1.1: Safety
Nothing should be more important to you than your own safety.
Think about Charlie Morecraft.
He had an industrial accident that left him in the hospital for about five years. Think about that. A shortcut in safety procedures left him in the hospital for five. Long. Years.Charlie had to open a valve on a petroleum line. It was a routine task he’d done “a thousand times” before. It was messy and difficult. The valves leaked and were sticky, but management had planned to replace the valves, and they put a safe procedure in place to ensure the valves could be operated without danger to the operator. Spoiler alert: Charlie didn’t follow the procedure.
Charlie worked on the valve without his safety glasses, and the valve leaked. Charlie didn’t bother trying to control the petroleum leak (“this will just take a minute”), and it got worse. When Charlie completely actuated the valve, it sprayed material into his face and soaked his clothes. Now blinded, he ran past his truck toward a nearby safety shower. Unfortunately, Charlie left the truck running (against procedure), and it ignited the vapors, engulfing Charlie in a ball of fire. When you think about taking shortcuts at work, remember Charlie .
The work we do is dangerous, and while I can never watch Mehdi Sadaghar (an electrical engineer who does unsafe things with electricity) without laughing, we have to remember that electricity can be fatal, and you never want to be involved in an arc flash incident.
Shock Current Path
As we’ve already learned, electricity requires a complete path (circuit) to continuously flow. This is why the shock received from static electricity is only a momentary jolt: the flow of electrons is necessarily brief when static charges are equalized between two objects. Shocks of self-limited duration like this are rarely hazardous.
Without two contact points on the body for current to enter and exit, respectively, there is no hazard of shock. This is why birds can safely rest on high-voltage power lines without getting shocked: they make contact with the circuit at only one point.
In order for electrons to flow through a conductor , there must be a voltage present to motivate them. Voltage, as you should recall, is always relative between two points . There is no such thing as voltage “on” or “at” a single point in the circuit, and so the bird contacting a single point in the above circuit has no voltage applied across its body to establish a current through it. Yes, even though they rest on two feet, both feet are touching the same wire, making them electrically common . Electrically speaking, both of the bird’s feet touch the same point, hence there is no voltage between them to motivate current through the bird’s body.
This might lend one to believe that its impossible to be shocked by electricity by only touching a single wire. Like the birds, if we’re sure to touch only one wire at a time, we’ll be safe, right? Unfortunately, this is not correct. Unlike birds, people are usually standing on the ground when they contact a “live” wire. Many times, one side of a power system will be intentionally connected to earth ground , and so the person touching a single wire is actually making contact between two points in the circuit (the wire and earth ground):
The ground symbol is that set of three horizontal bars of decreasing width located at the lower-left of the circuit shown, and also at the foot of the person being shocked. In real life the power system ground consists of some kind of metallic conductor buried deep in the ground for making maximum contact with the earth. That conductor is electrically connected to an appropriate connection point on the circuit with thick wire. The victim’s ground connection is through their feet, which are touching the earth.
A few questions usually arise at this point in the mind of the student:
- If the presence of a ground point in the circuit provides an easy point of contact for someone to get shocked, why have it in the circuit at all? Wouldn’t a ground-less circuit be safer?
- The person getting shocked probably isn’t bare-footed. If rubber and fabric are insulating materials, then why aren’t their shoes protecting them by preventing a circuit from forming?
- How good of a conductor can dirt be? If you can get shocked by current through the earth, why not use the earth as a conductor in our power circuits?
In answer to the first question, the presence of an intentional “grounding” point in an electric circuit is intended to ensure that one side of it is safe to come in contact with. Note that if our victim in the above diagram were to touch the bottom side of the resistor, nothing would happen even though their feet would still be contacting ground:
Because the bottom side of the circuit is firmly connected to ground through the grounding point on the lower-left of the circuit, the lower conductor of the circuit is made electrically common with earth ground. Since there can be no voltage between electrically common points, there will be no voltage applied across the person contacting the lower wire, and they will not receive a shock. For the same reason, the wire connecting the circuit to the grounding rod/plates is usually left bare (no insulation), so that any metal object it brushes up against will similarly be electrically common with the earth.
Circuit grounding ensures that at least one point in the circuit will be safe to touch. But what about leaving a circuit completely ungrounded? Wouldn’t that make any person touching just a single wire as safe as the bird sitting on just one? Ideally, yes. Practically, no. Observe what happens with no ground at all:
Despite the fact that the person’s feet are still contacting ground, any single point in the circuit should be safe to touch. Since there is no complete path (circuit) formed through the person’s body from the bottom side of the voltage source to the top, there is no way for a current to be established through the person. However, this could all change with an accidental ground, such as a tree branch touching a power line and providing connection to earth ground:
Such an accidental connection between a power system conductor and the earth (ground) is called a ground fault . Ground faults may be caused by many things, including dirt buildup on power line insulators (creating a dirty-water path for current from the conductor to the pole, and to the ground, when it rains), ground water infiltration in buried power line conductors, and birds landing on power lines, bridging the line to the pole with their wings. Given the many causes of ground faults, they tend to be unpredictable. In the case of trees, no one can guarantee which wire their branches might touch. If a tree were to brush up against the top wire in the circuit, it would make the top wire safe to touch and the bottom one dangerous—just the opposite of the previous scenario where the tree contacts the bottom wire:
With a tree branch contacting the top wire, that wire becomes the grounded conductor in the circuit, electrically common with earth ground. Therefore, there is no voltage between that wire and ground, but full (high) voltage between the bottom wire and ground. As mentioned previously, tree branches are only one potential source of ground faults in a power system. Consider an ungrounded power system with no trees in contact, but this time with two people touching single wires:
With each person standing on the ground, contacting different points in the circuit, a path for shock current is made through one person, through the earth, and through the other person. Even though each person thinks they’re safe in only touching a single point in the circuit, their combined actions create a deadly scenario. In effect, one person acts as the ground fault which makes it unsafe for the other person. This is exactly why ungrounded power systems are dangerous: the voltage between any point in the circuit and ground (earth) is unpredictable, because a ground fault could appear at any point in the circuit at any time. The only character guaranteed to be safe in these scenarios is the bird, who has no connection to earth ground at all! By firmly connecting a designated point in the circuit to earth ground (“grounding” the circuit), at least safety can be assured at that one point. This is more assurance of safety than having no ground connection at all.
In answer to the second question, rubber-soled shoes do indeed provide some electrical insulation to help protect someone from conducting shock current through their feet. However, most common shoe designs are not intended to be electrically “safe,” their soles being too thin and not of the right substance. Also, any moisture, dirt, or conductive salts from body sweat on the surface of or permeated through the soles of shoes will compromise what little insulating value the shoe had to begin with. There are shoes specifically made for dangerous electrical work, as well as thick rubber mats made to stand on while working on live circuits, but these special pieces of gear must be in absolutely clean, dry condition in order to be effective. Suffice it to say, normal footwear is not enough to guarantee protection against electric shock from a power system.
Research conducted on contact resistance between parts of the human body and points of contact (such as the ground) shows a wide range of figures (see end of chapter for information on the source of this data):
- Hand or foot contact, insulated with rubber: 20 MΩ typical.
- Foot contact through leather shoe sole (dry): 100 kΩ to 500 kΩ
- Foot contact through leather shoe sole (wet): 5 kΩ to 20 kΩ
As you can see, not only is rubber a far better insulating material than leather, but the presence of water in a porous substance such as leather greatly reduces electrical resistance.
In answer to the third question, dirt is not a very good conductor (at least not when its dry!). It is too poor of a conductor to support continuous current for powering a load. However, as we will see in the next section, it takes very little current to injure or kill a human being, so even the poor conductivity of dirt is enough to provide a path for deadly current when there is sufficient voltage available, as there usually is in power systems.
Some ground surfaces are better insulators than others. Asphalt, for instance, being oil-based, has a much greater resistance than most forms of dirt or rock. Concrete, on the other hand, tends to have fairly low resistance due to its intrinsic water and electrolyte (conductive chemical) content.
Safe Practices
If at all possible, shut off the power to a circuit before performing any work on it. You must secure all sources of harmful energy before a system may be considered safe to work on. In industry, securing a circuit, device, or system in this condition is commonly known as placing it in a Zero Energy State . The focus of this lesson is, of course, electrical safety. However, many of these principles apply to non-electrical systems as well.
Securing something in a Zero Energy State means ridding it of any sort of potential or stored energy, including but not limited to:
- Dangerous voltage
- Spring pressure
- Hydraulic (liquid) pressure
- Pneumatic (air) pressure
- Suspended weight
- Chemical energy (flammable or otherwise reactive substances)
- Nuclear energy (radioactive or fissile substances)
Voltage by its very nature is a manifestation of potential energy. In the first chapter I even used elevated liquid as an analogy for the potential energy of voltage, having the capacity (potential) to produce current (flow), but not necessarily realizing that potential until a suitable path for flow has been established, and resistance to flow is overcome. A pair of wires with high voltage between them do not look or sound dangerous even though they harbor enough potential energy between them to push deadly amounts of current through your body. Even though that voltage isn’t presently doing anything, it has the potential to, and that potential must be neutralized before it is safe to physically contact those wires.
