Kinetic Theory Of Gases Derivation

Kinetic Theory of Gases

The theory relates the macroscopic property of the gas, like- Temperature, Pressure, Volume to the microscopic property of the gas, like- speed, momentum, position. The gases are made up of a large number of molecules and they are flying in a random direction with a certain speed. By knowing their speed or position, one can figure out the macroscopic properties. In other words, by knowing the value of temperature or pressure one should be able to figure out the velocity or internal energy of gas molecules.

Assumptions of Kinetic theory

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

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

Derivation of Kinetic Theory of Gases

Consider a cubic box of length l filled with the gas molecule of

Mass m

Moving along the x-axis with velocity vx,

Therefore its momentum is mvx

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

It again collides with wall 2, the momentum is reversed. I.e., -mvx

Thus the change in the momentum is 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 –

\(Time=\frac{Distance\;travelled}{velocity}=\frac{2l}{v_{x}}\)—-(2)

These continuous collisions carry a force, given by-

\(Force=\frac{Change\;in\;momentum}{Change\;in\;Time}=\frac{\Delta p}{\Delta t}\)——(3)

Thus, substituting the values from equation(1) and (2),

\(Force=\frac{2mv_{x}}{\frac{2l}{v_{x}}}\) \(Force(F)=\frac{mv_{x}^{2}}{l}\) ——(4)

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

\(Pressure=\frac{Force}{Area}=\frac{F}{A}\) \(Pressure(P)=\frac{\left ( \frac{mv_{x}^{2}}{l} \right )}{l^{2}}\) \(P=\frac{mv_{x}^{2}}{l^{3}}\)——(5)

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

\(P=\frac{m}{l^{3}}\left (v_{x1}^{2} +v_{x2}^{2}+v_{x3}^{2}+…..+v_{xN}^{2} \right )\) \(\Rightarrow P=\frac{m}{l^{3}}\left (N\bar{v}_{x}^{2} \right )\) \(\Rightarrow P=\frac{mN\bar{v}_{x}^{2}}{V}\)——(6)

Where,

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

And l3 = V the volume

On extending the above equation to three dimensions we get-

\(v^{2}= v_{x}^{2}+v_{y}^{2}+v_{z}^{2}\) \(\bar{v}^{2}= \bar{v}_{x}^{2}+\bar{v}_{y}^{2}+\bar{v}_{z}^{2} \)

For a large number of gas molecules:

\(\bar{v}_{x}^{2}=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}\)

or

\(\bar{v}^{2}=3\bar{v}_{x}^{2}\) \(\Rightarrow \bar{v}_{x}^{2}=\frac{1}{3}\bar{v}^{2}\)—–(7)

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

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

On rearranging,

\(\Rightarrow PV= \frac{1}{3}Nm\bar{v}^{2}\) ——(8)

Average Kinetic Energy of a gas molecule

We know that PV=nRT —(9)

Thus equating equation (8) and (9) we get-

\(nRT=\frac{1}{3}Nmv^{2}\)

Multiplying and Dividing by 2 we get-

\(nRT=\frac{2}{3}N\left ( \frac{1}{2}mv^{2} \right )\) —–(10)

Rearranging equation (10) we get-

\(\frac{1}{2}mv^{2}=\frac{3}{2}\frac{nRT}{N}\) \(=\frac{3}{2}\frac{RT}{\left ( \frac{N}{n} \right )}\) —-(11)

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.

\(\frac{N}{n}=N_{A}\) , Avogadro’s number.

Avogadro number

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),

\((11)\Rightarrow \frac{1}{2}mv^{2}=\frac{3}{2}\frac{RT}{N_{A}}\) —–(12)

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

\(\Rightarrow K.E=\frac{3}{2}kT\)

Here, \(K.E= \frac{1}{2}mv^{2}\) and \(k=\frac{R}{N_{A}}\) a Boltzmann constant

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 are you 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.

Kinetic theory of gases formula

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

\(E_{total}=\frac{3}{2}PV\) [∵\(N\left ( \frac{1}{2}mv^{2} \right )=N(K.E)=E_{total}=\frac{3}{2}PV\)] \(E_{total}=\frac{3}{2}NkT\) \(E_{total}=\frac{3}{2}nRT\)

Physics related links:

Stay tuned with Byju’s for more such interesting derivations. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics related videos and an unlimited academic assist.


Practise This Question

A galvanometer having a resistance of 8Ω is shunted by a wire of resistance 2Ω. If the total current is 1 ampere, the part of it passing through the shunt will be