19.2: Basic Concepts
Ageism
Gerontology is the study of the social, cultural, psychological, cognitive, and biological aspects of aging. There are many stereotypes and negative attitudes about aging adults that persist in the US and around the world. This bias can be linked to a general lack of knowledge about the aging process and misunderstandings about older adults. Because of these influences, many individuals have anxiety about aging that can lead to negative stereotypes of older individuals called ageism . [1]
Ageism among nurses and other health care professionals puts older people at risk. Research has demonstrated that ageism in health care negatively impacts older adults’ overall health, well-being, and quality of care received. Ageism results in increased risks of mortality, poor functional health, and slower recovery times from illness. Negative perceptions about aging can also lead to poor mental health and depression. [2] As you read this chapter, think about your own attitudes about aging and how these beliefs may impact the care you provide.
Integrity Versus Despair
Aging individuals must continually adjust to changes in health and physical strength, lifestyle changes as a result of retirement, the loss of significant others, and changing roles and relationships with family members and friends. As a result, older individuals may find it difficult to accept the changes associated with aging. Nurses can support older adults in maintaining a positive self-image and outlook by considering Erikson’s theory of development. Erikson’s theory of development describes the stage of older adulthood as “Integrity versus Despair.” This stage begins at approximately age 65 and ends at death. During this stage, older adults reflect on their accomplishments and the person they have become. If they feel they have led a successful life, they often feel satisfied and develop a sense of integrity. Conversely, individuals who feel unsuccessful or do not feel they achieved their life goals often feel unsatisfied and may experience hopelessness and despair that can lead to depression. Nurses can assist older adults in developing a sense of integrity by encouraging the patient to reminiscence about previous positive life events and relationships and cultivate a positive mindset of guiding the next generation. [3]
Many older adults, especially those with declining health due to chronic disease, acknowledge that changes in their health status and mobility threaten the autonomy and independence they previously experienced throughout adulthood. As a result, many older adults strive to be autonomous so they are not overly reliant on others for their daily care. They often engage in self-management activities in response to changes in their health and physical strength, ranging from simple daily tasks, such as medication management, to more complex tasks, such as relocating to new residences that are better suited to their changes in physical and mental health. Research has found that when older adults are faced with declines in their physical health and/or cognitive abilities, they often draw upon experiences and skills acquired in earlier adulthood for the purpose of self-managing their new conditions. They reflect on their resilience used to overcome significant challenges faced in earlier adulthood and then apply skills and knowledge gained through previously productive activities to managing their new health changes. However, not all older adults have sufficient personal and external resources to devote towards successful self-management of their health conditions. Nurses can assist older adults by personalizing health self-management strategies that emphasize their existing skill sets and knowledge. [4]
Other Considerations
Retirement
In addition to the physiological changes that occur with aging, older adults vary in their level of activity. For example, many older adults continue working into their seventies and beyond. Individuals may choose to continue to work because of their sense of purpose or because of a need for income. Some older individuals experience a loss of identity when they retire because their work role was an important aspect of their life. Retirement can bring a sense of freedom and adventure, as well as a need to find new identity and purpose.
Social Isolation
Retirement and the loss of daily interaction with coworkers, as well as death of family members and friends, can lead to social isolation in the aging population. Social support impacts a person’s health and quality of life and should be included as part of the assessment. It is helpful for nurses to be familiar with community resources that provide socialization opportunities and provide referrals for patients in need of additional services.
Modified Living Environment
Although many aging adults live in assisted living facilities or skilled nursing centers, many older adults prefer to live at home. Modifications may be needed to the home environment to promote safety and independence. For example, grab bars, elevated toilet seats, and other modifications may be needed in the bathroom, along with good lighting, minimization of clutter, and removal of rugs throughout the home. Assessment of the home environment for safety and ease of mobility is an important aspect of home care nursing.
If an older adult requires more care than family members are able to provide at home, nurses provide valuable information about available care options and make referrals to social workers and case managers. There are a wide variety community-based resources to enhance care for older adults. Local aging and disability resource centers (ADRCs) can help facilitate referrals based on specific needs of the older adult. Examples of other resources include adult day centers, home health agencies that provide personal care and nursing assistance, community-based residential facilities (CBRFs), and residential care apartment complexes (RCACs). If an older adult requires 24-hour nursing care, placement in a nursing home (also referred to as a skilled nursing facility) may be required. Use the following hyperlink to read more information about nursing home resources provided by the Centers for Medicare and Medicaid (CMS).
Learn more about nursing home resources by reviewing the Nursing Home Resource Center provided by the Centers for Medicare and Medicaid (CMS).
- Merz, C. C., Stark, S. L., Morrow-Howell, N. L., & Carpenter, B. D. (2016). When I'm 64: Effects of an interdisciplinary gerontology course on first-year undergraduates' perceptions of aging. Gerontology & Geriatrics Education, 39 (1), 35-45. https://doi.org/10.1080/02701960.2016.1144600 ↵
- Burnes, D., Sheppard, C., Henderson, C. R., Wassel, M., Cope, R., Barber, C., & Pillemer, K. (2019). Interventions to reduce ageism against older adults: A systematic review and meta-analysis. American Journal of Public Health, 109 (8), e1-e9. https://doi.org/10.2105/AJPH.2019.305123 ↵
- This work is a derivative of StatPearls by Orenstein and Lewis and is licensed under CC BY 4.0 ↵
- Perry, T. E., Ruggiano, N., Shtompel, N., & Hassevoort, L. (2014). Applying Erikson's wisdom to self-management practices of older adults: Findings from two field studies. Research on Aging, 37 (3), 253-274. https://doi.org/10.1177/0164027514527974 ↵


1.18: Assignment- Science of Geology
You have been tasked with creating a brochure on geology to hand out at a high school career day event. The goal is to inform students what geology is, what types of careers geologists use and why they need a good foundation in science if they want to be a geologist.
Basic Requirements (assignment criteria):
Prepare a brochure or flyer that clearly and accurately informs the students about the field of geology. The brochure of flyer should include:
- An introduction with a clear understanding and definition of geology
- A section describing what geologists can do or the various fields (at least 3) geologists can be a part of
- Conclude with a discussion as to why a solid understanding of science and the scientific method is needed
Be creative with your brochure. You can use the following formats: Word, PDF, PPT or Publisher files.
Make sure you cite and reference any sources that you use. Remember to not only list references but cite WITHIN the text itself.
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Geology
Define geology as a science and understand the basic methods of science used in geology and related scientific disciplines
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Contributors and Attributions
- Science of Geology Assessment. Authored by : Kimberly Schulte. Provided by : SBCTC. Located at : http://www.columbiabasin.edu . License : CC BY: Attribution


6.10.2: Endemic Diseases
Some diseases reach a level of regularity or seasonality, meaning they stay at a particular level consistently or only increase and decrease in a predictable pattern. Essentially, these illnesses remain with us at a consistent background rate, or cycle with predictable peaks every year. For example, we may become accustomed to “cold and flu season” that typically peaks between December and February in the United States (CDC, 2022). Since the numbers of cases increase up to the peak and then decrease in a predictable pattern over the course of those months, we would call the common cold and flu “endemic” to the U.S. Other pathogens are endemic to specific areas of the world, like malaria. Since mosquitos are the vector which transmits malaria to humans, malaria is endemic in many tropical countries where mosquitoes abound. Many virologists predict that COVID-19 will become endemic if it has not already (Jared, 2023).


22.5C: Muscularis
The muscularis is responsible for the segmental contractions and peristaltic movements in the gastrointestinal (GI) tract.
- Identify the function of muscularis in the GI tract
Key Points
- The muscularis, or muscularis externa, consists of an inner circular muscular layer and a longitudinal outer muscular layer. The coordinated contractions of these layers is called peristalsis, which propels the food through the GI tract.
- Between the two muscle layers is the myenteric or Auerbach’s plexus, which controls peristalsis.
- In the colon, the muscularis externa is much thicker because the feces are large and heavy, requiring more force to push along.
- The stomach has a third layer of muscularis externa: the inner oblique layer. This helps churn the chyme in the stomach.
- Peristaltic activity in the muscularis externa is regulated by the enteric nervous system and the autonomic nervous system.
Key Terms
- muscularis externa : A region of muscle in many organs in the vertebrate body, adjacent to the submucosa membrane. It is responsible for gut movement such as peristalsis.
- oblique layer : This layer is responsible for creating the motion that churns and physically breaks down the food.
- tiniae coli : These are three, separate longitudinal ribbons of smooth muscle on the outside of the ascending, transverse, descending, and sigmoid colons.
The gastrointestinal (GI) tract is composed of four layers of tissue, known as tunics. Each layer has different structures and functions. From the inside out they are called the mucosa, submucosa, muscularis externa, and serosa.
Structure of the Muscularis Externa
Muscularis mucosa of the submucosa : The muscularis mucosa is adjacent to the submucosa, and should not be confused with the muscularis externa.
The muscularis externa is responsible for segmental contractions and peristaltic movement in the GI tract. These muscles cause food to move and churn together with digestive enzymes down the GI tract. The muscularis externa consists of an inner circular layer and a longitudinal outer muscular layer. It should not be confused with a thin layer of muscle known as the muscularis mucosa, which lies within the submucosa, a layer of tissue adjacent to the muscularis externa. The muscularis mucosa is made up of smooth muscle, and is most prominent in the stomach.
Within the muscularis externa, the circular muscle layer prevents food from traveling backward, while the longitudinal layer shortens the tract. The layers are not truly longitudinal or circular, rather the layers of muscle are helical with different pitches. The inner circular is helical with a steep pitch and the outer longitudinal is helical with a much shallower pitch.
The coordinated contractions of these layers is called peristalsis. Between the two muscle layers is the myenteric or Auerbach’s plexus, which controls peristalsis. Peristaltic activity is regulated by these nerve cells, and the rate of peristalsis can be modulated by the rest of the autonomic nervous system.
The thickness of muscularis externa varies in each part of the tract. In the colon, for example, the muscularis externa is much thicker because the feces are large and heavy, and require more force to push along. The outer longitudinal layer of the colon thins out into three discontinuous longitudinal bands known as tiniae coli (bands of the colon). This is one of the three features helping to distinguish between the large and small intestine.
General Structure of the gut wall : General structure of the gut wall—the muscularis externa is labeled circular muscle and longitudinal muscle here.
Occasionally in the large intestine (two to three times a day), there will be mass contraction of certain segments, moving a lot of feces along. This is generally when one gets the urge to defecate.
The pylorus of the stomach has a thickened portion of the inner circular layer: the pyloric sphincter. Alone among the GI tract, the stomach has a third layer of muscularis externa. This is the inner oblique layer, and helps churn the chyme in the stomach.


