Avogadro Example

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The Extensions Menu is a catalog of computational plugins equipped with Avogadro. These plugins can interact with molecules, generate input file dialogs for quantum codes, and create molecule property dialogs.

Avogadro’s Law Examples in Real Life Avogadro’s law tells about the relationship between the volume of a gas and the number of molecules possessed by it. It was formulated by an Italian scientist Amedeo Avogadro in the year 1811. Avogadro's Number Example Chemistry Problem - Water in a Snowflake Finding Number of Molecules in a Known Mass (Water in a Snowflake) Use Avogadro's number to determine quantity of molecules in a known mass, such as the number of water molecules in a single snowflake. Edward Kinsman / Getty Images. After selecting “Gaussian” from the Extensions menu, the dialog box depicted below will appear. You can edit the dialog box and add specific keywords to utilize these features in Avogadro. For example, typing “freq” in the dialog box will compute force constants and vibrational frequencies.

Animation

Selecting “Animation” will open the animate trajectory dialog box shown below. From here you can load a file, view and edit the animation, as well as save the file in a PC compatible format.

Optimize Geometry

“Optimize Geometry” provides a quick, realistic rendition of a molecule using molecular mechanics.

Molecular Mechanics

“Molecular Mechanics” allows you to edit the geometry optimization of a molecule, so that it best suits your purposes.

Setup Force Field…

A dialog box will open when “Setup Force Field…” is selected. This dialog box provides you with the ability to choose the type of force field, and algorithm that can best optimize your molecular parameters, and preferences.

Calculate Energy

“Calculate Energy” determines the amount of energy per the amount of material (kJ/mol), and displays this number in a pop up dialog box.

Conformer Search

“Conformer Search” is a way to easily search for conformers within a molecule (dialog box shown below). A more detailed outline on how to perform a conformer search is found in the “Optimizing Geometry” section of this manual. Avogadro only renders staggered conformations, and does not calculate ring conformers.

Constraints

“Constraints” is a way to ensure atom stability in various selections (dialog box depicted below). The constraints that can be applied to a molecule include Ignore Atom, Fix Atom, Fix Atom X, Fix Atom Y, Fix Atom Z, Distance, Angle, and Torsion Angle. A detailed outline on how to use the constraints feature is found in the “Optimizing Geometry” section of this manual.

Ignore Selection

“Ignore Selection” allows you to select a specific part of a molecule to omit during a geometry optimization.

Fix Selected Atoms

“Fix Selected Atoms” also allows you to set a certain part of a molecule to fix during optimization.

Avogadro Extensions–Plugins

Avogadro provides you with the ability to interface your molecules with other dialog based plugins. These extensions interact with a molecule to provide further molecular information, and additional computation abilities. These plugins include but aren’t limited to GAMESS, Abinit, Dalton, GAMESS-UK, Gaussian, MOLPRO, MOPAC, NWChem, PSI4, Q-Chem, and LAMMPS.

General “How To” for Plugins

Avogadro (as you will see below) can be used to display molecular orbitals, QTAIM, spectra, as well as create surfaces. However, many of these features can not be used to their full potential without first running one of the plugins listed in the section above. Gaussian is one of the most common plugins used, due to it’s wide range of basis sets/functions.

Running Gaussian

After selecting “Gaussian” from the Extensions menu, the dialog box depicted below will appear. You can edit the dialog box and add specific keywords to utilize these features in Avogadro. For example, typing “freq” in the dialog box will compute force constants and vibrational frequencies. More information on keywords for Gaussian can be found at the Gaussian website (http://www.gaussian.com/g_tech/g_ur/l_keywords09.htm). Then clicking generate will let you save the file to your computer, so you can run the file in external software.

Once the file has been run through the external software, you will have a .g03 or .g09 file that will open the keyword selection in a toolbar on the right hand side of the screen. “Freq” will open the vibrations toolbar shown below.

Molecular Orbitals

The “Molecular Orbitals” selection will display the molecular orbitals for orbitals with full status bars. The quality of the orbitals can be adjusted an reconfigured if need be. This feature only works by running gaussian extension files (.fchk, .g03, .g09, etc.).

