Kinetic Theory of Gases

The kinetic theory of gases relates the macroscopic property of the gas, like – Temperature, Pressure, and Volume to the microscopic property of the gas, like – speed, momentum, and position. In this model, the atoms and molecules are continually in random motion, constantly colliding with one another and the walls of the container within which the gas is enclosed. It is this motion that results in physical properties such as heat and pressure. In this article, let us delve deeper into the kinetic theory of gases.

Table of Contents

What is the Kinetic Theory of Gases?

The gases are made up of a large number of molecules and they are flying in a random direction at a certain speed. By knowing their speed or position, one can figure out the macroscopic properties. In other words, by knowing the value of velocity or the internal energy of gas molecules, one should be able to figure out the temperature or pressure.

Assumptions of Kinetic theory

The Kinetic theory of gases makes some basic assumptions. They are as follows:

Kinetic Theory Assumptions
  • The molecules do not interact with each other.
  • The collision of molecules with themselves or walls will be an elastic collision.
  • The momentum is conserved.
  • Kinetic energy will be conserved.

Read More: Kinetic Theory of Gases Assumptions

How Is the Kinetic Theory of Gases Derived?

Consider a cubic box of length l filled with the gas molecule of mass m, moving along the x-axis with velocity vTherefore its momentum is mvx.

The gas molecules collide with the walls. At wall 1, it collides and gains momentum mvx.

Similarly, the molecules collide wall 2, reversing the momentum i.e., -mvx.

Thus, the change in the momentum is given by

Δp = mvx-(-mvx) = 2mvx—–(1)

After the collision, the molecule travels a distance of 2l before colliding again with wall 1.

Thus, the time taken is given by

\(\begin{array}{l}Time=\frac{Distance\;travelled}{velocity}=\frac{2l}{v_{x}}\,\,\textup{—-(2)}\end{array} \)

These continuous collisions carry a force, given by

\(\begin{array}{l}Force=\frac{Change\;in\;momentum}{Change\;in\;Time}=\frac{\Delta p}{\Delta t}\,\,\textup{——(3)}\end{array} \)

Thus, by substituting the values from equations (1) and (2), we get

\(\begin{array}{l}Force=\frac{2mv_{x}}{\frac{2l}{v_{x}}}\end{array} \)

\(\begin{array}{l}Force(F)=\frac{mv_{x}^{2}}{l}\,\,\textup{——(4)}\end{array} \)

The continuous collisions also exert pressure on the wall given by-

\(\begin{array}{l}Pressure=\frac{Force}{Area}=\frac{F}{A}\end{array} \)

\(\begin{array}{l}Pressure(P)=\frac{\left ( \frac{mv_{x}^{2}}{l} \right )}{l^{2}}\end{array} \)

\(\begin{array}{l}P=\frac{mv_{x}^{2}}{l^{3}}\,\,\textup{——(5)}\end{array} \)

We know that the gas is made of N numbers of molecules and moves in all possible directions. Thus, the pressure exerted on wall 1 by the collision of N number of gas molecules is given by-

\(\begin{array}{l}P=\frac{m}{l^{3}}\left (v_{x1}^{2} +v_{x2}^{2}+v_{x3}^{2}+…..+v_{xN}^{2} \right )\end{array} \)

\(\begin{array}{l}\Rightarrow P=\frac{m}{l^{3}}\left (N\bar{v}_{x}^{2} \right )\end{array} \)

\(\begin{array}{l}\Rightarrow P=\frac{mN\bar{v}_{x}^{2}}{V}\textup{——(6)}\end{array} \)


\(\begin{array}{l}\bar{v}_{x}^{2}= v_{x1}^{2} +v_{x2}^{2}+v_{x3}^{2}+…..+v_{xN}^{2}\end{array} \)
is the average velocity (or velocity component) of all gas molecules colliding with wall 1 along x-direction.