All properly designed circuits have “disconnect” switch mechanisms for securing voltage from a circuit. Sometimes these “disconnects” serve a dual purpose of automatically opening under excessive current conditions, in which case we call them “circuit breakers.” Other times, the disconnecting switches are strictly manually-operated devices with no automatic function. In either case, they are there for your protection and must be used properly. Please note that the disconnect device should be separate from the regular switch used to turn the device on and off. It is a safety switch, to be used only for securing the system in a Zero Energy State:
With the disconnect switch in the “open” position as shown (no continuity), the circuit is broken and no current will exist. There will be zero voltage across the load, and the full voltage of the source will be dropped across the open contacts of the disconnect switch. Note how there is no need for a disconnect switch in the lower conductor of the circuit. Because that side of the circuit is firmly connected to the earth (ground), it is electrically common with the earth and is best left that way. For maximum safety of personnel working on the load of this circuit, a temporary ground connection could be established on the top side of the load, to ensure that no voltage could ever be dropped across the load:
With the temporary ground connection in place, both sides of the load wiring are connected to ground, securing a Zero Energy State at the load.
Since a ground connection made on both sides of the load is electrically equivalent to short-circuiting across the load with a wire, that is another way of accomplishing the same goal of maximum safety:
Either way, both sides of the load will be electrically common to the earth , allowing for no voltage (potential energy) between either side of the load and the ground people stand on. This technique of temporarily grounding conductors in a de-energized power system is very common in maintenance work performed on high voltage power distribution systems.
A further benefit of this precaution is protection against the possibility of the disconnect switch being closed (turned “on” so that circuit continuity is established) while people are still contacting the load. The temporary wire connected across the load would create a short-circuit when the disconnect switch was closed, immediately tripping any overcurrent protection devices (circuit breakers or fuses) in the circuit, which would shut the power off again. Damage may very well be sustained by the disconnect switch if this were to happen, but the workers at the load are kept safe.
It would be good to mention at this point that overcurrent devices are not intended to provide protection against electric shock. Rather, they exist solely to protect conductors from overheating due to excessive currents. The temporary shorting wires just described would indeed cause any overcurrent devices in the circuit to “trip” if the disconnect switch were to be closed, but realize that electric shock protection is not the intended function of those devices. Their primary function would merely be leveraged for the purpose of worker protection with the shorting wire in place.
Since it is obviously important to be able to secure any disconnecting devices in the open (off) position and make sure they stay that way while work is being done on the circuit, there is need for a structured safety system to be put into place. Such a system is commonly used in industry and it is called Lock-out/Tag-out .
A lock-out/tag-out procedure works like this: all individuals working on a secured circuit have their own personal padlock or combination lock which they set on the control lever of a disconnect device prior to working on the system. Additionally, they must fill out and sign a tag which they hang from their lock describing the nature and duration of the work they intend to perform on the system. If there are multiple sources of energy to be “locked out” (multiple disconnects, both electrical and mechanical energy sources to be secured, etc.), the worker must use as many of his or her locks as necessary to secure power from the system before work begins. This way, the system is maintained in a Zero Energy State until every last lock is removed from all the disconnect and shutoff devices, and that means every last worker gives consent by removing their own personal locks. If the decision is made to re-energize the system and one person’s lock(s) still remain in place after everyone present removes theirs, the tag(s) will show who that person is and what it is they’re doing.
Even with a good lock-out/tag-out safety program in place, there is still need for diligence and common-sense precaution. This is especially true in industrial settings where a multitude of people may be working on a device or system at once. Some of those people might not know about proper lock-out/tag-out procedure, or might know about it but are too complacent to follow it. Don’t assume that everyone has followed the safety rules!
After an electrical system has been locked out and tagged with your own personal lock, you must then double-check to see if the voltage really has been secured in a zero state. One way to check is to see if the machine (or whatever it is that’s being worked on) will start up if the Start switch or button is actuated. If it starts, then you know you haven’t successfully secured the electrical power from it.
Additionally, you should always check for the presence of dangerous voltage with a measuring device before actually touching any conductors in the circuit. To be safest, you should follow this procedure of checking, using, and then checking your meter:
- Check to see that your meter indicates properly on a known source of voltage.
- Use your meter to test the locked-out circuit for any dangerous voltage.
- Check your meter once more on a known source of voltage to see that it still indicates as it should.
While this may seem excessive or even paranoid, it is a proven technique for preventing electrical shock. I once had a meter fail to indicate voltage when it should have while checking a circuit to see if it was “dead.” Had I not used other means to check for the presence of voltage, I might not be alive today to write this. There’s always the chance that your voltage meter will be defective just when you need it to check for a dangerous condition. Following these steps will help ensure that you’re never misled into a deadly situation by a broken meter.
Finally, the electrical worker will arrive at a point in the safety check procedure where it is deemed safe to actually touch the conductor(s). Bear in mind that after all of the precautionary steps have taken, it is still possible (although very unlikely) that a dangerous voltage may be present. One final precautionary measure to take at this point is to make momentary contact with the conductor(s) with the back of the hand before grasping it or a metal tool in contact with it. Why? If, for some reason there is still voltage present between that conductor and earth ground, finger motion from the shock reaction (clenching into a fist) will break contact with the conductor. Please note that this is absolutely the last step that any electrical worker should ever take before beginning work on a power system, and should never be used as an alternative method of checking for dangerous voltage. If you ever have reason to doubt the trustworthiness of your meter, use another meter to obtain a “second opinion.”


4.3: Logarithmic Functions
In this section, you will:
- Convert from logarithmic to exponential form.
- Convert from exponential to logarithmic form.
- Evaluate logarithms.
- Use common logarithms.
- Use natural logarithms.
Figure \( \PageIndex{ 1 } \) Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)
In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings, like those shown in Figure \( \PageIndex{ 1 } \). Even though both caused substantial damage, the earthquake in 2011 was \( 100 \) times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale . The Haitian earthquake registered a \( 7.0 \) on the Richter Scale whereas the Japanese earthquake registered a \( 9.0 \).
The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude \( 8 \) is not twice as great as an earthquake of magnitude \( 4 \). It is \(10^{8−4} = 10^4 = 10,000\) times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.
Converting from Logarithmic to Exponential Form
In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were \( 500 \) times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is \(10^x = 500\), where \(x\) represents the difference in magnitudes on the Richter Scale. How would we solve for \(x\)?
We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve \(10^x =500\). We know that \(10^2 = 100\) and \(10^3 = 1000\), so it is clear that \(x\) must be some value between \( 2 \) and \( 3 \), since \(y = 10^x\) is increasing. We can examine a graph, as in Figure \( \PageIndex{ 2 } \), to better estimate the solution.
Figure \( \PageIndex{ 2 } \)
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure \( \PageIndex{ 2 } \) passes the Horizontal Line Test. The exponential function \( y = b^x \) is one-to-one, so its inverse, \( x = b^y \) is also a function. As is the case with all inverse functions, we simply interchange \(x\) and \(y\) and solve for \(y\) to find the inverse function. To represent \(y\) as a function of \(x\), we use a logarithmic function of the form \( y = \log_b(x) \). The base \(b\) logarithm of a number is the exponent by which we must raise \(b\) to get that number.
We read a logarithmic expression as, "The logarithm with base \(b\) of \(x\) is equal to \(y\)," or, simplified, "log base \(b\) of \(x\) is \(y\)." We can also say, " \(b\) raised to the power of \(y\) is \(x\)," because logs are exponents. For example, the base \( 2 \) logarithm of \( 32 \) is \( 5 \), because \( 5 \) is the exponent we must apply to \( 2 \) to get \( 32 \). Since \(2^5 = 32\), we can write \(\log_2(32) = 5\). We read this as "log base \( 2 \) of \( 32 \) is \( 5 \)."
We can express the relationship between logarithmic form and its corresponding exponential form as follows:\[\log_b( x ) = y \iff b^y = x, b > 0, b \neq 1. \nonumber \]Note that the base \(b\) is always positive.
Because the logarithm is a function, it is most correctly written as \(\log_b(x)\), using parentheses to denote function evaluation, just as we would with \(f(x)\). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as \(\log_b x\). Note that many calculators require parentheses around the \(x\).
We can illustrate the notation of logarithms as follows:
Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means \( y = \log_b(x) \) and \( y = b^x \) are inverse functions.
A logarithm base \(b\) of a positive number \(x\) satisfies the following definition.
For \(x > 0, b > 0, b \neq 1\),\[y =\log_b( x ) \text{ is equivalent to } b^y = x, \nonumber \]where,
- we read \( \log_b(x) \) as, "the logarithm with base \(b\) of \(x\)," or the "log base \(b\) of \(x\)."
- the logarithm \(y\) is the exponent to which \(b\) must be raised to get \(x\).
Also, since the logarithmic and exponential functions switch the \(x\) and \(y\) values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,
- the domain of the logarithm function with base \(b\) is \((0, \infty )\).
- the range of the logarithm function with base \(b\) is \((-\infty , \infty )\).
Can we take the logarithm of a negative number?
No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.
Write the following logarithmic equations in exponential form.
- \(\log_6 ( \sqrt{ 6 } ) = \frac{1}{2} \)
- \(\log_3 ( 9 ) = 2\)
- Solution
-
First, identify the values of \(b\), \(y\), and \(x\). Then, write the equation in the form \( b^y = x \).
- Here, \(b=6\), \(y = \frac{1}{2}\), and \(x=\sqrt{6}\). Therefore, the equation \(\log_6 ( \sqrt{ 6 } ) = \frac{1}{2}\) is equivalent to\[ 6^{1/2} = \sqrt{6}. \nonumber \]
- Here, \(b=3\), \(y=2\), and \(x=9\). Therefore, the equation \(\log_3(9)=2\) is equivalent to \(3^2 =9\).
