lab assignment

Computer

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CHEM 4331 Computer Experiment 1: Mathematics of Physical Chemistry

“Computer! Computer?… Hello, computer.”

Lieutenant Commander Montgomery “Scotty” Scott

Star Trek IV: The Voyage Home (1986)

Contents:

page #

1. Objectives and Notes on Completing the Exercises 1

2. Introduction to Mathematica 3

3. Solutions of Algebraic Equations 19

4. Differential Calculus 32

5. Geometrical Interpretation of Derivatives 45

6. Integral Calculus 54

7. References 69

1. Objectives and Notes on Completing the Exercises

The purpose of this first computer Physical Chemistry laboratory is to help you review [and,

perhaps, somewhat extend] the mathematical techniques learned in the Pre-Calculus and Calculus

courses, and apply those to solving simple problems in already familiar basic Physics and

Chemistry. In the process you will also be introduced to one of [if not the!] best computer algebra

software Mathematica [1,2,3] designed by Dr. Stephen Wolfram [4] (PhD in Particle Physics from

Caltech) and being developed by Wolfram Research [5]. The main goal of this lab is to help you

better understand the material presented in CHEM 4330, and, perhaps, demonstrate that

performing mathematical manipulations can be actually exciting and not very time-consuming!

The laboratory requires you to complete in Mathematica ten (10) exercises included in th

is

handout. Each exercise is worth 10 points. Mathematica notebooks for all exercises must be

submitted to a D2L dropbox by the date and time indicated in your syllabus.

Warning! Mathematica notebooks must be uploaded to D2L in the Mathematica notebook

format (which is the default):

http://reference.wolfram.com/language/ref/format/NB.html

Uploading a Mathematica notebook to D2L in the plain text format with the resulting loss of

all formatting:

http://reference.wolfram.com/language/ref/format/Text.html

will incur an automatic deduction of 50% per each file. Please, do not do that!

http://reference.wolfram.com/language/ref/format/NB.html

http://reference.wolfram.com/language/ref/format/Text.html

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Each exercise contains a set of objectives listed as (a), (b), (c) etc.. A successful completion of

each objective includes obtaining

1) performing all calculations in Mathematica, and

1) obtaining the correct numerical value(s) and/or formula/equation(s), and

2) printing out the correct units (where applicable).

Even if one of those is incorrect or missing (for example, units!), no credit shall be given for the

objective. When in doubt, go back to the worked-out examples, and if it does not help, ask your

lab instructor.

Please note that all intermediate and final numerical values must be evaluated (calculated)

in Mathematica. That is, you are not allowed to simply copy answers to your Mathematica

notebook – it is a violation of the MTSU Academic Integrity policy!

In each Mathematica notebook, you must clearly indicate/label objectives (answers and graphs)

with symbols (a), (b), (c) … as instructed in the Exercises. For example:

(𝐚) 𝑣(𝑡) = 4.4  + 0.6𝑡2(m/s)

(𝐛) 𝑥(𝑡) = 1.4  + 4.4𝑡 + 0.2𝑡3 (m)

….

(𝐞) 𝑎(2 s) = 2.4 (m/s2)

….

Unlabeled answers will be ignored, i.e. no points shall be given for unlabeled answers.

When working on the exercises, it is permitted to consult with your classmates. In fact, we

encourage discussions that promote a creative learning environment. However, copying each

other’s code and/or using someone else’s file(s) is a violation of the MTSU Academic Integrity

policy, and is strictly prohibited! You absolutely must write your own code following the

guidelines and style of the worked-out examples provided in the handout. As such, you will have

to come up with your own comments and names for the variables/parameters used in your

calculations. Of course, make sure that those names are meaningful (see worked-out examples

provided in this handout). This is your chance to be creative – make good use of it!

A couple of [hopefully, useful] suggestions. When working with Mathematica notebooks, save

them often as it is not uncommon for Windows OS to crash. When done for the day, do not forget

to either copy all your files from the computer to your USB flash drive, or email the files to your

own email (home, college etc.) account. Alternatively, you can store files in the “cloud” – for

more information contact the MTSU IT department (ITD). This, of course, applies to ALL CHEM

4331 labs (not just this one)

Good luck!

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2. Introduction to Mathematica

Mathematica is a an extremely powerful computer algebra software for numerical and

symbolic calculations. It is extensively used as a research tool in many scientific, engineering,

mathematical and computing disciplines [3].

Mathematica is split into two parts, the kernel and the front end [3]. The kernel interprets

expressions (Mathematica code) and returns result expressions [3]. The front end provides a GUI,

(graphical user interface) which allows the creation and editing of Notebook documents containing

program code with prettyprinting, formatted text together with results including typeset

mathematics, graphics, GUI components, tables, and sounds [3]. All contents and formatting can

be generated algorithmically or interactively edited [3]. Most standard word processing

capabilities are supported [3]. It includes a spell-checker but does not spell check automatically

as you type [3].

Documents can be structured using a hierarchy of cells, which allow for outlining and

sectioning of a document and support automatic numbering index creation [3]. Documents can be

presented in a slideshow environment for presentations [3]. Notebooks and their contents are

represented as Mathematica expressions that can be created, modified or analyzed by Mathematica

programs [3]. This allows conversion to other formats such as TeX or XML [3].

The front end includes development tools such as a debugger, input completion and

automatic syntax coloring [3].

To start Mathematica, click on the Mathematica icon located on your desktop

or go through the Windows “Start” menu: “Start” → “All Programs” → “Wolfram Mathematica”

→ “Wolfram Mathematica”:

Note that in this tutorial we will be using Mathematica 10.1, while you may be using Mathematica

9, 11 or 12. Not to worry – the interface and functionality of these versions are almost the same!

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After the program is loaded, you will see a welcoming screen

Note that because we have already worked with Mathematica, we have several projects listed

under the “Open Recent” menu. Now, left-click on the “New Document” button (or choose the

“Notebook” option in the “New Document”

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Whatever the case, a new Mathematica notebook will open.

The Mathematica fully-interactive notebook interface looks almost like a text editor, and is quite

intuitive. There is a menu bar, and an empty [for now] notebook canvas. Several short tutorials

on how to start working with a notebook are available on the Mathematica website at:

https://reference.wolfram.com/language/tutorial/UsingANotebookInterface.html

https://reference.wolfram.com/language/guide/MenuItems.html

https://reference.wolfram.com/language/guide/NotebookBasics.html

https://reference.wolfram.com/language/tutorial/WorkingWithTheNotebookInterfaceOverview.html

Let’s perform a couple of very simple calculations. In your empty notebook, type in “2+2”

(without the double quotes, of course) and hit the key combination to perform a

computation:

Note how Mathematica labels the input (In[n]) commands and the output (Out[n]), where n is a

greater-than-zero integer number. Usually, for each input command, there is an output. On the

right side of the notebook canvas there are cell (inner) and group (outer) brackets. Each input cell

can contain one or more commands. At this time, there is only one input command in the first cell

(i.e. “2+2” in cell In[1]). If you wish to add another command, for example “2-2”, you can add it

to the first input cell (In[1]) by positioning the mouse cursor inside that cell and typing the text.

When done, hit to get Mathematica to do a computation:

https://reference.wolfram.com/language/tutorial/UsingANotebookInterface.html

https://reference.wolfram.com/language/guide/MenuItems.html

https://reference.wolfram.com/language/guide/NotebookBasics.html

https://reference.wolfram.com/language/tutorial/WorkingWithTheNotebookInterfaceOverview.html

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Alternatively, you can open a new cell by positioning the cursor below the last output cell, and

type in the command (of course, do not forget the “magic” key combination!):

If at any time you wish to delete all your output cells, you can do so by using the “Delete All

Output” option under the “Cell” menu:

If you wish to delete a particular cell (input or output), simply left-click the bracket of the cell you

wish to get rid of, and hit the “Delete” button on your keyboard. Now, delete all output and the

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bottom set of cells, and, using the “File” menu, save the notebook to the Windows desktop as

“Test-1.nb”. Now select all previously entered input and delete all cells to leave a blank notebook.

Mathematica obviously can do much more than a simple addition and subtraction. To

multiply the two numbers, use the “*”symbol. To divide the two numbers, use the “/”symbol.

Note the difference between the results of the two divisions. Division of 6 by 3 gives 2 which is

an exact number. Why didn’t Mathematica do the same for the 2/3 operation? This is NOT

because Mathematica does not know how to divide 2 by 3, but rather because it knows that the

result is 0.6666666666. Thus, rather than truncating the number at some arbitrary decimal, it keeps

it as a fraction. If you want to get a numerical value, use the N[…] operator:

https://reference.wolfram.com/language/ref/N.html?q=N

By the way, to make input easier, you can use the good old copy-and-paste option. First, select

the text you wish to copy using the “left-click and drag” technique. Then, select “Copy” from the

“Edit” menu (or simply hit “Ctrl+C” on the keyboard), and paste using either the “Paste” option

from the “Edit” menu or the “Ctrl+V” key combination. Trust me, it makes working with

Mathematica a lot easier!

https://reference.wolfram.com/language/ref/N.html?q=N

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Note how the N[2 3⁄ ] command evaluates the fraction to some default number of decimals. The

N[2 3⁄ , 20] command evaluates the same to 20 decimal points. N[2 3⁄ , 80] gives the result to 80

decimals. How about that?!

When fractions are entered using the “/” symbol, the code is not very readable. The Wolfram

System’s notebook interface is a very powerful typesetting system that allows you to enter

formulas as they are written in mathematical literature, using two-dimensional notation such as

superscripts, subscripts, and so on. For example, try entering the “2/3” fraction using the

key combination:

https://reference.wolfram.com/language/tutorial/EnteringInputInNotebooks.html

Looks much better, doesn’t it?

Another useful option in Mathematica is its ability to store the result of a mathematical operation

(numerical or symbolic) in a named variable.

For example,

WARNING!!! Do not use subscripts, superscripts, and special symbols such as but not

limited to ( ) [ ] & % $ # @ etc., in the names of the Mathematica variables because it can

easily lead to program errors and wrong results !

https://reference.wolfram.com/language/tutorial/EnteringInputInNotebooks.html

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To see what is stored in a given variable, simply type in its name, or use the Print[]

command:

https://reference.wolfram.com/language/ref/Print.html

Note how in the last statement (Print[“variable c = “, c]) we have supplemented the printout with

text. This type of output can be very helpful when writing a code, and will be required when

processing the exercises.

If you wish to clear the value or definition of a given variable, use the Clear[] command:

Clearing the declared values as they are no longer needed is definitely a good habit [6]. Beginning

a new Mathematica session, like starting with a clean slate, for each problem is a good practice

[6].

https://reference.wolfram.com/language/ref/Print.html

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An alternative, almost as good as starting with a clean slate, is to execute the command

Clear[“Global` ∗ “] which clears the values for all symbols in the Global` context, except those that

are protected [6]. Note that symbol “ ` ” is not an apostrophe, it is the so-called “grave accent”:

https://en.wikipedia.org/wiki/Grave_accent

which can be located in your keyboard as shown below [7]:

Here is an example that shows how to use the Clear[“Global` ∗ “] statement:

If you wish to execute the entire notebook automatically (one cell after another), use the “Evaluate

Notebook” option in the “Evaluation” menu. Keep in mind, that Mathematica keeps definitions

and assignments of all variables in memory as it executes notebook(s). Sometimes, you want to

start over, and erase all those assignments. In this case, use option “Quit Kernel” → “Local” in

the same “Evaluation” menu, and re-evaluate the cells of interest, or use the “Evaluate Notebook”

option to evaluate all cells in your notebook. To clear the output only, use the “Cell” → “Delete

All Output” option.

https://en.wikipedia.org/wiki/Grave_accent

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Now, let’s try some other familiar mathematical operations. To raise a number or a variable

(including a symbolic variable) to a power, you can either use the “^” symbol or . To take

square root, you can use the Sqrt[] function, or .

