6.1: The Global Perspective
Learning Objectives
By the end of this section, you will be able to:
- Describe the components of Earth’s interior and explain how scientists determined its structure
- Specify the origin, size, and extent of Earth’s magnetic field
Earth is a medium-size planet with a diameter of approximately 12,760 kilometers (Figure \(\PageIndex{1}\)). As one of the inner or terrestrial planets, it is composed primarily of heavy elements such as iron, silicon, and oxygen—very different from the composition of the Sun and stars, which are dominated by the light elements hydrogen and helium. Earth’s orbit is nearly circular, and Earth is warm enough to support liquid water on its surface. It is the only planet in our solar system that is neither too hot nor too cold, but “just right” for the development of life as we know it. Some of the basic properties of Earth are summarized in Table \(\PageIndex{1}\).
| Property | Measurement |
|---|---|
| Semimajor axis | 1.00 AU |
| Period | 1.00 year |
| Mass | 5.98 × 10 24 kg |
| Diameter | 12,756 km |
| Radius | 6378 km |
| Escape velocity | 11.2 km/s |
| Rotational period | 23 h 56 m 4 s |
| Surface area | 5.1 × 10 8 km 2 |
| Density | 5.514 g/cm 3 |
| Atmospheric pressure | 1.00 bar |
Earth’s Interior
The interior of a planet—even our own Earth—is difficult to study, and its composition and structure must be determined indirectly. Our only direct experience is with the outermost skin of Earth’s crust, a layer no more than a few kilometers deep. It is important to remember that, in many ways, we know less about our own planet 5 kilometers beneath our feet than we do about the surfaces of Venus and Mars.
Earth is composed largely of metal and silicate rock (see the Composition and Structure of Planets section). Most of this material is in a solid state, but some of it is hot enough to be molten. The structure of material in Earth’s interior has been probed in considerable detail by measuring the transmission of seismic waves through Earth. These are waves that spread through the interior of Earth from earthquakes or explosion sites.
Seismic waves travel through a planet rather like sound waves through a struck bell. Just as the sound frequencies vary depending on the material the bell is made of and how it is constructed, so a planet’s response depends on its composition and structure. By monitoring the seismic waves in different locations, scientists can learn about the layers through which the waves have traveled. Some of these vibrations travel along the surface; others pass directly through the interior. Seismic studies have shown that Earth’s interior consists of several distinct layers with different compositions, illustrated in Figure \(\PageIndex{1}\). As waves travel through different materials in Earth’s interior, the waves—just like light waves in telescope lenses—bend (or refract) so that some seismic stations on Earth receive the waves and others are in “shadows.” Detecting the waves in a network of seismographs helps scientists construct a model of Earth’s interior, showing liquid and solid layers. This type of seismic imaging is not unlike that used in ultrasound, a type of imaging used to see inside the body.
The top layer is the crust , the part of Earth we know best (Figure \(\PageIndex{3}\)). Oceanic crust covers 55% of Earth’s surface and lies mostly submerged under the oceans. It is typically about 6 kilometers thick and is composed of volcanic rocks called basalt . Produced by the cooling of volcanic lava, basalts are made primarily of the elements silicon, oxygen, iron, aluminum, and magnesium. The continental crust covers 45% of the surface, some of which is also beneath the oceans. The continental crust is 20 to 70 kilometers thick and is composed predominantly of a different volcanic class of silicates (rocks made of silicon and oxygen) called granite . These crustal rocks, both oceanic and continental, typically have densities of about 3 g/cm 3 . (For comparison, the density of water is 1 g/cm 3 .) The crust is the easiest layer for geologists to study, but it makes up only about 0.3% of the total mass of Earth.
The largest part of the solid Earth, called the mantle , stretches from the base of the crust downward to a depth of 2900 kilometers. The mantle is more or less solid, but at the temperatures and pressures found there, mantle rock can deform and flow slowly. The density in the mantle increases downward from about 3.5 g/cm 3 to more than 5 g/cm 3 as a result of the compression produced by the weight of overlying material. Samples of upper mantle material are occasionally ejected from volcanoes, permitting a detailed analysis of its chemistry.
Beginning at a depth of 2900 kilometers, we encounter the dense metallic core of Earth. With a diameter of 7000 kilometers, our core is substantially larger than the entire planet Mercury. The outer core is liquid, but the innermost part of the core (about 2400 kilometers in diameter) is probably solid. In addition to iron, the core probably also contains substantial quantities of nickel and sulfur, all compressed to a very high density.
The separation of Earth into layers of different densities is an example of differentiation, the process of sorting the major components of a planet by density. The fact that Earth is differentiated suggests that it was once warm enough for its interior to melt, permitting the heavier metals to sink to the center and form the dense core. Evidence for differentiation comes from comparing the planet’s bulk density (5.5 g/cm 3 ) with the surface materials (3 g/cm 3 ) to suggest that denser material must be buried in the core.
Magnetic Field and Magnetosphere
We can find additional clues about Earth’s interior from its magnetic field. Our planet behaves in some ways as if a giant bar magnet were inside it, aligned approximately with the rotational poles of Earth. This magnetic field is generated by moving material in Earth’s liquid metallic core. As the liquid metal inside Earth circulates, it sets up a circulating electric current. When many charged particles are moving together like that—in the laboratory or on the scale of an entire planet—they produce a magnetic field.
Earth’s magnetic field extends into surrounding space. When a charged particle encounters a magnetic field in space, it becomes trapped in the magnetic zone. Above Earth’s atmosphere, our field is able to trap small quantities of electrons and other atomic particles. This region, called the magnetosphere , is defined as the zone within which Earth’s magnetic field dominates over the weak interplanetary magnetic field that extends outward from the Sun (Figure \(\PageIndex{4}\)).
Where do the charged particles trapped in our magnetosphere come from? They flow outward from the hot surface of the Sun; this is called the solar wind . It not only provides particles for Earth’s magnetic field to trap, it also stretches our field in the direction pointing away from the Sun. Typically, Earth’s magnetosphere extends about 60,000 kilometers, or 10 Earth radii, in the direction of the Sun. But, in the direction away from the Sun, the magnetic field can reach as far as the orbit of the Moon, and sometimes farther.
The magnetosphere was discovered in 1958 by instruments on the first US Earth satellite, Explorer 1 , which recorded the ions (charged particles) trapped in its inner part. The regions of high-energy ions in the magnetosphere are often called the Van Allen belts in recognition of the University of Iowa professor who built the scientific instrumentation for Explorer 1 . Since 1958, hundreds of spacecraft have explored various regions of the magnetosphere. You can read more about its interaction with the Sun in a later chapter.
Key Concepts and Summary
Earth is the prototype terrestrial planet. Its interior composition and structure are probed using seismic waves. Such studies reveal that Earth has a metal core and a silicate mantle. The outer layer, or crust, consists primarily of oceanic basalt and continental granite. A global magnetic field, generated in the core, produces Earth’s magnetosphere, which can trap charged atomic particles.
Glossary
- basalt
- igneous rock produced by the cooling of lava; makes up most of Earth’s oceanic crust and is found on other planets that have experienced extensive volcanic activity
- core
- the central part of the planet; consists of higher density material
- crust
- the outer layer of a terrestrial planet
- granite
- a type of igneous silicate rock that makes up most of Earth’s continental crust
- magnetosphere
- the region around a planet in which its intrinsic magnetic field dominates the interplanetary field carried by the solar wind; hence, the region within which charged particles can be trapped by the planetary magnetic field
- mantle
- the largest part of Earth’s interior; lies between the crust and the core
- seismic wave
- a vibration that travels through the interior of Earth or any other object; on Earth, these are generally caused by earthquakes
Contributors and Attributions
-
Andrew Fraknoi (Foothill College), David Morrison (NASA Ames Research Center), Sidney C. Wolff (National Optical Astronomy Observatory) with many contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/astronomy ).


2.4: The Attention Economy
“Attention is a resource—a person has only so much of it” (Crawford 11).
“How do you focus your attention? How do you protect it? How do you apply it productively and strategically, and avoid giving it to bad actors or dubious sources? And how do you do that in a world where decisions about what to engage with are made in seconds, not minutes or hours?” (Caulfield)
Overview
In a world where information is abundant, where we are overloaded and overwhelmed with information on a daily basis, attention is the scarcity (Caulfield).
Algorithms are being used by social media platforms and news organizations to keep our attention focused on their websites. These and other digital platforms have discovered that the best way to keep us engaged with their websites is to promote sensational, divisive, or outrage-inducing content. Emotional responses are what keep us clicking, liking, commenting, and sharing—when trying to capture attention, our anger, fear and disgust are a signal in the noise. “People tend to react more to inputs that land low on the brainstem. Fear and anger produce a lot more engagement and sharing than joy. The result is that the algorithms favor sensational content over substance” (McNamee).
“Engagement”
“Many of the biggest problems we face in this moment as a society result from decisions being made by the hidden creators of our digital world—the designers, developers, and editors that create and curate the media we consume….They do this by focusing on one over-simplified metric, one that supports advertising as its primary source of revenue. This metric is called engagement, and emphasizing it—above all else—has subtly and steadily changed the way we look at the news, our politics, and each other” (Rose-Stockwell, “This Is How Your Fear and Outrage Are Being Sold for Profit”).
Tobias Rose-Stockwell, a writer and technologist focused on ethical design, explains how emotional reactions are strong indicators of engagement, and how divisive content that captures our attention will be shown in our feeds first. We then share our moral judgments with our own followers, creating what he calls “outrage cascades” that dominate our conversations online.
Indeed, in their report on “Information Disorder,” researchers Wardle and Derakhshan emphasize that: “Social networks are driven by the sharing of emotional content. The architecture of these sites is designed such that every time a user posts content—and it is liked, commented upon or shared further— their brain releases a tiny hit of dopamine (13).
The most notable example of this idea? Most of our content feeds and timelines are no longer sorted chronologically or by relevance. Instead, the decision about which content to show us is instead based on how likely we are to engage with it .
Sources
Caulfield, Mike. “ Attention Is the Scarcity .” Hapgood , 4 Feb. 2019.
Crawford, Matthew. The World Beyond Your Head: On Becoming an Individual in an Age of Distraction . Farrar, Straus, and Giroux, 2016.
Graphics: “This Will Make You Cry,” “Engagement” definitions, “Chronological vs. Engagement Feed,” “Basic Sorting Algorithm” by Tobias Rose-Stockwell are licensed under CC BY-NC 4.0
Image: “Boy Mobile Phone” by Andi_Graf on Pixabay
McNamee, Roger. “ How to Fix Facebook – Before It Fixes Us .” Washington Monthly, Jan/Feb/Mar. 2018.