1.2.3: Polynomial Expressions
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We begin with the definition of a term.
Term
A term is either a single number or the product of a number and one or more variables.
For example, each of the following are terms.
\[ -5 \text{,}\quad-3 x^{2} \text{,}\quad 12 y^{2} z^{3} \text{,}\quad 13 a^{2} b c^{3} \nonumber \]
Note how the first term is a single number, while the remaining terms are products of a number and one or more variables. For example, \(−3x^2\) is the product of \(−3\), \(x\), and \(x\). The term that is a single number is called a constant term .
Coefficient
When a term is a product of a number and one or more variables, the number is called the coefficient of the term. In the case of a term that is a single number, the number itself is called the coefficient of the term.
The coefficients of the terms \[-5 \text{,}\quad-3 x^{2} \text{,}\quad 12 y^{2} z^{3} \text{,}\quad 13 a^{2} b c^{3} \nonumber \]are \(−5\), \(−3\), \(12\), and \(13\), respectively.
Degree
The degree of a term is the sum of the exponents on each variable of the term. A constant term (single number with no variables) has degree zero.
The degree of the terms \[-5\text{,}\quad-3 x^{2} \text{,}\quad 12 y^{2} z^{3} \text{,}\quad 13 a^{2} b c^{3} \nonumber \]are \(0\), \(2\), \(5\), and \(6\), respectively. In the last example, note that \(13a^2bc^3\) is equivalent to \(13a^2b^1c^3\), so adding exponents, we get:
\[\begin{aligned} \text { Degree of } 13 a^{2} b c^{3} &=\text { Degree of } 13 a^{2} b^{1} c^{3} \\ &=2+1+3 \\ &=6 \end{aligned} \nonumber \]
Monomial
The words monomial and term are equivalent.
Thus, \[-5 \text{,}\quad-3 x^{2} \text{,}\quad 12 y^{2} z^{3} \text{,}\quad 13 a^{2} b c^{3} \nonumber \]are monomials.
Binomial
A binomial is a mathematical expression containing exactly two terms, separated by plus or minus signs.
For example, each of the mathematical expressions \[2 x+3 y \text{,}\quad-3 a^{2}-3 b^{2} \text{,}\quad x y+7 \text{,}\quad-3 x^{2} y+5 x y^{2} \nonumber \]is a binomial. Each expression has exactly two terms.
Trinomial
A trinomial is a mathematical expression containing exactly three terms, separated by plus or minus signs.
For example, each of the mathematical expressions \[2 x^{2}+3 x+7 \text{,}\quad a^{2}+2 a b+b^{2} \text{,}\quad x^{4}-2 x^{2} y^{2}+3 y^{4} \nonumber \]is a trinomial. Each expression has exactly three terms.
A bicycle has two wheels, a binomial has two terms. A tricycle has three wheels, a trinomial has three terms. But once we get past three terms, the assignment of special names ceases and we use the generic word polynomial , which means “many terms.”
Polynomial
A polynomial is a many-termed mathematical expression, with terms separated by plus or minus signs. The coefficients of a polynomial are the coefficients of its terms.
Each of the previous expressions, \[12 y^{2} z^{3} \text{,}\quad-3 a^{2}-3 b^{2} \text{,}\quad x^{4}-2 x^{2} y^{2}+3 y^{4} \nonumber \]though assigned the particular names monomial, binomial, and trinomial, respectively, are also “many-termed” expressions and can also be called polynomials. However, because the word polynomial means “many terms,” we can also use the word polynomial to describe mathematical expressions with more than three terms, such as: \[x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4} \nonumber \]The coefficients of \(x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}\) are \(1\), \(−4\), \(6\), \(−4\), and \(1\).
Descending Order
When asked to simplify a polynomial expression, we should combine any like terms we find, and when possible, arrange the answer in descending order.
Example \(\PageIndex{1}\)
Simplify the following polynomial expression, arranging your answer in the descending order of \(x\). \[2 x^{3}+7 x-3 x^{2}+11 x+8 x^{2}+11+15 x \nonumber \]
Solution
In order to arrange our answer in descending powers of \(x\), we want to place the term with the highest power of \(x\) first and the term with the lowest power of \(x\) last. We use the commutative and associative properties to change the order and regroup, then we combine like terms.
\[\begin{aligned} 2 x^{3}+7 x &-3 x^{2}+11 x+8 x^{2}+11+15 x \\ &=2 x^{3}+\left(-3 x^{2}+8 x^{2}\right)+(7 x+11 x+15 x)+11 \\ &=2 x^{3}+5 x^{2}+33 x+11 \end{aligned} \nonumber \]
Note how the powers of \(x\) start at \(3\), then go down in order.
You Try \(\PageIndex{1}\)
Simplify the following polynomial, and arrange your answer in descending order of \[3 x^{2}-5 x^{3}+8 x+9 x^{2}-7 x+2 x^{3} \nonumber \]
- Answer
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\(-3 x^{3}+12x^{2}+x\)
When we have a polynomial in a single variable, such as the polynomial in Example \(\PageIndex{1}\), arranging the terms in descending order is fairly straightforward. However, a polynomial in two or more variables is a bit more difficult. We typically write in alphabetical order if they are multiple terms with the same degree, unless specified otherwise.
Example \(\PageIndex{2}\)
Simplify the following polynomial expression, then arrange your answer in descending order of \(x\).\[x^{3}+2 x y^{2}-6 x^{2} y+y^{3}-3 x y^{2}+4 x^{2} y \nonumber \]
Solution
We’ll again use the commutative and associative properties to change the order and regroup, putting the terms with the highest powers of \(x\) first, then follow with terms containing lower powers of \(x\) in order.
\[\begin{aligned} x^{3}+2 x y^{2} &-6 x^{2} y+y^{3}-3 x y^{2}+4 x^{2} y \\ &=x^{3}+\left(-6 x^{2} y+4 x^{2} y\right)+\left(2 x y^{2}-3 x y^{2}\right)+y^{3} \\ &=x^{3}-2 x^{2} y-x y^{2}+y^{3} \end{aligned} \nonumber \]
Note that this is a very natural order, the powers of \(x\) decrease while simultaneously the powers of \(y\) increase.
You Try \(\PageIndex{2}\)
Simplify the following polynomial, and arrange your answer in descending order of \(x\): \[-4 x^{2} y^{2}+3 x y^{3}+6 x^{3} y-x y^{3}+2 x^{2} y^{2} \nonumber \]
- Answer
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\(6 x^{3} y-2 x^{2} y^{2}+2 x y^{3}\)
Not all examples will have nice ordering presented in Example \(\PageIndex{2}\), with the powers of one variable descending while the powers of the other variable simultaneously ascends.
Example \(\PageIndex{3}\)
Simplify the following polynomial expression, then arrange your answer in descending order for variable \(a\). \[a^{3} b^{3}+2 a^{2} b-3 a^{2} b^{3}+4 a^{3} b^{3}+5 a^{4}+3 a^{2} b+b^{5} \nonumber \]
Solution
Let’s try to arrange the terms so that the powers of a descend. Again, we use the commutative and associative properties to change the order and regroup.
\[\begin{aligned} a^{3} b^{3}+2 a^{2} b &-3 a^{2} b^{3}+4 a^{3} b^{3}+5 a^{4}+3 a^{2} b+b^{5} \\ &=5 a^{4}+\left(a^{3} b^{3}+4 a^{3} b^{3}\right)+\left(2 a^{2} b+3 a^{2} b\right)-3 a^{2} b^{3}+b^{5} \\ &=5 a^{4}+5 a^{3} b^{3}+5 a^{2} b-3 a^{2} b^{3}+b^{5} \end{aligned} \nonumber \]
Note that in our final arrangement, the powers of \(a\) descend, but the powers of \(b\) bounce up and down, but at least we have the powers of \(a\) descending. That should help us spot if we’ve missed a term while simplifying the given problem.
You Try \(\PageIndex{3}\)
Simplify the following polynomial, and arrange your answer in descending order of \(b\): \[5 a^{3} b^{2}+4 a b^{3}-2 a^{2} b+3 a^{3} b^{2}-a b^{3} \nonumber \]
- Answer
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\(3ab^{3}+8a^{2}b-2a^{2}b\)
The Degree and Leading Coefficient of a Polynomial
To find the degree and leading coefficient of a polynomial, locate the term of the polynomial having the highest degree.
Degree of a Polynomial
The degree of a polynomial is the degree of the term having the highest degree.
Leading Coefficient of a Polynomial
The leading coefficient of a polynomial is the coefficient of the term with the highest degree.
Example \(\PageIndex{4}\)
What is the degree and leading coefficient of the polynomial \(x^{3}-4 x^{2}+5-6 x+2 x^{7}\)?
Solution
First, let’s arrange the polynomial in descending powers of x.
\[2 x^{7}+x^{3}-4 x^{2}-6 x+5 \nonumber \]
Arranging the polynomial in descending powers of \(x\) makes it easier to see that the term of the polynomial with the highest degree is \(2x^7\). Therefore, the degree of the polynomial is \(7\) and the leading coefficient is 2.
You Try \(\PageIndex{4}\)
What is the degree and leading coefficient of the polynomial \(2 x^{3}+8 x^{2}- x^{4}+2 x+10\)?
- Answer
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Degree is \(4\) and leading coefficient is \(-1\).
Finding the degree and leading coefficient of a polynomial of more than one variable is a little bit trickier.
Example \(\PageIndex{5}\)
What is the degree and leading coefficient of the polynomial \(x^{4}-2 x^{3} y^{7}+y^{5}\)?
Solution
Note that the polynomial is already arranged in descending powers of \(x\), an arrangement that is probably as good as we are going to get. In the following table, we list the degree of each term. Remember, the degree of any term is found by summing the exponents on its variables.
\[\begin{array}{cc}{\text { Term }} & {\text { Degree }} \\ \hline x^{4} & {4} \\ {-2 x^{3} y^{7}} & {10} \\ {y^{5}} & {5} \\ \hline\end{array} \nonumber \]
Hence, the term with the highest degree is \(-2 x^{3} y^{7}\), making \(10\) the degree of the polynomial. The leading coefficient is \(-2\).
You Try \(\PageIndex{5}\)
What is the degree and leading coefficient of the polynomial \(x^{2} y^{4}-6 x^{2} y^{2}+5 x^{2} y^{5}-2 x y\)?
- Answer
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The degree is \(7\) and the leading coefficient is 5.
Polynomial Functions
First, we define what we mean by a polynomial function.