Write the following logarithmic equations in exponential form.
- \(\log_{10}(1,000,000)=6\)
- \(\log_5(25)=2\)
Converting from Exponential to Logarithmic Form
To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base \(b\), exponent \(x\), and output \(y\). Then we write \( x = \log_b(y) \).
Write the following exponential equations in logarithmic form.
- \( 2^3 = 8 \)
- \( 5^2 = 25 \)
- \( 10^{-4} = \frac{1}{10,000} \)
- Solution
-
First, identify the values of \( b \), \( y \), and \( x \). Then, write the equation in the form \( x = \log_b(y) \).
- Here, \(b=2\), \(x=3\), and \(y=8\). Therefore, the equation \( 2^3 = 8 \) is equivalent to \( \log_2(8) = 3 \).
- Here, \(b=5\), \(x=2\), and \(y=25\). Therefore, the equation \( 5^2 = 25 \) is equivalent to \( \log_5(25) = 2 \).
- Here, \(b=10\), \(x=−4\), and \( y = \frac{1}{10,000} \). Therefore, the equation \( 10^{-4} = \frac{1}{10,000} \) is equivalent to \( \log_{10} \left( \frac{1}{10,000} \right) = -4 \).
Write the following exponential equations in logarithmic form.
- \( 3^2 = 9 \)
- \( 5^3 = 125 \)
- \( 2^{-1} = \frac{1}{2} \)
Evaluating Logarithms
Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider \( \log_2 8 \). We ask, "To what exponent must \(2\) be raised in order to get \( 8 \)?" Because we already know \( 2^3 = 8 \), it follows that \( \log_2 8 = 3 \).
Now consider simplifying \( \log_7 49 \) and \( \log_3 27 \) mentally.
- We ask, "To what exponent must \( 7 \) be raised in order to get \( 49 \)?" We know \( 7^2 = 49 \). Therefore, \( log_7 49 = 2\)
- We ask, "To what exponent must \( 3 \) be raised in order to get \( 27 \)?" We know \( 3^3 = 27 \). Therefore, \( log_3 27 = 3\)
Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate \( \log_{2/3} \frac{4}{9} \) mentally.
- We ask, "To what exponent must \( \frac{2}{3} \) be raised in order to get \( \frac{4}{9} \)?" We know \( 2^2 = 4 \) and \( 3^2 = 9 \), so \( \left( \frac{2}{3} \right)^2 = \frac{4}{9} \). Therefore, \( \log_{2/3} \left( \frac{4}{9} \right) = 2\).
Solve \( y = \log_4 (64) \) without using a calculator.
- Solution
-
First we rewrite the logarithm in exponential form: \(4^y =64\). Next, we ask, "To what exponent must \( 4 \) be raised in order to get \( 64 \)?"
We know\[4^3 = 64. \nonumber \]Therefore,\[ \log_4 ( 64 ) = 3. \nonumber \]
Solve \(y = \log_{121} (11)\) without using a calculator.
Evaluate \(y = \log_3 \left( \frac{1}{27} \right)\) without using a calculator.
- Solution
-
First we rewrite the logarithm in exponential form: \(3^y = \frac{1}{27} \). Next, we ask, "To what exponent must \( 3 \) be raised in order to get \( \frac{1}{27} \)? "
We know \( 3^3 = 27 \), but what must we do to get the reciprocal, \( \frac{1}{27} \)? Recall from working with exponents that \(b^{−a} = \frac{1}{b^a} \). We use this information to write\[3^{−3} = \dfrac{1}{3^3} = \dfrac{1}{27}. \nonumber \]Therefore, \( \log_3 \left( \frac{1}{27} \right) = −3\).
Evaluate \(y = \log_2 \left( \frac{1}{32} \right)\) without using a calculator.
Using Common Logarithms
Sometimes we may see a logarithm written without a base. In this case, we assume that the base is \( 10 \). In other words, the expression \( \log(x)\) means \( \log_{10}(x) \). We call a base-10 logarithm a common logarithm . Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars, decibel levels of sounds, and the pH of acids and bases also use common logarithms.
A common logarithm is a logarithm with base \(10\). We write \( \log_{10}(x) \) simply as \( \log(x) \). The common logarithm of a positive number \(x\) satisfies the following definition.
For \(x > 0\),\[y = \log(x) \text{ is equivalent to } 10^y = x. \nonumber \]We read \( \log(x) \) as, "the logarithm with base \(10\) of \(x\) " or "log base \( 10 \) of \(x\)."
The logarithm \(y\) is the exponent to which \(10\) must be raised to get \(x\).
Evaluate \(y= \log(1000)\) without using a calculator.
- Solution
-
First we rewrite the logarithm in exponential form: \( 10^y = 1000 \). Next, we ask, "To what exponent must \(10\) be raised in order to get \( 1000 \)?" We know\[10^3 = 1000. \nonumber \]Therefore, \( \log(1000)=3\).
Evaluate \(y= \log(1,000,000)\).
Evaluate \(y= \log(321)\) to four decimal places using a calculator.
- Solution
-
- Press [LOG] .
- Enter 321, followed by ) .
- Press [ENTER] .
Rounding to four decimal places, \( \log(321) \approx 2.5065\).
We can perform a quick "logic check" on our answer to Example \( \PageIndex{ 6 } \). Note that \( 10^2 = 100 \) and that \( 10^3 = 1000 \). Since \( 321 \) is between \( 100 \) and \( 1000 \), we know that \( \log(321)\) must be between \( \log(100) = 2\) and \(\log(1000) = 3\).
Evaluate \(y= \log(123)\) to four decimal places using a calculator.
The amount of energy released from one earthquake was \( 500 \) times greater than the amount of energy released from another. The equation \( 10^x = 500 \) represents this situation, where \(x\) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?
- Solution
-
We begin by rewriting the exponential equation in logarithmic form.\[ \begin{array}{rrclcl} & 10^x & = & 500 & & \\[6pt] \implies & \log(500) & = & x & \quad & \left( \text{definition of the common logarithm} \right) \\[6pt] \implies & x & \approx & 2.699 & \quad & \left( \text{using a calculator} \right) \\[6pt] \end{array} \nonumber \]The difference in Richter Scale magnitudes was about \(2.699\).
The amount of energy released from one earthquake was 8,500 times greater than the amount of energy released from another. The equation \(10^x = 8500\) represents this situation, where \(x\) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?
Using Natural Logarithms
The most frequently used base for logarithms is \(e\). Base \(e\) logarithms are important in Calculus and some scientific applications; they are called natural logarithms . The base \(e\) logarithm, \( \log_e(x) \), has its own notation, \( \ln(x) \).
Most values of \( \ln(x) \) can be found only using a calculator. The major exception is that, because the logarithm of \( 1 \) is always \( 0 \) in any base, \( \ln 1 = 0\). For other natural logarithms, we can use the \(\mathrm{ln}\) key that can be found on most scientific calculators. We can also find the natural logarithm of any power of \(e\) using the inverse property of logarithms.
A natural logarithm is a logarithm with base \(e\). We write \( \log_e(x) \) simply as \( \ln(x) \). The natural logarithm of a positive number \(x\) satisfies the following definition.
For \(x > 0\),\[ y = \ln(x) \text{ is equivalent to } e^y = x. \nonumber \]We read \( \ln(x) \) as, "the logarithm with base \(e\) of \(x\)" or "the natural logarithm of \(x\)."
The logarithm \(y\) is the exponent to which \(e\) must be raised to get \(x\).
Since the functions \( y = e^x \) and \( y = \ln( x )\) are inverse functions, \( \ln(e^x) = x\) for all \(x\) and \(e^{\ln(x)} = x\) for \(x > 0\).
Evaluate \(y = \ln(500)\) to four decimal places using a calculator.
- Solution
-
- Press [LN] .
- Enter \(500\), followed by ) .
- Press [ENTER] .
Rounding to four decimal places, \( \ln(500) \approx 6.2146\)
Evaluate \(\ln(−500)\).
Access this online resource for additional instruction and practice with logarithms.


15.1: Discovering Antimicrobial Drugs
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- Ying Liu
- City College of San Francisco
Learning Objectives
- Compare and contrast natural, semisynthetic, and synthetic antimicrobial drugs
- Describe the chemotherapeutic approaches of ancient societies
- Describe the historically important individuals and events that led to the development of antimicrobial drugs
Most people associate the term chemotherapy with treatments for cancer. However, chemotherapy is actually a broader term that refers to any use of chemicals or drugs to treat disease. Chemotherapy may involve drugs that target cancerous cells or tissues, or it may involve antimicrobial drugs that target infectious microorganisms. Antimicrobial drugs typically work by destroying or interfering with microbial structures and enzymes, either killing microbial cells or inhibiting of their growth. But before we examine how these drugs work, we will briefly explore the history of humans’ use of antimicrobials for the purpose of chemotherapy.
