11.R: Sequences, Probability and Counting Theory (Review)
11.1 Sequences and Their Notations
1) Write the first four terms of the sequence defined by the recursive formula \(a_1=2, a_n=a_{n-1}+n\).
- Answer
-
\(2,4,7,11\)
2) Evaluate \(\dfrac{6!}{(5-3)!3!}\).
3) Write the first four terms of the sequence defined by the explicit formula \(a_n=10^n+3\).
- Answer
-
\(13,103,1003,10003\)
4) Write the first four terms of the sequence defined by the explicit formula \(a_n=\dfrac{n!}{n(n+1)!}\).
11.2 Arithmetic Sequences
1) Is the sequence \(\dfrac{4}{7},\dfrac{47}{21},\dfrac{82}{21},\dfrac{39}{7},\ldots\) arithmetic? If so, find the common difference.
- Answer
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The sequence is arithmetic. The common difference is \(d=\dfrac{5}{3}\).
2) Is the sequence \(2,4,8,16,\ldots\) arithmetic? If so, find the common difference.
3) An arithmetic sequence has the first term \(a_1=18\) and common difference \(d=8\). What are the first five terms?
- Answer
-
\(18,10,2,-6,-14\)
4) An arithmetic sequence has terms \(a_3=11.7\) and \(a_8=-14.6\). What is the first term?
5) Write a recursive formula for the arithmetic sequence \(-20,-10,0,10,\ldots\)
- Answer
-
\(a_1=-20, a_n=a_{n-1}+10\)
6) Write a recursive formula for the arithmetic sequence \(0,-\dfrac{1}{2},-1,-\dfrac{3}{2},\ldots\), and then find the \(31^{st}\) term.
7) Write an explicit formula for the arithmetic sequence \(\dfrac{7}{8},\dfrac{29}{24},\dfrac{37}{24},\dfrac{15}{8},\ldots\)
- Answer
-
\(a_n=\dfrac{1}{3}n+\dfrac{13}{24}\)
8) How many terms are in the finite arithmetic sequence \(12,20,28,\ldots ,172\)?
11.3 Geometric Sequences
1) Find the common ratio for the geometric sequence \(2.5, 5, 10, 20,\ldots\)
- Answer
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\(r=2\)
2) Is the sequence \(4, 16, 28, 40,\ldots\) geometric? If so find the common ratio. If not, explain why.
3) A geometric sequence has terms \(a_7=16,384\) and \(a_9=262,144\). What are the first five terms?
- Answer
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\(4,16,64,256,1024\)
4) A geometric sequence has the first term \(a_1=-3\) and common ratio \(r=12\). What is the \(8^{th}\) term?
5) What are the first five terms of the geometric sequence \(a_1=3, a_n=4\cdot a_{n-1}\)?
- Answer
-
\(3, 12, 48, 192, 768\)
6) Write a recursive formula for the geometric sequence \(1,\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{27},\ldots\)
7) Write an explicit formula for the geometric sequence \(-\dfrac{1}{5},-\dfrac{1}{15},-\dfrac{1}{45},-\dfrac{1}{135},\ldots\)
- Answer
-
\(a_n=-\dfrac{1}{5}\cdot \left (\dfrac{1}{3} \right )^{n-1}\)
8) How many terms are in the finite geometric sequence \(-5,-\dfrac{5}{3},-\dfrac{5}{9},\ldots ,-\dfrac{5}{59,049}\)?
11.4 Series and Their Notations
1) Use summation notation to write the sum of terms \(\dfrac{1}{2}m+5\) from \(m=0\) to \(m=5\).
- Answer
-
\(\displaystyle \sum_{m=0}^{5}\left (\dfrac{1}{2}m+5 \right )\)
2) Use summation notation to write the sum that results from adding the number \(13\) twenty times.
3) Use the formula for the sum of the first \(n\) terms of an arithmetic series to find the sum of the first eleven terms of the arithmetic series \(2.5, 4, 5.5,\ldots \)
- Answer
-
\(S_{11}=110\)
4) A ladder has \(15\) tapered rungs, the lengths of which increase by a common difference. The first rung is \(5\) inches long, and the last rung is \(20\) inches long. What is the sum of the lengths of the rungs?
5) Use the formula for the sum of the first n terms of a geometric series to find \(S_9\) for the series \(12,6,3,\dfrac{3}{2},\ldots\)
- Answer
-
\(S_9\approx 23.95\)
6) The fees for the first three years of a hunting club membership are given in the Table below. If fees continue to rise at the same rate, how much will the total cost be for the first ten years of membership?
| Year | Membership Fees |
|---|---|
| 1 | $1500 |
| 2 | $1950 |
| 3 | $2535 |
7) Find the sum of the infinite geometric series \(\textstyle \sum_{k=1}^{\infty }45\cdot \left ( -\frac{1}{3} \right )^{k-1}\).
- Answer
-
\(S=\dfrac{135}{4}\)
8) A ball has a bounce-back ratio of \(35\) the height of the previous bounce. Write a series representing the total distance traveled by the ball, assuming it was initially dropped from a height of \(5\) feet. What is the total distance? ( Hint: the total distance the ball travels on each bounce is the sum of the heights of the rise and the fall.)
9) Alejandro deposits \(\$80\) of his monthly earnings into an annuity that earns \(6.25\%\) annual interest, compounded monthly. How much money will he have saved after \(5\) years?
- Answer
-
\(\$5,617.61\)
10) The twins Sarah and Scott both opened retirement accounts on their \(21^{st}\) birthday. Sarah deposits \(\$4,800.00\) each year, earning \(5.5\%\) annual interest, compounded monthly. Scott deposits \(\$3,600.00\) each year, earning \(8.5\%\) annual interest, compounded monthly. Which twin will earn the most interest by the time they are \(55\) years old? How much more?
11.5 Counting Principles
1) How many ways are there to choose a number from the set \(\left \{ -10,-6,4,10,12,18,24,32 \right \}\) that is divisible by either \(4\) or \(6\)?
- Answer
-
\(6\)
2) In a group of \(20\) musicians, \(12\) play piano, \(7\) play trumpet, and \(2\) play both piano and trumpet. How many musicians play either piano or trumpet?
3) How many ways are there to construct a \(4\)-digit code if numbers can be repeated?
- Answer
-
\(10^4=10,000\)
4) A palette of water color paints has \(3\) shades of green, \(3\) shades of blue, \(2\) shades of red, \(2\) shades of yellow, and \(1\) shade of black. How many ways are there to choose one shade of each color?
5) Calculate \(P(18,4)\).
- Answer
-
\(P(18,4)=73,440\)
6) In a group of \(5\) freshman, \(10\) sophomores, \(3\) juniors, and \(2\) seniors, how many ways can a president, vice president, and treasurer be elected?
7) Calculate \(C(15,6)\).
- Answer
-
\(C(15,6)=5005\)
8) A coffee shop has \(7\) Guatemalan roasts, \(4\) Cuban roasts, and \(10\) Costa Rican roasts. How many ways can the shop choose \(2\) Guatemalan, \(2\) Cuban, and \(3\) Costa Rican roasts for a coffee tasting event?
9) How many subsets does the set \(\left \{ 1, 3, 5, \ldots , 99 \right \}\) have?
- Answer
-
\(2^{50}=1.13\times 10^{15}\)
10) A day spa charges a basic day rate that includes use of a sauna, pool, and showers. For an extra charge, guests can choose from the following additional services: massage, body scrub, manicure, pedicure, facial, and straight-razor shave. How many ways are there to order additional services at the day spa?
11) How many distinct ways can the word DEADWOOD be arranged?
- Answer
-
\(\dfrac{8!}{3!2!}=3360\)
12) How many distinct rearrangements of the letters of the word DEADWOOD are there if the arrangement must begin and end with the letter \(D\)?
11.6 Binomial Theorem
1) Evaluate the binomial coefficient \(\dbinom{23}{8}\).
- Answer
-
\(490,314\)
2) Use the Binomial Theorem to expand \(\left ( 3x+\dfrac{1}{2}y \right )^6\).
3) Use the Binomial Theorem to write the first three terms of \((2a+b)^{17}\).
- Answer
-
\(131,072a^{17}+1,114,112a^{16}b+4,456,448a^{15}b^2\)
4) Find the fourth term of \(\left ( 3a^2-2b \right )^{11}\) without fully expanding the binomial.
11.7 Probability
For the exercises 1-7, assume two die are rolled.
1) Construct a table showing the sample space.
- Answer
-
1 2 3 4 5 6 1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
2) What is the probability that a roll includes a \(2\)?
3) What is the probability of rolling a pair?
- Answer
-
\(\dfrac{1}{6}\)
4) What is the probability that a roll includes a \(2\) or results in a pair?
5) What is the probability that a roll doesn’t include a \(2\) or result in a pair?
- Answer
-
\(\dfrac{5}{9}\)
6) What is the probability of rolling a \(5\) or a \(6\)?
7) What is the probability that a roll includes neither a \(5\) nor a \(6\)?
- Answer
-
\(\dfrac{4}{9}\)
For the exercises 8-11, use the following data: An elementary school survey found that \(350\) of the \(500\) students preferred soda to milk. Suppose \(8\) children from the school are attending a birthday party. (Show calculations and round to the nearest tenth of a percent.)
8) What is the percent chance that all the children attending the party prefer soda?
9) What is the percent chance that at least one of the children attending the party prefers milk?
- Answer
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\(1-\dfrac{C(350,8)}{C(500,8)}\approx 94.4\%\)
10) What is the percent chance that exactly \(3\) of the children attending the party prefer soda?
11) What is the percent chance that exactly \(3\) of the children attending the party prefer milk?
- Answer
-
\(\dfrac{C(150,3)C(350,5)}{C(500,8)}\approx 25.6\%\)
Practice Test
1) Write the first four terms of the sequence defined by the recursive formula \(a=-14, a_n=\dfrac{2+a_{n-1}}{2}\)
- Answer
-
\(-14,-6,-2,0\)
2) Write the first four terms of the sequence defined by the explicit formula \(a_n=\dfrac{n^2-n-1}{n!}\).
3) Is the sequence \(0.3, 1.2, 2.1, 3,\ldots\) arithmetic? If so find the common difference.
- Answer
-
The sequence is arithmetic. The common difference is \(d=0.9\).
4) An arithmetic sequence has the first term \(a_1=-4\) and common difference \(d=-\dfrac{4}{3}\). What is the \(6^{th}\) term?
5) Write a recursive formula for the arithmetic sequence \(-2,-\dfrac{7}{2},-5,-\dfrac{13}{2},\ldots\) and then find the \(22^{nd}\) term.
- Answer
-
\(a_1=-2,a_n=a_{n-1}-\dfrac{3}{2};a_{22}=-\dfrac{67}{2}\)
6) Write an explicit formula for the arithmetic sequence \(15.6, 15, 14.4, 13.8,\ldots\) and then find the \(32^{nd}\) term.
7) Is the sequence \(-2,-1,-\dfrac{1}{2},-\dfrac{1}{4},\ldots\) geometric? If so find the common ratio. If not, explain why.