Polynomial Function
A polynomial function is a function defined by a rule that assigns to each domain object a range object defined by a polynomial expression.
Advanced courses, such as multivariate calculus, frequently use polynomial functions of more than one variable such as \(f(x, y)=x^{2}+y^{2}\). However, in this course, our focus will be on polynomial functions of a single variable, such as \(p(x)=3-4 x-9 x^{2}\) and \(q(x)=x^{3}-9 x^{2}+11\).
Example \(\PageIndex{6}\)
Given the polynomial function \(p(x)=x^{3}-8 x-11\), evaluate \(p(−3)\).
Solution
To evaluate \(p(−3)\), first restate the function definition, then replace each occurrence of the variable \(x\) with open parentheses.
\[\begin{aligned} p(x) &= x^{3}-8 x-11 \quad \color {Red} \text { Original function definition. } \\ p(\;\;) &= (\;\;)^{3}-8(\;\;)-11 \quad \color {Red} \text { Replace each occurrence of } x \text { with open parentheses. } \end{aligned} \nonumber \]
Next, substitute \(−3\) for \(x\) in the open parentheses prepared in the last step.
\[\begin{aligned} p(-3) &= (-3)^{3}-8(-3)-11 \quad \color {Red} \text { Substitute }-3 \text { for } x \text { in the open parentheses positions.} \\ p(-3) &= -27-8(-3)-11 \quad \color {Red} \text { Exponent first: }(-3)^{3}=-27 \\ p(-3) &= -27+24-11 \quad \color {Red} \text { Multiply: }-8(-3)=24 \\ p(-3) &= -14 \quad \color {Red} \text { Add. } \end{aligned} \nonumber \]
Hence, \(p(−3) = −14\). You can easily check this result on your calculator (see Figure \(\PageIndex{1}\)).
You Try \(\PageIndex{6}\)
Given the polynomial function \(p(x)=-3 x^{2}+7 x+4\) , evaluate \(p(2)\).
- Answer
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\(6\)
The Graph of a Polynomial Function
A very important polynomial function in all of mathematics and science is the polynomial having degree two.
Quadratic Polynomial
The second degree polynomial having the form \[p(x)=a x^{2}+b x+c \nonumber \] is called a quadratic polynomial . The graph of this polynomial is called a parabola .
The parabola is approximately U-shaped. Some open upwards, some open downwards, depending on the sign of the leading term.
In Figure \(\PageIndex{2}\), the leading term of the parabola \(p(x)=2 x^{2}-8 x+6\) has positive two as its coefficient, so it opens upward.
In Figure \(\PageIndex{3}\), the leading term of the parabola \(p(x)=-2 x^{2}-8 x-6\) has negative two as its coefficient, so it opens downward.
Note
The sign of the leading term of \(p(x)=ax^2 +bx+c\) determines whether the parabola opens up or down.
- If \(a>0\), the parabola opens upward.
- If \(a<0\), the parabola opens downward.
The turning point of a parabola has a special name.
Vertex
The graph of the second degree polynomial \(p(x)=ax^2+bx+c\) has a single turning point, called the vertex of the parabola.
Example \(\PageIndex{7}\)
Use your graphing calculator to sketch the graph of the quadratic polynomial \(p(x)=−3x^2 + 12x + 25\).
Solution
The degree of the polynomial \(p(x)=−3x^2 + 12x + 25\) is two, so it is a quadratic polynomial and its graph is a parabola. Moreover, its leading term has negative three as its coefficient, so we know that the parabola opens downward. Enter \(y = −3x^2 + 12x + 25\) as \(Y 1=-3 * X \wedge 2+12 * X+25\) in the Y= menu (see the first image in Figure \(\PageIndex{4}\)), then select 6:ZStandard from the ZOOM menu to produce the third image in Figure \(\PageIndex{4}\).
Note that the graph in Figure \(\PageIndex{4}\) appears to have the U-shape of a parabola that opens downwards. Its vertex (turning point) is not visible, but one would surmise that it lies off the top of the screen. We need to adjust the WINDOW parameters so that the vertex of the parabola is visible in the viewing screen. After some experimentation, we settle on the parameters shown in the first image in Figure \(\PageIndex{5}\), then push the GRAPH button to produce the second image in Figure \(\PageIndex{5}\).
You Try \(\PageIndex{7}\)
Use your graphing calculator to sketch the graph of the quadratic polyno mial \(p(x)=2x^2 −5x−4\).
- Answer
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When the degree of the polynomial is larger than two, the number of turning points of the graph might increase. This makes for some very interesting curves. You will be introduced to a variety of techniques that will help you determine appropriate viewing windows for the graphs of these higher degree polynomials in precalculus. However, in this introductory section, we will assist you by suggesting a good viewing window for each polynomial, one that will allow you to see all of the turning points of the graph of the polynomial.
Example \(\PageIndex{8}\)
Use your graphing calculator to sketch the graph of the polynomial function \(p(x)=x^{4}-37 x^{2}+24 x+180\). Set your window parameters as follows: \(\mathbf{X} \min =-10, \mathbf{X} \max =10, \mathbf{X} \operatorname{scl}=1, \mathbf{Y} \min =-1000, \mathbf{Y} \max = 1000,\) and \(\mathbf{Y} \operatorname{scl} =100\).
Solution
Enter the polynomial function in \(\mathbf{Y} \mathbf{1}\) of the Y= menu , then enter the suggested window parameters in the WINDOW menu (see Figure \(\PageIndex{6}\)).
Push the GRAPH button on the top row of your calculator to produce the graph of the polynomial function shown in Figure \(\PageIndex{7}\).
Sweet-looking curve!
You Try \(\PageIndex{8}\)
Use your graphing calculator to sketch the graph of the quadratic polynomial p(x)=x^3 −14x^2 + 20x + 60. Set your window parameters as follows: \(\mathbf{X} \min =-10, \mathbf{X} \max =20, \mathbf{X} \operatorname{scl}=1, \mathbf{Y} \min =-200, \mathbf{Y} \max = 200,\) and \(\mathbf{Y} \operatorname{scl} =20\).
- Answer
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5.2: Blackbody Radiation
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Learning Objectives
By the end of this section you will be able to:
- Apply Wien’s and Stefan’s laws to analyze radiation emitted by a blackbody
- Explain Planck’s hypothesis of energy quanta
All bodies emit electromagnetic radiation over a range of wavelengths. In an earlier chapter, we learned that a cooler body radiates less energy than a warmer body. We also know by observation that when a body is heated and its temperature rises, the perceived wavelength of its emitted radiation changes from infrared to red, and then from red to orange, and so forth. As its temperature rises, the body glows with the colors corresponding to ever-smaller wavelengths of the electromagnetic spectrum. This is the underlying principle of the incandescent light bulb: A hot metal filament glows red, and when heating continues, its glow eventually covers the entire visible portion of the electromagnetic spectrum. The temperature ( T ) of the object that emits radiation, or the emitter , determines the wavelength at which the radiated energy is at its maximum. For example, the Sun, whose surface temperature is in the range between 5000 K and 6000 K, radiates most strongly in a range of wavelengths about 560 nm in the visible part of the electromagnetic spectrum. Your body, when at its normal temperature of about 300 K, radiates most strongly in the infrared part of the spectrum.
Radiation that is incident on an object is partially absorbed and partially reflected. At thermodynamic equilibrium, the rate at which an object absorbs radiation is the same as the rate at which it emits it. Therefore, a good absorber of radiation (any object that absorbs radiation) is also a good emitter. A perfect absorber absorbs all electromagnetic radiation incident on it; such an object is called a blackbody .
Although the blackbody is an idealization, because no physical object absorbs 100% of incident radiation, we can construct a close realization of a blackbody in the form of a small hole in the wall of a sealed enclosure known as a cavity radiator, as shown in Figure \(\PageIndex{1}\).  At thermodynamic equilibrium (at temperature T ), the cavity walls absorb exactly as much radiation as they emit. Furthermore, inside the cavity, the radiation entering the hole is balanced by the radiation leaving it. The emission spectrum of a blackbody can be obtained by analyzing the light radiating from the hole. Electromagnetic waves emitted by a blackbody are called blackbody radiation .
The intensity \(I(\lambda, T)\) of blackbody radiation depends on the wavelength \(\lambda\) of the emitted radiation and on the temperature T of the blackbody (Figure \(\PageIndex{2}\)). The function \(I(\lambda, T)\) is the power intensity that is radiated per unit wavelength; in other words, it is the power radiated per unit area of the hole in a cavity radiator per unit wavelength. According to this definition, \(I(\lambda, T)d\lambda\) is the power per unit area that is emitted in the wavelength interval from \(\lambda\) to \(\lambda + d\lambda\). The intensity distribution among wavelengths of radiation emitted by cavities was studied experimentally at the end of the nineteenth century. Generally, radiation emitted by materials only approximately follows the blackbody radiation curve (Figure \(\PageIndex{3}\)); however, spectra of common stars do follow the blackbody radiation curve very closely.
Two important laws summarize the experimental findings of blackbody radiation: Wien’s displacement law and Stefan’s law . Wien’s displacement law is illustrated in Figure \(\PageIndex{2}\) by the curve connecting the maxima on the intensity curves. In these curves, we see that the hotter the body, the shorter the wavelength corresponding to the emission peak in the radiation curve. Quantitatively, Wien’s law reads
\[\lambda_{max}T = 2.898 \times 10^{-3} m \cdot K \label{Wien} \]
where \(\lambda_{max}\) is the position of the maximum in the radiation curve. In other words, \(\lambda_{max}\) is the wavelength at which a blackbody radiates most strongly at a given temperature T . Note that in Equation \ref{Wien}, the temperature is in kelvins. Wien’s displacement law allows us to estimate the temperatures of distant stars by measuring the wavelength of radiation they emit.
Example \(\PageIndex{1}\): Temperatures of Distant Stars
On a clear evening during the winter months, if you happen to be in the Northern Hemisphere and look up at the sky, you can see the constellation Orion (The Hunter). One star in this constellation, Rigel, flickers in a blue color and another star, Betelgeuse, has a reddish color, as shown in Figure \(\PageIndex{4}\). Which of these two stars is cooler, Betelgeuse or Rigel?
Strategy
We treat each star as a blackbody. Then according to Wien’s law, its temperature is inversely proportional to the wavelength of its peak intensity. The wavelength \(\lambda_{max}^{(blue)}\) of blue light is shorter than the wavelength \(\lambda_{max}^{(red)}\) of red light. Even if we do not know the precise wavelengths, we can still set up a proportion.