Use of Antimicrobials in Ancient Societies
Although the discovery of antimicrobials and their subsequent widespread use is commonly associated with modern medicine, there is evidence that humans have been exposed to antimicrobial compounds for millennia. Chemical analyses of the skeletal remains of people from Nubia 1 (now found in present-day Sudan) dating from between 350 and 550 AD have shown residue of the antimicrobial agent tetracycline in high enough quantities to suggest the purposeful fermentation of tetracycline-producing Streptomyces during the beer-making process. The resulting beer, which was thick and gruel-like, was used to treat a variety of ailments in both adults and children, including gum disease and wounds. The antimicrobial properties of certain plants may also have been recognized by various cultures around the world, including Indian and Chinese herbalists (Figure \(\PageIndex{1}\)) who have long used plants for a wide variety of medical purposes. Healers of many cultures understood the antimicrobial properties of fungi and their use of moldy bread or other mold-containing products to treat wounds has been well documented for centuries. 2 Today, while about 80% of the world’s population still relies on plant-derived medicines, 3 scientists are now discovering the active compounds conferring the medicinal benefits contained in many of these traditionally used plants.
The First Antimicrobial Drugs
Societies relied on traditional medicine for thousands of years; however, the first half of the 20th century brought an era of strategic drug discovery. In the early 1900s, the German physician and scientist Paul Ehrlich (1854–1915) set out to discover or synthesize chemical compounds capable of killing infectious microbes without harming the patient. In 1909, after screening more than 600 arsenic-containing compounds, Ehrlich’s assistant Sahachiro Hata (1873–1938) found one such “magic bullet.” Compound 606 targeted the bacterium Treponema pallidum , the causative agent of syphilis. Compound 606 was found to successfully cure syphilis in rabbits and soon after was marketed under the name Salvarsan as a remedy for the disease in humans (Figure \(\PageIndex{2}\)). Ehrlich’s innovative approach of systematically screening a wide variety of compounds remains a common strategy for the discovery of new antimicrobial agents even today.
A few decades later, German scientists Josef Klarer, Fritz Mietzsch, and Gerhard Domagk discovered the antibacterial activity of a synthetic dye, prontosil, that could treat streptococcal and staphylococcal infections in mice. Domagk’s own daughter was one of the first human recipients of the drug, which completely cured her of a severe streptococcal infection that had resulted from a poke with an embroidery needle. Gerhard Domagk (1895–1964) was awarded the Nobel Prize in Medicine in 1939 for his work with prontosil and sulfanilamide, the active breakdown product of prontosil in the body. Sulfanilamide, the first synthetic antimicrobial created, served as the foundation for the chemical development of a family of sulfa drugs. A synthetic antimicrobial is a drug that is developed from a chemical not found in nature. The success of the sulfa drugs led to the discovery and production of additional important classes of synthetic antimicrobials, including the quinolines and oxazolidinones.
A few years before the discovery of prontosil, scientist Alexander Fleming (1881–1955) made his own accidental discovery that turned out to be monumental. In 1928, Fleming returned from holiday and examined some old plates of staphylococci in his research laboratory at St. Mary’s Hospital in London. He observed that contaminating mold growth (subsequently identified as a strain of Penicillium notatum ) inhibited staphylococcal growth on one plate. Fleming, therefore, is credited with the discovery of penicillin , the first natural antibiotic , (Figure \(\PageIndex{3}\)). Further experimentation showed that penicillin from the mold was antibacterial against streptococci, meningococci, and Corynebacterium diphtheriae , the causative agent of diphtheria.
Fleming and his colleagues were credited with discovering and identifying penicillin, but its isolation and mass production were accomplished by a team of researchers at Oxford University under the direction of Howard Florey(1898–1968) and Ernst Chain (1906–1979) (Figure \(\PageIndex{3}\)). In 1940, the research team purified penicillin and reported its success as an antimicrobial agent against streptococcal infections in mice. Their subsequent work with human subjects also showed penicillin to be very effective. Because of their important work, Fleming, Florey, and Chain were awarded the Nobel Prize in Physiology and Medicine in 1945.
In the early 1940s, scientist Dorothy Hodgkin (1910–1994), who studied crystallography at Oxford University, used X-rays to analyze the structure of a variety of natural products. In 1946, she determined the structure of penicillin, for which she was awarded the Nobel Prize in Chemistry in 1964. Once the structure was understood, scientists could modify it to produce a variety of semisynthetic penicillins . A semisynthetic antimicrobial is a chemically modified derivative of a natural antibiotic. The chemical modifications are generally designed to increase the range of bacteria targeted, increase stability, decrease toxicity, or confer other properties beneficial for treating infections.
Penicillin is only one example of a natural antibiotic. Also in the 1940s, Selman Waksman (1888–1973) (Figure \(\PageIndex{4}\)), a prominent soil microbiologist at Rutgers University, led a research team that discovered several antimicrobials, including actinomycin, streptomycin, and neomycin. The discoveries of these antimicrobials stemmed from Waksman’s study of fungi and the Actinobacteria, including soil bacteria in the genus Streptomyces , known for their natural production of a wide variety of antimicrobials. His work earned him the Nobel Prize in Physiology and Medicine in 1952. The actinomycetes are the source of more than half of all natural antibiotics 4 and continue to serve as an excellent reservoir for the discovery of novel antimicrobial agents. Some researchers argue that we have not yet come close to tapping the full antimicrobial potential of this group. 5
Query \(\PageIndex{1}\)
Key Concepts and Summary
- Antimicrobial drugs produced by purposeful fermentation and/or contained in plants have been used as traditional medicines in many cultures for millennia.
- The purposeful and systematic search for a chemical “magic bullet” that specifically target infectious microbes was initiated by Paul Ehrlich in the early 20th century.
- The discovery of the natural antibiotic , penicillin, by Alexander Fleming in 1928 started the modern age of antimicrobial discovery and research.
- Sulfanilamide, the first synthetic antimicrobial , was discovered by Gerhard Domagk and colleagues and is a breakdown product of the synthetic dye, prontosil.
Footnotes
- 1 M.L. Nelson et al. “Brief Communication: Mass Spectroscopic Characterization of Tetracycline in the Skeletal Remains of an Ancient Population from Sudanese Nubia 350–550 CE.” American Journal of Physical Anthropology 143 no. 1 (2010):151–154.
- 2 M. Wainwright. “Moulds in Ancient and More Recent Medicine.” Mycologist 3 no. 1 (1989):21–23.
- 3 S. Verma, S.P. Singh. “Current and Future Status of Herbal Medicines.” Veterinary World 1 no. 11 (2008):347–350.
- 4 J. Berdy. “Bioactive Microbial Metabolites.” The Journal of Antibiotics 58 no. 1 (2005):1–26.
- 5 M. Baltz. “Antimicrobials from Actinomycetes: Back to the Future.” Microbe 2 no. 3 (2007):125–131.


13.10: The Need for Ethics in the Workplace
- Explain the need for ethics in the workplace.
- Describe the costs to a company’s health of unethical behavior.
The Association of Certified Fraud Examiners estimates that fraud costs US companies about 5 percent of their revenues. In 2016, this meant roughly $900 billion was lost to fraud. [1] This includes the identity and credit card thefts that concern everyone and the massive corporate frauds that bring down companies. This estimate includes only reported fraud. We don’t know the cost of unreported fraud and unethical behavior, but we can assume it’s significant. This is a continuing battle that managers cannot delegate to HR or the legal department.
Just how much can unethical behavior cost a business? In some cases, billions. For example, in 2004 and 2005, General Motors engineers “misdiagnosed [an ignition switch problem] as a customer satisfaction issue and not a safety issue,” according to GM CEO Mary Barra, and they hid information from regulatory agencies and the public. Their unethical decisions led to 124 deaths. “A series of mistakes were made over a period of time . . . . This had tragic consequences,” said Barra. [2] GM has paid settlements, penalties, and legal costs close to $2.5 billion, and it could face additional lawsuits. [3]
The improper actions of board members, corporate-suite (C-suite) executives, mid-level managers, and nonmanagerial employees can create negative consequences ranging in intensity from petty annoyances to global recessions. Those affected by the consequences can include stakeholders, customers, and the general public. When the stakes are so high, it’s imperative that the leaders of an organization understand ethical behavior, be committed to making ethical decisions, and encourage and empower all stakeholders to take ethical actions.
Ethics is defined as principles of conduct or moral behavior. Business ethics is ethics as it relates to the conduct or behavior of the members of a business organization. The need for ethics in business is twofold. First, it prevents the external constituents of an organization (i.e., the public, other organizations, and the environment) from being harmed. Second, it benefits the organization internally by helping to ensure its success.
Good business ethics involves, but is not limited to, adhering to laws, regulations, and standards related to fair employment, product safety and quality, truthful advertisement, and environmental responsibility. Companies do suffer for bad ethical behavior and gain from good ethical behavior.
The reputation of a company reflects the ethical or unethical decisions of managers, and consumers often purchase the products or services of companies with a good reputation. If a business has a good reputation, consumers may prefer that company over similar businesses that offer the same products or services, even at higher prices. The companies Tom’s Shoes and Patagonia both advertise their ethical practices. And wise executives go to great lengths to repair their company’s reputation. For example, British Petroleum (BP) has extended cleanup and conservation efforts following oil spills, and Toyota has issued several major recalls to address potential problems associated with a faulty accelerator pedal.