As you can see, the readability of the notebook improves dramatically when using the Mathematica

typesetting system. Consider the following example.

Recall that another way to take a square root is to use the power of ½ , for example, 9½.

Alternatively, 9½ can be entered as 9^(1/2) which, despite being more a cumbersome expression,

gives the same result:

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However, sometimes students forget about the order of the mathematical operations (

https://en.wikipedia.org/wiki/Order_of_operations ) and type in “9^1/2”, which naturally gives a

very different result (91/2 = 9/2 = 4.5):

In order to avoid mishaps like that, we strongly suggest that you use the Mathematica typesetting

system.

Note that a variable need not have an assigned numerical value for Mathematica to apply a certain

mathematical operation to it. This comes in very handy when doing symbolic computations (we

will get to that in just a moment).

Perhaps you have noted that Mathematica always prints out variables and parameters as they are

defined. If you do not want Mathematica to do that, terminate a command with the semicolon

symbol (“;”). For example, compare the two seemingly identical sequences:

https://en.wikipedia.org/wiki/Order_of_operations

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In this lab, we require your Mathematica notebooks to have user-friendly printouts of parameters

using the Print[] option. In the following, we show how to do that:

Note how the values of parameters a, b, and c are not printed out as they are defined because each

command is terminated with the semicolon. Indeed, rather than printing them out one by one

without annotation, it makes much more sense to print out their values in a separate Print[]

statement with accompanying [self-explanatory] text.

It is often desirable to separate two printout statements by a blank line. While one can use the

]” Print[” option to print out a blank line, a more elegant solution is to include the newline

character “\n” in the second Print[] statement.

https://reference.wolfram.com/language/tutorial/StringsAndCharacters.html#21585

For example,

By default, Mathematica prints out all numbers in some default output format. Only very large

and very small numbers are given in the scientific notation:

Note that we used the curly brackets to collect several numerical values in a list:

https://reference.wolfram.com/language/howto/WorkWithLists.html

https://reference.wolfram.com/language/tutorial/StringsAndCharacters.html#21585

https://reference.wolfram.com/language/howto/WorkWithLists.html

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To force Mathematica to print out numbers in scientific notation, use the ScientificForm[] option:

There are many other number output formats available. For more information, refer to

https://reference.wolfram.com/language/guide/DisplayOfNumbers.html

Note how all the Mathematica built-in functions (Print[], Clear[], N[], Sqrt[] etc. are spelled with

the first character in upper case followed by a set of square brackets “[]”, while all characters that

follow are in lower case. If you misspell a name of the function and omit the square brackets,

the function will not work! At the same time, the name of the unrecognized function will be

highlighted in blue:

Also note that when entering mathematical expressions, do not use the square and curly

brackets, i.e. “[]” and “{}”, as these are reserved for functions and lists, respectively. As such,

regardless of how the mathematical equation is written on paper, always use the round brackets

“()”, also called parentheses. Let’s consider an example. Suppose we wish to code in Mathematica

the following equation:

𝑓 = [
(𝑎 + 𝑏){𝑐 + 𝑑}

𝑔
]

1/

2

If you use the square and curly brackets in the expression typed in Mathematica, you will get an

error:

https://reference.wolfram.com/language/guide/DisplayOfNumbers.html

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The issue is easily fixed by replacing the square and curly brackets in mathematical expressions

with parentheses:

When it comes to using the Mathematica typesetting system, there is no need to remember all the

special key combinations for symbols of mathematical operations – you can always use the

“Palettes” option. Left-click on the “Palettes” menu and choose the “Basic Math Assistant” option

(actually, it is not very basic at all!):

You will immediately get a new “Basic Math Assistant” palette, which has two options” “Basic”

and “Advanced”. Try switching between those to see the difference. Note that in addition to a

number of algebraic and trigonometric functions,

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it has many Calculus-related options, such as derivatives, limits, sums, and integrals:

If you ever need to use the Greek letters in Mathematica, for example:

you may find the following tutorial useful:

https://reference.wolfram.com/language/guide/GreekLetters.html

Of course, constants, such as , i, and e can also be entered using the “Basic Math Assistant”

palette. Alternatively, you can use the “Special Characters” palette under the “Palettes” menu.

Finally, before you can start working on the exercises, let’s go over a couple of other very useful

functions.

Functions Simplify[], FullSimplify[], and Refine[] :

http://reference.wolfram.com/language/ref/Simplify.html

http://reference.wolfram.com/language/ref/FullSimplify.html

https://reference.wolfram.com/language/ref/Refine.html

request simplification of the specified equation(s). These functions are extremely powerful

because Mathematica knows how to handle a lot of mathematical identities.

For example, here is a familiar simplification:

https://reference.wolfram.com/language/guide/GreekLetters.html

http://reference.wolfram.com/language/ref/Simplify.html

http://reference.wolfram.com/language/ref/FullSimplify.html

https://reference.wolfram.com/language/ref/Refine.html

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Perhaps, something a little more complicated involving a trigonometric function:

If you need to learn more about the use of trigonometric functions in Mathematica, see the

following guide:

https://reference.wolfram.com/language/guide/TrigonometricFunctions.html

In the following example, we illustrate a more general use of the Simplify[] function:

1) define function f that depends on variables 𝑥, 𝑦, and 𝑧, which we usually write as 𝑓(𝑥, 𝑦, 𝑧),

2) differentiate 𝑓(𝑥, 𝑦, 𝑧) with respect to (w.r.t.) 𝑥, and

3) simplify the resulting expression:

Note that taking a derivative of function 𝑓(𝑥, 𝑦, 𝑧) w.r.t. variable 𝑥 is very easy in Mathematica.

Alternatively, we could have written:

However, from our point of view the first way is more efficient because one may want to do

something else with function 𝑓(𝑥, 𝑦, 𝑧), i.e. re-use the definition of function 𝑓. Perhaps, you know

that when a function of several variables is differentiated w.r.t. one of the variables, the result is

called a partial derivative of the function with respect to the variable. For example, when

differentiating function 𝑓(𝑥, 𝑦, 𝑧) w.r.t. parameter x, we get partial derivative

𝜕𝑓(𝑥,𝑦,𝑧)

𝜕𝑥
. Because

function 𝑓(𝑥, 𝑦, 𝑧) depends on three parameters, there will be three partial derivatives. The

remaining two partial derivatives are
𝜕𝑓(𝑥,𝑦,𝑧)

𝜕𝑦
and

𝜕𝑓(𝑥,𝑦,𝑧)

𝜕𝑧
. The total derivative (aka differential

https://reference.wolfram.com/language/guide/TrigonometricFunctions.html

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change or total differential) of function 𝑓(𝑥, 𝑦, 𝑧), denoted usually as d𝑓(𝑥, 𝑦, 𝑧), is the sum of all

of its partial derivatives:

d𝑓(𝑥, 𝑦, 𝑧) =

𝜕𝑓(𝑥, 𝑦, 𝑧)

𝜕𝑥
d𝑥 +

𝜕𝑓(𝑥, 𝑦, 𝑧)

𝜕𝑦
d𝑦 +

𝜕𝑓(𝑥, 𝑦, 𝑧)

𝜕𝑧
d𝑧

where each of d𝑥, d𝑦, and d𝑧 is called simply a differential.

Ok, now that you know how to do some basic [and not very basic] things in Mathematica, let’s

see how you can use it to perform familiar scientific computations.

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3. Solutions of Algebraic Equations

Before diving into Calculus (which is what Physical Chemistry is all about), let’s learn how to

solve equations with Mathematica. In this section we limit ourselves to solving [relatively simple]

algebraic equations: quadratic, cubic, and quartic. For example, we wish to solve a standard

quadratic equation in the form

𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 (1)

Of course, everyone knows that the two solutions for x are

the equilibrium constant 𝐾 (𝐾𝑝) is equal to 977. In this particular example, the equilibrium

constant is given by

𝑥 =
−𝑏 ± √𝑏2 − 4𝑎𝑐

2

𝑎

(2)

Let’s see if Mathematica can do that. In this case, we use the Solve[] option:

http://reference.wolfram.com/language/ref/Solve.html

Yes, it can! Note how we have used the “= =” symbol

https://reference.wolfram.com/language/ref/Equal.html

to define the equality telling Mathematica that the right had side of the equation (rhs) is equal to

the left-hand side (lhs).

Actually, according to the Mathematica documentation,

“[b]uilt into the Wolfram Language is the world’s largest collection of both numerical and

symbolic equation solving capabilities—with many original algorithms, all automatically accessed

through a small number of exceptionally powerful functions. The Wolfram Language’s symbolic

architecture allows both equations and their solutions to be conveniently given in symbolic form,

and immediately integrated into computations and visualizations.”

http://reference.wolfram.com/language/guide/EquationSolving.html

http://reference.wolfram.com/language/ref/Solve.html

https://reference.wolfram.com/language/ref/Equal.html

http://reference.wolfram.com/language/guide/EquationSolving.html

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Example 3.1. Let’s consider the following exercise taken from your Physical Chemistry textbook

[8]. At a certain temperature, for reaction

N2(𝑔) + 3H2(𝑔) ↔ 2NH3(𝑔) (3)

the equilibrium constant 𝐾 (𝐾𝑝) is equal to 977. In this particular example, the equilibrium
constant is given by

𝐾 = 𝐾𝑝 =
𝑝NH3

2

𝑝𝑁2 𝑝𝐻2
3

(4)

where 𝑝J is the partial pressure of substance J divided by the standard pressure of 1 bar,

i.e.

𝑝J ← 𝑝J/

𝑝

o (5)

where 𝑝o= 1 bar. Note that the equilibrium constant 𝐾 is a dimensionless (unitless) quantity which

is given by the value of the reaction quotient Q at equilibrium. Thus, for a general chemical

reaction of the form

[8]:

𝑎 A + 𝑏 B ↔ 𝑐 C + 𝑑 D (6)

the equilibrium constant is given by [8]

𝐾 = 𝑄equilibrium = (
𝛼C

𝑐 𝛼D

𝑑

𝛼A
𝑎𝛼B

𝑏
)

equilibrium

(7)

where 𝑎J is the activity of species J [8]. For perfect gases, 𝛼J = 𝑝J/𝑝
o, the partial pressure of J

relative to the standard pressure 𝑝o= 1 bar [8]. For solutes in very dilute solutions, 𝛼J = [J]/𝑐
o,

the molar concentration (molarity) of J relative to the standard value 𝑐o= 1 mol L-1 (M) [8]. The

activities of pure solids at 1 bar and pure liquids at 1 bar are all taken to be equal to 1 [8].

Table. Activities and standard states * [9].

If one wishes to discuss a gas-phase reaction in terms of molar concentrations (the amount of gas

molecules in moles divided by the volume of the container, [J] = 𝑛J/𝑉), rather than partial

pressures, the equilibrium constant 𝐾𝑝 is replaced with 𝐾𝑐 [8]:

𝐾𝑐 =
[C]𝑐 [D]𝑑

[A]𝑎[B]𝑏
(8)

where each molar concentration (or, rather its numerical value) is raised to a power equal to the

stoichiometric coefficient of the species in the chemical equation [8]. As usual, we have replaced

[J]/𝑐o by [J] itself, which represents the numerical value of the molar concentration of the gas J

[8].