Solution
Writing Wien’s law for the blue star and for the red star, we have
\[\begin {align*} \lambda_{max}^{(red)}T_{(red)} &= 2.898 \times 10^{-3} m \cdot K \\[5pt] &= \lambda_{max}^{(blue)} T_{(blue)} \end{align*} \nonumber \]
When simplified, this gives
\[T_{(red)} = \dfrac{\lambda_{max}^{(blue)}}{\lambda_{max}^{(red)}}T_{(blue)} < T_{(blue)} \nonumber \]
Therefore, Betelgeuse is cooler than Rigel.
Significance
Note that Wien’s displacement law tells us that the higher the temperature of an emitting body, the shorter the wavelength of the radiation it emits. The qualitative analysis presented in this example is generally valid for any emitting body, whether it is a big object such as a star or a small object such as the glowing filament in an incandescent lightbulb.
Exercise \(\PageIndex{1}\)
The flame of a peach-scented candle has a yellowish color and the flame of a Bunsen’s burner in a chemistry lab has a bluish color. Which flame has a higher temperature?
- Answer
The second experimental relation is Stefan’s law, which concerns the total power of blackbody radiation emitted across the entire spectrum of wavelengths at a given temperature. In \(\PageIndex{2}\) , this total power is represented by the area under the blackbody radiation curve for a given T . As the temperature of a blackbody increases, the total emitted power also increases. Quantitatively, Stefan’s law expresses this relation as
\[P(T) = \sigma AT^4 \nonumber \]
where \(A\) is the surface area of a blackbody, \(T\) is its temperature (in kelvins), and \(σ\) is the Stefan–Boltzmann constant , \(\sigma = 5.670 \times 10^{-8} W/(m^2 \cdot K^4)\). Stefan’s law enables us to estimate how much energy a star is radiating by remotely measuring its temperature.
Example \(\PageIndex{2}\): Power Radiated by Stars
A star such as our Sun will eventually evolve to a “red giant” star and then to a “white dwarf” star. A typical white dwarf is approximately the size of Earth, and its surface temperature is about \(2.5 \times 10^4 K\). A typical red giant has a surface temperature of \(3.0 \times 10^3 K\) and a radius ~100,000 times larger than that of a white dwarf. What is the average radiated power per unit area and the total power radiated by each of these types of stars? How do they compare?
Strategy
If we treat the star as a blackbody, then according to Stefan’s law, the total power that the star radiates is proportional to the fourth power of its temperature. To find the power radiated per unit area of the surface, we do not need to make any assumptions about the shape of the star because P / A depends only on temperature. However, to compute the total power, we need to make an assumption that the energy radiates through a spherical surface enclosing the star, so that the surface area is \(A = 4\pi R^2\), where R is its radius.
Solution
A simple proportion based on Stefan’s law gives
\[\dfrac{P_{dwarf}/A_{dwarf}}{P_{giant}/A_{giant}} = \dfrac{\sigma T_{dwarf}^4}{\sigma T_{giant}^4} = \left(\dfrac{T_{dwarf}}{T_{giant}}\right)^4 = 4820 \label{6.5} \]
The power emitted per unit area by a white dwarf is about 5000 times that the power emitted by a red giant.  Despite its relatively lower temperature, the overall power radiated by a red giant far exceeds that of the white dwarf because the red giant has a much larger surface area. To estimate the absolute value of the emitted power per unit area, we again use Stefan’s law. For the white dwarf, we obtain
\[\dfrac{P_{dwarf}}{A_{dwarf}} = \sigma T_{dwarf}^4 = 5.670 \times 10^{-8} \dfrac{W}{m^2 \cdot K^4} (2.5 \times 10^4 \, K)^4 = 2.2 \times 10^{10} \dfrac{W}{m^2} \label{6.7} \]
The analogous result for the red giant is obtained by scaling the result for a white dwarf:
\[\dfrac{P_{giant}}{A_{giant}} = \dfrac{2.2 \times 10^{10}}{4.82 \times 10^3} \dfrac{W}{m^2} = 4.56 \times 10^6 \dfrac{W}{m^2} \cong 4.6 \times 10^{-6} \dfrac{W}{m^2} \nonumber \]
Significance
To estimate the total power emitted by a white dwarf, in principle, we could use Equation \ref{6.7}. However, to find its surface area, we need to know the average radius, which is not given in this example. Therefore, the solution stops here. The same is also true for the red giant star.
Exercise \(\PageIndex{2A}\)
An iron poker is being heated. As its temperature rises, the poker begins to glow—first dull red, then bright red, then orange, and then yellow. Use either the blackbody radiation curve or Wien’s law to explain these changes in the color of the glow.
- Answer
-
The wavelength of the radiation maximum decreases with increasing temperature.
Exercise \(\PageIndex{2B}\)
Suppose that two stars, \(α\) and \(β\), radiate exactly the same total power. If the radius of star \(α\) is three times that of star \(β\), what is the ratio of the surface temperatures of these stars? Which one is hotter?
- Answer
-
\(T_{\alpha}/T_{\beta} = 1/\sqrt{3} \cong 0.58\), so the star \(\beta\) is hotter.
The term “blackbody” was coined by Gustav R. Kirchhoff in 1862. The blackbody radiation curve was known experimentally, but its shape eluded physical explanation until the year 1900. The physical model of a blackbody at temperature T is that of the electromagnetic waves enclosed in a cavity (Figure \(\PageIndex{1}\)) and at thermodynamic equilibrium with the cavity walls. The waves can exchange energy with the walls. The objective here is to find the energy density distribution among various modes of vibration at various wavelengths (or frequencies). In other words, we want to know how much energy is carried by a single wavelength or a band of wavelengths. Once we know the energy distribution, we can use standard statistical methods (similar to those studied in a previous chapter) to obtain the blackbody radiation curve, Stefan’s law, and Wien’s displacement law. When the physical model is correct, the theoretical predictions should be the same as the experimental curves.
In a classical approach to the blackbody radiation problem, in which radiation is treated as waves (as you have studied in previous chapters), the modes of electromagnetic waves trapped in the cavity are in equilibrium and continually exchange their energies with the cavity walls. There is no physical reason why a wave should do otherwise: Any amount of energy can be exchanged, either by being transferred from the wave to the material in the wall or by being received by the wave from the material in the wall. This classical picture is the basis of the model developed by Lord Rayleigh and, independently, by Sir James Jeans. The result of this classical model for blackbody radiation curves is known as the Rayleigh–Jeans law . However, as shown in Figure \(\PageIndex{5}\), the Rayleigh–Jeans law fails to correctly reproduce experimental results. In the limit of short wavelengths, the Rayleigh–Jeans law predicts infinite radiation intensity, which is inconsistent with the experimental results in which radiation intensity has finite values in the ultraviolet region of the spectrum. This divergence between the results of classical theory and experiments, which came to be called the ultraviolet catastrophe , shows how classical physics fails to explain the mechanism of blackbody radiation.
The blackbody radiation problem was solved in 1900 by Max Planck . Planck used the same idea as the Rayleigh–Jeans model in the sense that he treated the electromagnetic waves between the walls inside the cavity classically, and assumed that the radiation is in equilibrium with the cavity walls. The innovative idea that Planck introduced in his model is the assumption that the cavity radiation originates from atomic oscillations inside the cavity walls, and that these oscillations can have only discrete values of energy. Therefore, the radiation trapped inside the cavity walls can exchange energy with the walls only in discrete amounts. Planck’s hypothesis of discrete energy values, which he called quanta , assumes that the oscillators inside the cavity walls have quantized energies . This was a brand new idea that went beyond the classical physics of the nineteenth century because, as you learned in a previous chapter, in the classical picture, the energy of an oscillator can take on any continuous value. Planck assumed that the energy of an oscillator (\(E_n\)) can have only discrete, or quantized, values:
\[E_n = nhf, \, where \, n = 1,2,3, \ldots \label{eq30} \]
In Equation \ref{eq30}, \(f\) is the frequency of Planck’s oscillator. The natural number \(n\) that enumerates these discrete energies is called a quantum number. The physical constant \(h\) is called Planck’s constant :
\[h = 6.626 \times 10^{-34} J \cdot s = 4.136 \times 10^{-15} eV \cdot s \label{6.10} \]
Each discrete energy value corresponds to a quantum state of a Planck oscillator . Quantum states are enumerated by quantum numbers. For example, when Planck’s oscillator is in its first \(n 1\) quantum state, its energy is \(E_1 = hf\); when it is in the \(n = 2\) quantum state, its energy is \(E_2 = 2hf\); when it is in the \(n = 3\) quantum state, \(E_3 = 3hf\); and so on.
Note that Equation \ref{eq30} shows that there are infinitely many quantum states, which can be represented as a sequence { hf , 2 hf , 3 hf ,…, ( n – 1) hf , nhf , ( n + 1) hf ,…}. Each two consecutive quantum states in this sequence are separated by an energy jump, \(\delta E = hf\). An oscillator in the wall can receive energy from the radiation in the cavity (absorption), or it can give away energy to the radiation in the cavity (emission). The absorption process sends the oscillator to a higher quantum state, and the emission process sends the oscillator to a lower quantum state. Whichever way this exchange of energy goes, the smallest amount of energy that can be exchanged is hf . There is no upper limit to how much energy can be exchanged, but whatever is exchanged must be an integer multiple of hf . If the energy packet does not have this exact amount, it is neither absorbed nor emitted at the wall of the blackbody.
PLANCK’S QUANTUM HYPOTHESIS
Planck’s hypothesis of energy quanta states that the amount of energy emitted by the oscillator is carried by the quantum of radiation, \(\Delta E\):
\[\Delta E = hf \label{planck} \]
Recall that the frequency of electromagnetic radiation is related to its wavelength and to the speed of light by the fundamental relation \(f\lambda = c\). This means that we can express Equation \ref{6.10} equivalently in terms of wavelength \(\lambda\). When included in the computation of the energy density of a blackbody, Planck’s hypothesis gives the following theoretical expression for the power intensity of emitted radiation per unit wavelength:
\[I(\lambda, T) = \dfrac{2\pi hc^2}{\lambda^5} \dfrac{1}{e^{hc/\lambda k_BT} - 1} \label{6.11} \]
where c is the speed of light in vacuum and kBkB is Boltzmann’s constant, \(k_B = 1.380 \times 10^{-23} J/K\). The theoretical formula expressed in Equation \ref{6.11} is called Planck’s blackbody radiation law . This law is in agreement with the experimental blackbody radiation curve (Figure \(\PageIndex{2}\)). In addition, Wien’s displacement law and Stefan’s law can both be derived from Equation \ref{6.11}. To derive Wien’s displacement law, we use differential calculus to find the maximum of the radiation intensity curve \(I(\lambda, T)\). To derive Stefan’s law and find the value of the Stefan–Boltzmann constant, we use integral calculus and integrate \(I(\lambda, T)\) to find the total power radiated by a blackbody at one temperature in the entire spectrum of wavelengths from \(\lambda = 0\) to \(\lambda = \infty\). This derivation is left as an exercise later in this chapter.