A broad survey of consumers shows that many factors affect reputation but that ethics is the primary concern. [4] Customers will switch to other brands when a company appears in headlines reporting ethics violations. BP gas stations suffered a boycott for a year after the Gulf of Mexico spill. The volume of sales did not recover to pre-spill levels in that time. [5]
A good reputation creates a buffer, a “halo” effect, where customers, suppliers, and regulators are slower to judge a company. Employees’ morale is better in ethical companies, and new employees are easier to hire. An ethics program also gives stock an edge. Ethisphere, an ethics consulting firm, found that share price of the publicly traded companies recognized as the 2016 World’s Most Ethical Companies consistently outperform other major indices, including performing 3.3 percent higher than the S&P 500 last year. Exhibiting good business ethics is essential to the survival and success of a company. [6]
Costs of Unethical Behavior
The costs of unethical behavior are varied and numerous. In addition to a poor reputation, these costs can include reduced customer loyalty and subsequent revenue loss, heavy fines, probation, criminal or civil prosecution, and the loss of needed employee talent. For example, Wells Fargo has paid penalties totaling $170 million and faces civil lawsuits for defrauding customers.
To reach high sales goals, Wells Fargo employees opened unauthorized customer accounts, forged client signatures, charged unnecessary fees on unwanted accounts, and misstated customer phone numbers to hinder customer-satisfaction surveys. Employees resorted to issuing illegal credit cards and lines of credit to avoid pressure from managers. Some employees lost their jobs after reporting the illegal tactics. A civil lawsuit filed in California in 2015 called for reimbursement of the fraudulent fees charged to customers, a fine of $2,500 for each violation of California’s Unfair Competition Law, and an injunction against unethical practices. [7]
As a result of the unethical behavior of executives and employees, there has been a strong drive to improve business ethics. Congress passed the Sarbanes-Oxley Act in 2002 to impose sanctions on executives who commit unethical acts in financial reporting and to protect employees who report fraud. In addition, business schools have increased their focus on informing students of ethical/unethical acts, and more individuals and organizations have filed lawsuits against companies involved in fraud, regardless of whether senior executives are aware of the unethical acts committed. [8]
- The Association of Certified Fraud Examiners, ACFE Report to the Nations on Occupational Fraud and Abuse, 2016, https://s3-us-west-2.amazonaws.com/acfepublic/2016-report-to-the-nations.pdf . ↵
- Margaret Cronin Fisk and Laurel Brubaker Calkins, “GM CEO Barra Tells Jury Company Flubbed Fatal Ignition Flaw,” August 15, 2016, https://www.bloomberg.com/news/articles/2016-08-15/gm-ceo-barra-tells-texas-jury-company-flubbed-fatal-switch-flaw . ↵
- Melissa Burden, “Lost Appeal Could Cost GM Billions,” April 24, 2017, http://www.detroitnews.com/story/business/autos/general-motors/2017/04/24/gm-ignition-switch/100837116/ . ↵
- Percy Marquina Feldman, Rolando Arellano Bahamonde, and Isabel Velasquez Bellido, “A New Approach for Measuring Corporate Reputation,” Revista de Administração de Empresas 54, no. 1 (2014): 53–66, http://www.scielo.br/pdf/rae/v54n1/a06v54n1.pdf . ↵
- Zhongmin Wang, Alvin Lee, and Michael Polonsky, “Egregiousness and Boycott Intensity: Evidence from the BP Deepwater Horizon Oil Spill” (discussion paper, Resources for the Future, February 2015), http://www.rff.org/files/sharepoint/WorkImages/Download/RFF-DP-15-06.pdf . ↵
- The Ethics Centre, “Investing in Ethics Pays Off Financially—Three Reasons to Care about Culture,” May 10, 2016. http://www.ethics.org.au/on-ethics/blog/may-2016/investing-in-ethics-pays-off-financially-%E2%80%93-three-r . ↵
- Zacks Equity Research, “Wells Fargo Faces LA Lawsuit for Unethical Conduct,” May 5, 2015. https://www.zacks.com/stock/news/173783/wells-fargo-faces-la-lawsuit-for-unethical-conduct . ↵
- Eugene Brigham and Joel Houston, Fundamentals of Financial Management, 12th ed. (Mason, OH: South-Western, Cengage Learning, 2009). ↵
Contributors and Attributions
- The Need for Ethics in the Workplace. Authored by : Talia Lambarki and Lumen Learning. License : CC BY: Attribution
- Image: Business Ethics. Authored by : Nick Youngson. Located at : http://www.picserver.org/b/business-ethics.html . License : CC BY-SA: Attribution-ShareAlike


1.4.0: The Dissolution Process
Learning Objectives
- Define solution and differentiate between the solute and the solvent in a solution.
- Predict the solubility of two substances based on their polarity.
- Explain“like dissolves like”.
Solutions are defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent . The other components of the solution present in relatively lesser concentrations are called solutes . Sugar is a covalent solid composed of sucrose molecules, \(\ce{C12H22O11}\). When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water:
\[\ce{C12H22O11 (s) ⟶ C12H22O11 (aq) } \label{Eq1}\]
The subscript “aq” in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to “settle out” over time.
Potassium dichromate, \(\ce{K_2Cr_2O_7}\), is an ionic compound composed of colorless potassium ions, \(\mathrm{K^+}\), and orange dichromate ions, \(\ce{Cr_2O_7^{2−}}\). When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure \(\PageIndex{1}\)), as indicated in this equation:
\[\ce{K2Cr2O7(s) ⟶ 2K^{+} (aq) + Cr2O7^{2-} (aq)} \label{Eq2}\]
As with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules.
Figure \(\PageIndex{1}\): When potassium dichromate (\(\ce{K_2Cr_2O_7}\)) is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott).
Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table \(\PageIndex{1}\) gives examples of several different solutions and the phases of the solutes and solvents.
| Solution | Solute | Solvent |
|---|---|---|
| air | O 2 ( g ) | N 2 ( g ) |
| soft drinks | CO 2 ( g ) | H 2 O( l ) |
| hydrogen in palladium | H 2 ( g ) | Pd( s ) |
| rubbing alcohol | H 2 O( l ) | C 3 H 8 O( l ) (2-propanol) |
| saltwater | NaCl( s ) | H 2 O( l ) |
| brass | Zn( s ) | Cu( s ) |
Solutions exhibit these defining traits:
- They are homogeneous; that is, after a solution is mixed, it has the same composition at all points throughout (its composition is uniform).
- The physical state of a solution—solid, liquid, or gas—is typically the same as that of the solvent, as demonstrated by the examples in Table \(\PageIndex{1}\).
- The components of a solution are dispersed on a molecular scale; that is, they consist of a mixture of separated molecules, atoms, and/or ions.
- The dissolved solute in a solution will not settle out or separate from the solvent.
- The composition of a solution, or the concentrations of its components, can be varied continuously, within limits.
The Formation of Solutions
The formation of a solution is an example of a spontaneous process , a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes we stir a mixture to speed up the dissolution process, but this is not necessary; a homogeneous solution would form if we waited long enough. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter’s discussion, it will suffice to consider two criteria that favor , but do not guarantee, the spontaneous formation of a solution:
- a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry)
- an increase in the disorder in the system (which indicates an increase in the entropy of the system, as you will learn about in the later chapter on thermodynamics)
In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in disorder always results when a solution forms.
When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution . A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions.
When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure \(\PageIndex{2}\)). The formation of this solution clearly involves an increase in disorder, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing.
Figure \(\PageIndex{2}\): Samples of helium and argon spontaneously mix to give a solution in which the disorder of the atoms of the two gases is increased.
Ideal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol (CH 3 OH) and ethanol (C 2 H 5 OH) form ideal solutions, as do mixtures of the hydrocarbons pentane, \(\ce{C5H12}\), and hexane, \(\ce{C6H14}\). Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure \(\PageIndex{2}\) will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how diffusion alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution.
Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solvent-solvent, and solute-solvent. As illustrated in Figure \(\PageIndex{3}\), the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation ). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation.
For example, cooking oils and water will not mix to any appreciable extent to yield solutions (Figure \(\PageIndex{4}\)). Water is a polar substance and will not mix with the nonpolar hydrocarbon molecules of cooking oils are not capable of hydrogen bonding, instead being held together by dispersion forces. Forming an oil-water solution would require overcoming the very strong hydrogen bonding in water, as well as the significantly strong dispersion forces between the relatively large oil molecules. And, since the polar water molecules and nonpolar oil molecules would not experience very strong intermolecular attraction, very little energy would be released by solvation.
On the other hand, a mixture of ethanol and water will mix in any proportions to yield a solution because they are both polar substances.
"Like Dissolves Like". In other words, ionic and polar covalent solutes dissolve in ionic and polar covalent solvents, and nonpolar covalent solutes dissolve in nonpolar covalent solvents.
As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate (NH 4 NO 3 ) is one such example and is used to make instant cold packs for treating injuries like the one pictured in Figure \(\PageIndex{5}\). A thin-walled plastic bag of water is sealed inside a larger bag with solid NH 4 NO 3 . When the smaller bag is broken, a solution of NH 4 NO 3 forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution.
Video \(\PageIndex{1}\) : Watch this brief video illustrating endothermic and exothermic dissolution processes.
Summary
A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy.
Footnotes
- If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution.
Glossary
- alloy
- solid mixture of a metallic element and one or more additional elements
- ideal solution
- solution that forms with no accompanying energy change
- solvation
- exothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established
- spontaneous process
- physical or chemical change that occurs without the addition of energy from an external source
Contributors and Attributions
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).


7.E: The Properties of Real Numbers (Exercises)
7.1 - Rational and Irrational Numbers
In the following exercises, write as the ratio of two integers.
- 6
- −5
- 2.9
- 1.8
In the following exercises, determine which of the numbers is rational.
- 0.42, 0.\(\overline{3}\), 2.56813…
- 0.75319…, 0.\(\overline{16}\), 1.95
In the following exercises, identify whether each given number is rational or irrational.