Rose-Stockwell, Tobias. “ How to Design Better Social Media .” Medium , 13 Apr. 2018. Licensed under CC BY-NC 4.0
Rose-Stockwell, Tobias. “ This Is How Your Fear and Outrage Are Being Sold for Profit. ” Medium , 14 July 2017. Licensed under CC BY-NC 4.0.
Wardle, Claire, and Hossein Derakhshan. “ Information Disorder: Toward an Interdisciplinary Framework for Research and Policy Making .” Council of Europe , 27 Sept. 2017.


7.5: Learning Activities
Learning Activities
(Answers to “Learning Activities” can be found in the “Answer Key” at the end of the book. Answers to interactive activity elements will be provided within the element as immediate feedback.)
Practice Activity
Data Collection/Assessment
- Working with a partner or a “simulated patient,” use the following questions and actions listed to perform a sensory function assessment. Use critical thinking to ask as-needed follow up questions.
Vision:
- Do you wear glasses or contact lenses?
- Do you have problems reading or doing close work? Do you have problems seeing far away objects?
- Do you have any floaters in your vision (spots)?
- When have you last had your eyes examined by an eye doctor?
- Is there a family history of glaucoma or other eye diseases?
- Have you ever had eye trauma or surgery?
- Read words from a book or newspaper.
Hearing:
- Do you currently have any ear pain, discharge, or hearing changes?
- Do you note yourself having trouble hearing in certain situations?
- Do you note any dizziness or ringing in your ears (tinnitus)?
- Do you work in an environment where you are exposed to loud noise on a regular basis? Do you wear protection on your ears from the noise?
- Have you taken any medications that came with a warning to report any changes in hearing?
- Have family members or friends mentioned that you seem not to hear?
- Perform the whisper test.
Smell:
- How is your sense of smell?
- Identify some scents like coffee or lemons with eyes closed.
Touch:
- Are you able to feel when someone is touching you?
- Do you have unusual sensations or numbness and tingling?
- Identify an object with eyes closed like a key.
Taste:
- Have you noted any changes in your ability to taste foods?
- Is your appetite “normal” for you? Have you noted a decrease?
- Taste and identify a food like sugar or salt.
Analysis and Care Planning
2. Create a nursing diagnoses based on your assessment findings.
3. Identify a patient-centered goal and SMART expected outcomes.
4. Outline nursing interventions to help the patient to meet the established goal and expected outcome.
Query \(\PageIndex{1}\)
Query \(\PageIndex{2}\)


3.3: Retention Plans
- Be able to discuss some of the theories on job satisfaction and dissatisfaction.
- Explain the components of a retention plan.
Effective high-performance work systems (HPWS) is the name given to a set of systematic HR practices that create an environment where the employee has greater involvement and responsibility for the success of the organization. A high-performance work system is a strategic approach to many of the things we do in HR, including retention. Generally speaking, a HPWS gets employees involved in conceiving, designing, and implementing processes that are better for the company and better for the employee, which increases retention. The Figure below gives an example of HR’s part in creating these systems.
Keeping HPWS in mind, we can begin to develop retention plans. The first step in this process is to understand some of the theories on job satisfaction and dissatisfaction. Next, we can gather data as to the satisfaction level of our current employees. Then we can begin to implement specific strategies for employee retention.
 While you may have learned about these theories in another class, such as organizational behavior, they are worth a review here to help us better understand employee satisfaction from an HR perspective.
Progression of Job Withdrawal
The first step to developing a retention plan is understanding some of the theories surrounding job satisfaction. One of the basic theories is the progression of job withdrawal theory, developed by Dan Farrell and James Petersen (Farrell & Petersen, 1984). It says that people develop a set of behaviors in succession to avoid their work situation. These behaviors include behavior change, physical withdrawal, and psychological withdrawal.
Within the behavior change area, an employee will first try to change the situation that is causing the dissatisfaction. For example, if the employee is unhappy with the management style, he or she might consider asking for a department move. In the physical withdrawal phase, the employee does one of the following:
- Leaves the job
- Takes an internal transfer
- Starts to become absent or tardy
If an employee is unable to leave the job situation, he or she will experience psychological withdrawal. They will become disengaged and may show less job involvement and commitment to the organization, which can create large costs to the organization, such as dissatisfied customers.
| Employee becomes dissatisfied |
|
|---|---|
| Behavior change |
|
| Physical withdrawal |
|
| Psychological withdrawal |
|
Hawthorne Studies
Between 1927 and 1932, a series of experiments were conducted by Elton Mayo in the Western Electric Hawthorne Works company in Illinois (Mayo, 1949; 2007). Mayo developed these experiments to see how the physical and environmental factors of the workplace, such as lighting and break times, would affect employee motivation.
This was some of the first research performed that looked at human motivation at work. His results were surprising, as he found that no matter which experiments were performed, worker output improved. His conclusion and explanation for this was the simple fact the workers were happy to receive attention from researchers who expressed interest in them. As a result, these experiments, scheduled to last one year, extended to five years to increase the knowledge base about human motivation.
The implication of this research applies to HR and managers even today. It tells us that our retention plans must include training and other activities that make the employee feel valued.
Maslow’s Hierarchy of Needs
In 1943, Abraham Maslow developed what was known as the theory of human motivation (Maslow, 1999). His theory was developed in an attempt to explain human motivation. According to Maslow, there is a hierarchy of five needs, and as one level of need is satisfied, it will no longer be a motivator. In other words, people start at the bottom of the hierarchy and work their way up. Maslow’s hierarchy consists of the following:
- Self-actualization needs
- Esteem needs
- Social needs
- Safety needs
- Physiological needs
Physiological needs are our most basic needs, including food, water, and shelter. Safety needs at work might include feeling safe in the actual physical environment, or job security. As humans, we have the basic need to spend time with others. Esteem needs refer to the need we have to feel good about ourselves. Finally, self-actualization needs are the needs we have to better ourselves.
The implications of his research tell us, for example, that as long as an employee’s physiological needs are met, increased pay may not be a motivator. Likewise, employees should be motivated at work by having all needs met. Needs might include, for example, fair pay, safety standards at work, opportunities to socialize, compliments to help raise our esteem, and training opportunities to further develop ourselves.
Herzberg Two-Factor Theory
In 1959, Frederick Herzberg published The Motivation to Work (Herzberg, et. al., 1993), which described his studies to determine which aspects in a work environment caused satisfaction or dissatisfaction. He performed interviews in which employees were asked what pleased and displeased them about their work. From his research, he developed the motivation-hygiene theory to explain these results.
The things that satisfied the employees were motivators, while the dissatisfiers were the hygiene factors. He further said the hygiene factors were not necessarily motivators, but if not present in the work environment, they would actually cause demotivation. In other words, the hygiene factors are expected and assumed, while they may not necessarily motivate.
His research showed the following as the top six motivation factors :
- Achievement
- Recognition
- The work itself
- Responsibility
- Advancement
- Growth
The following were the top six hygiene factors :
- Company policies
- Supervision
- Relationship with manager
- Work conditions
- Salary
- Relationship with peers
The implication of this research is clear. Salary, for example, is on the hygiene factor list. Fair pay is expected, but it doesn’t actually motivate someone to do a better job. On the other hand, programs to further develop employees, such as management training programs, would be considered a motivator. Therefore, our retention plans should be focused on the area of fair salary of course, but if they take the direction of Herzberg’s motivational factors, the actual motivators tend to be the work and recognition surrounding the work performed.
McGregor
Douglas McGregor proposed the X-Y theory in his 1960 book called The Human Side of Enterprise (McGregor, 2006). McGregor’s theory gives us a starting point to understanding how management style can impact the retention of employees. His theory suggests two fundamental approaches to managing people. Theory X managers , who have an authoritarian management style, have the following fundamental management beliefs:
- The average person dislikes work and will avoid it.
- Most people need to be threatened with punishment to work toward company goals.
- The average person needs to be directed.
- Most workers will avoid responsibility.
Theory Y managers , on the other hand, have the following beliefs:
- Most people want to make an effort at work.
- People will apply self-control and self-direction in pursuit of company objectives.
- Commitment to objectives is a function of expected rewards received.
- People usually accept and actually welcome responsibility.
- Most workers will use imagination and ingenuity in solving company problems.
As you can see, these two belief systems have a large variance, and managers who manage under the X theory may have a more difficult time retaining workers and may see higher turnover rates. As a result, it is our job in HR to provide training opportunities in the area of management, so our managers can help motivate the employees. Training is a large part of the retention plan. This will be addressed in more detail in “Implementing Retention Strategies” .
Human Resource Recall
What are the disadvantages of taking a theory X approach with your employees?
Carrot and Stick
It is unknown for sure where this term was first used, although some believe it was coined in the 1700s during the Seven Years’ War. In business today, the stick approach refers to “poking and prodding” to get employees to do something. The carrot approach refers to the offering of some reward or incentive to motivate employees. Many companies use the stick approach, as in the following examples:
- If you don’t increase your sales by 10 percent, you will be fired.
- Everyone will have to take a pay cut if we don’t produce 15 percent more than we are currently producing.
As you can see, the stick approach takes a punitive look at retention, and we know this may motivate for a short period of time, but not in the long term.
The carrot approach might include the following:
- If you increase sales by 10 percent, you will receive a bonus.
- If production increases by 15 percent, the entire team will receive an extra day off next month.
The carrot approach takes a much more positive approach to employee motivation but still may not be effective. For example, this approach can actually demotivate employees if they do not feel the goal is achievable. Also, if organizations use this as the only motivational technique, ignoring physiological rewards such as career growth, this could be a detriment as well. This approach is used as a retention method, usually as part of a compensation plan.
All the employee satisfaction theories we have discussed have implications for the development of our retention plans and reduction of turnover. These theories can be intertwined into the specific retention strategies we will implement.
Sources of Employee Satisfaction Data
After we have an understanding of why employees leave and employee satisfaction theories, research is our next step in developing a retention plan that will work for your organization. There isn’t a “one size fits all” approach to retention planning, so the research component is essential to formulate a plan that will make a difference in turnover rates.
Research can be performed in two ways. First, exit interviews of employees who are leaving the organization can provide important retention information. An exit interview is an interview performed by HR or a manager that seeks information as to what the employee liked at the organization and what they see should be improved. Exit interviews can be a valuable way to gather information about employee satisfaction and can serve as a starting point for determining any retention issues that may exist in the organization. However, the exit survey data should be reviewed over longer periods of time with several employees, so we can be sure we are not making retention plans based on the feedback of only a few people.
Sample Exit Interview Questions
- What is your primary reason for leaving?
- What did you like most about your job?
- What did you like least about your job?
- Did you feel there was room for growth in your job?
- What incentives did you utilize while at our company?
- Which incentives would you change and why?
- Did you have enough training to do your job effectively?