Example \(\PageIndex{3}\): Planck’s Quantum Oscillator
A quantum oscillator in the cavity wall in Figure \(\PageIndex{1}\) is vibrating at a frequency of \(5.0 \times 10^{14} Hz\). Calculate the spacing between its energy levels.
Strategy
Energy states of a quantum oscillator are given by Equation \ref{eq30}. The energy spacing \(\Delta E\) is obtained by finding the energy difference between two adjacent quantum states for quantum numbers n + 1 and n .
Solution
We can substitute the given frequency and Planck’s constant directly into the equation:
\[\begin {align*} \Delta E &= E_{n + 1} − En = (n + 1)hf − nhf \\[5pt] &= hf \\[5pt] &= (6.626 \times 10^{−34} \, J \cdot s)(5.0 \times 10^{14} \, Hz) \\[5pt] &= 3.3 \times 10^{− 19} \, J \end{align*} \nonumber \]
Significance
Note that we do not specify what kind of material was used to build the cavity. Here, a quantum oscillator is a theoretical model of an atom or molecule of material in the wall.
Exercise \(\PageIndex{3}\)
A molecule is vibrating at a frequency of \(5.0 \times 10^{14}\, Hz\). What is the smallest spacing between its vibrational energy levels?
- Answer
-
\(3.3 \times 10^{-19} J\)
Example \(\PageIndex{4}\): Quantum Theory Applied to a Classical Oscillator
A 1.0-kg mass oscillates at the end of a spring with a spring constant of 1000 N/m. The amplitude of these oscillations is 0.10 m. Use the concept of quantization to find the energy spacing for this classical oscillator. Is the energy quantization significant for macroscopic systems, such as this oscillator?
Strategy
We use Equation \ref{planck} as though the system were a quantum oscillator, but with the frequency f of the mass vibrating on a spring. To evaluate whether or not quantization has a significant effect, we compare the quantum energy spacing with the macroscopic total energy of this classical oscillator.
Solution
For the spring constant, \(k = 1.0 \times 10^3 N/m\), the frequency f of the mass, \(m = 1.0 \, kg\), is
\[f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{1.0 \times 10^3 N/m}{1.0 \, kg}} \simeq 5.0 \, Hz \nonumber \]
The energy quantum that corresponds to this frequency is
\[\Delta E = hf = (6.626 \times 10^{-34} J \cdot s)(5.0 \, Hz) = 3.3 \times 10^{-33} J \nonumber \]
When vibrations have amplitude \(A = 0.10 \, m\), the energy of oscillations is
\[E = \dfrac{1}{2} kA^2 = \dfrac{1}{2}(1000 \, N/m)(0.1 \, m)^2 = 5.0 \, J \nonumber \]
Significance
Thus, for a classical oscillator, we have \(\Delta E/E \approx 10^{-34}\). We see that the separation of the energy levels is immeasurably small. Therefore, for all practical purposes, the energy of a classical oscillator takes on continuous values. This is why classical principles may be applied to macroscopic systems encountered in everyday life without loss of accuracy.
Exercise \(\PageIndex{4}\)
Would the result in Example \(\PageIndex{4}\) be different if the mass were not 1.0 kg but a tiny mass of 1.0 µ g, and the amplitude of vibrations were 0.10 µ m?
- Answer
-
No, because then \(\Delta E /E \approx 10^{-21}\)
When Planck first published his result, the hypothesis of energy quanta was not taken seriously by the physics community because it did not follow from any established physics theory at that time. It was perceived, even by Planck himself, as a useful mathematical trick that led to a good theoretical “fit” to the experimental curve. This perception was changed in 1905 when Einstein published his explanation of the photoelectric effect, in which he gave Planck’s energy quantum a new meaning: that of a particle of light.


7.5: Ethics
Ethics or moral philosophy is a branch of philosophy that involves systematizing, defending, and recommending concepts of right and wrong conduct. The term ethics derives from the Ancient Greek word ἠθικός ethikos , which is derived from the word ἦθος ethos (habit, "custom"). The branch of philosophy axiology comprises the sub-branches of ethics and aesthetics, each concerned with values.
As a branch of philosophy, ethics investigates the questions "What is the best way for people to live?" and "What actions are right or wrong in particular circumstances?" In practice, ethics seeks to resolve questions of human morality, by defining concepts such as good and evil, right and wrong, virtue and vice, justice and crime. As a field of intellectual enquiry, moral philosophy also is related to the fields of moral psychology, descriptive ethics, and value theory.
Three major areas of study within ethics recognised today are:
- Meta-ethics, concerning the theoretical meaning and reference of moral propositions, and how their truth values (if any) can be determined
- Normative ethics, concerning the practical means of determining a moral course of action
- Applied ethics, concerning what a person is obligated (or permitted) to do in a specific situation or a particular domain of action
Defining ethics
Rushworth Kidder states that "standard definitions of ethics have typically included such phrases as 'the science of the ideal human character' or 'the science of moral duty'". Richard William Paul and Linda Elder define ethics as "a set of concepts and principles that guide us in determining what behavior helps or harms sentient creatures". The Cambridge Dictionary of Philosophy states that the word ethics is "commonly used interchangeably with 'morality' ... and sometimes it is used more narrowly to mean the moral principles of a particular tradition, group or individual." Paul and Elder state that most people confuse ethics with behaving in accordance with social conventions, religious beliefs and the law and don't treat ethics as a stand-alone concept.
The word "ethics" in English refers to several things. It can refer to philosophical ethics or moral philosophy—a project that attempts to use reason in order to answer various kinds of ethical questions. As the English philosopher Bernard Williams writes, attempting to explain moral philosophy: "What makes an inquiry a philosophical one is reflective generality and a style of argument that claims to be rationally persuasive." And Williams describes the content of this area of inquiry as addressing the very broad question, "how one should live" Ethics can also refer to a common human ability to think about ethical problems that is not particular to philosophy. As bioethicist Larry Churchill has written: "Ethics, understood as the capacity to think critically about moral values and direct our actions in terms of such values, is a generic human capacity." Ethics can also be used to describe a particular person's own idiosyncratic principles or habits. For example: "Joe has strange ethics."
The English word ethics is derived from an Ancient Greek word êthikos , which means "relating to one's character". The Ancient Greek adjective êthikos is itself derived from another Greek word, the noun êthos meaning "character, disposition".
Meta-ethics
Meta-ethics asks how we understand, know about, and what we mean when we talk about what is right and what is wrong. An ethical question fixed on some particular practical question—such as, "Should I eat this particular piece of chocolate cake?"—cannot be a meta-ethical question. A meta-ethical question is abstract and relates to a wide range of more specific practical questions. For example, "Is it ever possible to have secure knowledge of what is right and wrong?" would be a meta-ethical question.
Meta-ethics has always accompanied philosophical ethics. For example, Aristotle implies that less precise knowledge is possible in ethics than in other spheres of inquiry, and he regards ethical knowledge as depending upon habit and acculturation in a way that makes it distinctive from other kinds of knowledge. Meta-ethics is also important in G.E. Moore's Principia Ethica from 1903. In it he first wrote about what he called the naturalistic fallacy . Moore was seen to reject naturalism in ethics, in his Open Question Argument. This made thinkers look again at second order questions about ethics. Earlier, the Scottish philosopher David Hume had put forward a similar view on the difference between facts and values.
Studies of how we know in ethics divide into cognitivism and non-cognitivism; this is similar to the contrast between descriptivists and non-descriptivists. Non-cognitivism is the claim that when we judge something as right or wrong, this is neither true nor false. We may for example be only expressing our emotional feelings about these things. Cognitivism can then be seen as the claim that when we talk about right and wrong, we are talking about matters of fact.
The ontology of ethics is about value-bearing things or properties, i.e. the kind of things or stuff referred to by ethical propositions. Non-descriptivists and non-cognitivists believe that ethics does not need a specific ontology, since ethical propositions do not refer. This is known as an anti-realist position. Realists on the other hand must explain what kind of entities, properties or states are relevant for ethics, how they have value, and why they guide and motivate our actions.
Normative Ethics
Normative ethics is the study of ethical action. It is the branch of ethics that investigates the set of questions that arise when considering how one ought to act, morally speaking. Normative ethics is distinct from meta-ethics because it examines standards for the rightness and wrongness of actions, while meta-ethics studies the meaning of moral language and the metaphysics of moral facts. Normative ethics is also distinct from descriptive ethics, as the latter is an empirical investigation of people's moral beliefs. To put it another way, descriptive ethics would be concerned with determining what proportion of people believe that killing is always wrong, while normative ethics is concerned with whether it is correct to hold such a belief. Hence, normative ethics is sometimes called prescriptive, rather than descriptive. However, on certain versions of the meta-ethical view called moral realism, moral facts are both descriptive and prescriptive at the same time.
Traditionally, normative ethics (also known as moral theory) was the study of what makes actions right and wrong. These theories offered an overarching moral principle one could appeal to in resolving difficult moral decisions.
At the turn of the 20th century, moral theories became more complex and are no longer concerned solely with rightness and wrongness, but are interested in many different kinds of moral status. During the middle of the century, the study of normative ethics declined as meta-ethics grew in prominence. This focus on meta-ethics was in part caused by an intense linguistic focus in analytic philosophy and by the popularity of logical positivism.
In 1971 John Rawls published A Theory of Justice , noteworthy in its pursuit of moral arguments and eschewing of meta-ethics. This publication set the trend for renewed interest in normative ethics.


2.5: Surface Tension and Line Tension
Cell membranes are composed of phospholipids (most abundant), proteins, and glycolipids (Figure 1) 1,2 . Generally, phospholipids in cell membranes will aggregate together in order to maximize hydrophobic interactions among their tails and hydrophilic interactions among their heads . Any in-plane stretching of cell membrane will disrupt the interaction between phospholipids. However, in-plane membrane tension (T m ), which is the force needed to stretch the membrane, will counteract any membrane deformation 3 . T m , which seems to be constant across the whole cell membrane 4 , depends on lipid composition and surface area of the cell membrane. Without changing the amount of lipid, increasing the surface area of cell membrane will increase membrane tension 5 . For example, cells will swell when they are in a hypotonic environment. Cell swelling will cause an increase in cell membrane surface area without changing the the amount of lipid in the membrane. In addition, cell membranes are connected to a cross-linked actin network named the cortex 6 . The cortex is demonstrated as a purple mesh under the membrane in Figure 1. Actin cytoskeleton supports cell membrane and helps cells maintain their shape 6 . Membrane surface tension, which consist of T m and membrane-cytoskeleton adhesion, is the cohesive force that keep cell membranes intact. In order to deform cell membrane, you need to overcome both T m and membrane-cytoskeleton adhesion 3 ,4 . Membrane surface tension helps cell maintain a small surface area.