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For example, for the chemical reaction shown in equation (3) [8]

𝐾𝑐 =
[NH3]

2

[N2][H2]
3
(9)

The conversion between 𝐾𝑐 and 𝐾𝑝 for reaction

𝑎 A + 𝑏 B ↔ 𝑐 C + 𝑑 D

is given by [8]:

𝐾𝑝

= 𝐾𝑐 (
𝑐o𝑅𝑇

𝑝o
)

(𝑐+𝑑)−(𝑎+𝑏)

= 𝐾𝑐 (
𝑐o𝑅𝑇
𝑝o
)

Δ𝜈gas

(10)

where 𝑇 is the temperature, 𝑅 is the ideal gas constant, and Δ𝜈gas is the difference in the

stoichiometric coefficients of the gas-phase species, products minus reactants [8]. Substituting the

values of 𝑐o, 𝑝o, and 𝑅 into equation (10) gives a simple and convenient form of this expression

[8]:

𝐾𝑝 = 𝐾𝑐 (
𝑇

12.027 K
)

Δ𝜈gas

(11)

where the temperature 𝑇 must be given in kelvin (K).

Now, let’s go back to our example. Suppose we mix in 1.00 bar of N2(𝑔) with 3.00 bar of H2(𝑔).

We need to find the equilibrium partial pressures of all three gases: (a) nitrogen, (b) hydrogen, and

(c) ammonia, and (d) verify our results by comparing the equilibrium constant calculated from the

equilibrium partial pressures with the one given in the question. After setting up the ICE table [8],

we get:

𝐾 =
(2𝑥)2

(1.00 − 𝑥)(3.00 − 3𝑥)3
(12)

Because the highest power of 𝑥 is 4, the resulting equation is quartic! Mathematica can easily

solve this equation:

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Note how we used the N[] function to get Mathematica to output numerical values. Because this

is a quartic equation, we get four solutions. Which one is correct? Well, mathematically all of

them are correct! However, according to Physical Chemistry, partial pressure (or any other

physically observable quantity) can not be a complex number (i.e. number which includes the

imaginary component with 𝑖 = √−1). Thus, solutions two and three are not physical, and should

be discarded! In fact, in Thermodynamics and Kinetics we are usually interested in real solutions

only

https://en.wikipedia.org/wiki/Real_number

If x is real, it must belong to a domain of real numbers. In the language of Mathematics, it can be

written as 𝑥 ∈ ℝ, where the symbol “∈” means “is an element of” and the symbol “ℝ” denotes the

set of ALL real numbers. How about that! Anyway, if we are interested only in real solutions of

a given equation, the Solve[] command can be modified as follows:

Note how we have reduced the number of solutions to two. But which one (or both) is (are)

physically meaningful? Recall that in our ICE table, x denotes the change in partial pressure. The

equilibrium partial pressures of N2(𝑔) and H2(𝑔) are (1.00-x) bar and (3.00-3x) bar. Solution x =

1.1173 bar would give us negative equilibrium pressures of N2(𝑔) and H2(𝑔), and thus is not

physically meaningful. This leaves us with only one solution: x = 0.895018 bar. Now we can

calculate the final partial pressures of all three gases. Here is our Mathematica code that does that:

https://en.wikipedia.org/wiki/Real_number

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Let me explain what we are doing in each cell:

In[1]: Solve equation, get numerical solutions, and store all solutions in a variable (to be more

precise, a list of variables) called “solutions”

In[2]: Extract the first solution (command “solutions[[1]]”), and put it in variable “x” using

command “/.”. For example, command a2/. a → 2 means: evaluate “a2” while replacing

“a” with 2 ( the symbol → can be typed in Mathematica as “->”; for more information

about this so-called “\[Rule]” option, see

https://reference.wolfram.com/language/ref/character/Rule.html

beware that \[Rule] is not the same as \[RightArrow] ! ). Naturally, the answer is 4.

To do the same with several variables is not much more difficult:

A detailed information on the “/.” command can be obtained here:

http://reference.wolfram.com/language/ref/ReplaceAll.html

In[3]: Calculate the final partial pressures of all three gases and store them in variables pN2, pH2,

and pNH3. Note how the names of variables are self-explanatory. This is VERY

important !

Note on using units in Mathematica: Each scientific calculation must include units. In general,

Mathematica can use units:

https://reference.wolfram.com/language/guide/Units.html

However, the interface is not very user-friendly. Thus, for now, we shall ignore this functionality

when doing calculations in Mathematica.

https://reference.wolfram.com/language/ref/character/Rule.html

https://reference.wolfram.com/language/ref/character/RightArrow.html

http://reference.wolfram.com/language/ref/ReplaceAll.html

https://reference.wolfram.com/language/guide/Units.html

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That said, in every single Mathematica notebook you absolutely must include comments that

describe your calculation steps, and the units of variables used. A comment in Mathematica is any

text entered between symbol sequences “(*” and “*)”

Note how we have also added a check for 𝐾: the numerical value of the reaction quotient at

equilibrium must be equal to the equilibrium constant (𝐾 = 977). The difference between the

equilibrium constant that was calculated using the equilibrium partial pressures of the three gases

and the given value is below 10-12, which means that the final equilibrium partial pressures must

be correct!

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Now we tweak the processing of variables (to make the code more transparent and portable, and

less cluttered), and add the project number, author name, date, the Clear[“Global` ∗ “] command,

and user-friendly output of calculated properties. See for yourself:

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Note how we suppressed the default Mathematica printout by terminating each command (except

for the Solve[] statement) with a semicolon. This eliminates unnecessary printout from the final

version of the code (we suggest you do so only in the final version of the code, after you have

made sure that it produces correct results, and when you are ready to upload it to D2L).

Finally, once you are satisfied with your code, go through the following checklist before

submitting your Mathematica notebook to D2L:

1) Make sure that the header of the Mathematica notebook includes the exercise number, your

name, and the creation date.

2) Make sure that the first executable statement in your Mathematica code is

Clear[“Global` ∗ “];

3) Make sure that all required answers are printed out using the Print[] statement.

4) Verify that each Print[] statement includes the appropriate label, i.e.

Print[“(a) …… “];

Print[“(b) …… “];

Print[“(c) …… “]; etc.

5) Suppress output from lines that do not contain printout of answers.

6) Remove all output cells (go to the “Cell” option of the notebook menu and select “Delete

All Output”)

7) Merge all input cells, which can be done using the following two steps:

a. go to the “Edit” option of the notebook menu and select “Select All” (or simply use

the “Ctrl+A” key combination)

b. go to the “Cell” option of the notebook menu and select “Merge Cells” (or simply

use the “Shift + Ctrl + M”

8) Save the notebook, and upload it to D2L.

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Here is the final version of our Mathematica notebook for Example 3.1:

Note that your Mathematica notebook for this example should look very similar!

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Perhaps now is a good time to also demonstrate how to use the Mathematica Solve[]

function

http://reference.wolfram.com/language/ref/Solve.html

to solve two equations for two unknowns simultaneously. For example,

{

𝑎𝑥 + 𝑦 = 7
𝑏𝑥 − 𝑦 =

1

where 𝑥 and 𝑦 are the unknowns, and 𝑎 and 𝑏 are constants:

Note how we used the “&&” symbol to specify the logical AND function. Suppose, we wish to

save the solutions to some variables. Here is how we save the solution for 𝑥 to a variable called

xs:

, and the same for 𝑦:

If numerical values for 𝑎 and 𝑏 are known, xs and ys can be easily evaluated. For example,

Not bad!

Mathematica can solve even more complicated equations. For example, determining the real

solutions for the following system of equations:

{
(𝑥2 − 𝑦2)𝑥𝑦 = 180

𝑥2 − 𝑥𝑦 − 𝑦2 = −11

is easy in

Mathematica:

http://reference.wolfram.com/language/ref/Solve.html

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You can then use the replacement rule

https://reference.wolfram.com/language/tutorial/TransformationRulesAndDefinitions.html

to extract numerical values of x and y for any or all of the solutions. For example,

This technique allows you to use variables x1, y1, x2, y2 etc. that contain the solutions in some

other calculations. For example,

How about a system of three equations for three unknowns? No problem! Here is an example

{

𝑥 + 𝑦 + 𝑧 = 13

𝑥2 + 𝑦2 + 𝑧2 = 61
𝑥𝑦 + 𝑥𝑧 = 2𝑦𝑧

, and its solution in Mathematica:

Extracting numerical values for any of the solutions is also straightforward:

Isn’t it nice?!

https://reference.wolfram.com/language/tutorial/TransformationRulesAndDefinitions.html

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Note that if you are also interested in the complex solutions, simply remove the “Reals” option:

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Now that we have shown you how to perform this kind of calculations, it is your turn to do some.

Exercise 3.1. In the same chemical reaction as discussed in Example 3.1 but at a different

temperature, the nitrogen, hydrogen, and ammonia gases are mixed at partial pressures of 4.56 bar,

9.87 bar, and 3.21 bar, respectively, and allowed to reach equilibrium under conditions when K =

86.4. Using the final notebook from Example 3.1 as a template, create a Mathematica notebook

to calculate the equilibrium partial pressures (bar) of all three substances: (a) N2(g), (b) H2(g), (c)

NH3(g), and (d) perform the check for 𝐾𝑝. Save the notebook as “Exercise_3.1.nb”. Warning!

All the calculation steps must be performed in your Mathematica notebook, and all items (a-

d) must be clearly labelled using the Print[] functions, or the score for this exercise will be

zero!

Answers: (a) … (b) 0.858 bar (c) … (d) …

Exercise 3.2. In a study of the equilibrium,

A2 + B2 ⇌ 2 AB

1.0 mol of A2 and 3.0 mol of B2 gave rise at equilibrium to x mol of AB. Addition of a further 2.0

mol of A2 gave an additional x mol of AB. In Mathematica, (a) determine the value of x (mol),

and (b,c) evaluate the equilibrium constant via each of the two expressions that you set up to

determine (a). Save the notebook as “Exercise_3.2.nb”. Warning! All the calculation steps

must be performed in your Mathematica notebook, and all items (a-c) must be clearly labelled

using the Print[] function, or the score for this exercise will be zero! Hint: two ICE tables is

twice as much fun…

Answers: (a) 1.5 (b) … (c) 4.0

Exercise 3.3. Consider the following gas-phase dissociation reaction:

A2(g) ⇌ 2 A(g)

In Mathematica, derive a symbolic expression for the degree of dissociation of A2(g) at equilibrium

{ratio of the change in the partial pressure of A2(g) to the initial partial pressure of A2(g)}, , as a

function of the volume of the reaction vessel, V (liters), temperature, T (kelvins), initial chemical

amount of A2(g), n (moles), the equilibrium constant, 𝐾𝑝 (dimensionless), and the gas constant, R

(J K-1 mol-1). Do not assign numerical value to the gas constant yet! Use your newly derived

formula to calculate the degree of dissociation of A2(g) in a (b) 5.0 L and (c) 20 L reaction vessel

when 0.10 mol of A2(g) is dissociating at 2000 K, and the equilibrium constant is 0.570. Save the

notebook as “Exercise_3.3.nb”. Warning! All the calculation steps must be performed in

your Mathematica notebook, and all items (a-c) must be clearly labelled using the Print[]

functions, or the score for this exercise will be zero! Hints: the easiest way to evaluate an

expression numerically without permanent assignment of the variables is to use the rule option;

for example, (a + b)2/. {a−> 2, b−> 3} and (a + b)2/. {a−> 14, b−> 15}.

Answers: (a)
10𝐾𝑝𝑉

5𝐾𝑝𝑉+√𝐾𝑝𝑉(4𝑛𝑅𝑇+25𝐾𝑝𝑉)
(b) … (c) 0.34

https://reference.wolfram.com/language/ref/Rule.html

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4. Differential Calculus

In this part of the lab, we are going to review some of the most fundamental physical concepts,

and at the same time refresh your knowledge of Differential Calculus.