The second way to perform research is through employee satisfaction surveys. A standardized and widely used measure of job satisfaction is the job descriptive index (JDI) survey. While JDI was initially developed in 1969 at Bowling Green State University, it has gone through extensive revisions, the most recent one in 2009. JDI looks at five aspects of job satisfaction, including present job, present pay, opportunities for promotion, supervision, and coworkers 1 . Each of the five facets contains nine or eighteen questions; the survey can be given in whole or measure only one facet. The value of the scale is that an HR manager can measure job satisfaction over a period of time and compare current results to past results and even compare job satisfaction at their company versus their industry. This allows the HR manager to consider changes in the organization, such as a change in compensation structure, and see how job satisfaction is impacted by the change.
Any type of survey can provide information on the employee’s satisfaction with their manager, workload, and other satisfaction and motivational issues. An example of a general employee satisfaction survey is shown in Figure \(\PageIndex{3}\). However, a few things should be considered when developing an employee satisfaction survey:
- Communicate the purpose and goal of the survey.
- Once the survey is complete, communicate what changes have been made as a result of the survey.
- Assure employees their responses will be anonymous and private.
- Involve management and leadership in the survey development.
- Ask clear, concise questions that get at the root of morale issues.
Once data have been gathered and analyzed, we can formulate our retention plans. Our plan should always be tied to the strategic goals of the organization and the HPWS previously developed, and awareness of motivational theories should be coupled with the plans. Here are the components of a retention plan:
- JDI survey results, other survey results, and exit interview findings
- Current retention plans, strengths, and weaknesses
- Goals of a retention plan (e.g., reduce turnover by 10 percent)
- Individual strategies to meet retention and turnover reduction goals.
- Budgeting. An understanding of how your retention plans will impact the payroll budget is important. See Video 7.2 for an example on how to calculate turnover costs and compare those to costs saved with an effective retention strategy.
In “Implementing Retention Strategies” , we will discuss the implementation of specific retention strategies.
- A high-performance work system (HPWS) is a set of systematic HR practices that create an environment where the employee has greater involvement and responsibility for the success of the organization. The overall company strategy should impact the HPWS HR develops in regard to retention.
- Retention plans are developed to address employee turnover, resulting in a more effective organization.
- The first step in developing a retention plan is to use exit interviews and/or surveys to find out the satisfaction level of employees. Once you have the data, you can begin to write the plan, making sure it is tied to the organizational objectives.
- A standardized and widely used measure of job satisfaction is the JDI survey, or the Job Descriptive Index . While JDI was initially developed in 1969 at Bowling Green State University, it has gone through extensive revisions, the most recent one in 2009. JDI looks at five aspects of job satisfaction, including present job, present pay, opportunities for promotion, supervision, and coworkers 1 .
- A retention plan normally consists of survey and exit interview analysis, any current plans and strengths and weaknesses of those plans, the goal of the retention plan, and finally, the specific strategies to be implemented.
- There are many motivation theories that attempt to explain people’s motivation or lack of motivation at work.
- The Hawthorne studies were a series of studies beginning in 1927 that initially looked at physical environments but found that people tended to be more motivated when they felt cared about. The implications to retention are clear, in that employees should feel cared about and developed within the organization.
- Maslow’s theory on motivation says that if someone already has a need met, giving them something to meet more of that need will no longer motivate. Maslow divided the needs into physiological, safety, social, esteem, and self-actualization needs. Many companies only motivate based on the low-level needs, such as pay. Development of training opportunities, for example, can motivate employees on high-level self-actualization needs.
- Herzberg developed motivational theories based on actual motivation factors and hygiene factors. Hygiene factors are those things that are expected in the workplace and will demotivate employees when absent but will not actually motivate when present. If managers try to motivate only on the basis of hygiene factors, turnover can be high. Motivation on both of his factors is key to a good retention plan.
- McGregor’s theory on motivation looked at managers’ attitudes toward employees. He found that theory X managers had more of a negative view of employees, while theory Y managers had a more positive view. Providing training to the managers in our organization can be a key retention strategy based on McGregor’s theory.
- The carrot-and-stick approach means you can get someone to do something by prodding or by offering some incentive to motivate them to do the work. This theory implies these are the only two methods to motivate, which of course, we know isn’t true. The implication of this in our retention plan is such that we must utilize a variety of methods to retain employees.
- What types of things will motivate you in your career? Name at least five things. Where would these fit on Maslow’s hierarchy of needs and Herzberg’s two-factor theory?
-
How can you apply each of these motivation techniques to motivation theories?
- Training
- Employee recognition programs
- Bonuses
- Management training for your current managers
- Profit sharing
1 “Job Descriptive Index,” JDI Research Group, Bowling Green State University, accessed July 29, 2011, www.bgsu.edu/departments/psych/io/jdi/page54706.html .
References
Farrell D., and James C. Petersen, “Commitment, Absenteeism and Turnover of New Employees: A Longitudinal Study,” Human Relations 37, no. 8 (August 1984): 681–692, accessed August 26, 2011, http://libres.uncg.edu/ir/uncg/f/J_Petersen_Commitment_1984.pdf .
Herzberg, F., Bernard Mausner, and Barbara Bloch Snyderman, The Motivation to Work (New Brunswick, NJ: Transaction Publishers, 1993).
Maslow, A., Toward a Psychology of Being , 3rd ed. (New York: Wiley, 1999).
Mayo, E., The Social Problems of an Industrial Civilization (1949; repr., New York: Arno Press, 2007).


14.A: Review Questions
Complete as directed.
Fill in the Blanks:
- Trenches are narrow excavations may below the surface of the ground. Generally, the depth of light trench is greater than the width, but the width of the trench is not greater than ________ feet.
- A means of exit or entry from a trench excavation shall be provided for trench excavation for ________ feet or more depth.
- A ________ is one who is capable of identifying existing in predictable hazards and who has the authority to take prompt corrective measures to eliminate.
- Ladders, stairways or ramps are permitted means of exit and entry and they must be installed so that any worker does not have to travel more than ________ feet in a lateral direction to reach the exit.
- Subpart P requires that where oxygen levels of less than ________ % are present or where such oxygen deficient conditions could reasonably be expected to exist, the atmosphere in the excavation or trench shall be tested prior to employees entering any excavation deeper than four feet.
- The designated ________ for an excavation operation does a daily inspection of the excavation, including adjacent areas and structures, protective systems, hazardous atmospheres and personal protective equipment.
Multiple Choice: (Circle the Correct Letter)
7. Soil cannot be classified as type A, if which of the following conditions exists:
a. The soil is fissured.
b. The soil is subject to vibration from heavy traffic, pile driving or other similar effects.
c. The soil has been previously disturbed.
d. Any of the above.
True or False: (Circle the Correct Answer)
8. T or F
The locations marked by the utilities are estimated location only. Hand holes should be dug first to determine the exact location of the lines or piping.
9. T or F
The fatality rate for excavation work is nearly twice that of regular construction.
10. T or F
Employees shall not work in excavations in which there is an accumulation of water or water is accumulating, unless adequate precautions have been taken.


2.6: Classes of Minerals
Minerals are classified according to their chemical properties. Except for the native element class, the chemical basis for classifying minerals is the anion, the negatively charged ion that usually shows up at the end of the chemical formula of the mineral. For example, the sulfides are based on the sufur ion, S 2– . Pyrite, for example, FeS 2 , is a sulfide mineral. In some cases, the anion is of a mineral class is polyatomic, such as (CO 3 ) 2– , the carbonate ion. The major classes of minerals are:
- silicates
- sulfides
- carbonates
- oxides
- halides
- sulfates
- phosphates
- native elements
SILICATES
Based on the polyatomic anion, (SiO 4 ) 4– , which has a tetrahedral shape. Most minerals in the earth’s crust and mantle are silicate minerals. All silicate minerals are built of silicon-oxygen tetrahedra (SiO 4 ) 4– in different bonding arrangements which create different crystal lattices. You can understand the properties of a silicate mineral such as crystal shape and cleavage by knowing which type of crystal lattice it has.
- In nesosilicates , also called island silicates, the silicate tetrahedra are separate from each other and bonded completely to non silicate atoms. Olivine is an island silicate.
- In sorosilicates or paired silicates, such as epidote, the silicate tetrahedra are bonded in pairs.
- In cyclosilicates , also called ring silicates, the silicate tetrahedra are joined in rings. Beryl or emerald is a ring silicate.
- In phyllosilicates or sheet silicates, the tetrahedra are bonded at three corners to form flat sheets. Biotite is a sheet silicate.
- In single-chain inosilicates the silicate tetrahedra are bonded in single chains. Pyroxenes are singele-chain inosilicates.
- In double-chain inosilicates the silicate tetrahedra are bonded in double chains. Amphiboles are double-chain inosilicates.
- In tectosilicates , also known as framework silicates, all corners of the silicate tetrahedra are bonded to corners of other silicate tetrahedra, forming a complete framework of silicate tetrahedra in all directions. Feldspar, the most common mineral in earth’s crust, and quartz are both framework silicates.
SULFIDES
These are based on the sulfide ion, S 2– . Examples include pyrite, FeS 2 , galena, PbS, and sphalerite, ZnS in its pure zinc form. Some sulfides are mined as sources of such metals as zinc, lead, copper, and tin.
CARBONATES
These are based on the carbonate ion, (CO 3 ) 2– . Calcite, CaCO 3 , and dolomite, CaMg(CO 3 )2, are carbonate minerals. Carbonate minerals tend to dissolve relatively easily in water, especially acid water, and natural rain water is slightly acid.
OXIDES
These are based on the oxygen anion, O 2– . Examples include iron oxides such as hematite, Fe 2 O 3 and magnetite, Fe 3 O 4 , and pyrolusite, MgO.
HALIDES
These have a halogen element as the anion, whether it be fluoride, F – , chloride, Cl – , bromide, Br – , iodide, I – , or astatide, At – . Halite, NaCl, is a halide mineral.
SULFATES
These have the polyatomic sulfate ion, (SO 4 ) 2– , as the anion. Anhydrite, CaSO 4 , is a sulfate.
PHOSPHATES
These have the polyatomic phosphate ion, (PO 4 ) 3– , as the anion. Fluorapatite, Ca 5 (PO 4 ) 3 F, which makes your teeth hard, is a phosphate mineral.
NATIVE ELEMENTS
These are made of nothing but a single element. Gold (Au), native copper (Cu), and diamond and graphite, which are made of carbon, are all native element minerals. Recall that a mineral is defined as naturally occurring. Therefore, elements purified and crystallized in a laboratory do not qualify as minerals, unless they have also been found in nature.
Follow this link to the minerals classification table.
REFLECTION QUESTIONS
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Contributors and Attributions
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Original content from Kimberly Schulte (Columbia Basin College) and supplemented by Lumen Learning . The content on this page is copyrighted under a Creative Commons Attribution 4.0 International license.