- Answer
-
The sequence is geometric. The common ratio is \(r=\dfrac{1}{2}\).
8) What is the \(11^{th}\) term of the geometric sequence \(-1.5,-3,-6,-12,\ldots\)?
9) Write a recursive formula for the geometric sequence \(1,-\dfrac{1}{2},\dfrac{1}{4},-\dfrac{1}{8},\ldots\)
- Answer
-
\(a_1=1,a_n=-\dfrac{1}{2}\cdot a_{n-1}\)
10) Write an explicit formula for the geometric sequence \(4,-\dfrac{4}{3},\dfrac{4}{9},-\dfrac{4}{27},\ldots\)
11) Use summation notation to write the sum of terms \(3k^2-\dfrac{5}{6}k\) from \(k=-3\) to \(k=15\).
- Answer
-
\(\displaystyle \sum_{k=-3}^{15}\left (3k^2-\dfrac{5}{6}k \right )\)
12) A community baseball stadium has \(10\) seats in the first row, \(13\) seats in the second row, \(16\) seats in the third row, and so on. There are \(56\) rows in all. What is the seating capacity of the stadium?
13) Use the formula for the sum of the first \(n\) terms of a geometric series to find \(\displaystyle \sum_{k=1}^{7}-0.2\cdot (-5)^{k-1}\)
- Answer
-
\(S_7=-2604.2\)
14) Find the sum of the infinite geometric series \(\displaystyle \sum_{k=1}^{\infty }\dfrac{1}{3}\cdot \left ( -\dfrac{1}{5} \right )^{k-1}\)
15) Rachael deposits \(\$3,600\) into a retirement fund each year. The fund earns \(7.5\%\) annual interest, compounded monthly. If she opened her account when she was \(20\) years old, how much will she have by the time she’s \(55\)? How much of that amount was interest earned?
- Answer
-
Total in account: \(\$140,355.75\) Interest earned: \(\$14,355.75\)
16) In a competition of \(50\) professional ballroom dancers, \(22\) compete in the fox-trot competition, \(18\) compete in the tango competition, and \(6\) compete in both the fox-trot and tango competitions. How many dancers compete in the fox-trot or tango competitions?
17) A buyer of a new sedan can custom order the car by choosing from \(5\) different exterior colors, \(3\) different interior colors, \(2\) sound systems, \(3\) motor designs, and either manual or automatic transmission. How many choices does the buyer have?
- Answer
-
\(5\times 3\times 2\times 3\times 2=180\)
18) To allocate annual bonuses, a manager must choose his top four employees and rank them first to fourth. In how many ways can he create the “Top-Four” list out of the \(32\) employees?
19) A rock group needs to choose \(3\) songs to play at the annual Battle of the Bands. How many ways can they choose their set if have \(15\) songs to pick from?
- Answer
-
\(C(15,3)=455\)
20) A self-serve frozen yogurt shop has \(8\) candy toppings and \(4\) fruit toppings to choose from. How many ways are there to top a frozen yogurt?
21) How many distinct ways can the word EVANESCENCE be arranged if the anagram must end with the letter \(E\)?
- Answer
-
\(\dfrac{10!}{2!3!2!}=151,200\)
22) Use the Binomial Theorem to expand \(\left (\dfrac{3}{2}x-\dfrac{1}{2}y \right )^5\).
23) Find the seventh term of \(\left (x^2-\dfrac{1}{2} \right )^{13}\) without fully expanding the binomial.
- Answer
-
\(\dfrac{429x^{14}}{16}\)
For the exercises 24-28, use the spinner in the Figure below.
24) Construct a probability model showing each possible outcome and its associated probability. (Use the first letter for colors.)
25) What is the probability of landing on an odd number?
- Answer
-
\(\dfrac{4}{7}\)
26) What is the probability of landing on blue?
27) What is the probability of landing on blue or an odd number?
- Answer
-
\(\dfrac{5}{7}\)
28) What is the probability of landing on anything other than blue or an odd number?
29) A bowl of candy holds \(16\) peppermint, \(14\) butterscotch, and \(10\) strawberry flavored candies. Suppose a person grabs a handful of \(7\) candies. What is the percent chance that exactly \(3\) are butterscotch? (Show calculations and round to the nearest tenth of a percent.)
- Answer
-
\(\dfrac{C(14,3)C(26,4)}{C(40,7)}\approx 29.2\%\)
Contributors and Attributions
-
Jay Abramson (Arizona 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 https://openstax.org/details/books/precalculus .


7.5: Assignment- Reading Notebook Entry #2
Now that you’ve got an initial topic and viewpoint in mind, find ONE source that gives you helpful information about this topic. This source can come from anywhere, be any length, and can be the same source you consulted for the 3 Research Topic Ideas task earlier, if you’d like.
Create an entry for your Reading Notebook that includes any or all of the following:
- questions you had while reading
- emotional reactions to the text
- key terms that seem important to you
- what you think the thesis or main idea of this source is
- what you think the intended audience for this source is
- how effective you think this source is
- anything else that comes to mind
Assignment Requirements
This is an informal assignment. Your writing can be in complete sentences, or bullet points or fragments, as you see appropriate. Editing isn’t vital for this work, though it should be proofread to the point that obvious typos or misspellings are addressed and corrected. Target word count is 150-300 words for this entry.
Please do include the web link URL to the source you’ve chosen. You do not have to cite it with MLA or APA citation, though you are welcome to if you want the practice.
Though it isn’t mandatory that this reading notebook entry be about the same source you choose for the Source Evaluation Essay, you may find it very useful later on if it is.
Contributors and Attributions
- Composition II. Authored by : Alexis McMillan-Clifton. Provided by : Tacoma Community College. Located at : http://www.tacomacc.edu . Project : Kaleidoscope Open Course Project. License : CC BY: Attribution
- Image of notebook. Authored by : OpenClips. Provided by : Pixabay. Located at : pixabay.com/p-147191/?no_redirect. License : Public Domain: No Known Copyright


6.E: Fundamentals of Thermochemistry (Exercises)
-
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These are homework exercises to accompany the Textmap created for
"
Principles of Modern Chemistry" by Oxtoby et al.
Complementary General Chemistry question banks can be found for other Textmaps and can be accessed
here
.
Q1A
A sample of \(\ce{O_2}\) gas is under an external pressure of 17 atm and contained in a cylinder with a volume of 50 L. The gas is cooled and the resulting volume is 25 L. Calculate the amount of work done on the \(\ce{O_2}\) gas.
- Solution
-
\[\begin{align*} w &=-P\Delta V \nonumber \\[5pt] &= -P_{\text{ext}}(V_2-V_1) \\[5pt] &=-17atm\ (25\,L-50\,L)=425\, L\ atm \\[5pt] &=425\ atm\ L\ (101.325\dfrac{J}{atm\ L}) \\[5pt] &= 43, 063\,J \\[5pt] & =43.06\, kJ \end{align*} \nonumber \]
Q1B
A system containing oxygen gas is heated at a constant pressure of 40.0 atm so that its volume increases 177 L to 458 L. Express the amount of work that the system did in kilo-joules.
- Solution
-
The formula for work from the expansion of a gas at constant pressure
\[w=-P_{ext} \Delta V\nonumber \]
\[w=-40\; atm \times (458\; L - 177\; L) = -11240\; L\; atm \nonumber \]
Convert from L atm to joules
\[-11240 \; L\; atm \times 101.325\; \dfrac{J}{L\ atm} = -1138893\; J \nonumber \]
Convert from J to kJ and round to get the final answer
\[ -1138893\; J \times \dfrac{1\; kJ}{1000\; J}=-1140\; kJ \nonumber \]
\[w=-1140\; kJ \nonumber \]
Q2
The gas mixture inside one of the cylinders of an airplane expands against a constant external pressure of 5.00 atm because of the growing altitude, from an initial volume of 500 mL (at the end of the compression stroke) to the final volume of 1200 mL. Calculate the work done on the gas mixture during this process and express it in Joules.
- Solution
-
\[w = -(5.00\ \mathrm{atm}) (1200\ \mathrm{mL} - 500\ \mathrm{mL}) (\dfrac{1\ \mathrm{L}}{1000 \, \mathrm{mL}}) (101.3 \, \mathrm{\dfrac{J}{L\times atm}})= -354.55 \, \mathrm{J} \nonumber\]
Q3
The following events are based on a true story . A butcher for the local Chinese restaurant needs to defrost a large chunk of beef, which weights \(50\,lb\) and is currently at \(0^{\circ}C\). He wants to accomplish this by is going out onto the street and repeatedly drop it onto the ground. Suppose the potential energy of the meat completely transforms into heat each time it hits the ground, and that energy can be calculated from
\[V = mg\Delta{h}\nonumber \]
where \(m\) is mass of the object, \(g\) is the acceleration of gravity, and \(Δh\) is change in height. If the man is \(5\, \text{ft}\) tall and he wants to get the meat to room temperature (\(25^{\circ}\text{C}\)), how many times does he have to drop the piece of meat? Assume the environment has no effect on the meat, and that it does not lose any heat. (Specific Heat of the Meat: \(0.25\mathrm{\dfrac{J}{g \, ^{\circ}C}}\)).
- Solution
-
There are two steps to solve this equation. First, to find the amount of heat absorbed by the meat from one drop. Second, to find the total amount of heat needed to warm the meat up from \(0^{\circ}C\) to \(25^{\circ}C\).
To find the total amount of heat required to bring the meat from \(0^{\circ}C\) to \(25^{\circ}C\), the equation for calculating heat has to be used:
\(Q = mC_{p}\Delta{T}\)
The mass of the meat can be calculated via conversion from \(lb\) to \(g\), and is as follows:
\(50lb\times{453.592\dfrac{g}{lb}} = 22679.6g\)
The change in temperature is just the final temperature minus the initial, so it is just \(25^{\circ}C\).
Lastly, \(C_{p}\) is given. Plugging in all of these values gives:
\[Q = 22679.6g\times{0.25\dfrac{J}{g^{\circ}C}}\times{25^{\circ}C} = 141747.5 J\nonumber \]
Thus, \(141747.5 J\) is the total amount of heat needed to heat up the meat to \(25^{\circ}C\).
Now to find the amount of heat transferred into the meat from one drop. This utilizes the potential energy equation. The question mentions how to calculate the potential energy, and all potential energy is translated into heat. Therefore.
\[Q= V= mg\Delta{h}\nonumber \]
The problem gives the mass, the acceleration of gravity is \(9.8\dfrac{m}{s^2}\), and height. Therefore, since the potential energy for one drop is equal to the heat for one drop, the heat can be calculated as such:
\[Q= 22.6796kg\times9.8\dfrac{m}{s^2}\times{1.524m}= 338.724\dfrac{kg\;m^2}{s^2} = 338.724 J\nonumber \]
From there, since no heat is lost, it carries over for each drop. Therefore, the number of times he needs to drop the meat is found by simple division:
\[141747.5 J\div{338.724 J}= 418.474 drops\nonumber \]
Thus, the drops needed is around 419.
