Maxwell Boltzmann Distribution Formula

Not all the air molecules surrounding us travel at the same speed. Some air molecules travel faster, some move at a moderate speed and few others will hardly move at all. Hence, instead of asking the speed of any particular gas molecule we ask the distribution of speed in gas at a particular temperature. James Maxwell and Ludwig Boltzmann came up with a theory to show how the speeds of the molecule are distributed for an ideal gas. In the next section, we shall be discussing the Maxwell Boltzmann distribution formula in detail.

Maxwell Boltzmann Distribution Equation

The average kinetic energy of the gas molecules is given by the equation

\(\begin{array}{l}E_k=\frac{3}{2}k_BT=\frac{3}{2}\frac{k}{N_A}T\end{array} \)

where E_k is the average kinetic energy of the gas molecules

kB is the Boltzmann’s constant which is equal to

\(\begin{array}{l}1.38\times10^{-23}\,JK^{-1}\end{array} \)

R is the universal gas constant which is equal to 8.314 J/K/mol

NA is the Avagadro’s constant which is equal to

\(\begin{array}{l}6.023\times10^{23}\,mol^{-1}\end{array} \)

Solved Example

Example:

  1. What is the average speed of hydrogen molecules on the earth?

Solution:

Let us consider the temperature on earth to be 300 K.

The mass of a hydrogen molecule is

\(\begin{array}{l}2\times1.67\times10^{-27}\,kg\end{array} \)

The kinetic energy of the gas molecules is given by the equation

\(\begin{array}{l}E_k=\frac{3}{2}k_BT\end{array} \)

Substituting the values in the equation, we get

\(\begin{array}{l}E_k=\frac{3}{2}\times1.38\times10^{-23}\times300=6.21\times10^{-21}\,J\end{array} \)

To find the velocity from the average kinetic energy we use the formula

\(\begin{array}{l}E_k=\frac{1}{2}mv^2\end{array} \)

By re-arranging the formula, we get

\(\begin{array}{l}v^2=\frac{2E_k}{m}\end{array} \)

 

V2=(2)(6.21×10-21)/(2×1.67×10-27)

V   = √(12.42 × 106)/3.34

V  = √3.71× 106

V = 1.92× 103

Therefore, the average velocity is Vrms= 1928 m/s.

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