1: Complex Numbers
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- 1.5: Representing Complex Numbers in Vector Space
- So far we have coded the complex number z=x+jy with the Cartesian pair (x,y) and with the polar pair (r∠θ) . We now show how the complex number z may be coded with a two-dimensional vector z and show how this new code may be used to gain insight about complex numbers.
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- 1.6: An Electric Field Computation
- We have established that vectors may be used to code complex numbers. Conversely, complex numbers may be used to code or represent the orthogonal components of any two-dimensional vector. This makes them invaluable in electromagnetic field theory, where they are used to represent the components of electric and magnetic fields.


9.2: Asymmetric Information
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Asymmetric information describes situations wherein one party to the transaction (e.g., the seller) has more information about the product or service in question than another party (e.g., the buyer). For example, in many cases, the producer will possess better information about the quality of raw materials used in the manufacture of a product than the buyer. The producer may present the product to the buyer as representing the finest quality. The buyer could inspect the product prior to purchase, but there may be ways that the producer could mask shoddy workmanship. In many cases, the buyer will be unable to distinguish between good and poor quality before the sale. The consumer may only learn whether the producer used substandard inputs after consumption of the product or through its extended use. In short, the producer could use better information (and the buyer’s lack of information) to his or her advantage.
If only as a savvy consumer, you can relate to the scenario just described. In a famous paper, Akerlof (1970) noted that asymmetric information could, in severe cases, cause markets for high quality products to fail. He described a “lemons market,” where only low-quality products are provided. In this context, the term “lemons” refers to used automobiles of poor quality, the example used in Akerlof’s (1970) paper. The lemons market occurs because no buyer would be willing to pay a premium for high quality because everyone understands that sellers have an incentive to pass off low-quality products at a high-quality price.
To consider further how asymmetric information could cause market failures, let us explore two problems that relate to asymmetric information. One is called adverse selection. The other is called moral hazard. Adverse selection refers to cases where asymmetric information makes it difficult for parties to enter into a mutually beneficial transaction because of concerns that one party could use its informational advantage to the detriment of the other. The lemons market is an example of adverse selection. Adverse selection problems are the primary focus of this chapter. Moral hazard, on the other hand, refers to cases where asymmetric information can create problems after two parties have entered into a business arrangement because of one party’s inability to accurately monitor the behavior or effort of the other. According to Milgrom and Roberts (1992), adverse selection and moral hazard arose out of insurance market terminology but have since taken on broader meanings. Given the origins in insurance markets, let us use insurance as an example to illustrate each of these two problems.
Adverse Selection
Adverse selection results when asymmetric information creates pre-contractual problems. For example, customers who buy dental insurance likely know more about their risks of needing costly dental procedures than do insurance providers. Let us take the hypothetical case of John Doe. John is a recent graduate and knows that he has a high likelihood of being predisposed to dental problems because they run in his family. Moreover, he has had many cavities up to this point in his life. John is fairly certain that there will be expensive dental procedures in his near future. When there is an opportunity to sign up for dental insurance through his employer, John Doe jumps at the chance. To him it seems like a great deal. Jane Doe (no relation to John), on the other hand, has the same opportunity to sign up for dental insurance. Jane has never had a cavity before and feels fortunate to have a good set of teeth. To Jane, the premiums for dental insurance seem high. She would like to have some insurance against the costs of expensive dental care should she need it, but she knows that the most she will pay her dentist in a normal year will be the cost of routine cleanings and checkups. She can pay those out-of-pocket for much less than what she would pay in dental insurance premiums. Because of this, Jane decides not to enroll in the dental insurance program.
In the end, one would expect customers with bad teeth to enroll in dental insurance and customers with good teeth not to enroll. This is adverse selection. Many people with good teeth would like to insure against the cost of dental care should it be needed. but a problem arises because the insurance company is unable to distinguish between high-risk and low-risk customers a priori . It is possible that dental insurance is lemons market in that only high-premium insurance catering to high-risk customers is available. Asymmetric information, in this example, could prevent the emergence of a market for lower-cost dental insurance tailored to lower-risk customers.
Moral Hazard
Moral hazard results when asymmetric information creates post-contractual problems that result from one party being unable to fully monitor the effort or actions of the other. Once a people buy insurance, they may be less careful because they are at least partially insured against a loss. Insurance companies cannot monitor customers all the time, so an asymmetric information problem emerges after an insurance policy has been sold. Moral hazard is partially mitigated by providing incentives in the form of deductibles, co-payments, and partial coverage. For example, a person who has dental insurance, will likely still practice good oral hygiene. Aside from large deductibles and copayments, corrective dental procedures such as fillings, root canals, and crowns can be painful and inconvenient. Such losses in utility are not covered by the insurance.
Market Solutions to Adverse Selection and Moral Hazard
In the case of moral hazard, a common market solution is to develop contracts that provide incentives. For example, your automobile insurance company probably gives you a discount on your premiums if you have not filed a claim in the past few years, have a good driving record (few or no arrests for traffic violations), and/or have good grades. Basically, the company that insures you would like you to drive carefully. Since you are often in a hurry and have coverage that at least partially offsets the costs you would incur in the event of an accident, it knows that you may, on occasion, not be as careful as it would like you to be. It is prohibitively costly for the company to have a representative follow you around and file detailed reports on the carefulness of your driving habits. Instead, the insurance company offers you a set of incentives, based on things that it can observe (like the number of traffic violations) and that it knows are correlated with the degree of risk it takes in insuring you.
Livestock production contracts in agriculture are good examples of incentive-based contracts designed to address moral hazard problems. In broiler chicken production, an integrator provides contract growers with chicks, feed, and other inputs. The integrator would like its contract growers to work hard to minimize mortality and increase feed conversion. The integrator could offer a fixed payment for each flock of birds grown. However, this would result in a moral hazard problem because contract growers would get the same payment regardless of whether they put lots of effort or minimal effort into growing the birds. To address moral hazard, broiler contracts are designed so that growers who do well relative to others in their settlement pool receive premiums while those who do less well receive penalties. Again, it is costly for the integrator to monitor the effort expended by each of its contract growers, but it can easily measure mortality and feed conversion rates. If these are correlated with grower effort, it can design contract provisions to provide incentives for growers to put the desired levels of effort into raising the birds.
A common market solution to adverse selection problems is to rely on an economic signal. A signal is a way for one party to send a believable message to another about the characteristics contained in the products. Advertising is an example of an economic signal. As you will see, advertising clearly has a multifaceted role and can address problems of costly information as well as problems of asymmetric information. The remainder of the chapter is comprised primarily of a discussion of advertising and other economic signals to overcome adverse selection problems resulting from asymmetric information.


4.4: The Tensor Transformation Laws
We may wish to represent a vector in more than one coordinate system, and to convert back and forth between the two representations. In general relativity, the transformation of the coordinates need not be linear, as in the Lorentz transformations; it can be any smooth, one-to-one function. For simplicity, however, we start by considering the one-dimensional case, and by assuming the coordinates are related in an affine manner,
\[x'^{\mu} = ax^{\mu} + b.\]
The addition of the constant \(b\) is merely a change in the choice of origin, so it has no effect on the components of the vector, but the dilation by the factor \(a\) gives a change in scale, which results in \(v'^{\mu} = av^{\mu}\) for a contravariant vector. In the special case where \(v\) is an infinitesimal displacement, this is consistent with the result found by implicit differentiation of the coordinate transformation. For a contravariant vector, \(v'_{\mu} = \frac{1}{a} v_{\mu}\). Generalizing to more than one dimension, and to a possibly nonlinear transformation, we have
\[v'^{\mu} = v^{\kappa} \frac{\partial x'^{\mu}}{\partial x^{\kappa}} \label{[1]}\]
\[v'_{\mu} = v_{\kappa} \frac{\partial x'^{\kappa}}{\partial x^{\mu}} \label{[2]}\]
Note the inversion of the partial derivative in one equation compared to the other. Because these equations describe a change from one coordinate system to another, they clearly depend on the coordinate system, so we use Greek indices rather than the Latin ones that would indicate a coordinate-independent equation. Note that the letter \(\mu\) in these equations always appears as an index referring to the new coordinates, \(\kappa\) to the old ones. For this reason, we can get away with dropping the primes and writing, e.g., \(v^{\mu} = \frac{v^{\kappa} \partial x'^{\mu}}{\partial x^{\kappa}}\) rather than \(v'\) counting on context to show that \(v^{\mu}\) is the vector expressed in the new coordinates,\( v^{\kappa}\) in the old ones. This becomes especially natural if we start working in a specific coordinate system where the coordinates have names. For example, if we transform from coordinates (t, x, y, z) to (a, b, c, d), then it is clear that \(v^t\) is expressed in one system and \(v^c\) in the other.
Exercise \(\PageIndex{1}\)
Recall that the gauge transformations allowed in general relativity are not just any coordinate transformations; they must be (1) smooth and (2) one-to-one. Relate both of these requirements to the features of the vector transformation laws above.
In Equation \ref{[2]}, \(\mu\) appears as a subscript on the left side of the equation, but as a superscript on the right. This would appear to violate our rules of notation, but the interpretation here is that in expressions of the form \(\frac{\partial}{\partial x^{i}}\) and \(\frac{\partial}{\partial x_{i}}\), the superscripts and subscripts should be understood as being turned upside-down. Similarly, Equation \ref{[1]} appears to have the implied sum over κ written ungrammatically, with both \(\kappa\)’s appearing as superscripts. Normally we only have implied sums in which the index appears once as a superscript and once as a subscript. With our new rule for interpreting indices on the bottom of derivatives, the implied sum is seen to be written correctly. This rule is similar to the one for analyzing the units of derivatives written in Leibniz notation, with, e.g., \(\frac{d^{2} x}{dt^{2}}\) having units of meters per second squared. That is, the flipping of the indices like this is required for consistency so that everything will work out properly when we change our units of measurement, causing all our vector components to be rescaled.
A quantity \(v\) that transforms according to Equations \ref{[1]} or \ref{[2]} is referred to as a rank-1 tensor, which is the same thing as a vector.
Example 17: The identity transformation
In the case of the identity transformation \(x'^{\mu} = x^{\mu}\), Equation \ref{[1]} clearly gives v' = v, since all the mixed partial derivatives \(\frac{\partial x'^{\mu}}{\partial x^{\kappa}}\) with \(\mu \neq \kappa\) are zero, and all the derivatives for \(\kappa = \mu\) equal 1.
In Equation \ref{[2]}, it is tempting to write
\[\frac{\partial x^{\kappa}}{\partial x'^{\mu}} = \frac{1}{\frac{\partial x'^{\mu}}{\partial x^{\kappa}}} \quad (wrong!),\]
but this would give infinite results for the mixed terms! Only in the case of functions of a single variable is it possible to flip derivatives in this way; it doesn’t work for partial derivatives. To evaluate these partial derivatives, we have to invert the transformation (which in this example is trivial to accomplish) and then take the partial derivatives.