Abstract: Calculate heat from \(Q = mC_{p}\Delta{T}\) and \(Q= V= mg\Delta{h}\). Divide the two.
Q4
Suppose you have a ball \((C_p= 0.85\dfrac{J}{g^oC})\) at 25 °C, what will its final temperature be if the amount of work equal to dropping it down from a height of 86.6m is done to it ? \((g=9.81\dfrac{m}{s^2})\)
- Solution
-
\[\begin{align} U &=mg\Delta h \\[5pt] &=m_{ball}\cdot 86.6m\cdot 9.81\dfrac{m}{s^{2}}\\[5pt] &=850m_{ball} \end{align} \nonumber \]
If we assuming all the kinetic energy at the time of collision converts to heat, then
\[\begin{align} \Delta T &=\dfrac{q}{Cp\cdot m_{ball}}\\[5pt] &=\dfrac{U}{C_p\cdot m_{ball}} \\[5pt] &= \dfrac{850 m_{ball}}{85\dfrac{J}{g^oC} \cdot m_{ball}} \dfrac{1\,kg}{1,000\,g} \\[5pt] &= 1.0^{\circ}C \end{align} \nonumber \]
\[\begin{align} T_{final} &=25^{\circ}C+1^{\circ}C\\[5pt] &=26^{\circ}C \end{align} \nonumber \]
Q7A
The rule of Dulong and Petit shows that the molar heat capacities of most metallic elements group around a certain value “X” \(\dfrac{J}{K\ mol}\) at 25 o C. By first calculating and showing the molar heat capacities of metals rhenium, silver, lead, tungsten, copper, molybdenum and hafnium (given that the respective specific heat capacities of these metals are 0.14\(\dfrac{J}{K\ g}\), 0.23\(\dfrac{J}{K\ g}\), 0.13\(\dfrac{J}{K\ g}\), 0.13\(\dfrac{J}{K\ g}\), 0.39\(\dfrac{J}{K\ g}\), 0.25\(\dfrac{J}{K\ g}\) and 0.14\(\dfrac{J}{K\ g}\) ), find the value of “X”. (hint: take the average value of the calculated molar heat capacities and round off value to nearest whole number).
- Solution
-
To first find the molar heat capacities of each metal, multiply the molar mass of each metal by their specific heat capacities:
Molar heat capacity of:
\(\ce{Re} = 0.14\dfrac{J}{K\ g}\times186.207\dfrac{g}{mol}= 26.068\dfrac{J}{K\ mol}\) \(\ce{Ag}= 0.23\dfrac{J}{K\ g}\times 107.868\dfrac{g}{mol}=24.809\dfrac{J}{K\ mol}\) \(\ce{Pb}= 0.13\dfrac{J}{K\ g}\times 207.2\dfrac{g}{mol}= 26.936\dfrac{J}{K\ mol}\)calculate the average molar heat capacity:
Average molar heat capacity=\(\dfrac{1}{7}(26.068\dfrac{J}{K\ mol} + 24.809\dfrac{J}{K\ mol} + 26.936\dfrac{J}{K\ mol} + 23.899\dfrac{J}{K\ mol} + 24.7829\dfrac{J}{K\ mol} + 23.99\dfrac{J}{K\ mol} + 24.988\dfrac{J}{K\ mol})=25.068\dfrac{J}{K\ mol}\)
\(\ce{W}= 0.13\dfrac{J}{K\ g}\times 183.84\dfrac{g}{mol}=23.899\dfrac{J}{K\ mol}\)
\(\ce{Cu}= 0.39\dfrac{J}{K\ g}\times 63.546\dfrac{g}{mol}=24.7829\dfrac{J}{K\ mol}\)
\(\ce{Mo}=0.25\dfrac{J}{K\ g}\times 95.96\dfrac{g}{mol}= 23.99\dfrac{J}{K\ mol}\)
\(\ce{Hf}= 0.14\dfrac{J}{K\ g}\times 178.49\dfrac{g}{mol}=24.988\dfrac{J}{K\ mol}\)
Q7B
The specific heat capacities of metals aluminum, bismuth, copper, lead, and silver at 25°C are 0.900, 0.123, 0.386, 0.128, and 0.233\(\dfrac{J}{g\ K}\) . Calculate the molar heat capacities of these metals. According to the rule of Dulong and Petit, the molar heat capacities of metallic elements, like these, are approximately 25\(\dfrac{J}{K\ mol}\).
- Solution
-
Multiply each of the specific heat capacities by its corresponding molar mass to obtain their molar heat capacities.
Aluminum:
\[0.900 \dfrac{J}{g\ K}\times 26.98 \dfrac{g}{mol} = 24.3 \dfrac{J}{mol\ K}\nonumber \]
Bismuth:
\[0.123 \dfrac{J}{g\ K}\times 208.98 \dfrac{g}{mol} = 25.7 \dfrac{J}{mol\ K}\nonumber \]
Copper:
\[0.386 \dfrac{J}{g\ K}\times 63.55 \dfrac{g}{mol} =24.5 \dfrac{J}{mol\ K}\nonumber \]
Lead:
\[0.128 \dfrac{J}{g\ K}\times 207.2 \dfrac{g}{mol} =26.5 \dfrac{J}{mol\ K}\nonumber \]
Silver:
\[0.223 \dfrac{J}{g\ K}\times 107.87 \dfrac{g}{mol} =25.1 \dfrac{J}{mol\ K}\nonumber \]
Q9
An undisclosed volume of water is tightly sealed in a microwave-safe container at room temperature before it is placed in an ice bath where it is cooled by a student.
- During the cooling process within the ice bath, state whether (\(\Delta U\)), Q , and W of the system are negative, zero, or positive. Explain your reasoning.
- After being cooled, the student decides that the water seems too cold so it is placed in the microwave where the container of water is heated back to room temperature. What are the new signs of (\(\Delta U\)), Q , and W during the heating process? Explain your reasoning.
- Now determine the signs of (\(\Delta U_1 + \Delta U_2\)), (\(Q_1 + Q_2\)) , and (\(W_1 + W_2\)), where possible, assuming that the cooling process was step 1 and the heating process was step 2.
- Solution
-
- W is zero , Q is negative , and (\(\Delta U\)) is negative. Work is zero for this step because neither volume nor pressure of the container is changing. Since the water is cooler than before heat has left the system so it is negative; and because (\(\Delta U\)) = Q + W and W = 0 then (\(\Delta U\)) = Q which is negative so, (\(\Delta U\)) is negative.
- W is zero , Q is positive , and (\(\Delta U\)) is positive. Work is zero for this step because neither volume nor pressure of the container is changing. Since the water is hotter than before heat has entered the system so it is positive and because (\(\Delta U\)) = Q + W and W = 0 then (\(\Delta U\)) = Q which is positive so, (\(\Delta U\)) is positive.
- Since W was equal to zero for both processes above ( W 1 + W 2 ) is equal to zero . ( (\(\Delta U\)) 1 + (\(\Delta U\)) 2 ) = (Q 1 + Q 2 ) where they are both equal to zero .
If the water is cooled from room temperature to a lower temperature and then heated directly back to room temperature, wouldn't ( (\(\Delta U\)) 1 + (\(\Delta U\)) 2 ) be 0? Q is the only factor actually changing here and q(heating to cooling) = -q(cooling to heating) provided the change in t is the same. So both ( (\(\Delta U\)) 1 + (\(\Delta U\)) 2 ) and q should be 0.
Q11A
Your lab partner slipped a sample of unknown hot metal that is 40.0g, which is initially at 130.0°C into a 100.0 g water that is initially at 50.0°C. A temperature probe indicates that equilibrium is reached at 60.15°C. Using the specific heat capacity of water 4.18\(\dfrac{J}{K}\), calculate the specific heat capacity of the unknown metal.
- Solution
-
Imagine two sub-systems: the metal and the water. If mixing the hot metal and cool water inside a well-insulated container (that prevent leaks of heat), then the heat absorbed by the system will equal zero. Since the system is the sum of the two sub-systems:
\[q_{\text{sys}}=0=q_{\text{metal}}+q_{\text{water}}\nonumber \]
For both sub-systems, the amount of heat gained is equal to the specific heat capacity times the mass times the temperature change:
\[q_{\text{metal}}+ q_{\text{water}}= m_{\text{water}} C_{s,\text{water}} \, \Delta T_{\text{metal}} = 0\nonumber \]
Solving for the specific heat capacity of the metal:
\[c_{s,\text{metal}}=\dfrac{-m_{\text{water}}c_{s,\text{water}} \, \Delta T_{\text{water}}}{m_{\text{metal}} \Delta T_{\text{metal}}}=\dfrac{-100.0\;\text{g} \times 4.18 \mathrm{\dfrac{J}{K\ g}} \times 10.15 \mathrm{^{\circ}C}} {40.0\; \text{g} \times -69.85 \mathrm{^{\circ}C}} = 1.52 \mathrm{\dfrac{J}{K\ g}} \nonumber \]
Tip: Do not convert Celsius to Kelvin. The Kelvin and Celsius scales differ only in their location of their zero points. Temperature change in Celsius is the same temperature change of Kelvin.
Q11B
Let's say you 34.5 grams of some hot metal that is initially at 75°C and you put that metal into 64.0 grams of water that is initially at 25°C. If the the two objects reach thermal equilibrium at 39°C, what is the specific heat capacity of the metal when the specific heat capacity of water is \(4.18 \dfrac {J}{K\ g}\).
- Solution
-
The formula used to solve this question is \[m_1c_1 \Delta T_1 = -m_2c_2 \Delta T_2\nonumber \]
Plug in your known values and solve for \(c_1\)
\((34.15 \; g)(c_1)(36°C) = -(64 \; g)(4.18 \; \dfrac {J}{K\ g})(-14°C)\)
\(c_1 = 3.02 \; \dfrac {J}{K\ g}\)
Q12
A 10.00 g sample of Aluminum at 60.0 °C and a 30.0 g sample of copper at a temperature of -20.0 °C were thrown simultaneously into a 50.0 g of water at a temperature of 25.0 °C. What will be the final temperature of the system consisting of the two metal samples and the water? Assuming that this system is completely isolated from the surroundings. Use the information below for your calculations.
- \(c_{s(\ce{Al})}=0.900 J/(K \cdot g)\)
- \(c_{s(\ce{Cu})}=0.385 J/(K \cdot g)\)
- \(c_{s(\ce{H_2 O})}=4.184 J/(K \cdot g)\)
- Solution
-
Since our system is isolated, the thermal energy lost by one component is transferred to the other components.
\[q_1=-q_2\nonumber \]
which is equivalent to
\[C_1 ∆T=-C_2 ∆T\nonumber \]
Noting that
\[C=m C_s\nonumber \]
To avoid the complexity of handling three components, we can utilize the fact that temperature is a state function; we will simplify our calculations by choosing a different path to arrive to our final state.
