 # 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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