The metric is a rank-2 tensor, and transforms analogously:
\[g_{\mu \nu} = g_{\kappa \lambda} \frac{\partial x^{\kappa}}{\partial x'^{\mu}} \frac{\partial x^{\lambda}}{\partial x'^{\nu}}\]
writing g rather than g' on the left, because context makes the distinction clear).
Exercise \(\PageIndex{2}\)
Self-check: Write the similar expressions for \(g^{\mu \nu}, g^{\mu}_{\nu}\), and \(g^{\nu}_{\mu}\), which are entirely determined by the grammatical rules for writing superscripts and subscripts. Interpret the case of a rank-0 tensor.
Example 18: An accelerated coordinate system?
Let’s see the effect on Lorentzian metric g of the transformation
\[t' = t \qquad x' = x + \frac{1}{2} at^{2} \ldotp\]
The inverse transformation is
\[t = t' \qquad x = x' - \frac{1}{2} at'^{2} \ldotp\]
The tensor transformation law gives
\[g'_{t' t'} = 1 - (at')^{2}\]
\[g'_{x' x'} = -1\]
\[g'_{x' t'} = -at' \ldotp\]
Clearly something bad happens at \(at' = ±1\), when the relative velocity surpasses the speed of light: the \(t'\) component of the metric vanishes and then reverses its sign. This would be physically unreasonable if we viewed this as a transformation from observer A’s Lorentzian frame into the accelerating reference frame of observer B aboard a spaceship who feels a constant acceleration. Several things prevent such an interpretation:
- B cannot exceed the speed of light.
- Even before B gets to the speed of light, the coordinate \(t'\) cannot correspond to B’s proper time, which is dilated.
- Due to time dilation, A and B do not agree on the rate at which B is accelerating. If B measures her own acceleration to be a', A will judge it to be a < a', with a → 0 as B approaches the speed of light.
There is nothing invalid about the coordinate system (t', x'), but neither does it have any physically interesting interpretation.
Example 19: Physically meaningful constant acceleration
To make a more physically meaningful version of example 18, we need to use the result of example 4 . The somewhat messy derivation of the coordinate transformation is given by Semay. 11 The result is
\[t' = \left(x + \dfrac{1}{a}\right) \sinh at\]
\[x' = \left(x + \dfrac{1}{a}\right) \cosh at\]
Applying the tensor transformation law gives ( problem 7 ):
\[g'_{t' t'} = (1 + ax')^{2}\]
\[g'_{x' x'} = -1\]
Unlike the result of example 18, this one never misbehaves. The closely related topic of a uniform gravitational field in general relativity is considered in problem 7 .
11 arxiv.org/abs/physics/0601179
Example 20: Accurate timing signals
The relation between the potential A and the fields E and B given in section 4.2 can be written in manifestly covariant form as
\[F_{ij} = \partial _{[i}A_{j]}\]
where F, called the electromagnetic tensor, is an antisymmetric rank-two tensor whose six independent components correspond in a certain way with the components of the E and B three-vectors. If F vanishes completely at a certain point in spacetime, then the linear form of the tensor transformation laws guarantees that it will vanish in all coordinate systems, not just one. The GPS system takes advantage of this fact in the transmission of timing signals from the satellites to the users. The electromagnetic wave is modulated so that the bits it transmits are represented by phase reversals of the wave. At these phase reversals, F vanishes, and this vanishing holds true regardless of the motion of the user’s unit or its position in the earth’s gravitational field. Cf. problem 17 .
Example 21: Momentum wants a lower index
In example 5 , we saw that once we arbitrarily chose to write ruler measurements in Euclidean three-space as \(\Delta\)x a rather than \(\Delta\)x a , it became natural to think of the Newtonian force threevector as “wanting” to be notated with a lower index. We can do something similar with the momentum 3- or 4-vector. The Lagrangian is a relativistic scalar, and in Lagrangian mechanics momentum is defined by \(p_{a} = \frac{\partial L}{\partial v^{a}}\). The upper index in the denominator on the right becomes a lower index on the left by the same reasoning as was employed in the notation of the tensor transformation laws. Newton’s second law shows that this is consistent with the result of example 5 .


13.2: Strong Monoprotic Acids and Bases
Make sure you thoroughly understand the following essential concepts that have been presented above.
- Estimate the pH of a solution of a strong acid or base, given its concentration.
- Explain the distinction between the concentration of a substance and its activity in a solution.
- Describe the origin and effects of ion-pairing in concentrated solutions of strong acids.
- Name and write formulas for the four major strong acids .
To a good approximation, strong acids, in the forms we encounter in the laboratory and in much of the industrial world, have no real existence; they are all really solutions of \(\ce{H3O^{+}}\). So if you think about it, the labels on those reagent bottles you see in the lab are not strictly true! However, if the strong acid is highly diluted, the amount of \(\ce{H3O^{+}}\) it contributes to the solution becomes comparable to that which derives from the autoprotolysis of water. Under these conditions, we need to develop a more systematic way of working out equilibrium concentrations.
At Moderate Concentrations, Forget About Equilibria
A strong acid , you will recall, is one that is stronger than the hydronium ion \(\ce{H3O^{+}}\). This means that in the presence of water, the proton on a strong acid such as HCl will "fall" into the "sink" provided by H 2 O, converting the latter into its conjugate acid \(\ce{H3O^{+}}\). In other words, \(\ce{H3O^{+}}\) is the strongest acid that can exist in aqueous solution .
As we explained in the preceding lesson, all strong acids appear to be equally strong in aqueous solution because there are always plenty of H 2 O molecules to accept their protons. This is called the "leveling effect".
This greatly simplifies our treatment of strong acids because there is no need to deal with equilibria such as for hydrochloric acid
\[\ce{HCl + H_2O → H_3O^{+} + Cl^{–}}\]
The equilibrium constants for such reactions are so overwhelmingly large that we can usually consider the concentrations of acid species such as "HCl" to be indistinguishable from zero. As we will see further on, this is not strictly true for highly-concentrated solutions of strong acids (Figure \(\PageIndex{3}\)).
Over the normal range of concentrations we commonly work with (indicated by the green shading on this plot), the pH of a strong acid solution is given by the negative logarithm of its concentration in mol L –1 . Note that in very dilute solutions, the plot levels off, showing that this simple relation breaks down; there is no way you can make the solution alkaline by diluting an acid!
What will be the pH of a 0.025 mol/l solution of hydrochloric acid?
Solution
If we assume all the hydronium concentration originates from the added acid. So
\[\ce{H3O^{+}}\]
and we just find the negative logarithm of the concentration
\[pH = -\log_{10} [\ce{H3o^{+}}] = –\log_{10} 0.025 = 1.6\]
Since this pH is so far away from 7, our assumption is reasonable.
Major Strong Acids
Mineral acids are those that are totally inorganic. Not all mineral acids are strong; boric and carbonic acids are common examples of very weak ones. However, in ordinary usage, the term often implies one of those described below. With the exception of perchloric acid, which requires special handling, these are all widely used in industry and are almost always found in chemistry laboratories. Most have been known since ancient times
| acid | name | pK a | base | pK b |
|---|---|---|---|---|
| HClO 4 | perchloric acid | ~ –7 | ClO 4 – | ~ 21 |
| HCl | hydrochloric acid | ~ –7 | Cl – | ~ 17 |
| H 2 SO 4 | sulfuric acid | ~ –7 | HSO 4 – | ~ 17 |
| HNO 3 | nitric acid | –1 | NO 3 – | 15 |
| H 3 O + | hydronium ion | H 2 O | 14 | |
You should know the names and formulas of all four of these widely-encountered strong mineral acids.
- Hydrochloric acid \(HCl\): In contrast to the other strong mineral acids, no pure compound "hydrochloric acid" exists. One of the most commonly used acids, hydrochloric acid is usually sold as a constant-boiling 37% solution of hydrogen chloride gas in water, making its concentration about 10 M . Although higher concentrations are possible, the high partial pressures of HCl in equilibrium with the solution makes them difficult to store, ship and work with. Commercial-grade hydrochloric acid is often called muriatic acid .
- Nitric acid \(HNO_3\): Concentrated nitric acid is a 68% constant-boiling solution of HNO 3 in water, corresponding to about 11M. Although the acid itself is colorless, its slow decomposition into NO 2 (especially in the presence of light) often results in a yellow or orange color. More concentrated solutions are sold as "fuming nitric acid"; one form, available as "white fuming nitric acid" or "anhydrous nitricacid" contains 97.5% HNO 3 and only 2% water; the remainder consists of dissolved NO 2 . Pure HNO 3 forms crystals that melt at –42°C. In addition to being a strong acid, nitric acid at high concentrations acts as a powerful oxidizing agent, a property that accounts for its use as a rocket fuel.
- Perchloric acid \(HClO_4\) : This is the strongest of the mineral acids, thanks to the electron-withdrawing action of the oxygen atoms which make it energetically easier for HClO 4 to lose its proton. Its usual commercial form is a constant-boiling 72.5% solution in water. Concentrated perchloric acid is a powerful oxidizing agent that can react explosively with organic materials and certain metals. Its use in the laboratory requires special handling. A 1947 explosion of a 1000-L vat of HClO 4 in a Los Angeles electroplating plant killed 17 people and damaged over 250 homes and other buildings.
- Sulfuric acid \(H_2SO_4\): Sulfuric acid is by far the most industrially important acid, and is also the most concentrated one available. "Concentrated" sulfuric acid contains 98% by weight of H 2 SO 4 ; its density of 1.83 kg/L and oil-like viscosity reflect this high concentration. "100% H 2 SO 4 " (which can be prepared However, is not stable) actually contains a variety of other species, all in equilibrium with H 2 SO 4 . These include H 2 S 2 O 7 , H 2 S 4 O 13 , and H 2 SO 4 's autoprotolysis products H 3 SO 4 – and HSO 4 + .
Super acids
There is a class of super acids that are stronger than some of the common mineral acids. According to the classical definition, a superacid is an acid with an acidity greater than that of 100% pure sulfuric acid. Some, like fluorosulfuric acid \(FSO_3H\) are commercially available. Strong superacids are prepared by the combination of a strong Lewis acid and a strong Brønsted acid. The strongest known super acid is fluoroantimonic acid (\(H_2FSbF_6\)). This acid is so corrosive that the fumes alone will dissolve entire fume hoods, glass and plastic beakers, human skin, bone and most synthetic compounds.