As shown in Figure 1, cell membranes consist of various types of lipid (e. g. cholesterol, saturated/unsaturated phospholipids, and glycosylated lipids) 2 . Interactions between lipids in the membrane lead to formation of ordered membrane regions that recruit other lipids and proteins 2 . Consequently, there is a wide variety membrane domains with different molecular composition and properties 7 . In addition, lateral segregation of lipids leads to phase separations within cell membranes. Phase separation within cell membrane depends on temperature, pressure, and structural properties (e. g. hydrocarbon tails length or degrees of unsaturation, and the composition of the head group) of various lipid within the membrane 1 . For more info, refer to the wiki pages for Phase Coexistence and Membrane Phase Transitions . At the edge of membrane domains or at the lipid phase separations, phospholipid height mismatch and steric interactions have an energetic cost that depends on the length of the phase/domain boundary 8 . Line tension is the interfacial energy at the edge of membrane domain or at the lipid phase separation in cell membrane 9 . In order to minimize line tension, membrane domains tend to assume circular shapes (Figure 2) 9 . The free energy penalty associated with the line tension is compensated by the decrease in free energy due to membrane domain formations. Membrane tension plays important role in many biological processes, such as cell motility, endocytosis/exocytosis, and viral infection. Many techniques, such as atomic force microscopy ( AFM) and micropipette aspiration, are used to study membrane tension.
Biological Relevance
Membrane Tension and Cell Movement
Cell motility involves continuous cycles of protrusion, adhesion, and contraction of the cell membrane 10 . Actin-polymerization-mediated protrusion of cell membrane called lamellipodia influence the direction of migration. In the region of cell with low membrane tension, multiple lamellipodia that point in different directions coexist (Figure 3, left) 11 . As a result, cell migration in any particular direction is not favored. Cells can migrate toward a particular direction by increasing the membrane tension at a specific region of the membrane. Increasing the membrane tension will decrease the number of lamellipodia (Figure 3, right) 11 . Cells tend to migrate in the direction with fewer lamellipodia 11 .
Membrane Tension and Endocytosis/Exocytosis
Endocytosis involves formation of cell membrane invaginations, which eventually detach from cell membrane and form vesicles 12 . Increasing membrane tension or membrane-cytoskeleton adhesion will hinder both invagination and neck formation step ( Figure 4 ) 12 . Both steps involves membrane deformation.
Exocytosis involves vesicles docking to cell membrane and fuse with it 12 . Unlike endocytosis, increasing membrane tension will favor vesicle and membrane fusion ( Figure 5 ) 12 . Unlike endocytosis, exocytosis requires minimal membrane deformation.
Membrane Tension and Viral Infections
Polyomaviruses (e. g. Simian virus 40 (SV40), Murine polyomavirus, and BK virus) are nonenveloped DNA viruses 13 . Capsid protein VP1 13 , a main component of polyomavirus capsid, allows polyomaviruses to overcome membrane tension and infect cells. VP1 binds to glycosphingolipids on cell surface. 13 Different types of virus will bind to different types of glycosphingolipids 13 . For example, VP1 from SV40 binds to GM1 glycosphingolipid 13 . VP1 binding to GM1 reduces membrane tension by triggering actin cytoskeleton breakdown 13 . As a result, an invagination of cell membrane is formed, which eventually lead to the endocytosis of the virus 13 . SV40 endocytosis is shown in Figure 6.
Methods Used to Study Membrane Tension and Line Tension
Methods Used to Measure Surface Tension:
Common methods used to measure membrane tension is shown in Figure 7. One assay for measuring membrane tension is parallel plate confinement assay. In this assay, cells are compressed between two parallel plates. The radii of the principal curvatures of the membrane are measured 3 . Since the amount of force used to compress the cell is known, you can calculate the membrane tension with the equation shown in Figure 7 (Eq.1) 3 . Another method of measuring membrane tension is micropipette aspiration. For this method, a small section of the cell membrane is pulled using a micropipette , and the membrane tension is calculated from the cell radius, micropipette tip radius, extracellular pressure, and the suction pressure (Figure 7b, Equation 2) 3 . Micropipette aspiration can also be used to measure membrane tension from membrane tethers. Since membrane tethers lack continuous cytoskeleton, membrane-cytoskeleton adhesion will have very little (if any) contribution to the membrane tension measured 3 . Generally, cells are hold in place with a micropipette, and then tethers can be extracted from the cell membrane (Figure 7c, Equation 3) 3 . Other methods of tether extraction include atomic force microscopy (AFM) and optical tweezers (Figure 7d) 3 . The cantilever used in AFM can also be used to extract a membrane tether 3 . To extract membrane tether, optical tweezers can be used to pull on a uncoated polystyrene bead that was attached to the cell membrane 14 .
Fluorescence resonance energy transfer (FRET) refers to the distance-dependent transfer of energy from one fluorophore to another fluorophore ( Figure 8 ) 15 . For a more detail explanation of FRET refer to reference 15.  Fluorescence emission spectra of the donor fluorophore must overlap with the fluorescence absorption spectra of the acceptor fluorophore. In practice, only the donor flourophore will be excited. Generally, the florescence intensity ratio of acceptor fluorophore to donor fluorophore is measured. The distance between the donor and acceptor fluorophores affects the FRET efficiency. FRET efficiency increases as the distance between the two fluorophores decreases. Grashoff et al. developed a FRET-base biosensor for membrane tension 16 . A brief description of the method used in reference 16 is presented here. The biosensor consists of a tension sensor module in-between two halves (Vh and Vt) of vinculin, a membrane-cytoskeletal protein ( Figure 9c ). The tension sensor module consists of two fluorophores, mTFP1 and venus(A206K), linked together by a spider silk protein flagelliform linker sequence, (GPGGA) 8 ( Figure 9a ). The linker acts as a spring that holds the FRET pair, mTFP1 and venus(A206K), together. Vh and Vt help localized the biosensor to the cell membrane. Once cells are transfected with this biosensor, you can measure the membrane tension by measuring FRET efficiency. As the membrane tension increases, the FRET efficiency decreases because the distance between the two fluorophores increases ( Figure 9b ). In order to convert FRET efficiency to force, you need to construct a calibration curve (FRET efficiency vs Force). Grashoff et al. attached a similar FRET-base biosensor to a polymer-coated glass surface and a microsphere. They can apply force to the sensor by pulling on the microsphere with an optical tweezer. They then construct the calibration curve by measuring the change in FRET efficiency in response to the amount of force applied to the sensor.
Calculating Line Tension from Domain Nucleation Rate
Blanchette et al. developed a way to calculate line tension from nucleation rate of membrane domain 17 . A summary of the method used in reference 17 is presented here. First, supported lipid bilayers are heated to 55℃ and then cool down slowly until domain nucleation occurs. Domain nucleation events can be imaged with AFM. Next, lipid bilayers are heated slowly until membrane domains disappear. Averaging the nucleation and melting temperatures will give you the liquidus or miscibility temperature (T trans ) for liquid-solid and liquid-liquid phase coexistence, respectively 18 . Then, lipid bilayers are kept 5℃ above phase transition temperature. After the lipid bilayers are homogenized, they are rapidly cool down to various temperatures below T trans . During the cooling process, domain nucleation events are imaged with AFM. The line tension ( γ ) is calculated by fitting the nucleation rate ( J ) vs 1/TΔT curve with equation 1 for symmetrical domain nucleation, which occurs on both side of the membrane, and equation 2 for asymmetrical domain nucleation, which only occurs on one side of the membrane.
\[ J = A \exp \left( \dfrac{-\pi \gamma^2 a_g T_{trans}}{2k_B \Delta H} \times \dfrac{1}{T \Delta T} \right) \tag{1}\]
\[ J = A \exp \left( \dfrac{-\pi \gamma^2 a_g T_{trans}}{k_B \Delta H} \times \dfrac{1}{T \Delta T} \right) \tag{2}\]
J (obtained with AFM) is number of nuclei that form per unit area per unit time. The molar area of the gel phase, a g , and the enthalpy of the phase transition, ΔH , depend on the lipid composition of the lipid bilayer. T is the temperature, and ΔT is T trans – T . A is a pre-exponential factor 18 .
References
- Kranenburg, M. & Smit, B. Phase Behavior of Model Lipid Bilayers. J. Phys. Chem. B 109 , 6553-6563 (2005).
- Sezgin, E., Levental, I., Mayor, S. & Eggeling, C. The mystery of membrane organization: composition, regulation and roles of lipid rafts. Nat. Rev. Mol. Cell. Biol. (2017).
- Diz-Muñoz, A., Fletcher, D.A. & Weiner, O.D. Use the force: membrane tension as an organizer of cell shape and motility. Trends Cell Biol. 23 , 47-53 (2013).
- Sheetz, M.P. Cell control by membrane–cytoskeleton adhesion. Nat. Rev. Mol. Cell. Biol. 2 , 392-396 (2001).
- Lieber, Arnon D., Yehudai-Resheff, S., Barnhart, Erin L., Theriot, Julie A. & Keren, K. Membrane Tension in Rapidly Moving Cells Is Determined by Cytoskeletal Forces. Curr. Biol. 23 , 1409-1417 (2013).
- Salbreux, G., Charras, G. & Paluch, E. Actin cortex mechanics and cellular morphogenesis. Trends Cell Biol. 22 , 536-545.
- Sonnino, S. & Prinetti, A. Membrane Domains and the “Lipid Raft” Concept. Curr. Med. Chem. 20 , 4-21 (2013).
- García-Sáez, A.J. & Schwille, P. Stability of lipid domains. FEBS Lett. 584 , 1653-1658 (2010).
- García-Sáez, A.J., Chiantia, S. & Schwille, P. Effect of Line Tension on the Lateral Organization of Lipid Membranes. J. Biol. Chem. 282 , 33537-33544 (2007).
- Vicente-Manzanares, M., Webb, D.J. & Horwitz, A.R. Cell migration at a glance. J. Cell. Sci. 118 , 4917-4919 (2005).
- Pierre, S. & Julie, P. Membrane tension and cytoskeleton organization in cell motility. J. Phys. Condens. Matter 27 , 273103 (2015).