- (a) 49 (b) 55
- (a) 72 (b) 64
In the following exercises, list the (a) whole numbers, (b) integers, (c) rational numbers, (d) irrational numbers, (e) real numbers for each set of numbers.
- −9, 0, 0.361...., \(\dfrac{8}{9}, \sqrt{16}\), 9
- −5, \(− 2 \dfrac{1}{4}, − \sqrt{4}, 0.\overline{25}, \dfrac{13}{5}\), 4
7.2 - Commutative and Associative Properties
In the following exercises, use the commutative property to rewrite the given expression.
- 6 + 4 = ____
- −14 • 5 = ____
- 3n = ____
- a + 8 = ____
In the following exercises, use the associative property to rewrite the given expression.
- (13 • 5) • 2 = _____
- (22 + 7) + 3 = _____
- (4 + 9x) + x = _____
- \(\dfrac{1}{2}\)(22y) = _____
In the following exercises, evaluate each expression for the given value.
-
If y = \(\dfrac{11}{12}\), evaluate:
- y + 0.7 + (− y)
- y + (− y) + 0.7
-
If z = \(− \dfrac{5}{3}\), evaluate:
- z + 5.39 + (− z)
- z + (− z) + 5.39
-
If k = 65, evaluate:
- \(\dfrac{4}{9} \left(\dfrac{9}{4} k\right)\)
- \(\left(\dfrac{4}{9} \cdot \dfrac{9}{4}\right) k\)
-
If m = −13, evaluate:
- \(− \dfrac{2}{5} \left(\dfrac{5}{2} m\right)\)
- \(\left(− \dfrac{2}{5} \cdot \dfrac{5}{2}\right) m\)
In the following exercises, simplify using the commutative and associative properties.
- 6y + 37 + (−6y)
- \(\dfrac{1}{4} + \dfrac{11}{15} + \left(− \dfrac{1}{4}\right)\)
- \(\dfrac{14}{11} \cdot \dfrac{35}{9} \cdot \dfrac{14}{11}\)
- −18 • 15 • \(\dfrac{2}{9}\)
- \(\left(\dfrac{7}{12} + \dfrac{4}{5}\right) + \dfrac{1}{5}\)
- (3.98d + 0.75d) + 1.25d
- −12(4m)
- 30\(\left(\dfrac{5}{6} q\right)\)
- 11x + 8y + 16x + 15y
- 52m + (−20n) + (−18m) + (−5n)
7.3 - Distributive Property
In the following exercises, simplify using the distributive property.
- 7(x + 9)
- 9(u − 4)
- −3(6m − 1)
- −8(−7a − 12)
- \(\dfrac{1}{3}\)(15n − 6)
- (y + 10) • p
- (a − 4) − (6a + 9)
- 4(x + 3) − 8(x − 7)
In the following exercises, evaluate using the distributive property.
-
If u = 2, evaluate
- 3(8u + 9) and
- 3 • 8u + 3 • 9 to show that 3(8u + 9) = 3 • 8u + 3 • 9
-
If n = 7 8 , evaluate
- 8\(\left(n + \dfrac{1}{4}\right)\) and
- 8 • n + 8 • \(\dfrac{1}{4}\) to show that 8\(\left(n + \dfrac{1}{4}\right)\) = 8 • n + 8 • \(\dfrac{1}{4}\)
-
If d = 14, evaluate
- −100(0.1d + 0.35) and
- −100 • (0.1d) + (−100)(0.35) to show that −100(0.1d + 0.35) = −100 • (0.1d) + (−100)(0.35)
-
If y = −18, evaluate
- −(y − 18) and
- −y + 18 to show that −(y − 18) = − y + 18
7.4 - Properties of Identities, Inverses, and Zero
In the following exercises, identify whether each example is using the identity property of addition or multiplication.
- −35(1) = −35
- 29 + 0 = 29
- (6x + 0) + 4x = 6x + 4x
- 9 • 1 + (−3) = 9 + (−3)
In the following exercises, find the additive inverse.
- −32
- 19.4
- \(\dfrac{3}{5}\)
- \(− \dfrac{7}{15}\)
In the following exercises, find the multiplicative inverse.
- \(\dfrac{9}{2}\)
- −5
- \(\dfrac{1}{10}\)
- \(− \dfrac{4}{9}\)
In the following exercises, simplify.
- 83 • 0
- \(\dfrac{0}{9}\)
- \(\dfrac{5}{0}\)
- 0 ÷ \(\dfrac{2}{3}\)
- 43 + 39 + (−43)
- (n + 6.75) + 0.25
- \(\dfrac{5}{13} \cdot 57 \cdot \dfrac{13}{5}\)
- \(\dfrac{1}{6}\) • 17 • 12
- \(\dfrac{2}{3} \cdot 28 \cdot \dfrac{3}{7}\)
- 9(6x − 11) + 15
7.5 - Systems of Measurement
In the following exercises, convert between U.S. units. Round to the nearest tenth.
- A floral arbor is 7 feet tall. Convert the height to inches.
- A picture frame is 42 inches wide. Convert the width to feet.
- Kelly is 5 feet 4 inches tall. Convert her height to inches.
- A playground is 45 feet wide. Convert the width to yards.
- The height of Mount Shasta is 14,179 feet. Convert the height to miles.
- Shamu weighs 4.5 tons. Convert the weight to pounds.
- The play lasted \(1 \dfrac{3}{4}\) hours. Convert the time to minutes.
- How many tablespoons are in a quart?
- Naomi’s baby weighed 5 pounds 14 ounces at birth. Convert the weight to ounces.
- Trinh needs 30 cups of paint for her class art project. Convert the volume to gallons.
In the following exercises, solve, and state your answer in mixed units.
- John caught 4 lobsters. The weights of the lobsters were 1 pound 9 ounces, 1 pound 12 ounces, 4 pounds 2 ounces, and 2 pounds 15 ounces. What was the total weight of the lobsters?
- Every day last week, Pedro recorded the amount of time he spent reading. He read for 50, 25, 83, 45, 32, 60, and 135 minutes. How much time, in hours and minutes, did Pedro spend reading?
- Fouad is 6 feet 2 inches tall. If he stands on a rung of a ladder 8 feet 10 inches high, how high off the ground is the top of Fouad’s head?
- Dalila wants to make pillow covers. Each cover takes 30 inches of fabric. How many yards and inches of fabric does she need for 4 pillow covers?
In the following exercises, convert between metric units.
- Donna is 1.7 meters tall. Convert her height to centimeters.
- Mount Everest is 8,850 meters tall. Convert the height to kilometers.
- One cup of yogurt contains 488 milligrams of calcium. Convert this to grams.
- One cup of yogurt contains 13 grams of protein. Convert this to milligrams.
- Sergio weighed 2.9 kilograms at birth. Convert this to grams.
- A bottle of water contained 650 milliliters. Convert this to liters.
In the following exercises, solve.
- Minh is 2 meters tall. His daughter is 88 centimeters tall. How much taller, in meters, is Minh than his daughter?
- Selma had a 1-liter bottle of water. If she drank 145 milliliters, how much water, in milliliters, was left in the bottle?
- One serving of cranberry juice contains 30 grams of sugar. How many kilograms of sugar are in 30 servings of cranberry juice?
- One ounce of tofu provides 2 grams of protein. How many milligrams of protein are provided by 5 ounces of tofu?
In the following exercises, convert between U.S. and metric units. Round to the nearest tenth.
- Majid is 69 inches tall. Convert his height to centimeters.
- A college basketball court is 84 feet long. Convert this length to meters.
- Caroline walked 2.5 kilometers. Convert this length to miles.
- Lucas weighs 78 kilograms. Convert his weight to pounds.
- Steve’s car holds 55 liters of gas. Convert this to gallons.
- A box of books weighs 25 pounds. Convert this weight to kilograms.
In the following exercises, convert the Fahrenheit temperatures to degrees Celsius. Round to the nearest tenth.
- 95°F
- 23°F
- 20°F
- 64°F
In the following exercises, convert the Celsius temperatures to degrees Fahrenheit. Round to the nearest tenth.
- 30°C
- −5°C
- −12°C
- 24°C
PRACTICE TEST
- For the numbers 0.18349…, 0.\(\overline{2}\), 1.67, list the (a) rational numbers and (b) irrational numbers.
- Is \(\sqrt{144}\) rational or irrational?
- From the numbers −4, \(− 1 \dfrac{1}{2}\), 0, \(\dfrac{5}{8}\), \(\sqrt{2}\), 7, which are (a) integers (b) rational (c) irrational (d) real numbers?
- Rewrite using the commutative property: x • 14 = _________
- Rewrite the expression using the associative property: (y + 6) + 3 = _______________
- Rewrite the expression using the associative property: (8 · 2) · 5 = ___________
- Evaluate \(\dfrac{3}{16} \left(\dfrac{16}{3} n\right)\) when n = 42.
- For the number \(\dfrac{2}{5}\) find the (a) additive inverse (b) multiplicative inverse.
In the following exercises, simplify the given expression.
