# 1B.4: Problem Solving by Dimensional Analysis

- Page ID
- 50463

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- Use dimensional analysis to solve problems

## Introduction

In the section on unit conversions we came up with conversion factors that were based on equivalence statements, and through multiplying a measured number by the conversion factor we could convert a measured number in one set of units to another. For example, if we wanted to know how many feet are in 36 inches, we could multiply it by the conversion factor \(\left ( \frac{1 \; foot}{12 \; inches} \right)\), which would tell us that there are 3 feet in 36 inches. Because the value in the numerator equals the value in the denominator, using the conversion factor as a multiplier did not change the length of the object as we converted it from 36 inches to 3 feet. We also note from the unit conversion section that there were three types of conversion factors that we used; those relating different prefixes for the same units, those relating one type of unit to another for the same type of measurement, and those that used physical constants that relate one type of measurement to another. The last conversion factor is fundamentally different than the others in that it depends on the substance (different substances can have different densities).

Dimensional analysis is based on the use of conversion factors to solve problems and the term "dimensional analysis" is probably a misnomer in that the name originates from its use to relate different fundamental dimensions of physical objects like mass, volume and length to each other based on equivalence statements, but it can be used with any equivalence statement, even ones that are not based on physical dimensions. What needs to noted here is that many equivalence statements are conditional, for example the price of gasoline is a conditional equivalence statement. One day it may be $2.49/gallon, when another day it may be $4.20/gallon, so clearly the price of a gallon of gas is not a constant, but when you go to the gas pump, you can use the equivalence statement to calculate how much you need to pay for a gallon of gas, and thus it can be used in dimensional analysis.

## Setting Up Equations

The Trick to Dimensional Analysis is to keep your units in your conversion factors, and set them up so all units cancel except the units you want in your answer. This is easiest seen by looking at a problem.

General Strategy:

- Write down what is given.
- You must have a conversion factor with the unit of the answer you want
- You must have two values (inital condition or conversion factor) for each unit you wish to cancel, and these will be set up so one of those is on the denominator and the other on the numerator.

- Start with either the given quantity or the conversion factor that has the value you want (in the numerator) and stepwise add conversion factors until the units are the units of the desired answer

Example \(\PageIndex{1}\)

How much does it cost to drive a car from Little Rock to Memphis (137 miles) if a car gets 21 mpg and the price of gas is $2.54/gal

**Solution**

**Given Information:**

There are two equivalence statements here, along with the distance needed to be travelled

\[137 \;miles \\21 miles = 1 gallon \\ $2.54 = 1 gallon \nonumber \]

Lets solve this two ways

**First Way:** Start with the distance you need to travel.

\[137 miles \nonumber\]

you need to get rid of miles, so you need a conversion factor with miles in the denominator

\[137 \textcolor{red}{ \cancel{miles}}\left ( \frac{1 \; gallon}{21 \;\textcolor{red}{ \cancel{miles}}} \right) \nonumber\]

Solving the above equation would tell you how many gallons you need, but you want to convert gallons to dollars, so you need a conversion factor that has gallons and dollars, and you need dollars in the numerator and gallons in the denominator.

\[137 \textcolor{red}{ \cancel{miles}}\left ( \frac{1 \;\textcolor{blue}{ \cancel{ gallon}}}{21 \;\textcolor{red}{ \cancel{miles}}} \right) \left ( \frac{$2.54}{\textcolor{blue}{ \cancel{gallon}}} \right) = $16.57\nonumber\]

**Second Way:** You want to know how much it costs and only one of the given data has cost, so start with that conversion factor, placing cost in the numerator, and cancel everything else out.

\[\left ( \frac{$2.54}{gallon} \right) \nonumber\]

get rid of gallons

\[\left ( \frac{$2.54}{\textcolor{red}{ \cancel{gallon}}} \right) \left ( \frac{1 \;\textcolor{red}{ \cancel{ gallon}}}{21 \;miles} \right) \nonumber\]

get rid of miles and solve

\[\left ( \frac{$2.54}{\textcolor{red}{ \cancel{gallon}}} \right) \left ( \frac{1 \;\textcolor{red}{ \cancel{ gallon}}}{21 \;\textcolor{blue}{ \cancel{miles}}} \right) \left ( 137\textcolor{blue}{ \cancel{miles}} \right ) = $16.57\nonumber\]

The advantage of the second method is sometimes you may be given more than one starting value and may not know what to start with. The important thing here is to keep units in the calculations.

