In the following, we want to establish a relationship between temperature and the kinetic energy of gas particles.
For this, we first define the particle density \(n\) (not to be confused with the amount of substance) as the number of particles per volume:
$$
n = \frac{N}{V}
$$
Next, consider a cubic container in which the gas is enclosed.
The number of particles that collide with one of the 6 walls within a certain time interval \(\Delta t\) is given by:
$$
N' = n_x v_x A \Delta t
$$
Here, for simplicity, the \(x\)-direction has been chosen.
At each collision, the particle transfers the momentum
$$
\Delta |\vec{p}| = 2m v_x
$$
to the wall.
Instead of \(p\), the notation \(|\vec{p}|\) is used for momentum to avoid confusion with pressure later.
The factor 2 arises because the particles do not stick to the wall but are reflected without friction.
The total force on the wall is then obtained from the number of colliding particles \(N'\) and the momentum transfer:
$$
F = N' \frac{\Delta |\vec{p}|}{\Delta t} = 2 m n_x A v_x^2
$$
Thus, the pressure is:
$$
p = \frac{F}{A} = m n_x v_x^2
$$
Since just as many particles move in the negative as in the positive \(x\)-direction, the right-hand side is multiplied by \(1/2\), which eliminates the factor 2.
To determine the total pressure in all directions, we assume that the velocities of all particles are distributed isotropically, meaning there is no preferred direction.
Therefore, the mean squares of the velocity components are identical:
$$
\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac{1}{3}\,\overline{v^2}
$$
Thus, the total pressure is:
$$
p = \frac{1}{3} m n \overline{v^2}
$$
Multiplying this expression by \(V\) and additionally introducing the factor \(1/2\), we obtain:
$$
pV = \frac{2}{3} N \frac{1}{2} m \overline{v^2}
$$
Comparing this expression with the kinetic energy yields:
$$
pV = \frac{2}{3} N \overline{E_\mathrm{kin}}
$$
This formulation states that the product of pressure and volume equals the average energy of all gas particles of mass \(m\).
This result makes sense, since we have already noted that the term \(pV\) can be interpreted as energy.
By substituting into the gas equation, we further obtain the relation:
$$
\frac{1}{2} m \overline{v^2} = \frac{3}{2} k_B T
$$
Thus, the temperature can be directly associated with a kinetic energy and hence the average velocity of the gas particles.
It is important to note that not all particles have the same velocity; rather, it is statistically distributed.
Therefore, one can only speak of an average velocity or an average kinetic energy.
Finally, this requires the choice of an absolute temperature scale with \(T=0\) K corresponding to \(v=0\).