QTAIM (Quantum Theory of Atoms in Molecules)

QTAIM displays the implicit bonding that is theorized to take place between the hydrogens of organic crystals (the implicit bonding is conveyed through dots). This display type is utilized by importing a .wfn file from the “QTAIM”, “Molecular Graph” selection under the “Extensions” menu. Selecting “Molecular Graph with Lone Pairs” or “Atomic Charge” will provide concurrent information about the molecule. More information can be found on this process in the Tutorial section of this manual.

Spectra…

Clicking on “Spectra…” will create a spectra visualization of a .g03, or .g09 file that has been run with the keyword “freq”. A spectral visualization can also be created through the vibrations toolbar by selecting “Show Spectra…”.

Create Surfaces…

“Create Surfaces…” allows you to view the Van der Waals, Electrostatic Potential, Electron Density, and Molecular Orbital Surfaces. The surface type options for viewing depend on what type of calculations have previously been run on the molecule. The type of file you open/create allows for more or less surface viewing options (generally discussed under Avogadro Extensions–Plugins). This feature also allows you to edit the color, resolution, and iso value to further enhance your surface.

Avogadro's Law:
Ten Examples

Boyle's LawCombined Gas Law
Charles' LawIdeal Gas Law
Gay-Lussac's LawDalton's Law
Diver's LawGraham's Law
No Name LawReturn to KMT & Gas Laws Menu

Discovered by Amedo Avogadro, of Avogadro's Hypothesis fame. The ChemTeam is not sure when, but probably sometime in the early 1800s.

Gives the relationship between volume and amount when pressure and temperature are held constant. Remember amount is measured in moles. Also, since volume is one of the variables, that means the container holding the gas is flexible in some way and can expand or contract.

If the amount of gas in a container is increased, the volume increases.

If the amount of gas in a container is decreased, the volume decreases.

Why?

Suppose the amount is increased. This means there are more gas molecules and this will increase the number of impacts on the container walls. This means the gas pressure inside the container will increase (for an instant), becoming greater than the pressure on the outside of the walls. This causes the walls to move outward. Since there is more wall space the impacts will lessen and the pressure will return to its original value.

The mathematical form of Avogadro's Law is:

V
––– = k
n

This means that the volume-amount fraction will always generate a constant if the pressure and temperature remain constant.

Let V1 and n1 be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n2, then the volume will change to V2.

We know this:

V1
––– = k
n1

And we know this:

V2
––– = k
n2

Since k = k, we can conclude:

V1V2
––– = –––
n1n2
Avogadro Example

This equation will be very helpful in solving Avogadro's Law problems. You will also see it rendered thusly:

V1 / n1 = V2 / n2

Sometimes, you will see Avogadro's Law in cross-multiplied form:

V1n2 = V2n1

Avogadro's Law is a direct mathematical relationship. If one gas variable (V or n) changes in value (either up or down), the other variable will also change in the same direction. The constant K will remain the same value.

Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

Solution:

I'll use V1n2 = V2nAvogadro1

(5.00 L) (1.80 mol) = (x) (0.965 mol)

x = 9.33 L (to three sig figs)

Example #2: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)

Solution:

1) Convert grams of He to moles:

2.00 g / 4.00 g/mol = 0.500 mol

2) Use Avogadro's Law:

V1 / n1 = V2 / n2

2.00 L / 0.500 mol = 2.70 L / x

x = 0.675 mol

3) Compute grams of He added:

0.675 mol − 0.500 mol = 0.175 mol

(0.175 mol) (4.00 g/mol) = 0.7 grams of He added

Example #3: A balloon contains a certain mass of neon gas. The temperature is kept constant, and the same mass of argon gas is added to the balloon. What happens?

(a) The balloon doubles in volume.
(b) The volume of the balloon expands by more than two times.
(c) The volume of the balloon expands by less than two times.
(d) The balloon stays the same size but the pressure increases.
(e) None of the above.

Solution:

We can perform a calculation using Avogadro's Law:

V1 / n1 = V2 / n2

Let's assign V1 to be 1 L and V2 will be our unknown.

Let us assign 1 mole for the amount of neon gas and assign it to be n1.