Displacement [10]. Pretty much everything around [and inside] us is in motion. Motion involves

the displacement of an object from one place in space and time to another. Describing motion

requires some convenient coordinate system and a specified origin. A frame of reference is a

choice of coordinate axes that defines the starting point for measuring any quantity, an essential

first step in solving virtually any problem in mechanics. The displacement Δ𝑥 of an object is

defined as its change in position and is given by

Δ𝑥 = 𝑥𝑓 − 𝑥𝑖 (13)

where 𝑥𝑖 is the coordinate of the initial position of the object and 𝑥𝑓 is the coordinate of the objects’

final position (here, indices 𝑖 and 𝑓 stand for initial and final, respectively). The SI unit of

displacement is a meter (m).

Because displacement has both a magnitude (size) and a direction, it is a vector quantity.

In general, a vector quantity is characterized by having both a magnitude and a direction. By

contrast, a scalar quantity has magnitude, but no direction. Scalar quantities such as mass and

temperature are completely specified by a numeric value with appropriate units; no direction is

involved.

Velocity and speed [10]. In everyday usage the terms speed and velocity are interchangeable. In

physics, however, there’s a clear distinction between them: speed is a scalar quantity, having only

magnitude, whereas velocity is a vector quantity, having both magnitude and direction. However,

the two quantities share the same SI unit: meter per second (m/s).

The average speed of an object over a given time interval is the length of the path it travels

divided by the total elapsed time:

timeelapsed

lengthpath
speed Average = (14)

Unlike average speed, average velocity is a vector quantity, having both a magnitude and a

direction. The average velocity �̅� during a time interval Δ𝑡 is the displacement, Δ𝑥, divided by

the elapsed time, Δ𝑡:

�̅� =

∆𝑥

∆𝑡
=

𝑥𝑓 − 𝑥𝑖

𝑡𝑓 −

𝑡𝑖

(15)

While the average speed is always non-negative, the average velocity of an object in one dimension

can be either zero, or positive or negative, depending on the sign of the displacement. (The time

interval Δ𝑡 is always positive.)

The instantaneous velocity v is the limit of the average velocity as the time interval Δ𝑡

becomes infinitesimally small:

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𝑣 = lim
Δ𝑡→

0

∆𝑥

∆𝑡
(16)

The notation lim
Δ𝑡→0

means that the ratio Δ𝑥/Δ𝑡 is repeatedly evaluated for smaller and smaller time

intervals Δ𝑡. Notice that the displacement Δ𝑥 approaches zero as ∆𝑡 approaches zero, so the ratio

looks like 0/0 [11]. While this ratio may appear to be difficult to evaluate, it does have a specific

value. As Δ𝑥 and ∆𝑡 become smaller and smaller, the ratio Δ𝑥/Δ𝑡 approaches a value equal to the

slope of the line tangent to the 𝑥-versus- 𝑡 curve, which is defined as the instantaneous velocity

(Figure 1).

Figure 1. The instantaneous x-velocity v at any given point equals the slope of the tangent to the

𝑥 − 𝑡 curve at that point [12].

In the Calculus notation, this limit is called the derivative of x with respect to (w.r.t.) t, written as

𝑑𝑥/𝑑𝑡 [11]

𝑣 = lim
Δ𝑡→0
∆𝑥
∆𝑡
=

𝑑𝑥

𝑑𝑡
(17)

The instantaneous speed of an object, which is a scalar quantity, is defined as the magnitude of

the instantaneous velocity. Like average speed, instantaneous speed (which we will usually call,

simply, “speed”) has no direction associated with it and hence carries no algebraic sign.

Acceleration, a [10]. The changing of an object’s velocity with time is called acceleration. The

average acceleration �̅� during the time interval ∆𝑡 is the change in velocity ∆𝑣 divided by ∆𝑡:

�̅� =

∆𝑣

∆𝑡
=

𝑣𝑓 −

𝑣𝑖

𝑡𝑓 − 𝑡𝑖
(18)

Acceleration is a vector quantity having dimensions of length divided by the time squared. In the

SI system of units, acceleration has dimensions of meters per second per second ( (m/s)/s, which

is usually written as m s-2 ). Acceleration is a vector quantity. For the case of motion in a straight

line, the direction of the velocity of an object and the direction of its acceleration are related as

follows: When the object’s velocity and acceleration are in the same direction, the speed of the

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object increases with time. When the object’s velocity and acceleration are in opposite directions,

the speed of the object decreases with time (this type of acceleration is often called deceleration).

a) b)

Figure 2. (a) A car, modeled as a particle, moving along the x axis from A to B, has velocity 𝑣𝑖
at 𝑡 = 𝑡𝑖 and velocity 𝑣𝑓 at 𝑡 = 𝑡𝑓 . (b) Velocity–time graph (red-brown) for the particle moving

in a straight line. [11]

The value of the average acceleration often differs in different time intervals, so it is useful to

define the instantaneous acceleration, which is analogous to the instantaneous velocity discussed

above. The instantaneous acceleration a is the limit of the average acceleration as the time interval

∆𝑡 goes to zero:

𝑎 = lim
Δ𝑡→0

∆𝑣

∆𝑡
(19)

Here again, the notation lim
Δ𝑡→0

means that the ratio ∆𝑣/∆𝑡 is evaluated for smaller and smaller

values of ∆𝑡. The closer ∆𝑡 gets to zero, the closer the ratio gets to a fixed number, which is the

instantaneous acceleration.

In the language of Calculus, instantaneous acceleration equals the derivative of the velocity

with respect to time:

𝑎 = lim
Δ𝑡→0
∆𝑣
∆𝑡
=

𝑑𝑣

𝑑𝑡
(20)

which, by definition, is the slope of the velocity–time graph [11]. The slope of the green line in

Figure 2b is equal to the instantaneous acceleration at point B. Notice that Figure 2b is a velocity–

time graph, not a position–time graph like Figure 1. Therefore, we see that just as the velocity of

a moving particle is the slope at a point on the particle’s 𝒙 − 𝑡 graph, the acceleration of a particle

is the slope at a point on the particle’s 𝒗 − 𝑡 graph. One can interpret the derivative of the velocity

with respect to time as the time rate of change of velocity. If 𝑎 is positive, the acceleration is in

the positive 𝑥 direction; if 𝑎 is negative, the acceleration is in the negative 𝑥 direction (often called

deceleration) [11].

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Because 𝑣 = 𝑑𝑥/𝑑𝑡, the acceleration can also be written as

𝑎 =
𝑑𝑣

𝑑𝑡
=

𝑑

𝑑𝑡
(𝑣) =

𝑑

𝑑𝑡
(

𝑑𝑥

𝑑𝑡
) =

𝑑2𝑥

𝑑

𝑡2

(21)

That is, in one-dimensional motion, the acceleration also equals the second derivative of x with

respect to time [11].

Figure 3 [11] illustrates how an

acceleration–time graph is related to a velocity–

time graph. The acceleration at any time is the

slope of the velocity–time graph at that time.

Positive values of acceleration correspond to those

points in Figure 3 where the velocity is increasing

in the positive 𝑥 direction. The acceleration

reaches a maximum at time 𝑡A, when the slope of

the velocity–time graph is a maximum. The

acceleration then goes to zero at time 𝑡B, when the

velocity is a maximum (that is, when the slope of

the 𝑣 − 𝑡 graph is zero). The acceleration is

negative when the velocity is decreasing in the

positive 𝑥 direction, and it reaches its most

negative value at time 𝑡C [11].

Figure 3 [11]. (a) The velocity–time

graph for a particle moving along the x

axis. (b) The instantaneous acceleration

can be obtained from the velocity–time

graph.

Example 4.1. Consider a particle moving along the x axis according to the equation

𝑥(𝑡) = 𝒜 + ℬ 𝑡 + 𝒞 𝑡2 (22)

where 𝑥 is in meters (m), 𝑡 is in seconds (s), and 𝒜, ℬ, and 𝒞 are some constants. Using

Mathematica, derive general expressions for (a) velocity, and (b) acceleration as a function of time

𝑡 and general coefficients 𝒜, ℬ, and 𝒞. If 𝒜 = 2.00 m, ℬ = 3.00 m/s, and 𝒞 = −1.00 m/s2,

calculate (c) the position of the particle, (d) its velocity, and (e) its acceleration at 𝑡 = 3.00 s. Plot

(f) 𝑥 vs 𝑡, (g) 𝑣 vs 𝑡 , and (h) 𝑎 vs 𝑡 for 𝑡 = 0…5 seconds.

Open a new Mathematica notebook, and name it “Example_4.1.nb”. After typing in the project

number and clearing all the variables, define function 𝑥(𝑡) given above:

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Note that in order to match the notation for constants 𝒜, ℬ, and 𝒞 in equation (22), we used the

Mathematica “script capital” option when entering in the names of the variables:

https://reference.wolfram.com/language/ref/character/ScriptCapitalA.html

https://reference.wolfram.com/language/ref/character/ScriptCapitalB.html

https://reference.wolfram.com/language/ref/character/ScriptCapitalC.html

To learn more about letters and letter-like forms in Mathematica, see

https://reference.wolfram.com/language/tutorial/MathematicalAndOtherNotation.html#19606

First, we derive a general expression for velocity at any 𝑣, i.e. find 𝑣(𝑡). Recall that

𝑣(𝑡) =
𝑑𝑥(𝑡)

𝑑𝑡
,

so we need to differentiate 𝑥(𝑡) with respect to (w.r.t.) 𝑡. In Mathematica, differentiation can be

performed using the command D[]

https://reference.wolfram.com/language/ref/D.html?q=D

or using the button in the section of the “Basic Math Assistant” palette:

https://reference.wolfram.com/language/ref/character/ScriptCapitalA.html

https://reference.wolfram.com/language/ref/character/ScriptCapitalB.html

https://reference.wolfram.com/language/ref/character/ScriptCapitalC.html

https://reference.wolfram.com/language/tutorial/MathematicalAndOtherNotation.html#19606

https://reference.wolfram.com/language/ref/D.html?q=D

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Here is my code that demonstrates how to use both options (of course, you should use only one,

whichever makes more sense!):

As expected, the two differentiation commands give the same result. Thus, 𝑣(𝑡) = ℬ + 2 𝒞 𝑡

Note how we used the Print[] statement to print out the answer for (a).

How about the acceleration? Well, according to equation (20), we get 𝑎(𝑡) by differentiating 𝑣(𝑡)

w.r.t. 𝑡 once, or by differentiating 𝑥(𝑡) w.r.t. 𝑡 twice: 𝑎(𝑡) =
𝑑𝑣(𝑡)

𝑑𝑡
=

𝑑2𝑥(𝑡)

𝑑𝑡2
. Both methods can be

easily implemented in Mathematica (of course, use only one, whichever makes more sense!):

Again, as expected, 𝑎 is the same regardless of the method used. Note that since 𝑎 does not depend

on 𝑡, we are dealing with a case of the constant acceleration, i.e., 𝑎 = 2 𝒞 m/s2 (units come directly

from the question) at any time 𝑡. To summarize,

𝑥(𝑡) = 𝒜 + ℬ 𝑡 + 𝒞 𝑡2, where 𝑡 is in seconds, and 𝑥 is in meters,

𝑣(𝑡) = ℬ + 2 𝒞 𝑡, where 𝑡 is in seconds, and 𝑣 is in m/s.

𝑎(𝑡) = 2 𝒞 at any 𝑡, where 𝑎 is in m/s2.

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Now we need to define the numerical values for constants 𝒜, ℬ, and 𝒞. There are several ways

one can do that. The first way is quite straightforward:

However, the second way, using the Mathematica list option,

https://reference.wolfram.com/language/howto/CreateLists.html

is more elegant and uses less space in the notebook:

Of course, you should use only one method, whichever makes more sense! Then, finding 𝑥, 𝑣,

and especially 𝑎 at 𝑡 = 3.00 s is easy:

Note how the numerical value of 𝑡 is defined in one place only which makes the code much more

portable. Also note how we use the Print[] statements to create a user-friendly output. Finally,

note that acceleration, 𝑎, is of course constant at -2.00 m/s2.

