Kinetic Molecular Theory

Chemistry: IB Diploma Higher Level, topic 1.1.2

Molecular Velocities and Kinetic Energy

In a gas sample, individual molecules have widely varying speeds. However, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant.

$Root mean squared$

Root mean squared

The kinetic energy of a particle of mass $m$ and speed $u$ is given by:
$KE = \frac{1}{2}mv^2$
To deal with a large number of gas molecules, we use averages for both speed and kinetic energy
The root mean square velocity of a particle, $v_{rms}$ , is defined as the square root of the average of the squares of the velocities:
$v_{rms} = \sqrt{\frac{v{_1}^2+v{_2}^2+v{_3}^2+...}{n}}$
Where $n$ is the number of particles.

Average kinetic energy

The average kinetic energy for a mole of particles, $KE_{avg}$ , is then equal to:
$KE_{avg} = \frac{3}{2}RT$
Where $R$ is the gas constant and $T$ is the kelvin temperature.
When used in this equation, the appropriate form of the gas constant is 8.314 J/mol⋅K.

$Average speed$

Average speed

Combining these equations we can write:
$\frac{1}{2}M{u_{rms}}^2 = \frac{3}{2}RT$
$u_{rms} = \sqrt{\frac{3RT}{M}}$
For $u_{rms}$ , M is measured in kg/mol rather than g/mol.

Temperature dependence

If the temperature of a gas increases, $KE_{avg}$ increases.
More molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts to the right.
If temperature decreases, $KE_{avg}$ decreases.
More molecules have lower speeds and fewer molecules have higher speeds, so the distribution shifts to the left.
This behavior is shown here for nitrogen gas.

Test yourself on Kinetic Molecular Theory

Which of the following is the correct formula for the average speed of a particle?

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