We can do this in two steps:
Ignore the Aluminum sample and treat the copper and water as the only components of our system. After finding the equilibrium temperature, we add the Aluminum sample to the water and copper system, thereby reaching the same final state.
Which is the same answer we got before. This should make intuitive sense because regardless of the path we take, we end up with the exact same amount of thermal energy in our system.
Step 1
\[m_{\ce{Cu}}×C_{s(\ce{Cu})}×(T_f-T_{i(\ce{Cu})})=-m_{\ce{H_2O}}C_{s(\ce{H_2 O})}(T_f-T_{i(\ce{H_2 O})})\nonumber \]
\[30.0\,g×0.385\, J/(°C \cdot g)×(T_f+20 °C)=-50.0\,g × 4.184 \,J/(°C \cdot g)×(T_f-25.0 °C)\nonumber \]
Solving for \(T_f\) for the copper and water system gives us
\[T_f=22.6 °C\nonumber \]
Step 2
\[C_{(\ce{Cu}+\ce{H_2 O})}×(T_f-T_{i(\ce{Cu}+\ce{H_2 O})})=-m_{\ce{Al}}×C_{s(\ce{Al})}×(T_f-T_{i(\ce{Al})})\nonumber \]
Noting that
\[C_{(\ce{Cu}+\ce{H_2 O})}=m_{\ce{H_2O}}×C_{s(\ce{H_2 O})}+m_{\ce{Cu}}×C_{s(Cu)}\nonumber \]
Combining the two equations and plugging the values gives us
\[[(30.0\,g×0.385 \,J/(°C \cdot g))+(50.0\,g × 4.184 \,J/(°C \cdot g))]×(T_f-22.6 °C)= -10.0\,g×0.900\, J/(K \cdot g)×(T_f-60 °C)\nonumber \]
Solving for \(T_f\):
\[T_f=24.1°C\nonumber \]
Which is the final temperature of the whole system.
Alternative Approach
We can also choose a path where we add the Aluminum first, and then we add the copper. Again, temperature is a state function and choosing a different path will not affect the final answer. We also show the calculation for this path for the sake of completion.
Step 1
\[m_{\ce{Cu}}×C_{s(\ce{Al})}×(T_f-T_{i(Al)})=-m_{(\ce{H_2 O})}×C_{s(\ce{H_2 O})}×(T_f-T_{i(\ce{H_2 O})})\nonumber \]
\[10.0\,g×0.900\, J/(°C \cdot g)×(T_f+20 °C)=-50.0\,g × 4.184 \,J/(°C \cdot g)×(T_f-25.0 °C)\nonumber \]
Solving for \(T_f\) for the aluminum and water system gives us
\[T_f=26.4 °C\nonumber \]
Step 2
\[[(m_{(\ce{H_2 O})}×C_{s(\ce{H_2O})})+(m_{\ce{Al}}×C_{s(\ce{Al})}) )]×(T_f-T_{i(\ce{Al}+\ce{H_2 O})})=-m_{\ce{Al}}×C_{s(\ce{Cu})}×(T_f-T_{i(\ce{Cu})})\nonumber \]
Plugging the numbers,
\[[(10.0\,g×0.900\, J/(°C \cdot g))+(50.0\,g × 4.184 \,J/(°C \cdot g)]×(T_f-26.4 °C)=-30.0\,g×0.385 \,J/(K \cdot g)×(T_f+20 °C)\nonumber \]
Solving for \(T_f\)
\[T_f=24.1°C\nonumber \]
Q15A
Calculated the heat required to melt 3.00 g of ice and the heat required to change the temperature of water from 0°C to 100°C. What is the proportionality of the heat necessary to melt ice compared to the heat required to change the temperature of water from 0°C to 100°C? Use values from Table S2 and assume \(\Delta H_{f}\) =334 J g -1 for calculations? Why is the heat positive instead of negative?
- Solution
-
First lets determine the amount of heat needed to melt ice:
\(q = m\Delta H_{f}\)
\(q = 1,002\ \mathrm{J}\)
Next lets determine the heat required to raise the temperature of water 100°C:
\(q = mC_s \Delta{T}\)
\(q = 1,254\ \mathrm{J}\)
So the proportionality was determined to be 4:5
The heat is positive because the heat is required. The heat is needed to make the states go from solid to liquid when melting and then liquid to gas when the temperature is raised from 0 to 100 degrees Celsius.
Q15B
An observation in the 18th century stated that the heat that raised a certain mass of water from its freezing point to boiling point is equal to four-thirds of the heat required to melt the same mass of ice. Using the theory behind the observation, estimate the heat required to melt 10 g of ice, know that the heat capacity of water is 4.18 J/ o C.
- Solution
-
Let the heat required to raise the temperature of water from its freezing point to its boiling point is \(q_1\):
\[q = m c \Delta T\]
\[m = \text{mass}\]
\[c = 4.18\dfrac{J}{g\ ^{o}C}\]
\[\Delta T = \text{freezing point} - \text{boiling point} = 100 - 0 = 100^{o}C\]
\[q_{1}=(10 g)(4.18\dfrac{J}{g\ ^{o}C})(100^{o}C)\]
\[q_{1} = 4180 \, \text{J} = 4.18 \, \text{kJ}\]
Let the heat required to melt 10 g of ice is \(q_2\):
\[q_{1}=\dfrac{4}{3}q_{2}\]
\[q_{2}=\dfrac{3}{4}q_{1}\]
\[q_{2}=\dfrac{3}{4}(4.18 \, \text{kJ})=3.135 \, \text{kJ}\]
Q17
For his birthday, his John's parents have given him a compressible oven filled with 0.250 mol argon. If he sets this compressible oven at 1.00 atm and 273 K and let it contract from a constant external pressure of 0.100 atm until the gas pressure reaches 10.00 atm and the temperature reaches 400 K, what is the work done on the gas, the internal energy change, and the heat absorbed by the gas?
- Solution
-
Using the ideal gas law \(P V = n R T\), we can see that:
\[V_o = \dfrac{n R T_o}{P_o} \nonumber \]
\[V_f = \dfrac{n R T_f}{P_f} \nonumber \]
n = 0.25 mol, R = 0.082\(\dfrac{atm\ L}{mol\ K}\) Po = 1 atm, Pf = 10 atm To= 273K Tf = 400K
where \(V_o\) and \(V_f\) represent the initial and final volume of the chamber. Now, the pressure clearly changes inside the chamber, but outside the chamber, the pressure is held constant, so from the perspective of the surroundings:
\[W= -P(V_f - V_o)\nonumber \]
where \(P\) represents the constant external pressure. Now substituting in V f and V o we can see that we've solved for work. Now, for any thermodynamic process:
\[\Delta{U}= \dfrac{3}{2} n R (T_f - T_o) \nonumber \]
So plugging in those values allows us to solve for ∆U. Now, for heat, we simply subtract W from ∆U, by the first law of thermodynamics.
First law of thermodynamics: the total energy of an isolated system is a constant, energy can be transformed from one form to another, but it can not be created or destroyed.
The solution is incomplete.
Q19
Take 4 moles of ideal, monatomic gas going through expansion processes. The gas was initially put at a pressure of 5.00 atm and a temperature of 30°C. The gas first goes through isothermal expansion until the volume doubled. An isochoric process follows as the pressure is halved. \(C_v = \dfrac{3}{2}R\)
- Build \(\Delta U\), W, and Q table for each process.
- Find the final temperature.
- Solution
-
Isothermal Process:
\[\Delta U = 0\]
\[W = Q\]
\[W = nRT\ln \left( \dfrac{V_f}{V_i} \right) =(4\,\text{mol})(8.314\, \mathrm{J \cdot K^{-1} mol^{-1}})(303.15\, \text{K}) \ln(2) = 6988 \, \text{J} = Q\]
Isochoric Process
\[W = 0\]
\[\Delta U = Q\]
First find the initial pressure of this process.
Use ideal law to relate pressure to volume. \(PV=nRT\), in which P is inversely proportional to V.
Thus, when volume is doubled, pressure is halved so \(P_f = 2.50 \,\text{atm}\). Now, find the final temperature of this process.
Once again, relate P to T using ideal gas law. T is found to be directly proportional to P. Thus, if the pressure is halved from 2.50 atm, the temperature must also be halved from 303.15K.
\(\dfrac{1}{2}(303.15 \, \text{K}) = 151.58 \, \text{K}\)
\[Q= nC_v\Delta T=(4.00 \, \text{mol})(\dfrac{3}{2})(8.314 \mathrm{\dfrac{J}{K\, mol}})(-151.58K)=-7561 \, \text{J} = \Delta U\]
Processes ΔU Q W Isothermal 0 6988J 6988J Isochoric -7561J -7561J 0
Q20
An apparatus is set up such that an ideal gas is released into vacuum by opening a stopcock, hence allowing it to freely expand (i.e., no external force is applied). Calculate \(\Delta{U}\) of the system, and prove that the free expansion process is adiabatic i.e no heat transfer.
- Solution
-
\[\Delta U = \dfrac{3}{2}nR\Delta T\]
As the expansion is isothermal, i.e. it occurs without a change in temperature:
\[\Delta T = 0\]
Hence,
\[\Delta U = 0\]
and
\[\Delta U = q_p+w\]
As the external force applied on the gas is zero, the work done by the gas is zero.
Hence,
\[0 = q_p + 0\]
\[q_p = 0\]
This proves that the process is adiabatic.
Q21
A 150 L vessel contains 8.00 moles of neon at 270 K is compressed adiabatically, so that there is no gain nor loss of any heat, and irreversibly until the final temperature is 470 K. Calculate the change in internal energy, the heat added to the gas, and the work done on the gas.
- Solution
-
Since Ne is a monatomic ideal gas and the volume of the vessel remains constant, the heat capacity of Ne can be expressed as:
\[C_{p} = \dfrac{3}{2} \times R = \dfrac{3}{2} \times 8.314 \mathrm{\dfrac{J}{K \cdot mol}} = 12.47 \mathrm{\dfrac{J}{K \cdot mol}} \]
\[\Delta U = nC_{p} \Delta T\]
\[\Delta U = 8.00 \times 12.47 \mathrm{\dfrac{J}{K \cdot mol}} \times (470-2700) \; \mathrm{K} = 19952 \; \mathrm{J} = 20.0 \; \mathrm{kJ}\]
Since the vessel is adiabatically compressed, no heat is added, therefore \(q = 0\)
\[\Delta U = q + w \]
\[20.0 \; \mathrm{kJ} = 0 + w \]
\[w = 20.0 \; \mathrm{kJ}\]
Q22
A gas expands at constant external pressure of 3.00 atm until its volume has increased by 9.00 to 15.00 L. During this process, it absorbs 800J of heat from the surroundings.
- Calculate the energy change of the gas, \(\Delta U\)
- Calculate the work, w , done on the gas in an irreversible adiabatic ( q = 0) process connecting the same initial and final state.