Now, let’s get to a fun part. We can easily plot 𝑥(𝑡), 𝑣(𝑡), and 𝑎(𝑡) in Mathematica using the

Plot[] function. In the following, we show how to do it for 𝑡 = 0…5 s:

First, we define the range for 𝑡 using variables that we are going to call tmin and tmax:

https://reference.wolfram.com/language/howto/CreateLists.html

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Alternatively, you can use a more conventional way to define tmin and tmax:

Now, we plot 𝑥(𝑡) in the range of 𝑡 from tmin to tmax:

While the plot looks nice, it is missing labels for the x- and y- axes, and the plot label (legend).

Those can be added to any Mathematica plot as follows:

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Much better! Note that we used Mathematica’s function SetOptions[Plot, BaseStyle →

{FontSize → 14}] to increase the font size in the plot, and italicized “𝑡” and “𝑥” in the AxisLabel

and PlotLabel commands.

The SetOptions[Plot … . ] command applies to all plots that follow it, so there is no need to repeat

it separately for each new plot. To make changes to the plot style, simply insert another

SetOptions[Plot … . ] command with a new set of options.

Finally, let’s plot 𝑣(𝑡) and 𝑎(𝑡):

Note that in all three graphs, we specify the range for 𝑡 using variables tmin and tmax that are

defined only once!

We also hope that you can see that because velocity changes uniformly with time (i.e. the 𝑣 − 𝑡

plot is a straight line), acceleration is constant, and vice versa.

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Here is our final notebook for Example 4.1:

Finally, it may be worth the time to demonstrate how easy it is to plot two or more function on the

same graph in Mathematica. Consider the following three functions

𝑦1 = 1 + 2𝑥, 𝑦2 = 10 − 3𝑥, 𝑦3 = −5 + 𝑥
2

that we wish to plot on the same graph for 50  x . First, we define the functions:

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Then, we define the font size for the plot and the minimum and maximum values for x:

Because in this example we have terminated each command with a semicolon, there is no output

from either one of them. Finally, we plot all three functions on the same plot using the same

command Plot[] which is used to plot a single function:

Note how we have included a sequence of functions that we want to plot in the curly brackets thus

forming a so-called “list”:

https://reference.wolfram.com/language/howto/CreateLists.html

Also note that in addition to already familiar options AxesLabel and PlotLabel, we added the

PlotLegends option that, as you can guess from its name, adds a legend to the graph.

https://reference.wolfram.com/language/howto/CreateLists.html

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Sometimes Mathematica does not plot the entire range for the 𝑥 and/or 𝑦 variable(s). For example,

To fix that, simply add the PlotRange→Full option to either the Plot[] or SetOptions[Plot…]

command:

Note how in all the plots, we always italicize symbols for variables and physical quantities,

i.e. 𝑥, 𝑦, 𝑣, 𝑡, 𝑎, etc.

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Ok, it is now your turn to perform a couple of calculations in Mathematica.

Exercise 4.1. An object moves along the x axis according to the equation

𝑥(𝑡) = −𝒜 + ℬ 𝑡2 − 𝒞 𝑡3

where 𝑥 is the distance in centimeters (cm), 𝑡 is the time in seconds (s), and 𝒜, ℬ, and 𝒞 are some

constants. Using the notebook from Example 4.1 as a template, write a Mathematica code to

derive expressions for (a) velocity and (b) acceleration as a function of time and general

coefficients 𝒜, ℬ and 𝒞. Suppose 𝒜 = −45. cm, ℬ = 0.67 cm s−2, and 𝒞 = 3.2×10−2 cm

s-3. Calculate (c) the distance, (d) velocity, and (e) acceleration at 𝑡 = 8.5 s. Plot (f) 𝑥 vs 𝑡, (g)

𝑣 vs 𝑡, and (h) 𝑎 vs 𝑡 for 𝑡 = 0 … 15 s. Save the notebook as “Exercise_4.1.nb”. Warning! All

the calculation steps must be performed in your Mathematica notebook, and all items (a-h)

must be clearly labelled in the Print[] functions and/or the PlotLabel option in the Plot[]

commands, or the score for this exercise

will be zero!

Answers: (a) … (b) …

(c) 74 cm (d) … (e) -0.29 cm s-2

(f,g,h) …

Exercise 4.2. An object moves along the x axis according to the equation

𝑥(𝑡) = (𝑔 +
√𝑡3

ℎ
) (1 − 𝑒−𝑘𝑔𝑡 ) −

𝑡2

𝑛

where 𝑥 is the distance in meters (m), 𝑡 is the time in seconds (s), 𝑒 is the base of the natural

logarithm, and 𝑔, ℎ, 𝑘, and 𝑛 are some constants. Using the notebook from Example 4.1 as a

template, write a Mathematica code to derive expressions for (a) velocity and (b) acceleration as

a function of time and general coefficients 𝑔, ℎ, 𝑘, and 𝑛. Suppose 𝑔 = 12. m , ℎ = 0.19 s3/2

m−1, 𝑘 = 1.6×10−2 m−1 s−1, and 𝑛 = 0.76 s2 m−1. Calculate (c) the distance, (d) velocity, and

(e) acceleration at 𝑡 = 6.7 s. Plot (f) 𝑥 vs 𝑡, (g) 𝑣 vs 𝑡, and (h) 𝑎 vs 𝑡 for 𝑡 = 0 … 11 s. Save

the notebook as “Exercise_4.2.nb”. Warning! All the calculation steps must be performed in

your Mathematica notebook, and all items (a-h) must be clearly labelled in the Print[]

functions and/or the PlotLabel option in the Plot[] commands, or the score for this exercise

will be zero!

Answers: (a) … (b) … (c) … (d) 2.6 m s-1 (e) …

(f,g,h) …

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5. Geometrical interpretation of derivatives [13]

In the previous section we introduced the idea that in the limiting case the derivative represents an

instantaneous rate of change of two variables [13]. Hence, if, for example, 𝑦 = 𝑓(𝑥) is plotted in

a two-dimensional Cartesian coordinate system, then 𝑑𝑦/𝑑𝑥 is the slope of the curve at any point

(𝑥, 𝑦) on the curve. With exception of the function 𝑦(𝑥) = constant, functions either increase or

decrease as the value of 𝑥 increases. By looking at the derivative (or slope) evaluated at the point

(𝑥, 𝑦), we can determine whether the function 𝑓(𝑥) is increasing or decreasing as 𝑥 increases

without having to graph the function. If 𝑑𝑦/𝑑𝑥is positive, then 𝑓(𝑥) increases as 𝑥 increases. If

𝑑𝑦/𝑑𝑥 is negative, then 𝑓(𝑥) decreases as 𝑥 increases

[13].

Certain functions, such as parabolas (Figure 4a), or functions of higher order, such as cubic

functions (Figure 4b), have either a maximum or a minimum value, or both [13]. Differential

calculus can be used to help us determine the point or points along the curve where maxima or

minima occur. Since the slope of the curve must be zero at these points, the first derivative also

must be zero [13].

a) 𝑦 = 2𝑥2 − 3𝑥 + 2

b) 𝑦 = 2𝑥3 − 6𝑥 + 2

Figure 4 [13]. Sample graphs of (a) quadratic and (b) cubic functions.

For example, the parabola shown in Figure 4a is described by the equation [13]

𝑦(𝑥) = 2𝑥2 − 3𝑥 + 2 (23)

Taking the first derivative gives [13]

𝑑𝑦

𝑑𝑥
= 4𝑥 − 3 (24)

Setting the first derivative equal to zero and solving for x, we have [13]

4𝑥 − 3 = 0 or 𝑥 =
3

4
(25)

Substituting 𝑥 =
3

4
into the equation 𝑦(𝑥), yields 𝑦 = 0.875, which indicates the minimum point

on the curve [13]. In Mathematica, the same calculation can be performed as follows:

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In order to determine whether the curve is a maximum or a minimum at a point without actually

plotting the curve, we can substitute values for x that are greater than the value at the point where

the derivative is zero and the values for 𝑥 that are smaller than the value at the point where the

derivative is zero (a.k.a. stationary point). Then we note the behavior of 𝑦. A simpler way to test

whether the function is a maximum or a minimum is to look at the second derivative of the function

[13].

If 0
2

2


dx

yd
, then the function is a maximum.

If 0
2
2


dx

yd
, then the function is a minimum.

If 0
2
2

=
dx

yd
, then the function is at a point of inflection – a point on the curve where the curve

changes from one that exhibits a maximum at some point (i.e., is “concave

downward”) to one that exhibits a minimum at some point (i.e., is “concave

upward”), or vice versa; note that inflections points may be stationary points (as

shown in Figure 5), but are not local maxima or local minima.

Figure 5 [14]. Example of an inflection point (𝑑2𝑦/𝑑𝑥2 = 0) which is also a stationary point

(𝑑𝑦/𝑑𝑥 = 0).

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Thus, in the parabola example, the second derivative at x = ¾ (and, incidentally, everywhere else)

is

This makes sense because parabola shown in Figure 4a is “concave upward” everywhere…

Now, consider the cubic function shown in Figure 4b [13]:

𝑦 = 2𝑥3 − 6𝑥 + 2 (26)

Taking the first derivative and setting it equal to zero yields [13]

𝑑𝑦

𝑑𝑥
= 6𝑥2 − 6 = 0; 𝑥2 − 1 = 0; 𝑥 = ±1 (27)

which indicates that there is maximum or a minimum at 𝑥 = +1 and 𝑥 = −1 [13]. Taking the

second derivative of this cubic equation gives

𝑑2𝑦

𝑑𝑥2
= 12𝑥 (28)

For 𝑥 = +1, 𝑑2𝑦/𝑑𝑥2 = 12, which indicates that the curve is a minimum at this point. For 𝑥 =

−1, 𝑑2𝑦/𝑑𝑥2 = −12, which indicates that the curve is a maximum at this point.

In Mathematica, these calculations can be done as follows:

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You may wonder at which point the sign of the curvature changes from negative (maximum) to

positive (minimum). Well, clearly it should be at a point where the second derivative vanishes,

i.e.
𝑑2𝑦

𝑑𝑥2
= 12𝑥 = 0 (29)

It follows that 𝑥 = 0 is an inflection point of this function. Mathematica can easily find this point:

Example 5.1. The mathematical expression that tells us the probability, P, that the molecules in

a sample of gas have a speed that lies in a particular range at any instant is called the distribution

of molecular speeds [15]. The precise form of the distribution was worked out by James Clerk

Maxwell (1831-1879) towards the end of the nineteenth century, and his expression is known as

the Maxwell distribution of speeds [15]. A sample Maxwell distribution of molecular speeds for

the argon gas at 500 K is shown in Fig 6 [16].

Figure 6 [16]. The Maxwell distribution of molecular speeds for the argon gas at 500 K.

According to Maxwell’s equation, the probability 𝑃(𝑣, 𝑣 + Δ𝑣) that the molecules have a speed

in a narrow range between 𝑣 and 𝑣 + Δ𝑣 (for example, between 300 m/s and 301 m/s,

corresponding to 𝑣 = 300 m/s and Δ𝑣 = 1 m/s) is [15]

𝑃(𝑣, 𝑣 + ∆𝑣) = 𝜌(𝑣)∆𝑣 with

𝜌(𝑣) = 4𝜋 (
𝑀

2𝜋𝑅

𝑇
)

3/2

𝑣2𝑒
−

𝑀𝑣2

2𝑅𝑇 (30)

where 𝑀 is the molar mass of the gas, 𝑅 is the (ideal) gas constant, and 𝑇 is the temperature. We

wish to derive an expression for the most probable speed (identified as 𝑣most prob in Figure 6).