- \(\dfrac{3}{4}\)(−29)\(\left(\dfrac{4}{3}\right)\)
- −3 + 15y + 3
- (1.27q + 0.25q) + 0.75q
- \(\left(\dfrac{8}{15} + \dfrac{2}{9}\right) + \dfrac{7}{9}\)
- −18\(\left(\dfrac{3}{2} n\right)\)
- 14y + (−6z) + 16y + 2z
- 9(q + 9)
- 6(5x − 4)
- −10(0.4n + 0.7)
- \(\dfrac{1}{4}\)(8a + 12)
- m(n + 2)
- 8(6p − 1) + 2(9p + 3)
- (12a + 4) − (9a + 6)
- \(\dfrac{0}{8}\)
- \(\dfrac{4.5}{0}\)
- 0 ÷ \(\left(\dfrac{2}{3}\right)\)
In the following exercises, solve using the appropriate unit conversions.
- Azize walked \(4 \dfrac{1}{2}\) miles. Convert this distance to feet. (1 mile = 5,280 feet).
- One cup of milk contains 276 milligrams of calcium. Convert this to grams. (1 milligram = 0.001 gram)
- Larry had 5 phone customer phone calls yesterday. The calls lasted 28, 44, 9, 75, and 55 minutes. How much time, in hours and minutes, did Larry spend on the phone? (1 hour = 60 minutes)
- Janice ran 15 kilometers. Convert this distance to miles. Round to the nearest hundredth of a mile. (1 mile = 1.61 kilometers)
- Yolie is 63 inches tall. Convert her height to centimeters. Round to the nearest centimeter. (1 inch = 2.54 centimeters)
- Use the formula F = \(\dfrac{9}{5}\)C + 32 to convert 35°C to degrees F.
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


12.9: End of Chapter Problems
Gases and Pressure
1. How many atmospheres are in 1,547 mm of Hg?
2. How many millimeters of mercury are in 9.65 atm?
3. Covert 9.2 psi to:
a. atm.
b. torr.
4. Explain how a barometer works?
5. What is systolic pressure and diastolic pressure?
6. What is STP?
7. If you go camping in the mountains, why does it take longer to cook rice than it does at sea level? Explain with the help of the vapor pressure concept.
8. Explain the working of the autoclave with the help of the vapor pressure concept.
Gas Law Applications
Identify the gas laws involved in the following problems:
1. What is the relationship between temperature and pressure that must be kept in mind when oxygen cylinders are in use at the hospital ?
2. Aerosol cans should be stored in a cool place. Why? Explain with the help of a gas law.
3. Early winter, your tires on your car look deflated. Why? Explain with the help of a gas law.
4. Rapid ascent in an elevator may cause the volume of intestinal gas to increase and cause intestinal cramps. Why? Explain with the help of a gas law.
5. In respiration (exhaling), explain how changes in pressure in chest cavity changes the volume of the lungs. What gas law applies?
6. In respiration (inhaling), explain how changes in pressure in chest cavity changes the volume of the lungs. What gas law applies?
7. Suppose you were diving rather deep in the ocean and had to surface quickly while leaving your tank at the bottom. What precaution would you take so as not to put pressure on your lungs ? Gas law?
8. During severe bleeding, ADH (a hormone released by the hypothalamus) causes vasoconstriction (shrinking of the blood vessels) to take place. What effect does a decrease in blood vessel volume have on blood pressure?
9. What happens to the solubility of carbon dioxide gas in water when:
a. the pressure of CO 2 over the solution is increased.
b. the temperature is increased.
10. Why is breathing 100% O 2 in a pressurized hyperbaric chamber more effective at treating gangrene than breathing 100% O 2 at atmospheric pressure?
11. What is Bends? Explain with the help of a solubility law.
12. Two unopened bottles of carbonated water are at the same temperature. If one is opened at the top of a mountain and the other at sea level, which will produce more bubbles? Explain
13. Some people with respiratory problems is very uncomfortable in low pressure air that is very moist. Why
Gas Laws Calculations
Identify the gas laws involved in the following problems:
1. The pressure of an oxygen tank containing 15.0 L oxygen is 965 torr at 55.0 o C. What will be the pressure when the tank is cooled to 16.0 o C.
2. A sample of CO 2 occupies 3.23 L volume at 25.0 o C. Calculate the volume of the gas at 50.0 o C if pressure and amount of gas do not change?
3. The pressure of a 1.32 L sample of SO 2 gas at 0.532 atm is increased to 1.231 atm. Calculate the new volume of the gas if the temperature and the quantity of the gas remain the same?
4. An oxygen tank holds 20.0 L of oxygen at a pressure of 10.0 atm. What is the final pressure when the gas is released and occupies a volume of 200. L?
5. A weather balloon contains 212 L of helium at 25.0∘C and 750. mmHg. What is the volume of the balloon when it ascends to an altitude where the temperature is −40.0∘C and 540. mmHg, assuming the quantity of gas remains the same?
6. A sample of a gas at a temperature of 37.0 o C has a pressure of 725 torr and occupies a volume of 3.44 L.
a. Calculate the new pressure is the temperature is held constant and the volume is increased to 6.17 L.
b. Calculate the new pressure if the volume is held constant and the temperature is decreased to 10.0 o C.
c. Calculate the new volume if the pressure is held constant and the temperature is increased to 150.0 o C.
d. Calculate the new pressure if the temperature is decreased to 25.0 o C and the volume is increased to 15.9 L.
Respiration and Gases
1. Explain with the concept of partial pressures how oxygen gas (O 2 ) moves from the lungs to the tissues.
2. Explain with the concept of partial pressures how carbon dioxide gas (CO 2 ) moves from the tissues to the lungs.
3. Some runners train at high altitudes. Why?
4. Why does O 2 have greater solubility in blood than in water?


5.2.E: Problems on Derivatives of Extended-Real Functions
Complete the missing details in the proof of Theorems \(1,2,\) and \(4,\) Corollary \(4,\) and Lemma 1.
\(\text { [Hint for converse to Corollary } 4(\mathrm{ii}) : \text { Use Lemma } 1 \text { for an indirect proof. }]\)
Do cases \(p \leq 0\) in Example \((\mathrm{A})\).
Show that Theorems \(1,2,\) and 4 and Corollaries 2 to 4 hold also if \(f\) is discontinuous at \(a\) and \(b\) but \(f\left(a^{+}\right)\) and \(f\left(b^{-}\right)\) exist and are finite. (In Corollary \(2,\) assume also \(f\left(a^{+}\right)=f\left(b^{-}\right) ;\) in Theorems 1 and 4 and Corollary \(2,\) finiteness is unnecessary.)
\(\text { [Hint: Redefine } f(a) \text { and } f(b) .]\)
Under the assumptions of Corollary \(3,\) show that \(f^{\prime}\) cannot stay infinite on any interval \((p, q), a \leq p<q \leq b .\)
\(\text { [Hint: Apply Corollary } 3 \text { to the interval }[p, q] .]\)
Justify footnote \(1 .\)
[Hint: Let
\[
f(x)=x+2 x^{2} \sin \frac{1}{x^{2}} \text { with } f(0)=0 .
\]
At \(0,\) find \(f^{\prime}\) from Definition 1 in §1. Use also Problem 8 of §1. Show that \(f\) is not \(\left.\text { monotone on any } G_{0}(\delta) .\right]\)
Show that \(f^{\prime}\) need not be continuous or bounded on \([a, b]\) (under the standard metric), even if \(f\) is differentiable there.
\(\text { [Hint: Take } f \text { as in Problem } 5 .]\)
With \(f\) as in Corollaries 3 and \(4,\) prove that if \(f^{\prime} \geq 0\left(f^{\prime} \leq 0\right)\) on \((a, b)\) and if \(f^{\prime}\) is not constantly 0 on any subinterval \((p, q) \neq \emptyset,\) then \(f\) is strictly monotone on \([a, b] .\)
Let \(x=f(t), y=g(t),\) where \(t\) varies over an open interval \(I \subseteq E^{1}\), define a curve in \(E^{2}\) parametrically. Prove that if \(f\) and \(g\) have derivatives on \(I\) and \(f^{\prime} \neq 0,\) then the function \(h=f^{-1}\) has a derivative on \(f[I]\), and the slope of the tangent to the curve at \(t_{0}\) equals \(g^{\prime}\left(t_{0}\right) / f^{\prime}\left(t_{0}\right)\).
\(\text { [Hint: The word "curve" implies that } f \text { and } g \text { are continuous on } I \text { (Chapter } 4, §10),\) so Theorems 1 and 3 apply, and \(h=f^{-1}\) is a function. Also, \(y=g(h(x)) .\) Use \(\text { Theorem } 3 \text { of } §1 .]\)
Prove that if \(f\) is continuous and has a derivative on \((a, b)\) and if \(f^{\prime}\) has a finite or infinite (even one-sided) limit at some \(p \in(a, b),\) then this limit equals \(f^{\prime}(p) .\) Deduce that \(f^{\prime}\) is continuous at \(p\) if \(f^{\prime}\left(p^{-}\right)\) and \(f^{\prime}\left(p^{+}\right)\) exist.
[Hint: By Corollary \(3,\) for each \(x \in(a, b),\) there is some \(q_{x}\) between \(p\) and \(x\) such that
\[
f^{\prime}\left(q_{x}\right)=\frac{\Delta f}{\Delta x} \rightarrow f^{\prime}(p) \text { as } x \rightarrow p .
\]
\(\left.\text { Set } y=q_{x}, \text { so } \lim _{y \rightarrow p} f^{\prime}(y)=f^{\prime}(p) .\right]\)
From Theorem 3 and Problem 8 in §1, deduce the differentiation formulas
\[
(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}} ;(\arccos x)^{\prime}=\frac{-1}{\sqrt{1-x^{2}}} ;(\arctan x)^{\prime}=\frac{1}{1+x^{2}} .
