Main Concept

The Maxwell-Boltzmann Distribution describes the distribution of the speed of particles in an idealized gas. It plays an important role in understanding the kinetic energy distribution of electrons and ions as well as in characterizing a particular gaseous substance. In classical physics, it was thought that the molecules of ideal gases bounced around with arbitrary velocities, never interacting with one another. However, scientists later discovered that an ideal gas is better modeled by taking intermolecular interactions into account. The Maxwell-Boltzmann distribution is a probability density function for the speed v of molecules with mass m in a gas at absolute temperature T. It is given by the following equation:

$f\left(v\right)\=4 \mathrm{\pi } v^2 {\ⅇ}^{-\frac{m v^2}{2 k T}}{\left(\frac{m}{2 \mathrm{\π} k T}\right)}^{\frac{3}{2}}$,

where k is Boltzmann's constant, 1.38 × ${10}^{-23}$ J/K. The probability that the molecule will have any speed between $v_1$ and $v_2$ is given by:

$P\left(v_1< v < v_2\right) \&equals; {\int }_{v_1}^{v_2}f\left(v\right) \&DifferentialD;v$.

The mean speed, $\bar{v}\&comma;$ can be calculated as follows:

$\bar{v}\&equals;{\int }_0^{\infty }\mathrm{vf}\left(v\right) \&DifferentialD;v\&equals;\sqrt{\frac{8 \mathrm{kT}}{\mathrm{\&pi;} m}}$.

The mean speed is always higher than the most probable speed as a result of the skewness of the distribution.

Example

Adjust the sliders to see how temperature and mass affect the Maxwell-Boltzmann Distribution.

What is the average speed of nitrogen gas molecules (each with a mass of $4.65 \times {10}^{-26}$ kg) when they are exposed to a temperature of 25 °C (298.15 K)?

Solution

You can use the formula given in the previous section to determine the average speed:   $\bar{v}\={\int }_0^{\infty }\mathrm{vf}\left(v\right) \ⅆv\=\sqrt{\frac{8 \mathrm{kT}}{\mathrm{\pi } m}}$   $\bar{v}\=\sqrt{\frac{8 \left(1.38 \times {10}^{-23}\mathrm{J}\/\mathrm{K}\right)\left(298.15 \mathrm{K}\right)}{\mathrm{\π} \left(4.65 \times {10}^{-26} \mathrm{kg}\right)}}$   $\bar{v}\=471\frac{m}{s}$   Therefore the average speed is 471 m/s.

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