Example Of Avogadro's Law In Real Life

  

Avogadro's Law:
Ten Examples

  • Gases laws with examples kinds of gas laws gay-lussacs law avogadros law finding the temperature in gas laws charles law practical situation in amontons law 5 example of daltons law examples of amontons law volume and temperature examples Examples on Amontons Law of Gases amontons law examples practical application of daltons law.
  • Avogadro’s law says the volume (V) is directly proportional to the number of molecules of gas (n) at the same temperature. This means the ratio of n to V is equal to a constant value. Since this constant never changes, the ratio will always be true for different amounts of gas and volumes. Where ni = initial number of molecules.
  • Italian Scientist Amedeo Avogadro Source: Google. This law is an experimental gas law. Specifically, this law is for an ideal gas. This law is not applicable to real gases because they show some variations. Understand Avogadro’s Law Examples, Ballons. Avogadro’s law states that –.
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Avogadro’s law (now known as Avogadro’s hypothesis) was first published in 1811 and is one of the main theories that helped to build the foundation for the ideal gas laws. These laws help to explain the relationship that gases have between the number of molecules and the volume of the container they fill.

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

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 = V2n1

(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 / nExample1 = 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.

Example Of Avogadro's Law In Real Life

Solution:

We can perform a calculation using Avogadro's Law:

V1 / n1 = V2 / n2

Example of avogadro

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.

Avogadro's Law Equation

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:

V2Law = 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):

PV = 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

Example Of Avogadro's Law In Real Life Worksheet

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.)

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.

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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.

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.

Avogadro

Solution:

Pdf

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

V1V2
––– = ––––
n1n2

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)

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

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

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

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.

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Distillation of Mixtures

Example Of Avogadro's Law In Real Lifelaw In Real Life

There are two general types of mixtures to consider,mixtures of miscible liquids and mixtures of immiscible liquids. Their behaviorin distillation is very different from one another. Miscible liquids are soluble in each otherin all ratios. Immiscible liquids do not dissolve in one another to any extent.Water is immiscible with most organic substances and will generally always be one of the components in a mixture of immiscible liquids. Mixturesobey Dalton's law of partial pressures which states that the vapor pressure of a mixture is equal to the sum of the vapor pressures of the individualcomponents. When a mixture of immiscible liquids is heated, it will boilat a temperature which is less than the boiling point of either of thecomponents. Both components will be present in the vapor. For a specificexample consider a mixture of limonene and water. At a little over 97°C thevapor pressure of water is 695 mmHg(mm of mercury) and the vapor pressure of limonene is 65 mmHg.Since the sum of the vapor pressures equals 760 mmHg, the mixture boils. Themixture will continue to boil at this temperature as long as any limonene ispresent in the mixture. When all of the limonene is gone, the boiling pointrises to 100°C, the boiling point of pure water. On condensing the vapor, thelimonene and water, being immiscible in one another, separate into two phases. Thistechnique is referred to as steam distillation and can be an effective methodfor isolating organic materials from complex mixtures. The earliest isolationsof organic substances from natural materials were done using steamdistillation. One distinct advantage of steam distillation is the lowertemperature required to isolate the organic substance. In the example withlimonene, limonene is volatilized at 97°C whereas its normal boiling point is 175°C. Refrence: www.chem.wisc.edu/courses/342/Fall2004/Distillation.pdf