\]
Prove that if \(f\) has a derivative at \(p,\) then \(f(p)\) is finite, provided \(f\) is not constantly infinite on any interval \((p, q)\) or \((q, p), p \neq q\).
[Hint: If \(f(p)=\pm \infty,\) each \(G_{p}\) has points at which \(\frac{\Delta f}{\Delta x}=+\infty,\) as well as those \(x\) \(\left.\text { with } \frac{\Delta f}{\Delta x}=-\infty .\right]\)


1.5.5: Chemical Potentials- Solutions- Partial Molar Properties
A given solution comprises \(\mathrm{n}_{1}\) moles of solvent, liquid chemical substance 1, and \(\mathrm{n}_{\mathrm{j}}\) moles of solute, chemical substance \(\mathrm{j}\). We ask ---- What contributions are made by the solvent and by the solute to the volume of the solution at defined \(\mathrm{T}\) and \(\mathrm{p}\)? In fact we can only guess at these contributions. This is disappointing. The best that we can do is to probe the sensitivity of the volume of a given solution to the addition of small amounts of solute and of solvent. This approach leads to a set of properties called partial molar. The starting point is the Gibbs energy of a solution. We develop an argument which places the Gibbs energy at the centre from which all other thermodynamic variables develop.
A given closed system comprises \(\mathrm{n}_{1}\) moles of solvent (e.g. water) and \(\mathrm{n}_{\mathrm{j}}\) moles of a simple solute j (e.g. urea) at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\). The Gibbs energy of the solution is defined by equation (a).
\[\mathrm{G}=\mathrm{G}\left[\mathrm{T}, \mathrm{p}, \mathrm{n}_{\mathrm{l}}, \mathrm{n}_{\mathrm{j}}\right] \label{a} \]
We introduce a partial derivative having the following form: \(\left(\frac{\partial G}{\partial n_{j}}\right)_{T, p, n_{1}}\). The latter partial differential describes the differential dependence of Gibbs energy \(\mathrm{G}\) on the amount of chemical substance \(\mathrm{j}\). By definition, the chemical potential of chemical substance \(\mathrm{j}\),
\[\mu_{\mathrm{j}}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}_{\mathrm{l}}} \nonumber \]
We also envisage that displacement of the system by adding \(\delta n_{j}\) moles of chemical substance \(\mathrm{j}\) from the original state to a neighbouring state produces a change in Gibbs energy at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\). In one class of displacements the system moves along a path of constant affinity for spontaneous reaction \(\mathrm{A}\). In another displacement the system moves along a path at constant organisation/composition, \(\xi\); i.e. frozen. These two pathways are related by the following equation. For the system at fixed \(\mathrm{T}\), \(\mathrm{p}\) and \(\mathrm{n}_{1}\)
\[\left[\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{\mathrm{j}}}\right]_{\mathrm{A}}=\left[\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{\mathrm{j}}}\right]_{\xi}-\left[\frac{\partial \mathrm{A}}{\partial \mathrm{n}_{\mathrm{j}}}\right]_{\xi} \,\left[\frac{\partial \xi}{\partial \mathrm{A}}\right]_{\mathrm{n}(\mathrm{j})} \,\left[\frac{\partial \mathrm{G}}{\partial \xi}\right]_{\mathrm{n}(\mathrm{j})} \nonumber \]
The conditions, constant \(\mathrm{T}\) and \(\mathrm{p}\), refer to intensive variables. We direct attention to a closed system at equilibrium where ‘\(\mathrm{A} = 0\)’ and the composition \(\xi=\xi^{\mathrm{eq}}\). Moreover at equilibrium, \((\partial \mathrm{G} / \partial \xi)_{\mathrm{T}, \mathrm{p}}\) is zero. Therefore the chemical potential of chemical substance \(\mathrm{j}\) in a system at equilibrium is defined by the following equation. Hence from equation (c),
\[\mu_{\mathrm{j}}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}_{1}, \mathrm{~A}=0}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}_{1},, \mathrm{G}^{c_{q}}} \nonumber \]
A similar argument in the context of chemical substance 1 shows that,
\[\mu_{1}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}_{\mathrm{j}}, \mathrm{A}=0}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}_{\mathrm{j}}, \zeta^{\mathrm{eq}}} \nonumber \]
Equations (d) and (e) are key results. Similarly for a closed system at equilibrium at fixed \(\mathrm{T}\) and fixed \(\mathrm{p}\) (at a minimum in \(\mathrm{G}\), \(\mathrm{A} = 0\), \(\xi=\xi^{\mathrm{eq}}\) ), for all \(i\)-substances,
\[V_{j}(A=0)=V_{j}\left(\xi^{e q}\right) \nonumber \]
\[\mathrm{S}_{\mathrm{j}}(\mathrm{A}=0)=\mathrm{S}_{\mathrm{j}}\left(\xi^{\mathrm{eq}}\right) \nonumber \]
\[\mathrm{H}_{\mathrm{j}}(\mathrm{A}=0)=\mathrm{H}_{\mathrm{j}}\left(\xi^{e q}\right) \nonumber \]
\[\mu_{j}(A=0)=\mu_{j}\left(\xi^{e q}\right) \nonumber \]
But in the case of, for example, isobaric expansions and isobaric heat capacities, \(\mathrm{E}_{\mathrm{pj}}(\mathrm{A}=0) \neq \mathrm{E}_{\mathrm{pj}}\left(\xi^{\mathrm{eq}}\right)\) and \(\mathrm{C}_{\mathrm{pj}}(\mathrm{A}=0) \neq \mathrm{C}_{\mathrm{pj}}\left(\xi^{e q}\right)\). The identifications, (f) to (i), arise because these variables are first derivatives of the Gibbs energy of a closed system at equilibrium where \((\partial \mathrm{G} / \partial \xi)\) at fixed \(\mathrm{T}\) and \(\mathrm{p}\) is zero.


8.2.3: Using ChooseMyPlate in Meal Planning
Creating a basic menu plan involves selecting a food from each food group. To help create well-balanced meals, it is helpful to follow the ChooseMyPlate food guidelines. ChooseMyPlate.gov suggests the following key points:
- Focus on fruits
- Vary your vegetables
- Make half your plate fruits and vegetables
- Make at least half your grains whole grains
- Go lean with protein
- Get your calcium rich foods
Dietary Guidelines
In addition to taking in well-balanced foods and creating healthy meal plans, it is important for Home Health Aides/Personal Care Aides to keep the following in mind when planning and preparing meals to serve their patients:
- Consume a variety of foods within and among the basic food groups. This ensures meals are well-balanced and interesting.
- Control caloric intake in order to maintain an optimal weight.
- Be physically active every day. Even people who have physical limitations can engage in exercise within the constraints of their limitations. For example, people who have limited mobility can do leg and arm exercises independently or with help, rock in a rocking chair, turn side to side, and do simple exercises. Follow the guidelines set forth in the Care Plan that will detail the kind of activity the patient can engage in.
- Increase daily intake of fruits, vegetables, whole grains and nonfat or low fat milk and milk products.

- Choose fats wisely.
- Choose carbohydrates wisely. Select complex carbohydrates rather than simple carbohydrates which tend to have more sugar content in them.
- Choose and prepare foods with little sodium (salt). Read labels to check for sodium content and avoid prepackaged and boxed foods.
- Switch to water and reduce consumption of sugary drinks.
- If the patient drinks alcohol, they should do so in moderation. Guidelines suggest that women should consume no more than 1 drink per day and men no more than 2 drinks per day.
- Keep foods safe to eat and follow good hygiene practices during food preparation.
Other Considerations
Home Health Aides/Personal Care Aides should involve their patients in meal planning. Use the communication skills learned throughout this course to educate patients about healthy food choices. When planning meals, they must take into consideration their patient’s food preferences, cultural and or religious background, eating habits, ability to chew and digest food, convenience of preparation, and calorie and special nutrient needs. Food preferences are determined by our family, culture , religious beliefs, foods we may choose not to eat, such as animal products for vegetarians, and the area we grew up or in which we live. There may be regional or ethnic differences in food preferences. Talk to the patient about their food preferences and beliefs. Watch the types of foods they eat. Make suggestions and ask for feedback . Ask them to share the foods they most enjoy. Always also assess for any allergies to foods, as these would be items to avoid purchasing or preparing in the patient’s home.
1. Home Health Aides/Personal Care Aides should ask their patient what their food preferences are and plan meals according to foods they enjoy and religious and cultural preferences.
True or False? ______________
- Answer
-
1. True
FEEDBACK :
Food preferences are determined by a variety of factors such as family, culture, religious beliefs, regional and ethnic differences, personal preference, and allergies. It is important to ask patients what their preferences are to ensure the patient’s beliefs and preferences are respected, and to increase the likelihood that they will eat meals prepared.
Cost of Food
It is also important to consider the cost of food and the family’s ability to afford various types of food. If Home Health Aides/Personal Care Aides have a concern that the family is not getting enough to eat, they should speak to a supervisor. Various community resources are available to help families who cannot afford enough food. Use coupons when shopping for food and look for sales. Assist patients to plan meals weekly. Use items that are on hand or that wi ll spoil sooner first. Make a shopping list and stick to it. Purchase items such as dried beans and canned fishes to help provide less expensive sources of protein. When possible, buy in bulk. Purchase produce when it is in season as it will be freshest and cheaper.