how to find the area of the circle is vital for solving not only this problem but many others in both geometry and physics.

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Most popular questions from this chapter

ECP Kinetic energy KE (Chapter 5 ) has dimensions \(\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}\). It can be written in terms of the momentum \(p\) (Chapter 6) and mass m as $$K E=\frac{p^{2}}{2 m}$$ (a) Determine the proper units for momentum using dimensional analysis. (b) Refer to Problem 5, Given the units of force, write a simple equation relating a constant force \(F\) exerted on an object, an interval of time \(t\) during which the force is applied, and the resulting momentum of the object, \(p\).

A rectangular plate has a length of \((21.3 \pm 0.2) \mathrm{cm}\) and a width of \((9.8 \pm 0.1) \mathrm{cm}\). Calculate the area of the plate, including its uncertainty.

A fathom is a unit of length, usually reserved for measuring the depth of water. A fathom is approximately \(6 \mathrm{ft}\) in length. Take the distance from Earth to the Moon to be 250000 miles, and use the given approximation to find the distance in fathoms.

A furlong is an old British unit of length equal to \(0.125 \mathrm{mi}\), derived from the length of a furrow in an acre of ploughed land. A fortnight is a unit of time corresponding to two weeks, or 14 days and nights. Find the speed of light in megafurlongs per fortmight. (One megafurlong equals a million furlongs.)

(a) Find a conversion factor to convert from miles per hour to kilometers per hour. (b) For a while, federal law mandated that the maximum highway speed would be 55 \(\mathrm{mi} / \mathrm{h}\). Use the conversion factor from part (a) to find the speed in kilometers per hour, (c) The maximum highway speed has been raised to \(65 \mathrm{mi} / \mathrm{h}\) in some places. In kilometers per hour, how much of an increase is this over the \(55-\mathrm{mi} / \mathrm{h}\) limit?
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