And l3 = V

On extending the above equation to three dimensions, we get-

\(\begin{array}{l}v^{2}= v_{x}^{2}+v_{y}^{2}+v_{z}^{2}\end{array} \)

\(\begin{array}{l}\bar{v}^{2}= \bar{v}_{x}^{2}+\bar{v}_{y}^{2}+\bar{v}_{z}^{2} \end{array} \)

For a large number of gas molecules:

\(\begin{array}{l}\bar{v}_{x}^{2}=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}\,\,\textup{or}\,\,\bar{v}^{2}=3\bar{v}_{x}^{2}\end{array} \)

\(\begin{array}{l}\Rightarrow \bar{v}_{x}^{2}=\frac{1}{3}\bar{v}^{2}\,\,\textup{—-(7)}\end{array} \)

Substituting equation (7) in equation (6), we get

\(\begin{array}{l}P=\frac{Nm\left ( \frac{1}{3} \bar{v}^{2}\right )}{V}\end{array} \)

On rearranging,

\(\begin{array}{l}\Rightarrow PV= \frac{1}{3}Nm\bar{v}^{2}\,\,\textup{——(8)}\end{array} \)

What Is the Average Kinetic Energy of a Gas Molecule?

We know that PV=nRT —(9)

Thus equating equations (8) and (9), we get-

\(\begin{array}{l}nRT=\frac{1}{3}Nmv^{2}\end{array} \)

Multiplying and Dividing by 2, we get-

\(\begin{array}{l}nRT=\frac{2}{3}N\left ( \frac{1}{2}mv^{2} \right )\,\,\textup{—–(12)}\end{array} \)

Rearranging equation (10), we get-

\(\begin{array}{l}\frac{1}{2}mv^{2}=\frac{3}{2}\frac{nRT}{N}\end{array} \)
\(\begin{array}{l}=\frac{3}{2}\frac{RT}{\left ( \frac{N}{n} \right )}\,\,\textup{—-(11)}\end{array} \)

Here, N is the total number of gas molecules in a given region of space

n is the number of moles of gas present in a given region of space.

\(\begin{array}{l}\frac{N}{n}=N_{A}\end{array} \)
Avagadro’s number in this context is the number of molecules present in the one mole of gas.
NA = 6.022140857 × 1023.

Substituting NA in equation (11),

\(\begin{array}{l}(11)\Rightarrow \frac{1}{2}mv^{2}=\frac{3}{2}\frac{RT}{N_{A}}\,\,\textup{—–(12)}\end{array} \)

Thus, Average Kinetic Energy of a gas molecule is given by-

\(\begin{array}{l}\Rightarrow K.E=\frac{3}{2}kT\end{array} \)
\(\begin{array}{l}K.E= \frac{1}{2}mv^{2}\,\, \textup{and} \,\,k=\frac{R}{N_{A}}\textup{a Boltzmann constant}\end{array} \)

For monoatomic molecules, the total internal energy is given by-

\(\begin{array}{l}E_{total}=\frac{3}{2}PV\end{array} \)
\(\begin{array}{l}∵\,\,N\left ( \frac{1}{2}mv^{2} \right )=N(K.E)=E_{total}=\frac{3}{2}PV\end{array} \)
\(\begin{array}{l}E_{total}=\frac{3}{2}NkT\end{array} \)
\(\begin{array}{l}E_{total}=\frac{3}{2}nRT\end{array} \)

Solved Numericals on Kinetic Theory of Gases

Watch the video below and learn to solve numericals based on kinetic theory of gases

What Is the Significance of the Kinetic Theory of Gases?

By knowing the Temperature we can directly figure out the average Kinetic energy of a gas molecule. No matter what gas you are considering. Unless and until it is an ideal gas.

Knowing the macroscopic parameters of gases like Pressure, Volume, Temperature etc., one can accurately calculate the microscopic parameters like Momentum, Velocity, Internal energy, Kinetic energy, Thermal energy etc. and Vise-Versa.

Related links:

Boyle’s law
The behaviour of gas molecules
Avogadro’s Number
Ideal Gas

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