Exercise \(\PageIndex{1}\)

What is the value of a gold bar with dimensions of 1.5cm x 2.5cm x 2.0cm if gold sells for $300/oz and has a density of 19.32g/ml? (16 oz = 1 lb)(1 lb = 453.59 g)(19.32 g = 1 mL)

**Answer**-
Note: you were not given the conversion factor to go from cm

^{3}to ml, and this is because you are expected to know that 1 cm^{3}= 1 mL.\[\left ( \frac{$300}{oz} \right )\left ( \frac{16oz}{1lb} \right )\left ( \frac{1lb}{453.59g} \right )\left ( \frac{19.32g}{1mL} \right )\left ( \frac{1mL}{1cm^{3}} \right )\left ( 1.5cm \right )\left ( 2.5cm \right )\left ( 2.0cm \right )= $1500\]

.

Exercise \(\PageIndex{2}\)

Give the volume in liters of a box which is 2.4 yards by 2.4 inches by 2.4 feet in size? (3 ft = 1 yd)(12 in = 1ft)

**Answer**- Note: in addition to knowing 1 cm
^{3}= 1 mL, you are also expected to know that 1 in = 2.54 cm. -
\[\left ( 2.4 yd \right )\left ( 2.4 in \right )\left ( 2.4 ft \right )\left ( \frac{3 ft}{1 yd} \right )\left ( \frac{12 in}{1 ft} \right )^{2}\left ( \frac{2.54 cm}{1 in} \right )^{3}\left ( \frac{1 mL}{1 cm^{3}} \right )\left ( \frac{10^{-3} L}{1 mL} \right )= 98 L\]

Be Aware: area has the units of length squared and so conversion factors involving area in the dimensions of length need to be squared, and volume is length cubed, and so conversions of volume expressed by length cubed must be cubed. Not mL is not a unit of length, but of volume.

.

The following video shows the solution of a more complex problem involving going from English to metric units and the use of physical constants.

Example \(\PageIndex{1}\)

What is the mass of 1.0000 cubic feet of water?

**Solution**

(28.28 kg)

Note, the physical constant density had significant Figures, and those were what determined the number of significant Figures in the final answer.

**Important **

1. Always Include Units in Calculations

2. Check All Solutions for Proper Dimensions

## Reverse Engineering Eqs from Constants

When you are given a constant you must be given the units, or it makes no sense. For example the density of water is 1.00 g/ml but also 62.4 lb/ft^{3}, so you can't have a density of water without units. From the units you can back calculate the equation, although sometimes you need to analyze it to make sure it makes sense, for example temperature and temperature change can have the same units, but mean different things. If your constant is a linear proportionality constant, than you can say

Y = mX + b, where m is your constant, and the units of m must be the units of \(\left ( \frac{Y}{X}\right) \)

so if Density has the units of \(\left ( \frac{g}{mL}\right) \), then

- Y must be something that has the unit of g, and thus mass would work
- X must be something that has the unit of mL, and thus volume would work

Of course you need to think about these, as sometimes different things can have the same units. For example, a change of one degree kelvin equals a change of 1 degree Celsius, but 1 degree kelvin = 273.15 degrees Celsius, and so you do not know if the equation is T or \(\Delta T\)/ Likewise time and frequency (Hz) have the units of time, where frequency if the cycles per second (and so has units of reciprocal time).

## Test Yourself

## Contributors and Attributions

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

- Ronia Kattoum (Learning Objectives)
- Elena Lisitsyna (H5P interactive modules)