- Solution
-
a) \[\Delta U = q+w\]
\[\Delta U= 800J + -((3.00 \, \mathrm{atm})(15.00 \mathrm{L} - 9.00 \mathrm{L}) (101.3 \mathrm{\dfrac{J}{L\ atm}})) =-1023.4 \mathrm{J}\]
b) \[q=0\]
\[\Delta U = w\]
\[\Delta U= -((3.00 \mathrm{atm})(15.00 \mathrm{L} - 9.00 \mathrm{L}) (101.3 \mathrm{\dfrac{J}{L\ atm}}))= -1823.4 \mathrm{J}\]
Q23
Using the theorem of equipartition of energy, calculate the specific heat capacity at constant pressure \(C_p\) at 25ºC and 1 atm for \(\ce{O_2}\) and \(\ce{CO}\). Compare the calculated values to the experimental data \((\ce{O_2}=29.36, \ce{CO}=29.1)\) and thus calculate the percent of the experimental value that results from vibrational motions. Answer: The percent of \(C_p\) due to vibrational motions is 0.897% for \(\ce{O_2}\) and 0.141% for \(\ce{CO}\).
- Solution
-
The equipartition theorem states that each degree of freedom in a molecule contributes to ½ RT to the molar internal energy of a gas. To solve this problem, the number of degrees of freedom \(DOF\) in each molecule must be identifies.
\(\ce{O2}\) is a linear diatomic particle, thus it has 3 translational degrees of freedom and 2 rotational degrees of freedom. Since \(\ce{CO}\) is a linear molecule it also has 3 translational and 2 rotational degrees of freedom. The molecules are at room temperature. Thus, it is assumed that there is no vibration in the bond (other than the zero point energy).
The low percentage of \(C_p\) due to vibrational modes in both molecules indicates that the vibrational motion is extremely small and can be neglected.
\[DOF({\ce{O2}}) = {f_t} + {f_r} = 3 + 2 = 5\nonumber \]
\[DOF(\ce{CO}) = {f_t} + {f_r} = 3 + 2 = 5\nonumber \]
Using the equation
\[U = \left( DOF\right) \cdot \left( \dfrac{1}{2}RT \right)\nonumber \]
find the internal energy of each gaseous molecule.
\[U \left( \ce{O_2} \right) = \left( 5 \right) \cdot \left( \dfrac{1}{2}RT \right)\nonumber \]
Since \(\ce{O_2}\) and \(\ce{CO}\) have the same degrees of freedom,
\[U\left( \ce{O_2} \right) = U\left( \ce{CO} \right) = \dfrac{5}{2}RT\nonumber \]
The internal energy is related to \(C_v\) by
\[C_v= \left( \dfrac{\partial U}{\partial T} \right) = \left( \dfrac{\partial \dfrac{5}{2}RT}{\partial T} \right) = \dfrac{5}{2}R\nonumber \]
\(C_v\) is related to \(C_p\) by
\[C_p = C_v + R = \dfrac{5}{2}R + R = \dfrac{7}{2}R = \dfrac{7}{2} \times 8.314 \dfrac{J}{mol\ K} = 29.099\dfrac{J}{mol\ K}\nonumber \]
The percent of the experimental value that results from vibration motion is the difference between calculated and experimental value as a percentage of the experimental value.
\[\% \text{vibrational motion}(\ce{O_2}) = \dfrac{\text{experimental value} - \text{calculated value}}{ \text{experimental value}} \times 100 \% = \dfrac{ 29.36 - 29.099 }{29.099} \times 100 \% = 0.897\% \nonumber \]
\[\% \text{vibrational motion} (\ce{CO}) = \dfrac{\text{experimental value} - \text{calculated value}}{ \text{experimental value}} \times 100 \% = \dfrac{ 29.14 - 29.099}{ 29.099} \times 100 \% = 0.141\% \nonumber \]
Q23B
Using the classical equipartition theorem, calculate the value of \(C_p\) at 298 K and 1 atm for \(\ce{HF (g)}\) and \(\ce{F2 (g)}\) , assuming that their pressure is constant. Then compare your calculation with the experiment values of \(29.13\mathrm{\dfrac{J}{K\ mol}}\) and \(31.30\mathrm{\dfrac{J}{K\ mol}}\)respectively. What is the per cent of the measured value that arises from vibrational motions?
- Solution
-
Diatomic molecules possess a total of 6 degrees of freedom:
- 3 degrees in translational motion
- 2 degrees in rotational motion
- 1 degree in their vibrational motions
\[6.11 \text{ for } \ce{F2 (g)} = 0.735R\]
The differences are due to the fact that some quantized energy levels are not available at certain temperatures, in this case room temperature. This means that \(\ce{HF_{(g)}}\) contributed only 0.004R of its experimental value of \(29.13\mathrm{\dfrac{J}{K\ mol}}\) to the vibrational degree of freedom, which is 0.014%. \(\ce{F_ {2(g)}}\) contributed 0.265R of its experimental value \(31.30\mathrm{\dfrac{J}{K\ mol}}\), which is 0.847%.
Each translational and rotational degree of freedom contributes \(\dfrac{R}{2}\) to the heat capacity of a gas, while each vibrational degree of freedom contributes \(\dfrac{2R}{2}\) (=R). Therefore the predicted heat capacity \((C_v)\) at a constant volume is:
\[C_v = 3\times (\dfrac{R}{2}) + 2\times (\dfrac{R}{2}) + 1\times (\dfrac{2R}{2}) = 7(\dfrac{R}{2})\]
Since the question is asking for the heat capacity of these gases that are under a constant pressure, the result is multiplied by an additional \(\dfrac{2R}{2}\) because the volume can change in order to keep the pressure constant. Therefore:
\[C_v= (\dfrac{2R}{2})+(\dfrac{7R}{2}) = \dfrac{9R}{2}\]
\[R = 8.3145\mathrm{\dfrac{J}{K\ mol}}\]
\[C_p=\dfrac{9(8.3145\mathrm{\dfrac{J}{K\ mol}})}{2}=37.41\mathrm{\dfrac{J}{K\ mol}}\]
This is the calculated value for both of the two gases. Since the experiment values were \(29.13\mathrm{\dfrac{J}{K\ mol}}\) for \(\ce{HF (g)}\) and \(31.30\mathrm{\dfrac{J}{K\ mol}}\) for \(\ce{F2(g)}\), the differences between the calculations were:
\[8.28 \text{ for } \ce{HF (g)} = 0.996R\]
Q25
- Calculate the change of enthalpy when a 69-grams sample of zinc is heated from 753 K to 927 K at a constant pressure of 1 atm.
- Calculate the change of enthalpy when 3 moles of butane are heated from 204 K to 258 K at a constant pressure of 1 atm.
- Solution
-
a)
\(q=mC_s\Delta T\)
\(m=69\ \mathrm{g}\)
\(\Delta T=927-753=174\ \mathrm{K}\)
\(C_{\text{s(zinc)}}=0.39 \mathrm{\dfrac{J}{g\ ^oC}}\)
\(q=69\times 0.39\times 174=4682.34\ \mathrm{J}=4.56\ \mathrm{kJ}\)
b)
\(q=nC_p\Delta T\)
\(n=3\)
\(\Delta T=258-204=54\ \mathrm{K}\)
\(C_p=132.42\)
\(q=3\times 132.42\times 54=21452.04\ \mathrm{J}= 21.4504\ \mathrm{kJ}\)
Q31
What is the change in enthalpy when 8.19 grams of ethane (\(\ce{C_2H_6}\)) vaporizes, assuming a normal boiling point and \(\Delta H_{\text{vap}}= 14.72\ \mathrm{\dfrac{kJ}{mol}}\).
- Solution
-
First, convert grams of ethane to moles:
\[(8.19\; \cancel{g\; \ce{C_{2}H_{6}}}) \left(\frac{1\: mol}{30.07\: \cancel{g\; \ce{C_{2}H_{6}}}} \right) = 0.272\; mol\nonumber \]
This step is to match quantity units of ethane to those in the given value of \(\ce{\Delta H_{vap}}\).
Then, given the \(\Delta H_{\text{vap}}\),
\[(0.272\; \cancel{mol \;\ce{C_{2}H_{6}}} ) \left (\frac{14.72\: kJ}{1\: \cancel{mol\; \ce{C_{3}H_{8}}}} \right ) = 4.004\; kJ\nonumber \]
\(\ce{\Delta H}=4.004\; kJ\)
This value is the change in enthalpy, or thermodynamic energy, when the specified amount of ethane undergoes the above reaction.
Q33
The heat capacity \(C_p\) of ice is \(38 \mathrm{\dfrac{J}{K\ mol}}\) and \(C_p\) of water is \(75 \mathrm{\dfrac{J}{K\ mol}}\). A 24.0 g ice cube at -15 o C is placed into 120 g of water at room temperature (25 o C). What is the temperature of the water when it reaches equilibrium?
- Solution
-
\[q_{\mathrm{ice}} = -q_{\mathrm{water}}\]
\[m_{\mathrm{ice}}C_{\mathrm{ice}}(T_{f}-T_{\mathrm{ice}})=-m_{\mathrm{water}}C_{\mathrm{water}}(T_{f}-T_{\mathrm{water}})\]
\[\left(\dfrac{24g\ \mathrm{ice}}{18.02\dfrac{g}{mol}}\right)\times (38\dfrac{J}{K\ mol})\times (T_f - (-15^oC)) = -\left(\dfrac{120g\ \mathrm{water}}{18.02\dfrac{g}{mol}}\right)\times (75\dfrac{J}{K\ mol})\times (T_f - 25^oC)\]
\[T_{f}=21.316^{o}C\]
Q35
Determine the change in enthalpy \(\left(\Delta H\right)\) for the following chemical reaction
\(\ce{4NH_{3(g)} + 5O_{2(g)} -> 4NO_{(g)} + 6H_2O_{(g)}}\)
given that:
\(\ce{O_{2(g)} + N_{2(g)} -> 2NO_{(g)}}\) \(\Delta H = +180.5\ \mathrm{\dfrac{kJ}{mol}}\)
\(\ce{2H_2O_{(g)} -> 2H_{2\, (g)} + O_{2\, (g)}}\) \(\Delta H = +483.64\ \mathrm{\dfrac{kJ}{mol}}\)
\(\ce{3H_{2(g)} + N_{2(g)} -> 2NH_{3(g)}}\) \(\Delta H = -92.22\ \mathrm{\dfrac{kJ}{mol}}\).
- Solution
-
Using Hess's Law , we can obtain the \(\ce{\Delta}{H}\) for
\(\ce{4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)}}\)
by adding and modifying the other equations provided until they resemble the equation above.
