**The Kinetic Theory of Gases**

The behavior of gas molecules such as its speed and kinetic energy, varies with respect to the change in pressure and/or temperature. This relationship between the macroscopic properties of gases to the microscopic properties of gas molecules is what is stated in the **kinetic theory of gases**.

**Avogadro’s Number**

The very first step to study in the kinetic theory of gases is about the measuring of the amount of gas present in a substance or a certain sample in which the Avogadro’s number is used in the equation.

The Avogadro’s number denoted as ** N_{A}** is determined experimentally to have a value of

**, read as**

*6.02 x 10*^{23}mol^{-1}**. It is named after the Italian scientist Amedeo Avogadro who suggested that all gases contain the same number of atoms or molecules if they are occupying the same volume under the same conditions of temperature and pressure.**

*per mole*Mole is one of the SI base units used in measuring samples in terms of the atoms and molecules they contain. One mole is defined as the number of atoms in a 12-g sample of carbon-12. Mole is abbreviated as mol.

The ratio of the number of molecules, ** N **to the number of molecules

**in 1 mol is equal to the number of moles,**

*N*_{A}**present in a sample of any substance. Mathematically, it can be written as**

*n*\(n = {N \over N_A}\).

The above equation can also be written in terms of the given mass of the sample, ** M_{sam} **and either the molar mas,

**which is equivalent to the mass of 1 mol, or the molecular mass,**

*M***which is the mass of one molecule. The number of molecules is the ratio of this masses, written as**

*m*\(n = {M_{sam} \over M}\) or \(n = {M_{sam} \over mN_A}\).

This indicates that the mass of 1 mol of a substance is equal to the product of the mass of one molecule and the number of molecules in 1 mol expressed as

\(M = mN_A\).

**Ideal Gas**

This topic tends to explain the different properties of gases such as the pressure, temperature, and volume in terms of the behavior of the molecules of gases. Experiments show that all real gases have the tendency to obey the following relationship under low enough densities. The relationship is written as an equation which is called as the **ideal gas law**:

\(pV = nRT\),

where ** p** is the absolute pressure,

**is the volume of the gas,**

*V***is the number of moles of gas present,**

*n***is the temperature, and**

*T***has a value of**

*R***and is known as the**

*8.31 J/mol\(\cdot\)K***gas constant**.

Ideal gas law equation can also be rewritten in terms of a constant called **Boltzmann constant, k** which is defined as

\(k = {R \over N_A} = {8.31\;J/mol \cdot K \over 6.02 \times 10^{23}\;mol^{-1}}=1.38 \times 10^{-23}\;J/K\).

Since, from the above equation \(R = kN_A\) and that \(N_A = {N \over n}\), then we can write the equation as \(nR = Nk\) and this then leads us to another expression for the ideal gas law which is

\(pV = NkT\).

**Example 1.**

What is the equivalent of a 7.8 x 10^{25} molecules of an ideal gas in moles?

Solution:

To solve for the equivalent of the number of molecues to moles, we use the formula for the number of moles f an ideal gas which is \(n = {N \over N_A}\), where N is the number of molecules equal to 7.8 x 10^{25} molecules and N_{A} is the Avogadro's number that has a value of **6.02 x 10 ^{23} mol^{-1.}**

\(n = {7.8 \times 10^{24}\; \not{\text{molecules}} \over 1} \times {1\;\text{mole} \over 6.02 \times 10^{23} \not{\text{molecules}}}=12.96\;\text{moles}\)

Thus, there are 12.96 moles of the ideal gas.

**Example 2.**

Oxygen gas is present in a 0.5 m^{3} container and is held at a temperature of 420 K and pressure of 10^{6} pascals. How many moles of oxygen are there?

Given:

V = 0.5 m^{3}

T = ^{ }420 K

P = 10^{6} pascals

Solution:

To solve the problem, we use the ideal gas law equation \(pV = nRT\) and this implies that the equation for the number of moles is

\(pV = nRT\) \(\implies\)\(n = {pV \over RT}\)

Substitute the given values,

\(n = {(10^6\; Pa)(0.5\;m^3) \over (8.31\;J/mol \cdot K)(420\;K)}\) = **143 moles**