- Dai, J. & Sheetz, M.P. Regulation of Endocytosis, Exocytosis, and Shape by Membrane Tension. Cold Spring Harb. Symp. Quant. Biol. 60 , 567-571 (1995).
- Ewers, H. & Helenius, A. Lipid-Mediated Endocytosis. Cold Spring Harb. Symp. Quant. Biol. 3 (2011).
- Pontes, B. , et al. Cell Cytoskeleton and Tether Extraction. Biophys. J. 101 , 43-52 (2011).
- Ishikawa-Ankerhold, H.C., Ankerhold, R. & Drummen, G.P.C. Advanced Fluorescence Microscopy Techniques—FRAP, FLIP, FLAP, FRET and FLIM. Molecules 17 , 4047 (2012).
- Grashoff, C. , et al. Measuring mechanical tension across vinculin reveals regulation of focal adhesion dynamics. Nature 466 , 263-266 (2010).
- Blanchette, C.D., Lin, W.-C., Orme, C.A., Ratto, T.V. & Longo, M.L. Using Nucleation Rates to Determine the Interfacial Line Tension of Symmetric and Asymmetric Lipid Bilayer Domains. Langmuir 23 , 5875-5877 (2007).
- Blanchette, C.D., Lin, W.-C., Orme, C.A., Ratto, T.V. & Longo, M.L. Domain Nucleation Rates and Interfacial Line Tensions in Supported Bilayers of Ternary Mixtures Containing Galactosylceramide. Biophys. J. 94 , 2691-2697 (2008).


5.1: Definition and Properties
The main idea is to Laplace transform the constant-coefficient differential equation for \(x(t)\) into a simpler algebraic equation for the Laplace-transformed function \(X(s)\), solve this algebraic equation, and then transform \(X(s)\) back into \(x(t)\). The correct definition of the Laplace transform and the properties that this transform satisfies makes this solution method possible.
An exponential ansatz is used in solving homogeneous constant-coefficient odes, and the exponential function correspondingly plays a key role in defining the Laplace transform. The Laplace transform of \(f(t)\), denoted by \(F(s) = \mathcal{L}\{ f(t)\}\), is defined by the integral transform \[\label{eq:1}F(s)\int_0^{\infty}e^{-st}f(t)dt.\]
The improper integral given by \(\eqref{eq:1}\) diverges if \(f(t)\) grows faster than \(e^{st}\) for large \(t\). Accordingly, some restriction on the range of \(s\) may be required to guarantee convergence of \(\eqref{eq:1}\), and we will assume without further elaboration that these restrictions are always satisfied.
The Laplace transform is a linear transformation. We have \[\begin{aligned}\mathcal{L}\{c_1f_1(t)+c_2f_2(t)\}&=\int_0^{\infty}e^{-st}(c_1f_1(t)+c_2f_2(t))dt \\ &=c_1\int_0^{\infty}e^{-st}f_1(t)dt+c_2\int_0^{\infty}e^{-st}f_2(t)dt \\ &=c_1\mathcal{L}\{f_1(t)\}+c_2\mathcal{L}\{f_2(t)\}.\end{aligned}\]
There is also a one-to-one correspondence between functions and their Laplace transforms. A table of Laplace transforms can therefore be constructed and used to find both Laplace and inverse Laplace transforms of commonly occurring functions. Such a table is shown in Table \(\PageIndex{1}\) (and this table will be distributed with the exams). In Table \(\PageIndex{1}\), \(n\) is a positive integer. Also, the cryptic entries for \(u_c(t)\) and \(\delta (t − c)\) will be explained later in §5.3 .
Table \(\PageIndex{1}\): Table of Laplace Transformers
| \(f(t)=\mathcal{L}^{-1}\{F(s)\}\) | \(F(s)=\mathcal{L}\{f(t)\}\) |
|---|---|
| 1. \(e^{at}f(t)\) | \(\displaystyle F(s-a)\) |
| 2. \(1\) | \(\displaystyle\frac{1}{s}\) |
| 3. \(e^{at}\) | \(\displaystyle\frac{1}{s-a}\) |
| 4. \(t^n\) | \(\displaystyle\frac{n!}{s^{n+1}}\) |
| 5. \(t^ne^{at}\) | \(\displaystyle\frac{n!}{(s-a)^{n+1}}\) |
| 6. \(\sin bt\) | \(\displaystyle\frac{b}{s^2+b^2}\) |
| 7. \(\cos bt\) | \(\displaystyle\frac{s}{s^2+b^2}\) |
| 8. \(e^{at}\sin bt\) | \(\displaystyle\frac{b}{(s-a)^2+b^2}\) |
| 9. \(e^{at}\cos bt\) | \(\displaystyle\frac{s-a}{(s-a)^2+b^2}\) |
| 10. \(t\sin bt\) | \(\displaystyle\frac{2bs}{(s^2+b^2)^2}\) |
| 11. \(t\cos by\) | \(\displaystyle\frac{s^2-b^2}{(s^2+b^2)^2}\) |
| 12. \(u_c(t)\) | \(\displaystyle\frac{e^{-cs}}{s}\) |
| 13. \(u_c(t)f(t-c)\) | \(\displaystyle e^{-cs}F(s)\) |
| 14. \(\delta (t-c)\) | \(\displaystyle e^{-cs}\) |
| 15. \(\overset{.}{x}(t)\) | \(\displaystyle sX(s)-x(0)\) |
| 16. \(\overset{..}{s}(t)\) | \(\displaystyle s^2X(s)-sx(0)-\overset{.}{x}(0)\) |
The rows of Table \(\PageIndex{1}\) can be determined by a combination of direct integration and some tricks. We first compute directly the Laplace transform of \(e^{at} f(t)\) (line 1):
\[\begin{aligned}\mathcal{L}\{e^{at}f(t)\}&=\int_0^{\infty}e^{-st}e^{at}f(t)dt \\ &=\int_0^{\infty}e^{-(s-a)t}f(t)dt \\ &=F(s-a).\end{aligned}\]
We also compute directly the Laplace transform of \(1\) (line 2):
\[\begin{aligned}\mathcal{L}\{1\}&=\int_0^\infty e^{-st}dt \\ &=\left.-\frac{1}{s}e^{-st}\right]_0^\infty \\ &=\frac{1}{s}.\end{aligned}\]
Now, the Laplace transform of \(e^{at}\) (line 3) may be found using these two results:
\[\begin{align}\mathcal{L}\{e^{at}\}&=\mathcal{L}\{e^{at}\cdot 1\}\nonumber \\ &=\frac{1}{s-a}.\label{eq:2}\end{align}\]
The transform of \(t^n\) (line 4) can be found by successive integration by parts. A more interesting method uses Taylor series expansions for \(e^{at}\) and \(1/(s − a)\). We have \[\begin{align}\mathcal{L}\{e^{at}\}&=\mathcal{L}\left\{\sum\limits_{n=0}^\infty \frac{(at)^n}{n!}\right\}\nonumber \\ &=\sum\limits_{n=0}^\infty\frac{a^n}{n!}\mathcal{L}\{t^n\}.\label{eq:3}\end{align}\]
Also, with \(s > a\), \[\begin{align}\frac{1}{s-a}&=\frac{1}{s(1-\frac{a}{s})}\nonumber \\ &=\frac{1}{s}\sum\limits_{n=0}^\infty\left(\frac{a}{s}\right)^n\label{eq:4} \\ &=\sum\limits_{n=0}^\infty\frac{a^n}{s^{n+1}}.\nonumber\end{align}\]
Using \(\eqref{eq:2}\), and equating the coefficients of powers of \(a\) in \(\eqref{eq:3}\) and \(\eqref{eq:4}\), results in line 4:
\[\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}.\nonumber\]
The Laplace transform of \(t^n e^{at}\) (line 5) can be found from line 1 and line 4:
\[\mathcal{L}\{t^ne^{at}\}=\frac{n!}{(s-a)^{n+1}}.\nonumber\]
The Laplace transform of \(\sin bt\) (line 6) may be found from the Laplace transform of \(e^{at}\) (line 3) using \(a = ib\):
\[\begin{aligned}\mathcal{L}\{\sin bt\}&=\text{Im }\{\mathcal{L}\{e^{ibt}\}\} \\ &=\text{Im }\left\{\frac{1}{s-ib}\right\} \\ &=\text{Im }\left\{\frac{s+ib}{s^2+b^2}\right\} \\ &=\frac{b}{s^2+b^2}.\end{aligned}\]
Similarly, the Laplace transform of \(\cos bt\) (line 7) is \[\begin{aligned}\mathcal{L}\{\cos bt\}&=\text{Re }\{\mathcal{L}\{e^{ibt}\}\} \\ &=\frac{s}{s^2+b^2}.\end{aligned}\]
The transform of \(e^{at} \sin bt\) (line 8) can be found from the transform of \(\sin bt\) (line 6) and line 1:
\[\mathcal{L}\{e^{at}\sin bt\}=\frac{b}{(s-a)^2+b^2};\nonumber\] and similarly for the transform of \(e^{at}\cos bt\):
\[\mathcal{L}\{e^{at}\cos bt\}=\frac{s-a}{(s-a)^2+b^2}.\nonumber\]
The Laplace transform of \(t \sin bt\) (line 10) can be found from the Laplace transform of \(te^{at}\) (line 5 with \(n = 1\)) using \(a = ib\):
\[\begin{aligned}\mathcal{L}\{t\sin bt\}&=\text{Im }\{\mathcal{L}\{te^{ibt}\}\} \\ &=\text{Im }\left\{\frac{1}{(s-ib)^2}\right\} \\ &=\text{Im }\left\{\frac{(s+ib)^2}{(s^2+b^2)^2}\right\} \\ &=\frac{2bs}{(s^2+b^2)^2}.\end{aligned}\]
Similarly, the Laplace transform of \(t \cos bt\) (line 11) is \[\begin{aligned}\mathcal{L}\{t\cos bt\}&=\text{Re }\{\mathcal{L}\{te^{ibt}\}\} \\ &=\text{Re }\left\{\frac{(s+ib)^2}{(s^2+b^2)^2}\right\} \\ &=\frac{s^2-b^2}{(s^2+b^2)^2}.\end{aligned}\]
We now transform the inhomogeneous constant-coefficient, second-order, linear inhomogeneous ode for \(x = x(t)\), \[a\overset{..}{x}+b\overset{.}{x}+cx=g(t),\nonumber\] making use of the linearity of the Laplace transform:
\[a\mathcal{L}\{\overset{.}{x}\}+b\mathcal{L}\{\overset{.}{x}\}+c\mathcal{L}\{x\}=\mathcal{L}\{g\}.\nonumber\]
To determine the Laplace transform of \(\overset{.}{x}(t)\) (line 15) in terms of the Laplace transform of \(x(t)\) and the initial conditions \(x(0)\) and \(\overset{.}{x}(0)\), we define \(X(s) =\mathcal{L}\{x(t)\}\), and integrate \[\int_0^{\infty}e^{-st}\overset{.}{x}dt\nonumber\] by parts. We let \[\begin{array}{ccc}u=e^{-st}&\quad &dv=\overset{.}{x}dt \\ du=-se^{-st}dt&\quad &v=x.\end{array}\nonumber\]
Therefore, \[\begin{aligned}\int_0^{\infty}e^{-st}\overset{.}{x}dt&=xe^{-st}]_0^{\infty}+s\int_0^{\infty}e^{-st}xdt \\ &=sX(s)-x(0),\end{aligned}\] where assumed convergence of the Laplace transform requires \[\underset{t\to\infty}{\lim}x(t)e^{-st}=0.\nonumber\]
Similarly, the Laplace transform of \(\overset{..}{x}(t)\) (line 16) is determined by integrating \[\int_0^{\infty}e^{-st}\overset{..}{x}dt\nonumber\] by parts and using the just derived result for the first derivative. We let \[\begin{array}{ccc}u=e^{-st}&\quad &dv=\overset{..}{x}dt \\ du=-se^{-st}dt&\quad&v=\overset{.}{x},\end{array}\nonumber\] so that \[\begin{aligned}\int_0^{\infty}e^{-st}\overset{..}{x}dt&=\overset{.}{x}e^{-st}]_0^{\infty}+s\int_0^{\infty}e^{-st}\overset{.}{x}dt \\ &=-\overset{.}{x}(0)+s(sX(s)-x(0)) \\ &=s^2X(s)-sx(0)-\overset{.}{x}(0),\end{aligned}\] where similarly we assume \(\lim_{t\to\infty}\overset{.}{x}(t)e^{-st}=0\).


