Avogadro's Law

Avogadro's Law was invented by Amedeo Carlo Avogadro in 1811. He was an Italian chemist and mathematician physicist. He said that, at constant temperature and pressure, two different ideal gases with similar volumes must contain an equal number of molecules. This law can be derived from the ideal gas equation.

In this article, we will discuss Avogadro's law, its formula, derivation, application, and other aspects facts related to it in detail.

What is Avogadro's Law?

Avogadro's law says that at constant temperature and pressure, gases with equal volume have an equal number of atoms or molecules. It means that the volume of gas is directly proportional to the number of moles in the gas. It is also called Avogadro's hypothesis or Avogadro's principle. Avogadro's law is suitable for ideal gases, but real gases behave a little differently from their ideal behavior. The law gives only approximation for the real gases. This law is convenient to measure the amount of gas present in a container.

👁 Avogadro's-Law

Avogadro Law Statement

At constant temperature and pressure, the total number of atoms or molecules of a gas, i.e., the amount of gaseous substance, is directly proportional to the volume occupied by the gas.

The law states that at constant temperature and pressure, the amount and volume of the gaseous substance in a gas is proportional to each other. To evaluate the number of particles in 1 mole gas, we use Avogadro's number. The value of Avogadro's number is = 6.023 × 10²³. It is used for the conversion between moles and grams.

Formula and Derivation of Avogadro's Law

At constant pressure and temperature, a particular gas with volume = V and a number of particles = n can be expressed by Avogadro's law as follows:

V ∝ n

⇒ V/n = k (k = Avogadro's constant)

When the number of particles is increased or decreased from n1 to n2, the volume is also increased or decreased from V1 to V2. This change in volume can be evaluated from:

V₁/n₁ = V₂/n₂

Derivation of Avogadro's Law from the Ideal Gas Equation

Avogadro's law can arise from the ideal gas equation. The process is discussed below:

We know, for ideal gases,

PV = nRT

where,

• ‘P’ is the gaseous pressure applied on the walls of its container
• ‘V’ is the volume preoccupied by the gas
• ‘n’ is the number of moles of gas
• ‘R’ is the universal gas constant
• ‘T’ is the temperature of the gas in kelvin

Now, we can obtain the following equation by reshuffling the ideal gas equation,

V/n = (RT)/P

V/n = k

Here, the value of (RT)/P = k or constant, because the temperature and pressure are kept constant and the product of two or more constants is always a constant.

Thus, it is proved that the volume occupied by the gas and the number of molecules present in the gas are proportionally related.

Graphical Representation of Avogadro's Law

In the graphical representation of Avogadro's law, the X axis represents the amount of substances and the Y axis represents the volume of the gas. The presentation is given below:

👁 Avogadro's-Law-Graph

In this graph, the straight line from the center implies that the increase in volume is proportional to the increase in substances or moles.

Molar Volumes of a Gas

The molar volume of all ideal gases at STP is 22.4 liters. The value of Avogadro's constant k can be derived from the following equation:

k = RT/P (at constant pressure and temperature)

At standard temperature and pressure, the value of T = 273 K, P = 101.325 kPa, P = 8.314 joule.mol^-1.K^-1. Thus the volume of one mole of a gas at STP is,

Volume of 1 mole of gas = (8.314 J.mol^-1.K^-1)×(273 K)/(101.325 kPa) = 22.4 liters

Moles to Gram

The following formula shows how to change from moles to gram, which is another common unit of measure:

Moles = grams/molar mass

Examples of Avogadro's Law

Below are some examples of Avogadro's law:

• Deflation of tyres: A common example of the use of Avogadro’s law is the deflation of tyres. When the air trapped inside a tyre is released, the amount of substances of air present in the tire is also decreased. This results in a decrease in the volume occupied by the gas, causing the tyre to lose its shape and deflate.
• Process of Respiration: It is one of the important examples of Avogadro's law. When a person breaths in, the amount of molar quantity of air increased in his/her lungs is proportional to the increase in the volume of his/her lungs.
• Balloon: When we add air(moles of gas) to a balloon, its volume increases.

Applications of Avogadro's Law

The applications of Avogadro's law are mentioned below:

• This law allows chemists to relate the volumes of gases involved in a reaction to the number of moles of reactants and products. By knowing the volumes of gases involved in a reaction and applying Avogadro's law, chemists can determine the molar ratios of reactants and products, facilitating the calculation of reaction yields.

• Avogadro's law is utilized in gas analysis techniques, such as gas chromatography and mass spectrometry, to determine the composition and concentration of gas mixtures.

• Avogadro's law is employed in the determination of gas densities and molar masses. By measuring the mass and volume of a gas sample at known temperature and pressure conditions, Avogadro's law allows scientists to calculate the number of moles of gas present and subsequently determine its molar mass.

Limitations of Avogadro's Law

The limitations of Avogadro's law is mentioned below:

• Avogadro's law is appropriate for ideal gases, but for real gases, it only gives an approximation. The divergence of real gases from ideal behavior increases with high temperature and low pressure.
• Gaseous molecules with lower molecular masses such as helium, hydrogen, etc. follow Avogadro's law more than gaseous molecules with heavier molecular masses.
• It is not applicable for solids and liquids, it is only applicable for gases.
• Avogadro's law doesn't work for highly dense or very light gases.

Also, Check

• Van Der Waals equation
• Ideal Gas Law
• Boyle's law
• Charles law

Solved Examples on Avogadro's Law

Example 1:A puncture takes away half of the volume of a tire with 10 moles of air and a 40-liter volume. How much air is left in a tire that has been deflated?

Solution:

We know, V₁/n₁ = V₂/n₂

Here, initial volume = V₁ = 40L

Initial number of moles = n₁ = 10mole

Final volume= V₂ = 20L

Final number of moles = n₂ = x mole

Now, putting these values we get,

40/10 = 20/x

⇒ x = (20 × 10)/40 = 200/40

⇒ x = 5mole

Example 2:5L of gas is known to contain 0.90 mol. If the amount of gas is increased to 1.80 mol, what will be the new volume(at a constant temperature and pressure)?

Solution:

We know V₁/n₁ = V₂/n₂

Initial volume = V₁ = 5L

Initial no. of moles = n₁ = 0.90mole

Final volume = V₂ = x L

Final no. of moles = n₂ = 1.80mole

Now, putting these values we get,

5 × 1.80 = x × 0.90

⇒x = 10L