\(\ce \left( {O_{2(g)} + N_{2(g)} \rightleftharpoons 2NO_{(g)}} \right) \times \left( 2 \right) \) \(\ce{\Delta}{H}\) = \(+180.5\ \dfrac{kJ}{mol} \times \left( 2 \right) \)
\(\ce \left( {2H_2O_{(g)} \rightleftharpoons 2H_{2\, (g)} + O_{2\, (g)}} \right) \times \left( -3 \right) \) \(\ce{\Delta}{H}\) = \(+483.64\ \dfrac{kJ}{mol} \left( -3 \right) \)
\(\ce \left( {3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)}} \right) \times \left(-2 \right) \) \(\ce{\Delta}{H}\) = \(-92.22\ \dfrac{kJ}{mol} \times \left(-2 \right) \)
The above modifications simplify out to
\(\ce {2O_{2(g)} + 2N_{2(g)} \rightleftharpoons 4NO_{(g)}} \) \(\ce{\Delta}{H}\) = \(+361.0\ \dfrac{kJ}{mol} \)
\(\ce {6H_{2\, (g)} + 3O_{2\, (g)} \rightleftharpoons 6H_2O_{(g)}} \) \(\ce{\Delta}{H}\) = \(-1450.92\ \dfrac{kJ}{mol} \)
\(\ce {4NH_{3\, (g)} \rightleftharpoons 6H_{2\, (g)} + 2N_{2\, (g)}} \) \(\ce{\Delta}{H}\) = \(184.4\ \dfrac{kJ}{mol} \)
Species that are common to both sides can be cancelled out.
\(\ce { 2O_{2(g)} + \require{cancel} \cancel{2N_{2(g)}} \rightleftharpoons 4NO_{(g)}} \) \(\ce{\Delta}{H}\) = \(+361.0\ \dfrac{kJ}{mol} \)
\(\ce { \require{cancel} \cancel{6H_{2\, (g)}} + 3O_{2\, (g)} \rightleftharpoons 6H_2O_{(g)}} \) \(\ce{\Delta}{H}\) = \(-1450.92\ \dfrac{kJ}{mol} \)
\(\ce { 4NH_{3\, (g)} \rightleftharpoons \require{cancel} \cancel{6H_{2\, (g)}} + \cancel{2N_{2\, (g)}}} \) \(\ce{\Delta}{H}\) = \(184.4\ \dfrac{kJ}{mol} \)
Finally, after adding the remaining specimens and the enthalpy change that occurs with them we have:
\(\ce{4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)}}\) with \(\ce{\Delta}{H}\) = \(-905.5\ \dfrac{kJ}{mol} \)
Q37
\(\Delta H\) of \(\ce{C6H6(g)}\) from \(\ce{C6H6(l)}\) to its gaseous form is +33.9 kJ/mol. Which of these two substances absorbs off more heat when 10 lbs of the substance is burned? Which of these two substances require less energy when condensed to solid form?
- Solution
-
An alternative way to view this is to inspect the reactions under comparison
\[\ce{2C6H6(g) + 15O2(g) -> 12CO2(g) + 6H2O(g)} \nonumber\]
vs.
\[\ce{2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(g)} \nonumber\]
Recognizing that the difference between these equations is
\[\ce{2C6H6(l) -> 2C6H6(g)} \nonumber\]
The problem states th at the \(∆H\) of the conversion of \(\ce{C6H6(g)}\) to \(\ce{C6H6(l)}\) is positive, and therefore, is an endothermic process. Consequently, ten pounds of \(\ce{C6H6(g)}\) contains more enthalpy than 10 pounds of \(\ce{C6H6(l)}\). Since these will both produce carbon dioxide and water when burned, their products are of the same final energy level. As a result, \(\ce{C6H6(g)}\) will give off more energy than \(\ce{C6H6(l)}\) when burned and \(\ce{C6H6(l)}\) will require less energy to form a solid.
Q39
Calculate the enthalpy change under standard conditions for the reaction:
\[\ce{C3H6O(l) + 4 O2 (g) → 3 CO2 (g) + 3 H2O(l)} \nonumber \]
given information below:
- ΔH° f, O 2 (g) : 0 kJ/mol
- ΔH° f, CO 2 (g) : -393.5 kJ/mol
- ΔH° f, H 2 O(l) : -285.83 kJ/mol
- ΔH° f, C 3 H 6 O(l) : -248 kJ/mol
- Solution
-
\[ΔH^o_{rxn} = \sum ΔH^o_{f,\, products} – \sum ΔH^o_{f,\, reactants} \nonumber\]
ΔH° rxn = [ 3 (-393.5 kJ/mol) + 3 (-285.83 kJ/mol) ] - [ (-248 kJ/mol) + 4 (0 kJ/mol) ] = -1790 kJ/mol
Q47
A 10 gram sample of pure hydrogen is burned completely with excess oxygen to generate liquid water in a constant volume calorimeter at 25 °C. The amount of heat evolved is 1,420 kJ.
- Write and balance the chemical equation for the combustion reaction.
- Calculate the standard change in internal energy for the combustion of 1.00 mole hydrogen to liquid water.
- Calculate the standard enthalpy change per mole of hydrogen for the same reaction as in part (a).
- Calculate the standard enthalpy of formation per mole of hydrogen, using data for the standard enthalpies of formation of liquid water ( Table T1 ).
- Solution
-
- \(\ce{2H2 (g) + O2 (g) -> 2 H2O (l)}\)
- \(ΔU^o= q_v = \dfrac{-1420 \,kJ}{10/(1.008 \times 2 ) g/mol} = -286.3 \,kJ/(mol\, of\, H_2) \)
- \(ΔH^o_{rxn}= ΔU^o + Δn_{gas} RT = -286.3 \,kJ/(mol\, of\, H_2) + (2 mol. \, of\, H_2)(-3) (8.314 J/{K/mol}) (298 \,K) = -580\, kJ/mol\)
- \(ΔH^o_rxn= 2 \Delta H_f^o \{\ce{H2} \} \) so \(\Delta H_f^o /2 \).
A consequence of the constant-volume condition is that the heat released corresponds to \(q_v\) and thus to the internal energy change \(ΔU_{sys}\) rather than to \(ΔH_{sys}\) under constant pressure conditions.
\[\Delta U_{sys} = q_v\nonumber \]
This comes from the definition of enthalpy
\[H_{sys} = U_{sys} + PV\nonumber \]
and associated change
\[\Delta H_{sys} = \Delta U_{sys} + \Delta (PV)\nonumber \]
or using the chain rule
\[\Delta H_{sys} = \Delta U_{sys} + \cancelto{0}{P\Delta V} + V \Delta P \label{Change in Bomb}\nonumber \]
which simplifies to
\[\Delta H_{sys} = \Delta U_{sys} + V \Delta P\nonumber \]
If we assuming ideal gas law for the gases
\[PV=nRT\nonumber \]
or
\[\Delta P = \dfrac{\Delta n R T}{V}\nonumber \]
substituting this into the equation for enthalpy change gives the enthalpy change under constant volume conditions in terms of changing number of moles
\[ΔH_{sys} = \Delta U_{sys} + Δn_gRT\nonumber \]
or
\[ΔH_{sys} = q_v + Δn_gRT\nonumber \]
where \(Δn_g\) is the change in the number of moles of gases in the reaction.
Q49
Carbon dioxide (\(\ce{CO_2}\)) is a common byproduct of the combustion of fossil fuels. Estimate the standard enthalpy of formation (\(\Delta H_f\)) of carbon dioxide at 25°C, use Table T1 .
- Solution
-
First write the equation:
\[\ce{C (s) + O2 (g) -> CO2 (g)}\nonumber \]
The bond enthalpies in Table T3 allows us to calculate \(\Delta H°\) because 2 C=O and 2 C-O bonds are formed:
\(\Delta H°\) ≈ 2 mol (-192.0 kJ mol -1 ) + 2 mol (- 85.5) = -555 kJ
Next we need to write the equations for "striping" the atoms from there standard state to single atoms. Each of the atomization has an enthalpy that was found in Table T1 :
\[\ce{C (s) -> C (g)}\nonumber \] \(\Delta H°\) = 1 mol (716.7 kJ mol -1 ) = 716.7 kJ
\[\ce{O2 (g) -> 2 O(g) }\nonumber \] \(\Delta H°\) = 2 mol (249.2 kJ mol -1 ) = 498.4 kJ
Combine the results of all three equations to calculate \(\Delta H_f\) of 1 mol of CO 2 :
-555 kJ + 716.7 kJ + 498.4 kJ = 660.1 kJ
This solution should be the same if finding the standard enthalpy of formation by using Table T1 which would be -393.5 kJ for one mole of CO 2 . Since this question is not clear, it is better if the students answering this question to just use Table T1 which will just use the standard heat of formation for and then have the sum of products minus the sum of reactants = the answer. So the answer would be -393.5 kJ.
Q51
Given the table of average bond enthalpies shown below, estimate the enthalpy change \(\Delta H°\) for the following reaction:
\[\ce{2H2 (g) + O2 (g) -> 2H2O (l)} \nonumber \]
| Bond | Bond Enthalpy (kJ/mol) |
|---|---|
| \(\ce{H-H}\) | \(436\) |
| \(\ce{H-O}\) | \(463\) |
| \(\ce{O=O}\) | \(498\) |
- Solution
-
To calculate the total enthalpy change of the reaction, we use Hess's Law. Thus, the total enthalpy of formation is equal to the sum of the enthalpy of formation of the reactants and the enthalpy of formation for the products.
Step 1. \(\Delta H_1\) = the bond enthalpies of \(\ce{2 H2 (g) + O2 (g)}\)
= \(2(\ce{H-H}) + (\ce{O=O})\)
= \(2(436) \; \mathrm{kJ/mol} + 498 \; \mathrm{kJ/mol}\)
= \(+1370 \; \mathrm{kJ/mol}\)
Step 2. \(\Delta H_2\) = the bond enthalpies of \(2 \ce{H2O} (l)\)
= \(2 \times 2 (\ce{H-O})\)
= \(4(463) \; \mathrm{kJ/mol}\)
= \(-1852 \; \mathrm{kJ/mol}\)
Step 3. \(\Delta H°\) = \(\Delta H_1\) + \(\Delta H_2\)
= \(1370 \; \mathrm{kJ/mol} - 1852 \; \mathrm{kJ/mol}\)
= \(-482 \; \mathrm{kJ/mol}\)
Note in Step 1, \(\Delta H_1\) is positive because the process of breaking the bonds into individual atoms is always an endothermic process while in Step 2, \(\Delta H_2\) is negative because the process of creating bonds out of the individual atoms is always an exothermic process.
Q59
Two quantum states are separated by an energy of 0.6 × 10 - 21 J . Calculate the relative populations of the two quantum states at a temperature of 33 °C.
- Solution
-
The equation for the relative populations of two quantum states, P 2 /P 1 is given by the equation
\[\dfrac{P_{2}}{P_{1}} = e^{-(E_{2} - E_{1}) / k_B T}\nonumber \]
Where E 2 -E 1 is the energy difference between the two states , \(k_B\) i s the Boltzmann constant, and T is the temperature.