Strong bases
The only strong bases that are commonly encountered are solutions of Group 1 hydroxides, mainly NaOH and KOH. Unlike most metal hydroxides, these solids are highly soluble in water, and can thus yield concentrated solutions of hydroxide ion, the strongest base that can exist in water — the ultimate aquatic proton sink. Sodium hydroxide is by far the most important of these strong bases; its common names "lye", "caustic soda" (or, in industry, often just "caustic"), reflect the diverse uses of NaOH.
Solid NaOH is usually sold in the form of pellets. When exposed to air, they become wet ( deliquescence ) and absorb CO 2 , becoming contaminated with sodium carbonate. NaOH is the most soluble of the Group 1 hydroxides, dissolving in less than its own weight of water (111 g / 100 ml) to form a 2.8 M/L solution at 20°C. However, as with strong acids, the pH of such a solution cannot be reliably calculated from such a high concentration.
Acids at High Concentrations
At higher concentrations, intermolecular interactions and ion-pairing can cause the effective concentration (known as the activity ) of \(\ce{H3O^{+}}\) to deviate from the value corresponding to the nominal or "analytical" concentration of the acid. Activities are important because only these work properly in equilibrium calculations. Also, pH is defined as the negative logarithm of the hydrogen ion activity, not its concentration. The relation between the concentration of a species and its activity is expressed by the activity coefficient \(\gamma\):
\[a = \gamma C\]
As a solution becomes more dilute, \(\gamma\) approaches unity. At ionic concentrations not exceeding about 2 M , concentrations of typical strong acids can generally be used in place of activities without serious error. Note that activities of single ions other than \(\ce{H3O^{+}}\) cannot be determined, so activity coefficients in ionic solutions are always the average, or mean , of those for the ionic species present. This quantity is denoted as \(\gamma_±\).
| molarity | \(\gamma_±\) |
|---|---|
| 0.0005 | 0.975 |
| 0.01 | 0.904 |
| 0.10 | 0.796 |
| 1 | 0.809 |
| 2 | 1.01 |
| 5 | 2.38 |
| 10 | 10.44 |
| 12 | 17.25 |
Because activities of single ions cannot be measured, these mean values are the closest we can get to \(\{H^+\}\) in solutions of strong acids.
Activity is Important for Concentrated Samples
Activity is a practical consideration when dealing with strong mineral acids which are available at concentrations of 10 M or greater.
In a 12 M solution of hydrochloric acid, for example, the mean ionic activity coefficient is 17.25. This means that under these conditions with [H 3 O + ] = 12 M, the activity {H 3 O + } = 12 × 17.25 = 207, corresponding to a pH of about –2.3, instead of –1.1 as might be predicted if concentrations were being used.
These very high activity coefficients also explain another phenomenon: why you can detect the odor of HCl (g) over a concentrated hydrochloric acid solution even though this acid is supposedly 100% dissociated. It turns out that the activity {HCl} (which represents the "escaping tendency" of HCl (g) from the solution) is almost 49,000 for a 10 M solution! The source of this gas is best described as the result of ion pairing.
Similarly, in a solution prepared by adding 0.5 mole of the very strong acid HClO 4 to sufficient water to make the volume 1 liter, freezing-point depression measurements indicate that the concentrations of hydronium and perchlorate ions are only about 0.4 M. This does not mean that the acid is only 80% dissociated; there is no evidence of HClO 4 molecules in the solution. What has happened is that about 20% of the \(\ce{H3O^{+}}\) and ClO 4 – ions have formed ion-pair complexes in which the oppositely-charged species are loosely bound by electrostatic forces (Figure \(\PageIndex{4}\)).
Ion-pairing reduces effective dissociation at high concentrations. If you have worked with concentrated hydrochloric acid in the lab, you will have noticed that the choking odor of hydrogen chloride gas is very apparent. How can this happen if this strong acid is really "100 percent dissociated" as all strong acids are said to be?
At very high concentrations, too few H 2 O molecules are available to completely fill the extended hydration shells that normally help keep the ions apart, reducing the fraction of "free" \(\ce{H3O^{+}}\) ions capable of acting independently. Under these conditions, the term "dissociation" begins to lose its meaning. Although the concentration of HCl (aq) is never very high, its own activity coefficient can be as great as 2000 (Table \(\PageIndex{2}\)), which means that its escaping tendency from the solution is extremely high, so that the presence of even a tiny amount is very noticeable.
Estimate the pH of a 10.0 M solution of hydrochloric acid in which the mean ionic activity coefficient - is 10.4.
Solution
pH = – log {H + } ≈ – log (10.4 × 10.0) = – log 104 = – 2.0
Compare this result with what you would get by using – log[H + ]
Systematic treatment of strong acids
In this section, we will derive expressions that relate the pH of solution of a strong acid to its concentration in a solution of pure water. We will use hydrochloric acid as an example. When HCl gas is dissolved in water, the resulting solution contains the ions H 3 O + , OH – , and Cl−, However, except in very concentrated solutions, the concentration of HCl is negligible; for all practical purposes, molecules of “hydrochloric acid”, HCl, do not exist in dilute aqueous solutions.
To specify the concentrations of the three species present in an aqueous solution of HCl, we need three independent relations between them. These relations are obtained by observing that certain conditions must always be true in any solution of HCl. These are:
1. The autoprotolysis equilibrium of water must always be satisfied:
\[[H_3O^+][OH^–] = K_w \label{4-1}\]
2. For any acid-base system, one can write a mass balance equation that relates the concentrations of the various dissociation products of the substance to its “nominal concentration”, which we designate here as C a . For a solution of HCl, this equation would be
\[[HCl] + [Cl^–] = C_a \label{4-2}\]
However, since HCl is a strong acid and therefore no "HCl" exists in the solution, we can neglect the first term, so the mass balance equation becomes simply
\[[Cl^–] = C_a \label{4-3}\]
3. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle .
\[[H_3O^+] = [OH^–] + [Cl^–] \label{4-4}\]
The next step is to combine these three equations into a single expression that relates the hydronium ion concentration to \(C_a\). This is best done by starting with an equation that relates several quantities, such as Equation \(\ref{4-4}\), and substituting the terms that we want to eliminate. Thus we can get rid of the [Cl−] term by substituting Equation \(\ref{4-3}\) into Equation \(\ref{4-4}\) :
\[[H_3O^+] = [OH^–] + C_a \label{4-5}\]
The [OH – ] term can be eliminated by use of Equation \(\ref{4-3}\):
\[[H_3O^+] = C_a + \dfrac{K_w}{[H_3O^+]} \label{4-6}\]
This equation tells us that the hydronium ion concentration will be the same as the nominal concentration of a strong acid as long as the solution is not very dilute. Notice that Equation \(\ref{4-6}\) is a quadratic equation. Recalling that K w = 10 –14 , it is apparent that the final term of the above equation will ordinarily be very small in comparison to the other terms, so it can be ordinarily be dropped, yielding the simple relation
\[[H_3O^+] \approx C_a \label{4-7}\]
Only in extremely dilute solutions, around 10 –6 M or below (where the plot curves), does this approximation become untenable. However, even then, the effect is tiny. After all, the hydronium ion concentration in a solution of a strong acid can never fall below 10 –7 M ; no amount of dilution can make the solution alkaline! So for almost all practical purposes, Equaton \(\ref{4-7}\) is all you will ever need for a solution of a strong acid.


16.5: Ring Homomorphisms and Ideals
In the study of groups, a homomorphism is a map that preserves the operation of the group. Similarly, a homomorphism between rings preserves the operations of addition and multiplication in the ring. More specifically, if \(R\) and \(S\) are rings, then a ring homomorphism is a map \(\phi : R \rightarrow S\) satisfying
\begin{align*} \phi( a + b ) & = \phi( a ) + \phi(b)\\ \phi( a b ) & = \phi( a ) \phi(b) \end{align*}
for all \(a, b \in R\text{.}\) If \(\phi : R \rightarrow S\) is a one-to-one and onto homomorphism, then \(\phi\) is called an isomorphism of rings.
The set of elements that a ring homomorphism maps to \(0\) plays a fundamental role in the theory of rings. For any ring homomorphism \(\phi : R \rightarrow S\text{,}\) we define the kernel of a ring homomorphism to be the set
\[ \ker \phi = \{ r \in R : \phi( r ) = 0 \}\text{.} \nonumber \]
Example \(16.20\)
For any integer \(n\) we can define a ring homomorphism \(\phi : {\mathbb Z} \rightarrow {\mathbb Z}_n\) by \(a \mapsto a \pmod{n}\text{.}\) This is indeed a ring homomorphism, since
Solution
\begin{align*} \phi( a + b ) & = (a + b) \pmod{n}\\ & = a \pmod{n} + b \pmod{n}\\ & = \phi( a ) + \phi(b) \end{align*}
and
\begin{align*} \phi( a b ) & = ab \pmod{n}\\ & = a \pmod{n}\cdot b \pmod{n}\\ & = \phi( a ) \phi(b)\text{.} \end{align*}
The kernel of the homomorphism \(\phi\) is \(n {\mathbb Z}\text{.}\)
Example \(16.21\)
Let \(C[a, b]\) be the ring of continuous real-valued functions on an interval \([a,b]\) as in Example 16.5. For a fixed \(\alpha \in [a, b]\text{,}\) we can define a ring homomorphism \(\phi_{\alpha} : C[a, b] \rightarrow {\mathbb R}\) by \(\phi_{\alpha} (f ) = f( \alpha)\text{.}\) This is a ring homomorphism since
Solution
\begin{gather*} \phi_{\alpha}( f + g ) = (f + g)( \alpha) = f(\alpha) + g(\alpha) = \phi_{\alpha}( f ) + \phi_{\alpha}(g )\\ \phi_{\alpha}( f g ) = (f g)( \alpha) = f(\alpha) g(\alpha) = \phi_{\alpha}( f ) \phi_{\alpha}(g )\text{.} \end{gather*}
Ring homomorphisms of the type \(\phi_{\alpha}\) are called evaluation homomorphisms .
In the next proposition we will examine some fundamental properties of ring homomorphisms. The proof of the proposition is left as an exercise.
Proposition \(16.22\)
Let \(\phi : R \rightarrow S\) be a ring homomorphism.
- If \(R\) is a commutative ring, then \(\phi(R)\) is a commutative ring.
- \(\phi( 0 ) = 0\text{.}\)
- Let \(1_R\) and \(1_S\) be the identities for \(R\) and \(S\text{,}\) respectively. If \(\phi\) is onto, then \(\phi(1_R) = 1_S\text{.}\)
- If \(R\) is a field and \(\phi(R) \neq \{ 0 \}\text{,}\) then \(\phi(R)\) is a field.