11.2: Behaviors of Non-Ideal Gases
Non-ideal gases (sometimes also referred to as “real gases”), do not behave as ideal gases because at least one of the assumptions in Definition: Ideal Gas is violated. What characterizes non-ideal gases is that there is no unique equation that we can use to describe their behavior. For this reason, we have a plethora of several experimental models, none of which is superior to the other. The van der Waals (vdW) equation is the only model that we will analyze in detail because of its simple interpretation. However, it is far from universal, and for several non-ideal gases, it is severely inaccurate. Other popular non-ideal gases equations are the Clausius equation, the virial equation, the Redlich–Kwong equation and several others.\(^1\)
The van der Waals equation
One of the simplest empirical equation that describes non-ideal gases was obtained in 1873 by Johannes Diderik van der Waals (1837–1923). The vdW equation includes two empirical parameters (\(a\) and \(b\)) with different values for different non-ideal gases. Each of the parameters corresponds to a correction for the breaking of one of the two conditions that define the ideal gas behavior ( Definition: Ideal Gas ). The vdW equation is obtained from the ideal gas equation performing the following simple substitutions:
\[ \begin{aligned} P & \;\rightarrow\;\left( P + \dfrac{a}{\overline{V}^2} \right)\\ \overline{V} & \;\rightarrow\;\left( \overline{V} - b\right),\\ \end{aligned} \label{11.2.1} \]
which results in:
\[ \begin{aligned} P\overline{V} &=RT \; \rightarrow \; \left( P + \dfrac{a}{\overline{V}^2} \right)\left( \overline{V} - b\right)=RT\\ P &=\dfrac{RT}{\overline{V} - b}-\dfrac{a}{\overline{V}^2}. \end{aligned} \label{11.2.2} \]
The parameter \(a\) accounts for the presence of intermolecular interactions, while the parameter \(b\) accounts for the non-negligible volume of the gas molecules. Despite the parameters having simple interpretations, their values for each gas must be determined experimentally. Values for these parameters for some significant non-ideal gas are reported below:
| \(a \left[ \dfrac{\mathrm{L}^2\mathrm{bar}}{\mathrm{mol}^2} \right]\) | \(b \left[ \dfrac{\mathrm{L}}{\mathrm{mol}} \right]\) | |
|---|---|---|
| Ammonia | 4.225 | 0.0371 |
| Argon | 1.355 | 0.03201 |
| Carbon dioxide | 3.640 | 0.04267 |
| Carbon monoxide | 1.505 | 0.03985 |
| Chlorine | 6.579 | 0.05622 |
| Freon | 10.78 | 0.0998 |
| Helium | 0.0346 | 0.0238 |
| Hydrogen | 0.2476 | 0.02661 |
| Mercury | 8.200 | 0.01696 |
| Methane | 2.283 | 0.04278 |
| Neon | 0.2135 | 0.01709 |
| Nitrogen | 1.370 | 0.0387 |
| Oxygen | 1.382 | 0.03186 |
| Radon | 6.601 | 0.06239 |
| Xenon | 4.250 | 0.05105 |
Joule–Thomson effect
We have already met William Thomson, also known as Lord Kelvin, and his seminal work on the second law of thermodynamics. In conjunction with that work, Thomson is famous for developing a sensitive method for measuring the temperature changes related to the expansion of a gas. These experiments improved on the earlier work by James Joule, and Lord Kelvin’s improved instrument depicted in Figure \(\PageIndex{4}\) is named the Joule–Thomson apparatus. The apparatus is composed of two chambers, each with its own mobile piston. The chambers are connected via a valve or a porous plug. The entire equipment is also thermally isolated from the surroundings. This instrument is a more sensitive version of the Joule expansion apparatus that we already described in section 3 (compare with Figure 3.1.1 ).
Thomson realized that a gas flowing through an obstruction experience a drop in pressure. If the entire apparatus is insulated, it will not exchange heat with its surroundings (\(Q=0\)), and each transformation will happen at adiabatic conditions. Let’s consider an initial condition with 1 mol of gas in the left chamber, occupying a volume \(V_l\), and a completely closed right chamber, for which \(V_r^i=0\). After the process completes, the volume of the right chamber will reduce to \(V_l^f=0\), while the volume of the right chamber will be \(V_r\). Using the first law of thermodynamics, we can write:
\[ \Delta U=U_r-U_l=\underbrace{Q}_{=0}+W=W_l+W_r, \label{11.2.3} \]
with:
\[ \begin{aligned} W_l &=-\int_{V_l}^0 P_l dV = P_l V_l\\ W_r &=-\int_0^{V_r} P_r dV = - P_r V_r. \end{aligned} \label{11.2.4} \]
Replacing \ref{11.2.4} into Equation \ref{11.2.3}, results in:
\[ \begin{aligned} U_r-U_l &=P_l V_l-P_r V_r \\ \underbrace{U_r+P_r V_r}_{H_r} &= \underbrace{U_l + P_l V_l}_{H_l}, \end{aligned} \label{11.2.5} \]
which, replacing the definition of enthalpy \(H=U+PV\), we obtain:
\[ \begin{aligned} H_r &=H_l \\ \Delta H &=0, \end{aligned} \label{11.2.6} \]
or, in other words, the process is isenthalpic . Using the total differential of \(H\):
\[ dH=\left(\dfrac{\partial H}{\partial T} \right)_P dT + \left(\dfrac{\partial H}{\partial P} \right)_T dP = C_P dT + \left(\dfrac{\partial H}{\partial P} \right)_T dP, \label{11.2.7} \]
we obtain:
\[ \Delta H=\int dH = \int C_P dT + \int \left(\dfrac{\partial H}{\partial P} \right)_T dP =0, \label{11.2.8} \]
or, in purely differential form:
\[ dH = C_P dT + \left(\dfrac{\partial H}{\partial P} \right)_T dP =0, \label{11.2.9} \]
From Equation \ref{11.2.9} we can define a new coefficient, called the Joule–Thomson coefficient , \(\mu_{\mathrm{JT}}\), that measures the rate of change of temperature of a gas with respect to pressure in the Joule–Thomson process:
\[ \mu_{\mathrm{JT}}=\left( \dfrac{\partial T}{\partial P} \right)_H=-\dfrac{1}{C_P} \left( \dfrac{\partial H}{\partial T} \right)_P \label{11.2.10} \]
The value of \(\mu_{\mathrm{JT}}\) depends on the type of gas, the temperature and pressure before expansion, and the heat capacity at constant pressure of the gas. The temperature at which \(\mu_{\mathrm{JT}}\) changes sign is called the “Joule–Thomson inversion temperature.” Since the pressure decreases during an expansion, \(\partial P\) is negative by definition, and the following possibilities are available for \(\mu_{\mathrm{JT}}\):
| Gas temperature: | \(\partial P\) | \(\mu_{\mathrm{JT}}\) | \(\partial T\) | The gas will: |
|---|---|---|---|---|
| Below the inversion temperature | – | + | – | cool |
| Above the inversion | temperature– | – | + | warm |
For example, helium has a very low Joule–Thomson inversion temperature at standard pressure \((T=45\;\text{K})\), and it warms when expanded at constant enthalpy at typical room temperatures. The only other gases that have standard inversion temperature lower than room temperature are hydrogen and neon. On the other hand, nitrogen and oxygen have high inversion temperatures (\(T=621\;\text{K}\) and \(T=764\;\text{K}\), respectively), and they both cool when expanded at room temperature. Therefore, it is possible to use the Joule–Thomson effect in refrigeration processes such as air conditioning.\(^2\) As we already discussed in chapter 3 , the temperature of an ideal gases stays constant in an adiabatic expansion, therefore its Joule–Thomson coefficient is always equal to zero.
- For more information on empirical equations for non-ideal gases see this Wikipedia page .
- Nitrogen and oxygen are the two most abundant gases in the air. A sequence of Joule–Thomson expansions are also used for the industrial liquefaction of air.


9: Microsoft PowerPoint Basic Skills Last updated Save as PDF Page ID 46385 Learning Objectives Create and save a new presentation Edit and organize a presentation Add objects to a presentation 9.1: Why It Matters- Microsoft PowerPoint Basic Skills 9.2: Introduction to Creating Presentations 9.3: New Presentations 9.4: Themes 9.5: Introduction to Using PowerPoint 9.6: Editing a Presentation 9.7: Organizing Slides 9.8: Views 9.9: Introduction to Objects in PowerPoint 9.10: Text Boxes 9.11: Images and Clip Art 9.12: SmartArt 9.13: Arranging Objects 9.14: Grouping Objects 9.15: Putting It Together- Microsoft PowerPoint Basic Skills 9.16: Assignment- Present Sales Information 9.17: Discussion- Infographics and Custom Shapes