The mass of argon now added is exactly equal to the neon, but argon has a higher gram-atomic weight (molar mass) than neon. Therefore less than 1 mole of Ar will be added. Let us use 1.5 mol for the total moles in the balloon (which will be n2) after the Ar is added. (I picked 1.5 because neon weighs about 20 g/mol and argon weighs about 40 g/mol.)

1 / 1 = x / 1.5

x = 1.5

answer choice (c).

Example #4: A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas. More gas is then added to the container until it reaches a final volume of 18.10 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Solution:

V1 / n1 = V2 / n2
5.120 L18.10 L
–––––––– = ––––––
8.500 molx

x = 30.05 mol <--- total moles, not the moles added

30.05 − 8.500 = 21.55 mol (to four sig figs)

Notice the specification in the problem to determine moles of gas added. The Avogadro Law calculation gives you the total moles required for that volume, NOT the moles of gas added. That's why the subtraction is there.

Example #5: If 0.00810 mol neon gas at a particular temperature and pressure occupies a volume of 214 mL, what volume would 0.00684 mol neon gas occupy under the same conditions?

Solution:

1) Notice that the same conditions are the temperature and pressure. Holding those two constant means the volume and the number of moles will vary. The gas law that describes the volume-mole relationship is Avogadro's Law:

V1V2
––– = ––––
n1n2

2) Substituting values gives:

214 mLV2
––––––––– = ––––––––––
0.00810 mol0.00684 mol

3) Cross-multiply and divide for the answer:

V2 = 181 mL (to three sig figs)

When I did the actual calculation for this answer, I used 684 and 810 when entering values into the calculator.

4) You may find this answer interesting:

Dividing PV1 = n1RT by PV2 = n2RT, we get

V1/V2 = n1/n2

V2 = V1n2/n1

V2 = [(214 mL) (0.00684 mol)] / 0.00810 mol

V2 = 181 mL

In case you don't know, PV = nRT is called the Ideal Gas Law. You'll see it a bit later in your Gas Laws unit, if you haven't already.

Example #6: A flexible container at an initial volume of 6.13 L contains 7.51 mol of gas. More gas is then added to the container until it reaches a final volume of 13.5 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Solution:

1) Let's start by rearranging the Ideal Gas Law (which you'll see a bit later or you can go review it right now):

Avogadro ExamplePV = nRT

V/n = RT / P

R is, of course, a constant.

2) T and P are constant, as stipulated in the problem. Therefore, we can write this:

k = RT / P

where k is some constant.

3) Therefore, this is true:

V/n = k

4) Given V and n at two different sets of conditions, we have:

V1 / n1 = k
V2 / n2 = k

5) Since k = k, we have this relation:

V1 / n1 = V2 / n2

6) Insert data and solve:

6.13 / 7.51 = 13.5 / n

(6.13) (n) = (13.5) (7.51)

n = [(13.5) (7.51)] / 6.13

n = 16.54 mol (this is not the final answer)

7) Final step:

16.54 − 7.51 = 9.03 mol (this is the number of moles of gas that were added)

Example #7: A container with a volume of 25.47 L holds 1.050 mol of oxygen gas (O2) whose molar mass is 31.9988 g/mol. What is the volume if 7.210 g of oxygen gas is removed from the container, assuming the pressure and temperature remain constant?

Solution #1:

1) Initial mass of O2:

(1.050 mol) (31.9988 g/mol) = 33.59874 g

2) Final mass of O2:

33.59874 − 7.210 = 26.38874 g

3) Final moles of O2:

26.38874 g / 31.9988 g/mol = 0.824679 mol

4) Use Avogadro's Law:

V1 / n1 = V2 / n2

25.47 L / 1.050 mol = V2 / 0.824679 mol

V2 = 20.00 L

Solution #2:

1) Let's convert the mass of O2 removed to moles:

7.210 g / 31.9988 g/mol = 0.225321 mol

2) Subtract moles of O2 that got removed:

1.050 mol − 0.225321 mol = 0.824679 mol

3) Use Avogadro's Law as above.