In group theory we found that normal subgroups play a special role. These subgroups have nice characteristics that make them more interesting to study than arbitrary subgroups. In ring theory the objects corresponding to normal subgroups are a special class of subrings called ideals. An ideal in a ring \(R\) is a subring \(I\) of \(R\) such that if \(a\) is in \(I\) and \(r\) is in \(R\text{,}\) then both \(ar\) and \(ra\) are in \(I\text{;}\) that is, \(rI \subset I\) and \(Ir \subset I\) for all \(r \in R\text{.}\)
Example \(16.23\)
Every ring \(R\) has at least two ideals, \(\{ 0 \}\) and \(R\text{.}\) These ideals are called
Solution
the trivial ideals .
Let \(R\) be a ring with identity and suppose that \(I\) is an ideal in \(R\) such that \(1\) is in \(I\text{.}\) Since for any \(r \in R\text{,}\) \(r1 = r \in I\) by the definition of an ideal, \(I = R\text{.}\)
Example \(16.24\)
If \(a\) is any element in a commutative ring \(R\) with identity, then the set
\[ \langle a \rangle = \{ ar : r \in R \} \nonumber \]
is an ideal in \(R\text{.}\)
Solution
Certainly, \(\langle a \rangle\) is nonempty since both \(0 = a0\) and \(a = a1\) are in \(\langle a \rangle\text{.}\) The sum of two elements in \(\langle a \rangle\) is again in \(\langle a \rangle\) since \(ar + ar' = a(r + r')\text{.}\) The inverse of \(ar\) is \(-ar = a (-r) \in \langle a \rangle\text{.}\) Finally, if we multiply an element \(ar \in \langle a \rangle\) by an arbitrary element \(s \in R\text{,}\) we have \(s(ar) = a(sr)\text{.}\) Therefore, \(\langle a \rangle\) satisfies the definition of an ideal.
If \(R\) is a commutative ring with identity, then an ideal of the form \(\langle a \rangle = \{ ar : r \in R \}\) is called a principal ideal .
Theorem \(16.25\)
Every ideal in the ring of integers \({\mathbb Z}\) is a principal ideal.
- Proof
-
The zero ideal \(\{ 0 \}\) is a principal ideal since \(\langle 0 \rangle = \{ 0 \}\text{.}\) If \(I\) is any nonzero ideal in \({\mathbb Z}\text{,}\) then \(I\) must contain some positive integer \(m\text{.}\) There exists a least positive integer \(n\) in \(I\) by the Principle of Well-Ordering. Now let \(a\) be any element in \(I\text{.}\) Using the division algorithm, we know that there exist integers \(q\) and \(r\) such that
\[ a = nq + r \nonumber \]
where \(0 \leq r \lt n\text{.}\) This equation tells us that \(r = a - nq \in I\text{,}\) but \(r\) must be \(0\) since \(n\) is the least positive element in \(I\text{.}\) Therefore, \(a = nq\) and \(I = \langle n \rangle\text{.}\)
Example \(16.26\)
The set \(n {\mathbb Z}\) is ideal in the ring of integers. If \(na\) is in \(n{\mathbb Z}\) and \(b\) is in \({\mathbb Z}\text{,}\)
Solution
then \(nab\) is in \(n {\mathbb Z}\) as required. In fact, by Theorem 16.25, these are the only ideals of \({\mathbb Z}\text{.}\)
Proposition \(16.27\)
The kernel of any ring homomorphism \(\phi : R \rightarrow S\) is an ideal in \(R\text{.}\)
- Proof
-
We know from group theory that \(\ker \phi\) is an additive subgroup of \(R\text{.}\) Suppose that \(r \in R\) and \(a \in \ker \phi\text{.}\) Then we must show that \(ar\) and \(ra\) are in \(\ker \phi\text{.}\) However,
\[ \phi(ar) = \phi(a) \phi(r) = 0 \phi(r) = 0 \nonumber \]
and
\[ \phi(ra) = \phi(r) \phi(a) = \phi(r)0 = 0\text{.} \nonumber \]
Remark \(16.28\)
In our definition of an ideal we have required that \(rI \subset I\) and \(Ir \subset I\) for all \(r \in R\text{.}\) Such ideals are sometimes referred to as two-sided ideals . We can also consider one-sided ideals ; that is, we may require only that either \(rI \subset I\) or \(Ir \subset I\) for \(r \in R\) hold but not both. Such ideals are called left ideals and right ideals , respectively. Of course, in a commutative ring any ideal must be two-sided. In this text we will concentrate on two-sided ideals.
Theorem \(16.29\)
Let \(I\) be an ideal of \(R\text{.}\) The factor group \(R/I\) is a ring with multiplication defined by
\[ (r + I)(s + I) = rs + I\text{.} \nonumber \]
- Proof
-
We already know that \(R/I\) is an abelian group under addition. Let \(r+I\) and \(s +I\) be in \(R/I\text{.}\) We must show that the product \((r + I)(s + I) = rs + I\) is independent of the choice of coset; that is, if \(r' \in r+I\) and \(s' \in s+I\text{,}\) then \(r's'\) must be in \(rs+I\text{.}\) Since \(r' \in r+I\text{,}\) there exists an element \(a\) in \(I\) such that \(r' = r + a\text{.}\) Similarly, there exists a \(b \in I\) such that \(s' = s + b\text{.}\) Notice that
\[ r' s' = (r+a)(s+b) = rs + as + rb + ab \nonumber \]
and \(as + rb + ab \in I\) since \(I\) is an ideal; consequently, \(r' s' \in rs + I\text{.}\) We will leave as an exercise the verification of the associative law for multiplication and the distributive laws.
The ring \(R/I\) in Theorem \(16.29\) is called the factor or quotient ring . Just as with group homomorphisms and normal subgroups, there is a relationship between ring homomorphisms and ideals.
Theorem \(16.30\)
Let \(I\) be an ideal of \(R\text{.}\) The map \(\phi : R \rightarrow R/I\) defined by \(\phi( r ) = r + I\) is a ring homomorphism of \(R\) onto \(R/I\) with kernel \(I\text{.}\)
- Proof
-
Certainly \(\phi : R \rightarrow R/I\) is a surjective abelian group homomorphism. It remains to show that \(\phi\) works correctly under ring multiplication. Let \(r\) and \(s\) be in \(R\text{.}\) Then
\[ \phi(r) \phi(s) = (r + I)(s+I) = rs + I = \phi(rs)\text{,} \nonumber \]
which completes the proof of the theorem.
The map \(\phi : R \rightarrow R/I\) is often called the natural or canonical homomorphism . In ring theory we have isomorphism theorems relating ideals and ring homomorphisms similar to the isomorphism theorems for groups that relate normal subgroups and homomorphisms in Chapter 11. We will prove only the First Isomorphism Theorem for rings in this chapter and leave the proofs of the other two theorems as exercises. All of the proofs are similar to the proofs of the isomorphism theorems for groups.
Theorem \(16.31\). First Isomorphism Theorem
Let \(\psi : R \rightarrow S\) be a ring homomorphism. Then \(\ker \psi\) is an ideal of \(R\text{.}\) If \(\phi : R \rightarrow R/\ker \psi\) is the canonical homomorphism, then there exists a unique isomorphism \(\eta: R/\ker \psi \rightarrow \psi(R)\) such that \(\psi = \eta \phi\text{.}\)
- Proof
-
Let \(K = \ker \psi\text{.}\) By the First Isomorphism Theorem for groups, there exists a well-defined group homomorphism \(\eta: R/K \rightarrow \psi(R)\) defined by \(\eta(r + K) = \psi(r)\) for the additive abelian groups \(R\) and \(R/K\text{.}\) To show that this is a ring homomorphism, we need only show that \(\eta( (r + K)(s + K) ) = \eta(r + K) \eta( s + K)\text{;}\) but
\begin{align*} \eta( (r + K)( s +K )) & = \eta(r s +K )\\ & = \psi(r s)\\ & = \psi(r) \psi(s)\\ & = \eta( r + K ) \eta( s + K )\text{.} \end{align*}
Theorem \(16.32\). Second Isomorphism Theorem
Let \(I\) be a subring of a ring \(R\) and \(J\) an ideal of \(R\text{.}\) Then \(I \cap J\) is an ideal of \(I\) and
\[ I / I \cap J \cong (I+ J) /J\text{.} \nonumber \]
Theorem \(16.33\). Third Isomorphism Theorem
Let \(R\) be a ring and \(I\) and \(J\) be ideals of \(R\) where \(J \subset I\text{.}\) Then
\[ R/I \cong \frac{R/J}{I/J}\text{.} \nonumber \]
Theorem \(16.34\). Correspondence Theorem
Let \(I\) be an ideal of a ring \(R\text{.}\) Then \(S \mapsto S/I\) is a one-to-one correspondence between the set of subrings \(S\) containing \(I\) and the set of subrings of \(R/I\text{.}\) Furthermore, the ideals of \(R\) containing \(I\) correspond to ideals of \(R/I\text{.}\)


15.7: Statistical Interpretation of Entropy and the Second Law of Thermodynamics- The Underlying Explanation
Learning Objectives
By the end of this section, you will be able to:
- Identify probabilities in entropy.
- Analyze statistical probabilities in entropic systems.
The various ways of formulating the second law of thermodynamics tell what happens rather than why it happens. Why should heat transfer occur only from hot to cold? Why should energy become ever less available to do work? Why should the universe become increasingly disorderly? The answer is that it is a matter of overwhelming probability. Disorder is simply vastly more likely than order.
When you watch an emerging rain storm begin to wet the ground, you will notice that the drops fall in a disorganized manner both in time and in space. Some fall close together, some far apart, but they never fall in straight, orderly rows. It is not impossible for rain to fall in an orderly pattern, just highly unlikely, because there are many more disorderly ways than orderly ones. To illustrate this fact, we will examine some random processes, starting with coin tosses.
Coin Tosses
What are the possible outcomes of tossing 5 coins? Each coin can land either heads or tails. On the large scale, we are concerned only with the total heads and tails and not with the order in which heads and tails appear. The following possibilities exist:
\[5 \, heads, \, 0 \, tails\]
\[4 \, heads, \, 1 \, tail\]
\[3 \, heads, \, 2 \, tails\]
\[2 \, heads, \, 3 \, tails\]
\[1 \, head, \, 4 \, tails\]
\[0 \, heads, \, 5 \, tails\]
These are what we call macrostates . A macrostate is an overall property of a system. It does not specify the details of the system, such as the order in which heads and tails occur or which coins are heads or tails.
