1.6 Dimensional Analysis

Throughout the text we use an approach called dimensional analysis as an aid in problem solving. In dimensional analysis we carry units through all calculations. Units are multiplied together, divided into each other, or "canceled." Dimensional analysis will help ensure that the solutions to problems yield the proper units. Moreover, dimensional analysis provides a systematic way of solving many numerical problems and of checking our solutions for possible errors.

The key to using dimensional analysis is the correct use of conversion factors to change one unit into another. A conversion factor is a fraction whose numerator and denominator are the same quantity expressed in different units. For example, 2.54 cm and 1 in. are the same length, 2.54 cm 1 in. This relationship allows us to write two conversion factors:

We use the first of these factors to convert inches to centimeters. For example, the length in centimeters of an object that is 8.50 in. long is given by

The units of inches in the denominator of the conversion factor cancel the units of inches in the given data (8.50 inches). The centimeters in the numerator of the conversion factor become the units of the final answer. Because the numerator and denominator of a conversion factor are equal, multiplying any quantity by a conversion factor is equivalent to multiplying by the number 1 and so does not change the intrinsic value of the quantity. The length 8.50 in. is the same as 21.6 cm.

In general, we begin any conversion by examining the units of the given data and the units we desire. We then ask ourselves what conversion factors we have available to take us from the units of the given quantity to those of the desired one. When we multiply a quantity by a conversion factor, the units multiply and divide as follows:

If the desired units are not obtained in a calculation, then an error must have been made somewhere. Careful inspection of units often reveals the source of the error.

SAMPLE EXERCISE 1.9

If a woman has a mass of 115 lb, what is her mass in grams? (Use the relationships between units given on the back inside cover of the text.)

Solution Because we want to change from lb to g, we look for a relationship between these units of mass. From the back inside cover we have 1 lb 453.6 g. In order to cancel pounds and leave grams, we write the conversion factor with grams in the numerator and pounds in the denominator:

The answer can be given to only three significant figures, the number of significant figures in 115 lb.

PRACTICE EXERCISE

By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0-mi automobile race.

Answer: 804.7 km

Using Two or More Conversion Factors

It is often necessary to use more than one conversion factor in the solution of a problem. For example, suppose we want to know the length in inches of an 8.00-m rod. The table on the back inside cover doesn't give the relationship between meters and inches. It does give the relationship between centimeters and inches (1 in. 2.54 cm), though, and from our knowledge of metric prefixes we know that 1 cm Thus, we can convert step by step, first from meters to centimeters, and then from centimeters to inches as diagrammed in the column.

Combining the given quantity (8.00 m) and the two conversion factors, we have

The first conversion factor is applied to cancel meters and convert the length to centimeters. Thus, meters are written in the denominator and centimeters in the numerator. The second conversion factor is written to cancel centimeters, so it has centimeters in the denominator and inches, the desired unit, in the numerator.

SAMPLE EXERCISE 1.10

The average speed of a nitrogen molecule in air at 25°C is 515 Convert this speed to miles per hour.

Solution To go from the given units, to the desired units, we must convert meters to miles and seconds to hours. From the relationships given on the back inside cover of the book, we find that 1 mi 1.6093 km. From our knowledge of metric prefixes we know that 1 km103 m. Thus, we can convert m to km and then convert km to mi. From our knowledge of time we know that 60 s 1 min and 60 min 1 hr. Thus, we can convert s to min and then convert min to hr.

Applying first the conversions for distance and then those for time, we can set up one long equation in which unwanted units are canceled:

Our answer has the desired units. We can check our calculation using the estimating procedure described in the previous "Strategies" box. The given speed is about 500 Dividing by 1000 converts m to km, giving 0.5 Because 1 mi is about 1.6 km, this speed corresponds to Multiplying by 60 gives about Multiplying again by 60 gives The approximate solution (about ) and the detailed solution () are reasonably close. The answer to the detailed solution has three significant figures, corresponding to the number of significant figures in the given speed in

PRACTICE EXERCISE

A car travels 28 mi per gallon of gasoline. How many kilometers per liter will it go?

Answer: 12 kmL

Conversions Involving Volume

The conversion factors previously noted convert from one unit of a given measure to another unit of the same measure, such as from length to length. We also have conversion factors that convert from one measure to a different one. The density of a substance, for example, can be treated as a conversion factor between mass and volume. Suppose that we want to know the mass in grams of two cubic inches (2.00 in.3) of gold, which has a density of 19.3 g cm3. The density gives us the following factors:

Because the answer we want is a mass in grams, we can see that we will use the first of these factors, which has mass in grams in the numerator. To use this factor, however, we must first convert cubic inches to cubic centimeters. The relationship between in.3 and cm3 is not given on the back inside cover, but the relationship between inches and centimeters is given: 1 in. 2.54 cm (exactly). Cubing both sides of this equation gives (1 in.)3 (2.54 cm)3 from which we write the desired conversion factor:

Notice that both the numbers and the units are cubed. Also, because 2.54 is an exact number, we can retain as many digits of (2.54)3 as we need. We have used four, one more than the number of digits in the density (19.3 gcm3). Applying our conversion factors, we can now solve the problem:

The final answer is reported to three significant figures, the same number of significant figures as is in 2.00 and 19.3.

SAMPLE EXERCISE 1.11

What is the mass in grams of 1.00 gal of water? The density of water is 1.00 gmL.

Solution Before we begin solving this exercise, we note the following:

  1. We are given 1.00 gal of water.

  2. We wish to obtain the mass in grams.

  3. We have the following conversion factors either given, commonly known, or available on the back inside cover of the text:

The first of these conversion factors must be used as written (with grams in the numerator) to give the desired result, whereas the last conversion factor must be inverted in order to cancel gallons. The solution is given by

The units of our final answer are appropriate, and we've also taken care of our significant figures. We can further check our calculation by the estimation procedure. We can round 1.057 off to 1. Focusing on the numbers that don't equal 1 then gives merely in agreement with the detailed calculation.

PRACTICE EXERCISE

  1. Calculate the mass of 1.00 qt of benzene if it has a density of 0.879 gmL.

  2. If the volume of an object is reported as 5.0 ft3, what is the volume in cubic meters?

Answers: (a) 832 g; (b) 0.14 m3