Solution #3:

1) This solution depends on seeing that the mass ratio is the same as the mole ratio. Allow me to explain by using Avogadro's Law:

V1V2
–––– = ––––
n1n2

2) Replace moles with mass divided by molar mass:

V1V2
–––––––––– = ––––––––––
mass1 / MMmass2 / MM

3) Since the molar mass is of the same substance (oxygen in this case), they cancel out leaving us with this:

V1V2
–––– = ––––
mass1mass2

4) Solve using the appropriate values

25.47 LV2
–––––––– = ––––––––
33.59874 g26.38874 g

V2 = 20.00 L

Example #8: What volume (in L) will 5.5 g of oxygen gas occupy if 2.2 g of the oxygen gas occupies 3.0 L? (Under constant pressure and temperature.)

Avogadro's Law Example Problem

Solution:

1) State the ideal gas law:

P1V1P2V2
––––– = –––––
n1T1n2T2

Note that it is the full version which includes the moles of gas. Usually a shortened version with the moles not present is used. Since grams are involved (which leads to moles), we choose to use the full version.

2) The problem states that P and T are constant:

V1V2
––– = –––
n1n2

3) Cross-multiply and rearrange to isolate V2:

V2n1 = V1n2

V2 = (V1) (n2 / n1)

4) moles = mass / molecular weight:

n = mass / mw

V2 = (V1) [(mass2 / mw) / (mass1 / mw)]

5) mw is a constant (since they are both the molecular weight of oxygen), which means it can be canceled out:

V2 = (V1) (mass2 / mass1)

6) Solve:

V2 = (3.0 L) (5.5 g / 2.2 g)

V2 = 7.5 L

Example #9: At a certain temperature and pressure, one mole of a diatomic H2 gas occupies a volume of 20 L. What would be the volume of one mole of H atoms under those same conditions?

Solution:

One mole of H2 molecules has 6.022 x 1023 H2 molecules.

One mole of H atoms has 6.022 x 1023 H atoms.

Avogadro Software Examples

The number of independent 'particles' in each sample is the same.

Therefore, the volumes occupied by the two samples are the same. The volume of the H atoms sample is 20 L.

By the way, I agree that one mole of H2 has twice as many atoms as one mole of H atoms. However, the atoms in H2 are bound up into one mole of molecules, which means that one molecule of H2 (with two atoms) counts as one independent 'particle' when considering gas behavior.

Example #10: A flexible container at an initial volume of 6.13 L contains 8.51 mol of gas. More gas is then added to the container until it reaches a final volume of 15.5 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Solution:

1) State Avogadro's Law in problem-solving form:

V1V2
––– = ––––
n1n2
Avogadro

2) Substitute values into equation and solve:

6.13 L15.5 L
––––––– = ––––––
8.51 molx

x = 21.5 mol

3) Determine moles of gas added:

21.5 mol − 8.51 mol = 13.0 mol (when properly rounded off)

Avogadro's Example Problems

Bonus Example: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.50 L? (The temperature was held constant.)

Solution:

1) The two variables are the volume and the amount of gas (temp and press are constant). The gas law that relates these two variables is Avogadro's Law:

V1V2
––– = ––––
n1n2

Avogadro Constant Example

2) We convert the grams to moles:

2.00 g / 4.00 g/mol = 0.500 mol

3) Now, we use Avogadro's Law:

2.00 L2.50 L
–––––––– = ––––––
0.500 molx

Avogadro's Principle Example

x = [(0.500 mol) (2.50 L)] / 2.00 L

Avogadro's Number Example

x = 0.625 mol <--- this is the ending amount of moles, not the moles of gas added

4) This is the total moles to create the 2.50 L. We need to convert back to grams:

(4.00 g/mol) (0.125 mol) = 0.500 g <--- this is the amount added.

Notice that I subtracted 0.500 mol from 0.625 mol and used 0.125 mol in the calculation. This is because I want the amount added, not the final ending amount.

Avogadro's Law Equation Example

Boyle's LawCombined Gas Law
Charles' LawIdeal Gas Law
Gay-Lussac's LawDalton's Law
Diver's LawGraham's Law
No Name LawReturn to KMT & Gas Laws Menu