Note that the most probable speed is the speed at which the probability is a maximum. Thus, we

must maximize the function 𝜌(𝑣),

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𝜌(𝑣) = 4𝜋 (
𝑀
2𝜋𝑅𝑇
)
3/2

𝑣 2𝑒
−

𝑀𝑣2

2𝑅𝑇 (31)

with respect to 𝑣, set the derivative to zero, and solve for 𝑣. This is a simple task for Mathematica.

We always start with the project title (and so should you!). Then, we define function 𝜌(𝑣):

Note that an exponential can be taken using the Exp[] function, or using the button in the

“Basic Math Assistant” palette

Naturally, the result is the same:

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Now, we calculate the derivative of 𝜌(𝑣) w.r.t. v,
𝑑𝜌(𝑣)

𝑑𝑣
, and store the result in variable d𝜌dv:

In order to make the final result more “visually appealing”, we can ask Mathematica to try to

simplify the expressions using command Simplify[], FullSimplify[], or Refine[]:

https://reference.wolfram.com/language/ref/Simplify.html

https://reference.wolfram.com/language/ref/FullSimplify.html

http://reference.wolfram.com/language/ref/Refine.html

When simplifying expressions, it is recommended to include (whenever possible) the Assuming[]

function

http://reference.wolfram.com/language/ref/Assuming.html

Consider the following example:

The symbol “&&” is the logical AND function:

http://reference.wolfram.com/language/ref/And.html

In the following, we simplify the expression for
𝑑𝜌(𝑣)

𝑑𝑣
stored in variable d𝜌dv:

https://reference.wolfram.com/language/ref/Simplify.html

https://reference.wolfram.com/language/ref/FullSimplify.html

http://reference.wolfram.com/language/ref/Refine.html

http://reference.wolfram.com/language/ref/Assuming.html

http://reference.wolfram.com/language/ref/And.html

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We set the derivative (stored in variable d𝜌dv) to zero, and use the Solve[] function to solve the

equation for 𝑣:

We can simply the resulting expressions by applying one of Mathematica’s “simplifying”

functions, for example FullSimplify[]:

Note how we have specified that 𝑅 (gas constant), 𝑀 (molar mass), and 𝑇 (temperature) are all

positive.

There are three solutions. While all of them are mathematically correct, the first two solutions are

not physically meaningful (we hope you see why…). Thus, the only acceptable solution is number

three, which we then copy to variable vmostprob, and print it out:

Now, compare the expression derived in Mathematica to that given in [almost] every single

Physical Chemistry textbook [8,9,14,15,16]:

𝑣most prob = (
2𝑅𝑇

𝑀
)

1/2

(32)

Not bad, eh?!

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Well, it is now your time to differentiate a couple of functions in Mathematica.

Exercise 5.1. Consider the following function:

𝑔(𝑥) = 4𝑥3 + 9𝑥2 − 12𝑥 + 5

Using the final notebook from Example 5.1 as a template, determine the numerical values of 𝑥 for

the stationary and inflection points in the function 𝑔(𝑥). Save the notebook as

“Exercise_5.1.nb”. Warning! All the calculation steps must be performed in your

Mathematica notebook, or the score for this exercise will be zero! Your Mathematica

notebook must include the following items in the printout, all clearly labelled using the

Print[] function or the PlotLabel option in the Plot[] command:

(a) define function 𝑔(𝑥);

(b) derive and simplify formula for the first derivative of 𝑔(𝑥) with respect to (w.r.t.) 𝑥,

𝑑𝑔/𝑑𝑥;

(c) locate the stationary points in 𝑔(𝑥) by determining the values of 𝑥 at which 𝑑𝑔/𝑑𝑥 = 0

(solutions: −2,
1

2
);

(d) derive and simplify formula for the second derivative of 𝑔(𝑥) w.r.t. 𝑥, 𝑑2𝑔/𝑑𝑥2;

(e,f) for each stationary point found at step (c), calculate numerical value of the second

derivative, 𝑑2𝑔/𝑑𝑥2, and decide whether 𝑔(𝑥) at this point is a minimum, a maximum, or

an inflection point;

(g) locate the inflection points by determining the values of 𝑥 at which 𝑑2𝑔/𝑑𝑥2 = 0 (solution:

−
3

4
);

(h) plot 𝑔(𝑥) vs 𝑥 for −4 ≤ 𝑥 ≤ 3 to confirm your conclusions for (e-g).

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Exercise 5.2. Suppose, the following equation of state for a non-ideal gas is proposed:

𝑝 =
𝑅𝑇

𝑉m − 𝑏
𝑒

−
𝑎

𝑅𝑇3/2𝑉m (5.2.a)

where 𝑝 is pressure, 𝑇 is temperature, 𝑅 is the (ideal) gas constant, 𝑉m is the molar volume (i.e.

volume per mole of substance, 𝑉m = 𝑉/𝑛 ) [8,14], 𝑎 and 𝑏 are constants characteristic of the gas,

and 𝑒 is the base of the natural logarithm. Note that if 𝑏 ≪ 𝑉m and 𝑎 ≪ 𝑅𝑇
3/2𝑉m, equation (5.2.a)

reduces to the ideal gas law:

𝑝 =
𝑅𝑇

𝑉m
(5.2.b)

If equation (5.2.a) is to be applicable to a real gas, is must also predict the existence of a critical

point. The critical point terminates the vapor-liquid phase boundary (see your Physical Chemistry

textbook [8,9,14,15]), and is characterized by the critical constants of a gas: 𝑉c, 𝑇c, and 𝑝c, which

must all be positive. Considering that the critical point is an inflection and stationary point in the

𝑝 − 𝑉 curve, use Mathematica to derive expressions for 𝑉c, 𝑇c, and 𝑝c in terms of 𝑎, 𝑏, and R. Use

the final notebook from Example 5.1 as a template. Save the notebook as “Exercise_5.2.nb”.

Warning! All the calculation steps must be performed in your Mathematica notebook, or the

score for this exercise will be zero! Your Mathematica notebook must include the following

items in the printout, all clearly labelled using the

Print[] function:

(a) define 𝑝 as a function of 𝑉m as given in equation (5.2.a);

(b) derive formula for the first derivative of 𝑝 with respect to 𝑉m, 𝑑𝑝/𝑑𝑉m;

(c) derive formula for the second derivative of 𝑝 with respect to 𝑉m, 𝑑
2𝑝/𝑑𝑉m

2;

(d) since at the critical point both the first and second derivatives are zero, set 𝑑𝑝/𝑑𝑉m = 0

and 𝑑2𝑝/𝑑𝑉m
2 = 0, and solve the two equations simultaneously for 𝑉m and 𝑇; the resulting

solutions should give you the expressions for 𝑉c and 𝑇c;

(e) select, simplify, and print out the appropriate solution for 𝑉c;

(f) select, simplify and print out the appropriate solution for 𝑇c;

(g) insert 𝑉c and 𝑇c into equation (5.2.a) defined at step (a), and simplify and print out the

expression for 𝑝c.

Answers: 𝑉c = 2𝑏 𝑇c = (
𝑎

4𝑏𝑅
)

2/3

𝑝c =
𝑅

𝑏
(

𝑎
4𝑏𝑅
)
2/3

𝑒−2

Hint: It seems that in this exercise, Refine[] works better than Simplify[] and FullSimplify[].

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6. Integral Calculus

In Section 4 we discussed that the velocity of a particle moving in a straight line can be obtained

if its position as a function of time is known [11]. Mathematically, the velocity equals the

derivative of the position with respect to time [11], i.e.

𝑣(𝑡) =
𝑑𝑥(𝑡)

𝑑𝑡

It is also possible to find the position of a particle if its velocity is known as a function of time

[11]. In calculus, the procedure used to perform this task is referred to either as integration or as

finding the antiderivative. Graphically, it is equivalent to finding the area under a curve [11].

Figure 7 [11]. Velocity versus time for a particle moving along the x axis. The total area under

the curve is the total displacement of the particle [11].

Suppose the 𝑣 − 𝑡 graph for a particle moving along the x axis is as shown in Figure 7 [11].

Let us divide the time interval 𝑡𝑓 − 𝑡𝑖 into many small intervals, each of duration ∆𝑡𝑛. From the

definition of average velocity, we see that the displacement of the particle during any small

interval, such as the one shaded in Figure 7, is given by ∆𝑥𝑛 = 𝑣𝑛,avg∆𝑡𝑛, where 𝑣𝑛,avg is the

average velocity in that interval [11]. Therefore, the displacement during this small interval is

simply the area of the shaded rectangle in Figure 7 [11]. The total displacement for the interval

𝑡𝑓 − 𝑡𝑖 is the sum of the areas of all the rectangles from 𝑡𝑖 to 𝑡𝑓

[11]:

Δ𝑥 =

∑ 𝑣𝑛,avgΔ𝑡𝑛
𝑛

(33)

Now, as the intervals are made smaller and smaller, the number of terms in the sum increases and

the sum approaches a value equal to the area under the curve in the velocity–time graph [11].

Therefore, in the limit 𝑛 → ∞, or Δ𝑡 → 0, the displacement is [11]

Δ𝑥 =

lim
Δ𝑡𝑛→0

∑ 𝑣𝑛,avgΔ𝑡𝑛
𝑛

(34)

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If we know the 𝑣 − 𝑡 graph for motion along a straight line, we can obtain the displacement during

any time interval by measuring the area under the curve corresponding to that time interval [11].

The limit of the sum shown in equation (34) is called a definite integral and is written [11]

lim
Δ𝑡𝑛→0
∑ 𝑣𝑛,avgΔ𝑡𝑛
𝑛

= ∫ 𝑣(𝑡)𝑑𝑡
𝑡𝑓

𝑡𝑖

(35)

where 𝑣(𝑡) denotes the velocity at any time 𝑡 [11]. If the explicit functional form of 𝑣(𝑡) is known

and the limits are given, the integral can be evaluated [11].

Sometimes the 𝑣 − 𝑡 graph for a moving particle has a shape much simpler than that shown

in Figure 7 [11]. For example, suppose an object is described with the particle under a constant

velocity model [11]. In this case, the 𝑣 − 𝑡 graph is a horizontal line as in Figure 8 and the

displacement of the particle during the time interval Δ𝑡 is simply the area of the shaded rectangle

[11]:

Δ𝑥 = 𝑣𝑖,avgΔ𝑡 (when 𝑣 = 𝑣𝑖 = constant) (36)

Figure 8 [11]. The velocity–time curve for a particle moving with constant velocity 𝑣𝑖 . The
displacement of the particle during the time interval 𝑡𝑓 − 𝑡𝑖 is equal to the area of the shaded

rectangle.