Plug in the given values into the equation to reach the final answer.
\[\dfrac{P_{2}}{P_{1}}=e^{-(0.6 \cdot 10^{-21} \, \mathrm{J})/1.38\cdot 10^{-23} \mathrm{K^{-1}} (298 \, \mathrm{K}))}\nonumber \]
\[\dfrac{P_{2}}{P_{1}}=0.86\nonumber \]
Q63
By utilizing the Harmonic Oscillator Model, calculate the relative population of the first energy state and the ground state, both at \(278.15 \, \text{K}\), for \(\ce{H2}\). The force constant for \(\ce{H2}\) is \(510 \mathrm{\dfrac{N}{m}}\)
- Solution
-
A few pieces of information need to be understood before going about solving this equation. First, the question asks for the “relative population” of the first energy state at a given temperature (\(278.15 \, \mathrm{K}\)) to the energy state at \(0 \, \mathrm{K}\). Recall that energy states are basically a way of saying that every chemical species (molecule, atom, etc.) can “have” a specific (discreet) amount of energy. Each of these values are a “state”, and the lowest- value state can be considered the “ground state”, and all states above are “excited states”. Also recall that the states can be named by quantum number; i.e if \(n\) were the quantum number, than the ground state would be \(n=0\) and the first state after that, an excited state, has a quantum number \(n=1\). The next step is to understand what “relative population is”. Basically, this refers to the probability that the species, in this case \(\ce{H2}\), will be found in one energy state versus another. The equation that describes the probability of a species being in energy level \(n\) is as follows:
\(P(n)= Ce^{-\varepsilon_{n}/k_{b}T}\)
Where C is a constant, n is quantum number (energy level), \(k_{b}\) is Boltzmann’s constant, T is temperature, and \(\varepsilon\) is the energy of the molecule, which can be determined as such:
\(\varepsilon_{n} = \left(n + \dfrac{1}{2}\right)hv\)
\(hv = \dfrac{h}{2\pi}\sqrt{\dfrac{k}{\mu}}\)
n, as already seen, is the quantum number. \(hv\) in the equation for \(\varepsilon\) is defined as the product of planck’s constant (\(h\)) divided by \(2\pi\) and the \(\sqrt{\dfrac{k}{\mu}}\), where k is the force constant, which is given in the problem, and the \(\mu\) is the reduced mass , which can be calculated as follows:
\(\mu = \dfrac{m_{1}m_{2}}{m_{1} + m_{2}}\)
Where \(m\) is just the mass of each of the atoms. Since the mass of both hydrogens are 1 amu (\(u\)),
\(\mu = \dfrac{1\times1}{1 + 1} = \dfrac{1}{2} u\)
\(\dfrac{1}{2} u = 8.3027\times10^{-28}kg\)
Because the question is only interested in relative population, the actual number of molecules there are doesn’t really matter, since relative means to take the ratio between the probability of one energy state to another. Since the two energy states in comparison for this problem is \(n=1, n=0\), (remember ground state is \(n=0\)), the equation is as follows.
\(\dfrac{P(1)= Ce^{-\varepsilon_{1}/k_{b} 278.15 \, \mathrm{K}}}{P(0)= Ce^{-\varepsilon_{0}/k_{b} 278.15 \, \mathrm{K}}}\)
By simplification,
\(= e^{\dfrac{\varepsilon_{1} - \varepsilon_{0}}{K_{b}T}}\)
And, substituting in the formula for energy(\(\varepsilon\)),
\(= e^{\dfrac{\left[\left(n + \dfrac{1}{2}\right)hv - \dfrac{1}{2}hv\right]}{K_{b}T}}\)
\(= e^{nhv/K_{b}T}\)
Here, \(n=1\) (because the 0 has already been taken into consideration) and \(hv\) is as follows:
\(hv = \dfrac{h}{2\pi}\sqrt{\dfrac{k}{\mu}}\)
\(hv = \dfrac{6.62607\times10^{-34}\dfrac{kgm^{2}}{s^2}}{2\pi}\sqrt{\dfrac{510\dfrac{N}{m}}{8.3027\times10^{-28}kg}}\)
\(hv = \dfrac{6.62607\times10^{-34}Js^{-1}}{2\pi}\sqrt{\dfrac{510\dfrac{N}{m}}{8.3027\times10^{-28}kg}}\)
\(hv= 8.2650\times 10^{-20} J\)
All the variables have been found, so all there is left is to substitute and solve.
\(= e^{\dfrac{-\left(1\right)\left(8.2650\times 10^{-20} J\right)}{1.3806\times10^{-23}\dfrac{m^{2}kg}{s^{2}K}\left(278.15K\right)}}\)
\(= e^{\dfrac{-\left(1\right)\left(8.2650\times 10^{-20} J\right)}{1.3806\times10^{-23}\dfrac{J}{K}\left(278.15K\right)}}\)
\(\approx e^{-21.5229}\)
\(\approx 4.495\times10^{-10}\)
Thus, the probability of the \(\ce{H2}\) molecules being in the 1st energy state when compared to the vibrational state of the lowest energy is \(\approx 4.495\times10^{-10}\). This extremely low probability makes sense, because intuition says that the compound will most likely be found at its lowest energy.
Abstract : Find difference in energy, \(\varepsilon\), divide by Boltzmann's constant (\(K_{b}\)) and Temperature. Plug as \(e^{x}\) and solve.
Q69
A container holds 2 L of gas under 5.00 atm and a ball floating in 10 L of NaOH. As the volume of the gas expands to 20 L, the ball is turned upside down. This turn is caused by the temperature increase of the NaOH after the gas expands. Assuming that no heat is lost, the density of NaOH is 2.13 g/cm 3 and the specific heat is 4.184 J/g K, calculate the increase in temperature of the NaOH.
- Solution
-
The "work" completed by the ball to turn equals the negative value of the work the gas absorbs.
w = -(-P (V 2 -V 1 ))
w = -(-5 atm) (20-2) L
w = 90 atm L (101.325 J/ 1 atm L)
w = 9119.25 J
The work carried out by the ball by the NaOH (in order to turn the ball) correlates to the increase in 10 L of NaOH temperature .
10 L of NaOH (1000 mL/1 L) = 10000 mL = 10000 cm 3 (2.13 g/cm 3 ) = 21300 g of NaOH
q = mC p ΔT
ΔT = (q NaOH) / (Specific Heat NaOH * g NaOH)
ΔT = (9119.25 J) / (4.184 J/K g)(21300 g)
ΔT = 0.102 K gained by the NaOH to turn the ball
Q69
Some gas in a piston expands against a constant pressure of 1.2 atm from a volume of 3 L to 18 L. The piston turns an egg beater submerged in 150 g of water. If the water was originally at 25°C, what is its temperature once the gas stops expanding? Assume that all of heat goes into the water and the specific heat capacity of water is 4.184 J K -1 g -1 .
- Solution
-
From the first law of thermodynamics
\[\Delta U = q + w\nonumber \]
In this case,\(\Delta U =0\), so
\[q =w = P_{ext}\Delta V = P_{ext} (V_f - V_i)\nonumber \]
\[ q = (1.2 \; \mathrm{atm} )(18\; \mathrm{L} - 3\; \mathrm{L}) = 14.4 \; \mathrm{L\; atm} \nonumber \]
Convert From L atm to joules using the conversion factor
\[14.4 \; \mathrm{L\; atm} \times 101.325\; \mathrm{J \; L^{-1} atm^{-1}} = 1459\; \mathrm{J} \nonumber \]
Now connect the heat transferred to the temperature increased via the specific heat \(c_{sp}\) via
\[ q = m c_{sp} \Delta T\nonumber \]
\[ \Delta T = \dfrac{q}{m c_{sp}}\nonumber \]
\[\Delta T = \dfrac{1459\; \mathrm{J}}{(150\; \mathrm{g})(4.184\; \mathrm{J\; K^{-1} g^{-1}})} = 2.325 \; \mathrm{K} \nonumber \]
Since initial temperatures is 25° = 298 K so
\[T_{f} = 298 \; K + 2.325 \; \mathrm{K}\nonumber \]
\[T_{f}=300.325\; K = 27.325 \mathrm{°C} \nonumber \]


1.11: Introduction to the Elements of Music
Before we begin our first historical periods, the Middle Ages and the Renaissance, it will be important that you become familiar with some of the basic elements of music. The next few readings will focus on those basic elements. This information is vital to your understanding of the genres and styles we’ll be studying over the course of the semester.
This section includes the following pages:
- Slideshow: The Elements of Music
- Slideshow: Form
- A Historical Approach to the Elements of Music
- Melody
- Rhythm
- Meter in Music
- Yale Lecture on Rhythm and Meter
- Texture
Contributors and Attributions
CC licensed content, Original
- Provided by : Lumen Learning. License : CC BY: Attribution
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu . License : CC BY: Attribution


4.3: Informatics
By the end of this section, you will be able to:
- Explore the advantages, disadvantages, and capabilities of an EHR
- Define meaningful use and list the main components of meaningful use legislation
- Identify standard regulations relating to HIE
The field of healthcare informatics , or the field that incorporates health care and computer and information sciences, has drastically transformed the way the healthcare profession interacts (HIMSS, 2021). Communication is now transmitted through smartphones and web conferencing. Implementation of the EHR is one example of how the healthcare industry took advantage of this new digital platform. The seamless flow of information created by EHRs has changed the way patient care is delivered and how provided services are compensated. A few examples of how informatics has improved healthcare delivery include improved patient care outcomes, increased patient participation in health management, and improved care coordination among healthcare teams (Reid et al., 2021).
Nurse:
Gerald, CNS
Clinical Setting:
Pediatric acute care
Years in practice:
28
Facility location:
Major metropolitan area in southern Louisiana
Nursing informatics today is more important than ever in the clinical nursing world. As clinicians, we must adapt to the ever-evolving EHR. From the bedside nurse to the executive leadership of a large hospital system, we use informatics to drive clinical practice and improve patient care outcomes. Often people wonder what the informatics nurse does behind the scenes.
As the Clinical Nurse Specialist of a large hospital system, let me tell you that informatics nurses are the bridge that connect hands-on clinical practice and information technology. They drive patient care outcomes and population health statistics and report data for government organizations, such as the Centers for Medicare and Medicaid Services .
One of the biggest drivers for me is patient satisfaction. Data gathered on real perceptions of patient experiences. The data are broken down into categories for improvement on a monthly basis. These experiences include things such as quiet at night, recommendation to others of the services provided by the hospital, education received about medications, and interactions with clinicians. Informatics nurses disseminate surveys to all the patients and then collect data to be analyzed. They then compare results to state and national averages. My department uses this information for quality improvement projects and unit-based policy changes.
Another important aspect of nursing informatics is improving workflow for the EHR based on users’ recommendations. Hospitals have regulatory bodies that complete accreditation visits in person on an annual basis, if not biannually. The accrediting bodies want data and the ability to review patient charts. As a Clinical Nurse Specialist, I would not have the ability to paint the full picture of what was done to the patient, their positive outcomes, and financial revenue without the use of informatics .