Using this nomenclature, a system of 5 coins has the 6 possible macrostates just listed. Some macrostates are more likely to occur than others. For instance, there is only one way to get 5 heads, but there are several ways to get 3 heads and 2 tails, making the latter macrostate more probable. Table \(\PageIndex{1}\) lists of all the ways in which 5 coins can be tossed, taking into account the order in which heads and tails occur. Each sequence is called a microstate —a detailed description of every element of a system.
| Individual microstates | Number of microstates | |
|---|---|---|
| 5 heads, 0 tails | HHHHH | 1 |
| 4 heads, 1 tail | HHHHT, HHHTH, HHTHH, HTHHH, THHHH | 5 |
| 3 heads, 2 tails | HTHTH, THTHH, HTHHT, THHTH, THHHT HTHTH, THTHH, HTHHT, THHTH, THHHT | 10 |
| 2 heads, 3 tails | TTTHH, TTHHT, THHTT, HHTTT, TTHTH, THTHT, HTHTT, THTTH, HTTHT, HTTTH | 10 |
| 1 head, 4 tails | TTTTH, TTTHT, TTHTT, THTTT, HTTTT | 5 |
| 0 heads, 5 tails | TTTTT | 1 |
| Total: 32 |
The macrostate of 3 heads and 2 tails can be achieved in 10 ways and is thus 10 times more probable than the one having 5 heads. Not surprisingly, it is equally probable to have the reverse, 2 heads and 3 tails. Similarly, it is equally probable to get 5 tails as it is to get 5 heads. Note that all of these conclusions are based on the crucial assumption that each microstate is equally probable. With coin tosses, this requires that the coins not be asymmetric in a way that favors one side over the other, as with loaded dice. With any system, the assumption that all microstates are equally probable must be valid, or the analysis will be erroneous.
The two most orderly possibilities are 5 heads or 5 tails. (They are more structured than the others.) They are also the least likely, only 2 out of 32 possibilities. The most disorderly possibilities are 3 heads and 2 tails and its reverse. (They are the least structured.) The most disorderly possibilities are also the most likely, with 20 out of 32 possibilities for the 3 heads and 2 tails and its reverse. If we start with an orderly array like 5 heads and toss the coins, it is very likely that we will get a less orderly array as a result, since 30 out of the 32 possibilities are less orderly. So even if you start with an orderly state, there is a strong tendency to go from order to disorder, from low entropy to high entropy. The reverse can happen, but it is unlikely.
| Macrostate | Number of Microstates | |
|---|---|---|
| Heads | Tails | ( W ) |
| 100 | 0 | 1 |
| 99 | 1 | 100 |
| 95 | 5 | \(7.5 \times 10^7\) |
| 90 | 10 | \(1.7 \times 10^{13}\) |
| 75 | 25 | \(2.4 \times 10^{23}\) |
| 60 | 40 | \(1.4 \times 10^{28}\) |
| 55 | 45 | \(6.1 \times 10^{28}\) |
| 51 | 49 | \(9.9 \times 10^{28}\) |
| 50 | 50 | \(1.0 \times 10^{29}\) |
| 49 | 51 | \(9.9 \times 10^{28}\) |
| 45 | 55 | \(6.1 \times 10^{28}\) |
| 40 | 60 | \(1.4 \times 10^{28}\) |
| 25 | 75 | \(2.4 \times 10^{23}\) |
| 10 | 90 | \(1.7 \times 10^{13}\) |
| 5 | 95 | \(7.5 \times 10^7\) |
| 1 | 99 | 100 |
| 0 | 100 | 1 |
| Total: | \(1.27 \times 10^{30}\) |
This result becomes dramatic for larger systems. Consider what happens if you have 100 coins instead of just 5. The most orderly arrangements (most structured) are 100 heads or 100 tails. The least orderly (least structured) is that of 50 heads and 50 tails. There is only 1 way (1 microstate) to get the most orderly arrangement of 100 heads. There are 100 ways (100 microstates) to get the next most orderly arrangement of 99 heads and 1 tail (also 100 to get its reverse). And there are \(1 \times 10^{29} ways to get 50 heads and 50 tails, the least orderly arrangement. Table \(\PageIndex{2}\) is an abbreviated list of the various macrostates and the number of microstates for each macrostate. The total number of microstates—the total number of different ways 100 coins can be tossed—is an impressively large \(1.27 \times 10^{30}\). Now, if we start with an orderly macrostate like 100 heads and toss the coins, there is a virtual certainty that we will get a less orderly macrostate. If we keep tossing the coins, it is possible, but exceedingly unlikely, that we will ever get back to the most orderly macrostate. If you tossed the coins once each second, you could expect to get either 100 heads or 100 tails once in \(2 \times 10^{2}\) years! This period is 1 trillion (\(10^{12}\) times longer than the age of the universe, and so the chances are essentially zero. In contrast, there is an 8% chance of getting 50 heads, a 73% chance of getting from 45 to 55 heads, and a 96% chance of getting from 40 to 60 heads. Disorder is highly likely.
Disorder in a Gas
The fantastic growth in the odds favoring disorder that we see in going from 5 to 100 coins continues as the number of entities in the system increases. Let us now imagine applying this approach to perhaps a small sample of gas. Because counting microstates and macrostates involves statistics, this is called statistical analysis . The macrostates of a gas correspond to its macroscopic properties, such as volume, temperature, and pressure; and its microstates correspond to the detailed description of the positions and velocities of its atoms. Even a small amount of gas has a huge number of atoms: \(1.0 \, cm^3\) of an ideal gas at 1.0 atm and \(0^oC\) has \(2.7 \times 10^{19}\) atoms. So each macrostate has an immense number of microstates. In plain language, this means that there are an immense number of ways in which the atoms in a gas can be arranged, while still having the same pressure, temperature, and so on.
The most likely conditions (or macrostates) for a gas are those we see all the time—a random distribution of atoms in space with a Maxwell-Boltzmann distribution of speeds in random directions, as predicted by kinetic theory. This is the most disorderly and least structured condition we can imagine. In contrast, one type of very orderly and structured macrostate has all of the atoms in one corner of a container with identical velocities. There are very few ways to accomplish this (very few microstates corresponding to it), and so it is exceedingly unlikely ever to occur. (See Figure \(\PageIndex{2}\)(b).) Indeed, it is so unlikely that we have a law saying that it is impossible, which has never been observed to be violated—the second law of thermodynamics.
The disordered condition is one of high entropy, and the ordered one has low entropy. With a transfer of energy from another system, we could force all of the atoms into one corner and have a local decrease in entropy, but at the cost of an overall increase in entropy of the universe. If the atoms start out in one corner, they will quickly disperse and become uniformly distributed and will never return to the orderly original state (Figure \(\PageIndex{2}\)(b)). Entropy will increase. With such a large sample of atoms, it is possible—but unimaginably unlikely—for entropy to decrease. Disorder is vastly more likely than order.
The arguments that disorder and high entropy are the most probable states are quite convincing. The great Austrian physicist Ludwig Boltzmann (1844–1906)—who, along with Maxwell, made so many contributions to kinetic theory—proved that the entropy of a system in a given state (a macrostate) can be written as \[S = klnW,\] where \(k = 1.38 \times 10^{-23} \, J/K\) is Boltzmann’s constant, and \(lnW\) is the natural logarithm of the number of microstates \(W\) corresponding to the given macrostate. \(W\) is proportional to the probability that the macrostate will occur. Thus entropy is directly related to the probability of a state—the more likely the state, the greater its entropy. Boltzmann proved that this expression for \(S\) is equivalent to the definition \(\Delta S = Q/T\) which we have used extensively.
Thus the second law of thermodynamics is explained on a very basic level: entropy either remains the same or increases in every process. This phenomenon is due to the extraordinarily small probability of a decrease, based on the extraordinarily larger number of microstates in systems with greater entropy. Entropy can decrease, but for any macroscopic system, this outcome is so unlikely that it will never be observed.
Example \(\PageIndex{1}\): Entropy Increases in a Coin Toss
Suppose you toss 100 coins starting with 60 heads and 40 tails, and you get the most likely result, 50 heads and 50 tails. What is the change in entropy?
Strategy
Noting that the number of microstates is labeled \(W\) in Table \(\PageIndex{2}\) for the 100-coin toss, we can use \(\Delta S = S_f - S_i = klnW_f - klnW_i\) to calculate the change in entropy.
Solution
The change in entropy is \[\Delta S = S_f - S_i = klnW_f - klnW_i,\] where the subscript i stands for the initial 60 heads and 40 tails state, and the subscript f for the final 50 heads and 50 tails state. Substituting the values for \(W\) from Table \(\PageIndex{2}\) gives
\[\Delta S = (1.38 \times 10^{-23} \, J/K)[ln(1.0 \times 10^{29}) - ln(1.4 \times 10^{28})]\]\[= 2.7 \times 10^{-23} \, J/K\]
Discussion
This increase in entropy means we have moved to a less orderly situation. It is not impossible for further tosses to produce the initial state of 60 heads and 40 tails, but it is less likely. There is about a 1 in 90 chance for that decrease in entropy \((-2.7 \times 10^{-23} \, J/K)\) to occur. If we calculate the decrease in entropy to move to the most orderly state, we get \(\Delta S = -92 \times 10^{-23} \, J/K\). There is about 1 in \(10^{30}\) chance of this change occurring. So while very small decreases in entropy are unlikely, slightly greater decreases are impossibly unlikely. These probabilities imply, again, that for a macroscopic system, a decrease in entropy is impossible. For example, for heat transfer to occur spontaneously from 1.00 kg of \(0^oC\) ice to its \(0^oC\) environment, there would be a decrease in entropy of \(1.22 \times 10^3 \, J/K\). Given that a \(\Delta S\) of \(10^{-21} \, J/K\) corresponds to about a 1 in \(10^{30}\) chance, a decrease of this size \((10^3 \, J/K)\) is an utter impossibility. Even for a milligram of melted ice to spontaneously refreeze is impossible.
PROBLEM-SOLVING STRATEGIES FOR ENTROPY
- Examine the situation to determine if entropy is involved.
- Identify the system of interest and draw a labeled diagram of the system showing energy flow.
- Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
- Make a list of what is given or can be inferred from the problem as stated (identify the knowns). You must carefully identify the heat transfer, if any, and the temperature at which the process takes place. It is also important to identify the initial and final states.
- Solve the appropriate equation for the quantity to be determined (the unknown). Note that the change in entropy can be determined between any states by calculating it for a reversible process.
- Substitute the known value along with their units into the appropriate equation, and obtain numerical solutions complete with units.
- To see if it is reasonable: Does it make sense? For example, total entropy should increase for any real process or be constant for a reversible process. Disordered states should be more probable and have greater entropy than ordered states.
Summary
- Disorder is far more likely than order, which can be seen statistically.
- The entropy of a system in a given state (a macrostate) can be written as \(s = KLNw,\) where \(k = 1.38 \times 10^{-23} \, J/K\) is Boltzmann’s constant, and \(lnW\) is the natural logarithm of the number of microstates \(W\) corresponding to the given macrostate.
Glossary
- macrostate
- an overall property of a system
- microstate
- each sequence within a larger macrostate
- statistical analysis
- using statistics to examine data, such as counting microstates and macrostates