Recall, that the acceleration is defined as

𝑎 =
𝑑𝑣
𝑑𝑡

First, we rewrite this equation as

𝑑𝑣 = 𝑎𝑑𝑡 (37)

We now integrate the two sides between the initial (𝑖) and final (𝑓) values:

∫ 𝑑𝑣
𝑣𝑓

𝑣𝑖

= ∫ 𝑎𝑑𝑡
𝑡𝑓

𝑡𝑖

(38)

The left hand side (lhs) is trivial as it evaluates to Δ𝑣:

∫ 𝑑𝑣
𝑣𝑓
𝑣𝑖

= 𝑣𝑓 − 𝑣𝑖 = Δ𝑣 (39)

In Mathematica, integration is easy. You can either type in the command Integrate[]

http://reference.wolfram.com/language/ref/Integrate.html

http://reference.wolfram.com/language/ref/Integrate.html

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or use the integral buttons in the section of the “Basic Math Assistant” palette:

Here is my Mathematica code that demonstrates how to use both options:

As expected the result is the same, and, of course, agrees with that given in equation (39). Now

we deal with the right hand side (rhs) of equation (38). Note that we integrate between the initial

time (ti) and final time (tf). By convention, the initial time is set to zero, i.e. 𝑡𝑖 = 0: recall that

when you time an event, you usually start at zero, i.e. 𝑡 = 0! On the contrary, the initial velocity

is not necessarily zero (the object can be moving when we start the timer), so 𝑣𝑖 = 𝑣0 (which may

or may not be zero). Then, we can drop the index (𝑓) on the final time, 𝑡 = 𝑡𝑓 , and velocity (𝑣 =

𝑣𝑓 ) and write:

Δ𝑣 = 𝑣 − 𝑣0 = ∫ 𝑎𝑑𝑡
𝑡

0

(40)

This shows that the change in the x-velocity is the time integral of the x-acceleration. In general,

acceleration may change with time, 𝑎 = 𝑎(𝑡), which gives:

𝑣 = 𝑣0 + ∫ 𝑎(𝑡)𝑑𝑡
𝑡

0

(41)

Using the same arguments, we can rewrite the equation for the displacement as

𝑥 = 𝑥0 + ∫ 𝑣(𝑡)𝑑𝑡
𝑡

0

(42)

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Here 𝑥 and 𝑣 are the position and 𝑥-velocity at time 𝑡. If we know the 𝑥-acceleration as a function

of time and we know the initial velocity we can use equation (41) to find the 𝑥-velocity at any

time, i.e., as a function of time. Once we know this function, and given the initial position we can

use equation (42) to find the position 𝑥 at any time.

Equations (41) and (42) are called the kinematic equations. Unfortunately, the right-hand

side of each equation is impossible to evaluate without the explicit knowledge of a time-

dependence of velocity or acceleration, i.e. 𝑣(𝑡) and 𝑎(𝑡).

In your basic Physics course, you considered the special case in which the acceleration is

constant, 𝑎 = constant. Recall that a constant can be moved out of an integral. As such, the

integral in equation (41) can be evaluated as

Which gives a very simple formula for the time-dependence of velocity v(t):

𝑣 = 𝑣0 + ∫ 𝑎𝑑𝑡
𝑡

0

= 𝑣0 + 𝑎 ∫ 𝑑𝑡
𝑡

0

= 𝑣0 + 𝑎(𝑡 − 0) = 𝑣0 + 𝑎𝑡 (43)

This powerful expression enables us to determine an object’s velocity at any time t if we know the

object’s initial velocity v0 and its (constant) acceleration 𝑎 [11]. A velocity–time graph for this

constant-acceleration motion is shown in Figure 9a [11]. The graph is a straight line, the slope of

which is the acceleration a; the (constant) slope is consistent with 𝑎 = 𝑑𝑣/𝑑𝑡 being a constant

[11]. Notice that the slope is positive, which indicates a positive acceleration [11]. If the

acceleration were negative, the slope of the line in Figure 9a would be negative [11]. When the

acceleration is constant, the graph of acceleration versus time (Figure 9b) is a straight line having

a slope of zero [11].

a)

b)

Figure 9 [11]. A particle under constant acceleration a moving along the x axis:

(a) the velocity–time graph, and (b) the acceleration–time graph [11].

Because we know how velocity depends on time, 𝑣 = 𝑣0 + 𝑎𝑡, we can use this expression to

evaluate the integral on the right-hand side of equation (42):

𝑥 = 𝑥0 + ∫ (𝑣0 + 𝑎𝑡)𝑑𝑡
𝑡

0

(44)

First we recall that the integral of a sum is a sum of integrals, i.e.

𝑥 = 𝑥0 +

∫ 𝑣0𝑑𝑡
𝑡

0

+ ∫ (𝑎𝑡)𝑑𝑡
𝑡

0

(45)

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The first integral is easy because the initial velocity (𝑣0) is a constant (naturally, it does not change

with time), so we get:

∫ 𝑣0𝑑𝑡
𝑡
0

= 𝑣0 ∫ 𝑑𝑡
𝑡

0

= 𝑣0(𝑡 − 0) = 𝑣0𝑡 (46)

The second integral is not much more complicated. Because acceleration is constant, we get:

∫ 𝑎𝑡𝑑𝑡
𝑡

0

= 𝑎 ∫ 𝑡𝑑𝑡
𝑡

0

= 𝑎 (
𝑡2

2
−

02

2
) =

1

2
𝑎𝑡2 (47)

Combining the results for the two integrals, we get:

𝑥 = 𝑥0 + 𝑣0𝑡 +
1

2
𝑎𝑡2 (48)

This equation provides the final position of the particle at time t in terms of the initial position, the

initial velocity, and the constant acceleration. Of course in Mathematica, equation (48) can be

obtained from equation (44) in one simple step:

Example 6.1. Suppose, a car is moving along a straight highway. At exactly 4 hours (h) after

beginning its journey, it is 200 km from the starting point (𝑥0), and the speed is 120 km/h (𝑣0). At

this point the driver begins to decrease the speed exponentially (with a constant acceleration) such

that at 𝑡 = 5 h the car’s speed is 16.240 km/h. (a) Calculate the numerical value of acceleration

Find expressions for (b) 𝑥-velocity 𝑣 and (c) position 𝑥 as functions of time after 𝑡 = 4 hours.

Compute the (d) velocity and (e) distance of the car from its starting point at 𝑡 = 6 hours. Plot (f)

𝑥 vs 𝑡, and (g) 𝑣 vs 𝑡 for 𝑡 = 4…7 hours

In the first cell, we specify the project number:

In the following statements we define 𝑥0 and 𝑣0 as

given in the question:

Note that because we do not wish to print those out, we terminate each statement with the “;”

symbol. Now there is a more difficult part. The question says that speed decreases exponentially

with time but acceleration is constant, i.e. 𝑒−𝑎𝑡, where 𝑡 is time (in h) 𝑎 is the acceleration (in

Computer Experiment 1: Mathematics of Physical Chemistry revision 1/10/22

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km/h2). Note that in this equation 𝑡 is being counted from 4 hours (because that is when

deceleration started). To bring 𝑡 to the absolute time scale (i.e. counting 𝑡 from the beginning of

the journey), we modify 𝑡 in 𝑒−𝑎𝑡 to 𝑡 − 4, i.e. 𝑒−𝑎(𝑡−4). Note that this equation shows the change

in speed, but the initial speed was 𝑣0, so we append it to 𝑒
−𝑎(𝑡−4) thus completing the expression

for 𝑣(𝑡):

𝑣(𝑡) = 𝑣0𝑒
−𝑎(𝑡−4) for 𝑡 ≥ 4 hours (49)

where 𝑣 is speed (in km/h) after 𝑡 = 4 hours, 𝑡 is time (in h), and 𝑎 is the acceleration (in km/h2).

Of course, this equation is valid only for 𝑡  4 hours, because we have no knowledge of how the

car was moving between 𝑡 = 0 and 𝑡 = 3.99999… hours. Now we enter this expression into

Mathematica and include a bunch of comments (so we do not forget what we are doing):

Note that in this expression acceleration, 𝑎, is not known. But we can certainly determine it

because we know that at 𝑡 = 5 h the speed (𝑣) was 16.240 km/h. All we need to do is insert these

values of 𝑡 and 𝑣 into the formula for 𝑣(𝑡) (equation 49) and solve it for 𝑎 :

Once 𝑎 is known, we can complete the expression for 𝑣(𝑡)

If 𝑣(𝑡) is known, we can determine 𝑥(𝑡) using equation (42):

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Note that 𝑥0 is given in the question. We can easily verify that our expressions for 𝑥(𝑡) and 𝑣(𝑡)

are correct, by using 𝑡 = 4 h and 𝑡 = 5 h and comparing the calculated values of 𝑥 and 𝑣 with those

given in the question:

They agree!!! Now we can calculate 𝑥 and 𝑣 at 𝑡 = 6 hours:

Finally, plotting 𝑥(𝑡) is straightforward:

Note that we, again, used Mathematica’s function SetOptions[Plot, BaseStyle → {FontSize →

14}] to increase the font size in the plot(s).

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Let’s see what the 𝑣(𝑡) plot looks like:

Now, that wasn’t too bad, was it?!

Integral Calculus is, in general, not easy, so we shall consider another example. This time it is

your [possibly] favorite subject – Chemistry!

Example 6.2. Consider the following chemical reaction [8,14,15]

PA ⎯→⎯
k

(50)

where A is a reactant, P is a product, and k is the rate constant [10,6]. Suppose the reaction follows

the first order mechanism, which means that the rate of consumption of A is proportional to the

first power of the concentration of A, [A]:

[A]

[A]

k
dt

d
=− (51)

We need to (a) derive the expression for [A] = f(t) (called the integrated rate law), and (b)

determine the expression for the parameter half-life, t1/2 at which the concentration [A] falls to half

of its original concentration [A]0.

Before we get to Mathematica, we need to rearrange equation (51) in such a way as to

collect terms related to the concentration of A (i.e. [A] and d[A]) on one side of the equation

(suppose, left hand side, lhs) and terms related to time (t) on the other side (suppose, right hand

side, rhs). It does not matter where you put constants such as k and (-1) because they always factor

out of integrals. To this purpose, we rearrange equation (51) as

kdtd −=

[A]
[A]

1
(52)

In the next step, we need to integrate both sides. Suppose, the left hand side is integrated between

the initial concentration [A]0 and concentration [A] at time t. Then, the right hand side is integrated

between the initial time (which we always consider as zero), and time t at which the concentration

of A falls to [A]. The following equation demonstrates that:

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 −=
t

kdtd
0

[A]

[A]0

[A]
[A]

1
(53)

Now, it is time to go to Mathematica. As always, the first cell contains the project number:

Then, we add a comment summarizing the preliminary steps:

We now integrate the left hand side:

Note that because Mathematica always thinks that a given symbolic parameter may be complex,

we had to give it additional information about the integrand and the limits of the integration.

Indeed, the initial concentration [A]0 must always be positive, while [A] is always less than [A]0.

Note that while in general, [A] = [A]0 at t = 0, in this particular case it is safer to use the relationship

shown above. As you can see, we make use of the Mathematica command Assumptions:

http://reference.wolfram.com/language/ref/Assumptions.html

As expected, the integral on rhs evaluates to 










0
[A]

[A]
ln . Note that in Mathematica the natural

logarithm is entered as Log[ ],

http://reference.wolfram.com/language/ref/Log.html

while the base-10 logarithm is written as Log10[]:

http://reference.wolfram.com/language/ref/Log10.html

http://reference.wolfram.com/language/ref/Assumptions.html

http://reference.wolfram.com/language/ref/Log.html

http://reference.wolfram.com/language/ref/Log10.html

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Ok, now we need to evaluate the right-hand side of equation (53). It is easy to do so in

Mathematica:

Now we combine the two sides, and solve for [A]:

Note how we told Mathematica that we are interested in Real solutions only (concentrations can

not be complex!). In the next step, we copy the solution to variable “A” and print it out:

This is exactly the same result as given in your Physical Chemistry textbook [8,14,15].

kt
e
−

=
0

[A][A] (54)

Now, all we need to do is solve this equation for half-life (t1/2). From the definition of half-life, it

is time (t) at which [A] falls to [A]0/2. We can certainly do that in Mathematica:

Compare our formula with that given in your Physical Chemistry textbook [8,14,15],

𝑡1/2 =
ln( 2)

𝑘
(55)

The agreement is perfect!

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Exercise 6.1. An object is moving along the 𝑥-axis with acceleration 𝑎(𝑡) given by the equation

𝑎(𝑡) = Φ(𝑡 + Ω)𝑒−Θ𝑡 (56)

where 𝑎 is in cm/s2, and 𝑡 is seconds (s). Using your Mathematica notebook from Example 6.1

as a template, determine the object’s (a) 𝑥-velocity and (b) 𝑥-position as a function of time, the

general constants Φ, Ω, and Θ, and some (yet unknown) initial position 𝑥0 and initial velocity 𝑣0.

