# Important Kinetic Theory of Gas Formulas for JEE

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## Kinetic Theory of Gas

Gases can be studied by considering two theories, either by considering the small scale activity of the individual molecules present in the gas or by considering the gas as a whole in a large scale activity. To study the small-scale activity of the molecules present in the gas, we must have to use a theoretical model. This theoretical model is known as the Kinetic theory of gases. It was founded by the British scientist James Clerk Maxwell and the Austrian physicist Ludwig Boltzmann in the 19th century.

This theory assumes that molecules are tiny in size in relation to the distance between the molecules and the molecules are in constant random motion and regularly collide with each other as well as the walls of the container in which they are stored. This is what you learn in the kinetic theory of gases for JEE.

The individual molecules present in gases possess the standard physical properties of an element, such as mass, momentum, energy. The sum of masses of all the molecules divided by the volume that the gas occupies gives the density of the gas. The measure of the linear momentum of the gas is considered as the pressure of the gas. When the molecules of the gas collide against the walls of the container they are in, the molecules impart some momentum on the walls that produce a force that can be measured. The temperature of a gas is a measure that shows the mean kinetic energy of the gas. The molecules of the gas are in constant motion and produce energy from that motion. The temperature is directly proportional to the motion. These are some important points covered under the topic of the kinetic theory of gases IIT JEE syllabus.

### Assumptions of Kinetic Theory of Gases

The kinetic theory of gases is based upon some assumptions. Let us discuss some of these assumptions.

• The molecules or atoms of the gases are in constant random motion; they do not remain still at any time.

• The molecules or atoms of the gas are repeatedly colliding against each other as well as the walls of the container the gas is stored in.

• The particles of gases are very small, and the total volume occupied by these particles is considered negligible if compared with the total volume of the container.

• There is a negligible force of attraction between the particles of the gases.

• The average kinetic energy of gas particles is directly proportional to the actual temperature of that particular gas, and all gases at the same temperature have equal kinetic energy.

### Kinetic Theory of Gas Formulas

1. Boltzmann’s Constant

kB = nR/N

kB is the Boltzmann’s constant

R is the gas constant

n is the number of moles

N is the number of particles in one mole (the Avogadro number)

1. Total Translational K.E of Gas

K.E = (3/2)nRT

n is the number of moles

R is the universal gas constant

T is the absolute temperature

1. Maxwell Distribution Law

Vrms ＞ V＞ Vp

Vrms is the RMS speed.

V is the Average speed.

Vp is the most probable speed.

1. RMS Speed (Vrms)

Vrms = $\sqrt{8kt/m} = \sqrt{3RT/M}$

R is the universal gas constant.

T is the absolute temperature.

M is the molar mass.

1. Average Speed

$\overrightarrow{v} = \sqrt{8kt/\pi m} = \sqrt{8RT/\pi M}$

1. Most Probable Speed (Vp)

$V _\rho = \sqrt{2kt/m} = \sqrt{2RT/M}$

1. The Pressure of Ideal Gas

$P = \frac{1}{3} V^{2} rms$

P is the density of molecules.

1. Equipartition of Energy

$K = \frac{1}{2} K_{B}T$ for each degree of freedom.

K = (f/2) KвT for molecules having f degrees of freedom.

KB is the Boltzmann’s constant.

T is the temperature of the gas.

1. Internal Energy

U = (f/2) nRT

For n moles of an ideal gas.

### Solved Examples

Example 1

What is the average velocity or root mean square velocity of a molecule in a sample of oxygen at 0-degree celsius?

Solution

R = 8.3145 (kg.m2/sec2)/ K.mol

T = Absolute temperature in kelvin

M = mass of the moles present in the gas considered in kilograms.

T = ℃ + 273

T = 0 + 273

T = 273 K

Molar mass of oxygen = 2 ｘ 16

Molar mass of oxygen = 32 g/mol

Convert this to kg/mol

Molar mass of oxygen = 32g/mol ｘ 1 kg/1000g

Molar mass of oxygen = 3.2 ｘ10-2kg/mol

Vrms = $\sqrt{3RT/M}$

= [ 3( 8.3145 (kg.m2/sec2)/ K.mol)(273 K) / 3.2 ｘ10-2kg/mol ]1/2

=( 2.128 ｘ 105m2/sec2)1/2

= 461 m/sec

RMS = 461 m/sec.