Capabilities of an Electronic Health Record
The ability to share real-time patient information means the physician can access critical data no matter their physical proximity to the patient. A provider can be in their offsite office and access the hospital’s EHR system remotely to analyze critical diagnostic results. The healthcare team can confidentially and securely share information about the patient’s status change or vital health history . A decreased likelihood of errors exists because all providers have access to the patient’s most current medication list. For example, if a patient is seeing one primary physician and three specialty physicians, they can all have access to all prescriptions, resulting in a decreased probability of duplicative treatment and medication errors. Patients receive a more tailored and individualized plan of care because of real-time data exchange. Fewer forms are needed if the patient is transferred between facilities, as the electronic version is constantly updated and transferred with the patient.
One of the newest capabilities of the EHR is patient engagement. Many providers give patients access to their EHR, allowing a more patient-centered approach to care. This permits patients the capability to answer their own questions immediately, instead of waiting for office personnel to return a call. Patient access to the EHR also allows for direct and private messaging to their providers.
The Technology Informatics Guiding Education Reform (TIGER) initiative created by the leaders in healthcare informatics (Healthcare Information and Management Systems Society) has helped transform nursing informatics through personalized interactive learning resources. The TIGER Virtual Learning Environment is a free tool where students can watch webinars, access the resource library, and even take a free self-paced certification course.
Advantages
According to the American Nurses Association, care coordination among interdisciplinary healthcare members can be enhanced with incorporating electronic health records (Nelson & Parker, 2019). All departments have access to “real-time data” of vital information, without the need to wait for data transfer between departments. Primary care providers can easily distribute patient information to referrals or specialty care providers, without the previous lags in delivery. Clinical reasoning is improved because of access to more patient data, allowing for more well-informed decision-making and reduced safety risks. Additional advantages include efficient documentation that is legible; ability to collect and store data to be used for analytics; and built-in features, such as clinical decision support tools, electronic prescribing, and real-time data transfer to patient portals.
Disadvantages
Among the widespread advantages of EHRs, there are a few disadvantages. Technical challenges required health-care entities to develop their own information technology departments to handle these issues when they arise. Incorporation of the EHR has brought forth the need for policy revisions, organizational restructuring, and financial investment. These challenges, while adaptable, can be overwhelming to healthcare institutions. One of the most notable disadvantages of the EHR is related to user-related issues and design flaws. While informatics leadership has said that improved training and healthcare clinicians incorporated in the design process can alleviate these issues, it is still worth mentioning as one of the most cited disadvantages (Green, 2019). EHRs may also be a barrier to patients, particularly those who are unfamiliar with technology.
Meaningful Use and the Health Information Technology for Economic and Clinical Health Act
The Health Information Technology for Economic and Clinical Health (HITECH) Act , enacted as part of the American Recovery and Reinvestment Act of 2009, was signed into law on February 17, 2009. The main purpose of the HITECH Act is to promote meaningful ways to incorporate health information technology while still protecting patients’ privacy rights. It is important to note the HITECH Act played a significant role in the adoption of EHRs. They developed five main goals to help achieve this objective:
- The first goal is aimed at improving how, why, and where information technology should be incorporated.
- The second goal focuses on patients being active team members in their own care.
- The third goal was designed to improve care coordination among various interdisciplinary team members.
- The fourth goal is focused on improving population health through health information technology.
- The fifth goal ensures privacy standards are incorporated through evaluation of security measures when using electronic health information.
Using electronic health information in a meaningful way to improve patient care is known as meaningful use . The U.S. government developed minimum standards defining ways in which healthcare information could be used. These standards outline how the exchange of information among providers, patients, and insurers should take place. There are five pillars of health outcomes that support meaningful use (Figure 4.11) (Particle Health, 2023).
Releasing Health Record Information and Health Information Exchanges
Release of information (ROI) in health care is critical for quality patient care and positive patent care outcomes. The ROI plays a role in reporting between entities, researching current best practice standards, and billing for services provided. The Department of Health and Human Services’ Office of Civil Rights (OCR) ensures HIPAA privacy laws are applied to the release of healthcare information and the surrounding usability. The term usability refers to the extent to which a product can be used by specified users to achieve specified goals with effectiveness, efficiency, and satisfaction in a specified context of use. A few of the steps involved in ROI include following quality control regulations between sender and receiver, validating data being released, and sending the information in a timely manner. Failure to follow the guidelines outlined by the OCR and incorrectly releasing health information can violate privacy policies and result in noncompliance penalties.
Health Information Exchange
A health information exchange (HIE) allows for the exchange of health-related information among all stakeholders involved. HIE allows patients and healthcare team members to exchange information safely and securely. The healthcare entities must be extremely careful when releasing a patient’s health information. Nurses should verify that consents to release information are signed by the patient. Nurses should pay particular attention to release requests involving anyone under the state’s designated age of majority .
There are currently three identified rationales for health information exchange: directed exchange, query-based exchange, and consumer-mediated exchange (ONC, 2019). Directed exchange is used for sending public data regarding immunizations to local and state public health organizations or for direct reporting to CMS for the purpose of quality control initiatives. Directed exchange is also used for sending pertinent health data between providers. Query-based exchange is used to research best practice standards and clinical sources by providers when delivering patient care. For example, if a provider is faced with a disease or condition they are not accustomed to treating, they can access the research databases on current practice guidelines. Consumer-mediated exchange allows the patient access to their own health information. This form of HIE allows the patient the ability to monitor their own health, identify any mistakes in the demographic information, and make sure billing information is current.


13.1: Properties of proteins
40
Harper College Chemistry Department
Proteins are one of the most abundant organic molecules in living systems and have the most diverse range of functions of all macromolecules. Proteins may be structural, regulatory, contractile, or protective. They may serve in transport, storage, or membranes; or they may be toxins or enzymes. Each cell in a living system may contain thousands of proteins, each with a unique function. Their structures, like their functions, vary greatly. They are all, however, amino acid polymers arranged in a linear sequence.
Types and Functions of Proteins
Enzymes , which living cells produce, are catalysts in biochemical reactions (like digestion) and are usually complex or conjugated proteins. Each enzyme is specific for the substrate (a reactant that binds to an enzyme) upon which it acts. The enzyme may help in breakdown, rearrangement, or synthesis reactions. We call enzymes that break down their substrates catabolic enzymes. Those that build more complex molecules from their substrates are anabolic enzymes, and enzymes that affect the rate of reaction are catalytic enzymes. Note that all enzymes increase the reaction rate and, therefore, are organic catalysts. An example of an enzyme is salivary amylase, which hydrolyzes its substrate amylose, a component of starch.
Hormones are chemical-signaling molecules, usually small proteins or steroids, secreted by endocrine cells that act to control or regulate specific physiological processes, including growth, development, metabolism, and reproduction. For example, insulin is a protein hormone that helps regulate the blood glucose level. Table 1 lists the primary types and functions of proteins.
| Protein Types and Functions | ||
|---|---|---|
| Type | Examples | Functions |
| Digestive Enzymes | Amylase, lipase, pepsin, trypsin | Help in food by catabolizing nutrients into monomeric units |
| Transport | Hemoglobin, albumin | Carry substances in the blood or lymph throughout the body |
| Structural | Actin, tubulin, keratin | Construct different structures, like the cytoskeleton |
| Hormones | Insulin, thyroxine | Coordinate different body systems’ activity |
| Defense | Immunoglobulins | Protect the body from foreign pathogens |
| Contractile | Actin, myosin | Effect muscle contraction |
| Storage | Legume storage proteins, egg white (albumin) | Provide nourishment in early embryo development and the seedling |
Denaturation and Protein Folding
Each protein has its own unique sequence and shape that chemical interactions hold together. If the protein is subject to changes in temperature, pH, or exposure to chemicals, the protein structure may change, losing its shape without losing its primary sequence in what scientists call denaturation. Denaturation is often reversible because the polypeptide’s primary structure is conserved in the process if the denaturing agent is removed, allowing the protein to resume its function. Sometimes denaturation is irreversible, leading to loss of function. One example of irreversible protein denaturation is frying an egg. The albumin protein in the liquid egg white denatures when placed in a hot pan. Not all proteins denature at high temperatures. For instance, bacteria that survive in hot springs have proteins that function at temperatures close to boiling. The stomach is also very acidic, has a low pH, and denatures proteins as part of the digestion process; however, the stomach’s digestive enzymes retain their activity under these conditions.
Protein folding is critical to its function. Scientists originally thought that the proteins themselves were responsible for the folding process. Only recently researchers discovered that often they receive assistance in the folding process from protein helpers, or chaperones (or chaperonins) that associate with the target protein during the folding process. They act by preventing polypeptide aggregation that comprise the complete protein structure, and they disassociate from the protein once the target protein is folded.
For an additional perspective on proteins, view this animation called “Biomolecules: The Proteins.”


2.9: Putting It Together- Economic Environment
Synthesis
In this module you learned about the fundamental economic principles that affect the environment in which businesses operate.  Might you need to pack a sweater or an umbrella or grab some sunscreen? Perhaps, like Dorothy in the Wizard of Oz , head for the nearest cellar? If you ignore the forecast, you can find yourself unprepared and caught in a storm. Of course economic forecasts aren’t totally reliable—sometimes there’s a freak weather event that no one saw coming. Nonetheless, having a basic understanding of how supply and demand work, how different economic systems function, and how the business cycle connects to the economy can help you make informed decisions—and make the best out of a rainy day.
Summary
In this module you learned about the fundamental principles of economics and how they shape the business environment. Below is a summary of the key points covered.
What Is Economics?
Economics focuses on the ways in which people, businesses, and governments make decisions when faced with scarce resources. Economists study the economy at either the microeconomic level (focus on individuals) or the macroeconomic level (focus on systems).
Economic Systems
Economic systems can be organized as traditional, planned, or market economies. Traditional systems are hunter-gatherer economies in which people consume what they produce. In command economies such as communism and socialism, the government exercises a high degree of control over production and pricing. In market economies such as capitalism, free-market supply and demand drives what is produced and consumed. The increasing complexity of the world has led to mixed economic systems that have characteristics of both command and market economies.
Demand
Demand is the amount that consumers are willing and able to purchase of a good or service at a given price. Quantity demanded is a specific quantity that will be supplied at a single point (price) on the demand curve.
Supply
Supply is the amount of a good or service that a business is willing to produce at a given price. Quantity supplied refers to a specific quantity that will be supplied at a single point (price) on the supply curve.
Equilibrium
Equilibrium is said to exist at the point where quantity supplied equals the quantity demanded, and therefore there is no excess or shortage in the market. The market is “in balance.” The equilibrium price is the price where the amount that consumers want to purchase is equal to the quantity that the producers are willing to supply. The equilibrium quantity is the quantity supplied and demanded at the equilibrium price .
Health of the Economy
Economists use several measures to evaluate the health of an economy. Among the most important are GDP (Gross Domestic Product), the unemployment rate, and the CPI (Consumer Price Index). These three key economic indicators are used to measure how well the economy is achieving the goals of growth, high employment, and price stability.
Economic Stages
The business environment is cyclical, meaning it goes through a cycle of stages, each of which is characterized by a different set of economic conditions. The four stages of the business environment are expansion, peak, contraction, and trough.