Suppose, Φ = 3.5 cm/s3, Ω = 1.2 s, and Θ = 0.47 s−1. Determine the numerical values for (c)

𝑥0 (cm) and (d) 𝑣0 (cm/s) if it is known that at 𝑡 = 1.2 s, the velocity of the object was 13. cm/s,

and it was located at +14. cm. Plot (e) 𝑥 vs 𝑡, (f) 𝑣 vs 𝑡, and (g) 𝑎 vs 𝑡 for 𝑡 = 0 … 15 s. Save the

notebook as “Exercise_6.1.nb”. Warning! All the calculation steps must be performed in

your Mathematica notebook, and all items (a-g) must be clearly labelled using the Print[]

and/or PlotLabel[] function, or the score for this exercise will be zero!

Answers: (a) … (b) …

(c) 1.8 cm (d) …

(e,f,g) …

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Exercise 6.2. For a closed system in the absence of non-expansion work and at constant

composition, the variation of the molar Gibbs energy, 𝐺m, with pressure, 𝑝, and temperature, 𝑇, is

given by the following equation:

𝑑𝐺m = 𝑉m𝑑𝑝 − 𝑆m𝑑𝑇 (6.2.a)

where 𝑉m is the molar volume (i.e. volume per mole of substance, 𝑉m = 𝑉/𝑛 ) and 𝑆m is the molar

entropy (i.e. entropy per mole of substance, 𝑆m = 𝑆/𝑛) [8,14]. To determine the change in the

molar Gibbs energy for the isothermal (𝑇 = const.) reversible expansion from initial pressure 𝑝i

to final pressure 𝑝f, we set 𝑑𝑇 = 0 which gives

𝑑𝐺m = 𝑉m𝑑𝑝 (6.2.b)

and integrate:

Δ𝐺m = 𝐺m(𝑝f) − 𝐺m(𝑝i) = ∫ 𝑉m𝑑𝑝

𝑝f

𝑝i

(6.2.c)

This expression is applicable to any phase of matter, but to evaluate it we need to know how the

molar volume, 𝑉m, depends on the pressure [8,14]. For an ideal gas, we rearrange 𝑝𝑉m = 𝑅𝑇 as

𝑉m = 𝑅𝑇/𝑝, and substitute it into the integral (6.2.c).

Δ𝐺m = ∫ 𝑉m𝑑𝑝
𝑝f

𝑝i

= ∫
𝑅𝑇

𝑝
𝑑𝑝

𝑝f
𝑝i

(6.2.d)

Because 𝑅𝑇 is a constant (the process is isothermal), we get:

Δ𝐺m = 𝑅𝑇 ∫
1

𝑝
𝑑𝑝
𝑝f
𝑝i

= 𝑅𝑇 ln
𝑝f
𝑝i

(6.2.e)

This very important expression, which applies to ideal gases, shows that when the pressure is

increased tenfold at room temperature, the molar Gibbs energy increases by 𝑅𝑇 ln 10 ≈

6 kJ mol−1 [8,14]. In Mathematica, the very same formula can be obtained using the following

code:

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Note that when integrating, we had to tell Mathematica that 𝑝i is positive (𝑝i>0), and that 𝑝f is

greater than 𝑝i (𝑝f>𝑝i) and thus, positive as well. Also note that the natural logarithm (ln) in

Mathematica is written as Log[], while the base 10 (common) logarithm, log10, is named Log10[].

Suppose, a certain real gas obeys the following equation of state:

𝑝𝑉m = 𝑅𝑇 {1 +
9

128
(

𝑝

𝑝𝑐
) (

𝑇𝑐

𝑇

) [1 − 6 (
𝑇𝑐
𝑇

)
2

]} (6.2.f)

where 𝑝𝑐 and 𝑇𝑐 are the critical pressure and temperature of the gas, respectively. Using the sample

code given above and the final notebook from Example 6.1 or Example 6.2 as a template, create a

Mathematica code to derive and simplify as much as possible an expression for Δ𝐺m for the

isothermal (𝑇 = const.) reversible expansion of the real gas that obeys equation (6.2.f) from initial

pressure 𝑝i to final pressure 𝑝f, and verify that your result agrees with the literature expression

Δ𝐺m = 𝑅𝑇 ln
𝑝f

𝑝i
+

9𝑅

128
(
𝑇𝑐

𝑝𝑐
) {1 − 6 (

𝑇𝑐
𝑇
)
2

} (𝑝f − 𝑝i). Save the resulting notebook as

“Exercise_6.2.nb”. Warning! All the calculation steps (see below) must be performed in

your Mathematica notebook, or the score for this exercise will be zero! Your Mathematica

notebook must include the following items in the printout, all clearly labelled using the
Print[] function:

(a) define the left-hand side (l.h.s) of equation (6.2.f);

(b) define the right-hand side (r.h.s) of equation (6.2.f);

(c) solve l.h.s = r.h.s. for 𝑉m;

(d) determine Δ𝐺m by evaluating the integral (6.2.c);

(e) verify that your answer is identical to the literature expression; Hint: recall that there are

two easy ways to show mathematically that the two expressions / formulas / numerical

values 𝑎 and 𝑏 are the same: 𝑎 − 𝑏 = 0 or 𝑎/𝑏 = 1.

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Exercise 6.3. The construction of many thermometers is based on the fact that a column of

mercury (Hg) changes its length when its temperature is changed. The length of a solid or the

volume of a gas at constant pressure are also used in some thermometers. Consider a relationship

between the length of the mercury column 𝑙 and centigrade temperature 𝜃. First, we assign its

length the value 𝑙100 when it is at thermal equilibrium with boiling water at 1 atm pressure. Then,

we assign its length the value 𝑙0 when it achieves a thermal equilibrium with melting ice exposed

to 1 atm pressure. Assuming a linear relationship between the temperature 𝜃 and the length, and

assuming 100 divisions between the fixed marks, allows us to write

𝜃 =
𝑙𝜃 − 𝑙0

𝑙100 − 𝑙0
× 100℃ (6.3.a)

where 𝑙𝜃 is the length at temperature 𝜃, and 𝑙0 and 𝑙100 are the lengths at the freezing and boiling

water temperatures, respectively.

Suppose you calibrate a mercury thermometer at 0°C and 100°C. The thermal expansion

coefficient, 𝛼, for mercury is

𝛼 =
1

𝑉0
(

𝜕𝑉

𝜕𝑇
)

𝑝
= 𝑎 + 𝑏 𝜃 + 𝑐 𝜃2 (6.3.b)

where 𝑉 is volume, 𝑇 is the absolute (kelvin) temperature, 𝜃 is the value of the Celsius temperature,

𝑉0 = 𝑉 at 𝜃 = 0℃, and 𝑎, 𝑏, and 𝑐 are the empirical parameters. The quantity (𝜕𝑉/𝜕𝑇)𝑝, which

is ordinarily read a partial of 𝑉 with respect to 𝑇 at constant 𝑝, is the derivative of 𝑉 in which

pressure and temperature are independent variables. As such, the thermal expansion coefficient 𝛼

gives the fractional increase of volume per unit rise of temperature at constant pressure. Switching

from length to volume modifies equation (6.3.a) as follows:

𝜃 =
𝑉𝜃 − 𝑉0

𝑉100 − 𝑉0
× 100℃ =

Δ𝑉𝜃
Δ𝑉100

× 100℃ (6.3.c)

where 𝑉𝜃 is the volume at temperature 𝜃, Δ𝑉100 is the change in volume that accompanies the

temperature increase from 0℃ to 100℃, and Δ𝑉𝜃 is the change in volume at temperature 𝜃 relative

to that at 0℃. Note that Δ𝑉𝜃 can be conveniently obtained by integrating the expression for

(𝜕𝑉/𝜕𝑇)𝑝 = 𝛼𝑉0 between 0℃ and 𝜃:

Δ𝑉𝜃 = 𝑉𝜃 − 𝑉0 = ∫ 𝛼𝑉0 𝑑𝜃
𝜃

0

(6.3.d)

Using your Mathematica notebook from Example 6.1 or Example 6.2 as a template, determine the

temperature that would appear on the mercury thermometer scale at 50°C given that for mercury,

𝑎 = 1.817 × 10−4, 𝑏 = 5.90 × 10−9 and 𝑐 = 3.45 × 10−10. Warning! All the calculation

steps (see below) must be performed in your Mathematica notebook, or the score for this

exercise will be zero! Your Mathematica notebook must include the following items in the

printout, all clearly labelled using the Print[] function:

(a) define the expansion coefficient  using the right-hand side (r.h.s) of equation (6.3.b); do

not assign numerical values to the coefficients 𝑎, 𝑏, and 𝑐 yet!

(b) obtain the analytical expression for Δ𝑉𝜃 by evaluating the integral (6.3.d);

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(c) assign the given numerical values to the parameters 𝑎, 𝑏, and 𝑐;

(d) using formula (6.3.c), the expression for Δ𝑉𝜃 derived at (b), and the numerical parameters

defined at (c), calculate the temperature 𝜃Hg (to three significant digits) that would appear

on the mercury scale at 50°C.

Answers: (a) … (b) … (c) … (d) 49.7°C

Believe it or not, but this concludes our brief Mathematica tutorial. We hope that we have been

able to convince you that this amazing software can provide a significant help in learning many

concepts in Mathematics, Physics, and Chemistry (recall that Physical Chemistry = Mathematics

+ Physics + Chemistry).

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7. References

1. Wolfram Research, Inc. Mathematica 9.0. Champaign, Illinois, USA,

2012.

2. http://www.wolfram.com/mathematica

3. https://en.wikipedia.org/wiki/Mathematica

4. https://en.wikipedia.org/wiki/Stephen_Wolfram

5. https://en.wikipedia.org/wiki/Wolfram_Research

6. Tam, P. T. A Physicist’s Guide to Mathematica, 2nd edition, Academic Press, 2008.

7. https://www.duolingo.com/comment/15264140/How-To-Type-Accented-Characters-On-An-

English-Keyboard-Windows-10-Tutorial

8. Atkins, P; de Paula, J. Elements of Physical Chemistry, 5th edition, W. H. Freeman, New York,

2009.

9. Atkins, P; de Paula, J. Physical Chemistry for the Life Sciences, 1st edition, W. H. Freeman,

New York, 2006

10. Serway, R. A., Vuille, C. College Physics, 9th edition, Cengage Learning, 2011.

11. Serway, R. A., Jewett, J. W. Physics for Scientists and Engineers with Modern Physics, 9th

edition, Brooks Cole, 2013.

12. Young, H. D., Freedman, R. A., Ford, A. L. Sears and Zemansky’s University Physics with

Modern Physics, 13th edition. Pearson Education, Inc., 2012.

13. Barrante, J. R. Applied Mathematics for Physical Chemistry, 3rd edition, Prentice Hall; 2003.

14. Atkins, P; de Paula, J. Physical Chemistry, 8th edition, W. H. Freeman and Company, New

York, 2006.

15. Atkins, P; de Paula, J. Elements of Physical Chemistry, 6th edition, W. H. Freeman, New York,

2012.

16. Ball, D. W. Physical Chemistry, 1st edition, Cengage Learning; 2002.

http://www.wolfram.com/mathematica

https://en.wikipedia.org/wiki/Mathematica

https://en.wikipedia.org/wiki/Stephen_Wolfram

https://en.wikipedia.org/wiki/Wolfram_Research

https://www.duolingo.com/comment/15264140/How-To-Type-Accented-Characters-On-An-English-Keyboard-Windows-10-Tutorial

https://www.duolingo.com/comment/15264140/How-To-Type-Accented-Characters-On-An-English-Keyboard-Windows-10-Tutorial