Let us consider a system of N ideal gas particles inside a cubic container of length d and volume V = d^{3}. Each particle has mass m. Let us neglect particle-particle collisions.

Now, let us apply Newtonian mechanics to work out what is the pressure exerted by these particles on the wall of the container.

**Change in momentum per particle on collision with wall.**

When a particle of velocity v(=(v_{x}, v_{y}, v_{z})) collides with the container wall that is parallel to the y-z plane, v_{x} is reversed while v_{y} and v_{z} remain unchanged. The change in momentum of the particle is: $\Delta p_{x} = \, – 2 mv_{x}$.

**Average force on wall per particle**

For the particle to collide with the same wall again, it must travel a round trip distance of 2d in the x direction. Assuming no collisions occur in-between, this takes time of $\left( \frac{2d}{v_{x}} \right)$. The average force exerted by this particle on the wall is equal to the average rate at which it transfers momentum to the wall: (Recall that $F_{ave} = \frac{\Delta p}{\Delta t}$)

$$F_{i} = \frac{2 m v_{x, i}}{\frac{2 d}{v_{x, i}}} = \frac{m v_{x,i}^{2}}{d}$$

The subscript i indicates that this force on the wall is exerted by particle i.

The total average force exerted by all N particles on the wall is given by the sum:

$$F = \sum\limits_{i} F_{i} = \frac{m}{d} \left( \sum\limits_{i = 1}^{N} v_{x,i}^{2} \right)$$

Recalling the definition of average: $\left< a \right> = \frac{1}{N} \sum\limits_{i=1}^{N} a_{i}$ so $\sum\limits_{i = 1}^{N} a_{i} = N \left< a \right>$. Let us denote:

$$\sum\limits_{i = 1}^{N} v_{x,i}^{2} = N \left< v_{x}^{2} \right>$$

where $\left< V_{x}^{2} \right>$ is the average of the square of $v_{x}$ of the particles.

Hence, The total force on the wall is:

$$F = \frac{N m}{d} \left< v_{x}^{2} \right>$$

Let us write in terms of $\left< v^{2} \right>$.

Since the velocity of the particle can be resolved into the x,y,z direction, Pythagoras theorem gives $v_{i}^{2} = v_{x,i}^{2} + v_{y,i}^{2} + v_{z,i}^{2}$.

Now, we sum the velocity of all the particles:

$$\begin{aligned} \sum\limits_{i = 1}^{N} v_{i}^{2} &= \sum\limits_{i=1}^{N} \left( v_{i,x}^{2} + v_{i,y}^{2} + v_{i,z}^{2} \right) \\ \frac{1}{N} \sum\limits_{i=1}^{N} v_{i}^{2} &= \frac{1}{N} \sum\limits_{i=1}^{N} \left( v_{i,x}^{2} + v_{i,y}^{2} + v_{i,z}^{2} \right) \end{aligned}$$

Hence:

$$\left< v^{2} \right> = \left< v_{x}^{2} \right> + \left< v_{y}^{2} \right> + \left< v_{z}^{2} \right>$$

Since, motion is completely random, we have:

$$\left< v_{x}^{2} \right> = \left< v_{y}^{2} \right> = \left<v_{z}^{2} \right>$$

Hence,

$$\left< v^{2} \right> = 3 \left< v_{x}^{2} \right> $$

**Thus the total force on the wall can be written as:**

$$F = \frac{Nm}{3s} \left< v^{2} \right>$$

The pressure on the wall is given by:

$$\begin{aligned} P &= \frac{F}{A} \\ &= \frac{F}{d^{2}} \\ &= \frac{Nm}{3d^{3}} \left< v^{2} \right> \\ &= \frac{Nm}{3V} \left< v^{2} \right> \end{aligned}$$

**Rearranging the above equation and recalling that kinetic energy, $E_{kin} = \frac{1}{2} m_{i} v_{i}^{2}$ so $\left< E_{kin} \right> = \left< \frac{1}{2} m v^{2} \right> = \frac{1}{2} m \left< v^{2} \right>$**

$$P = \frac{2}{3} \frac{N}{V} \left( \frac{1}{2} m \left< v^{2} \right> \right) $$

,where

- $\frac{N}{V}$ is the number density of the particles
- $\left( \frac{1}{2} m \left< v^{2} \right> \right)$ is the average particle translational kinetic energy

Therefore the pressure of the ideal gas is proportional to the number of particles per unit volume and the average translational kinetic energy of the particles.

The pressure is given by two-thirds of the translational kinetic energy density of the gas.

Experimentally, it is known that PV = Nk_{B}T

**Therefore, the average kinetic energy:**

$$\frac{1}{2} m \left< v^{2} \right> = \frac{3}{2} k_{B} T$$

Since the average translational kinetic energy of the particles gives its temperature, this temperature is also called the kinetic temperature of the particles.

Later, we will find that this result can be derived separately from the Maxwell-Boltzmann distribution, and also from the energy equipartition theorem. Therefore, the ideal gas model does correctly predict the “ideal” behaviour of real gases in the limit of low pressures and high temperatures.

Next: Root-mean-square speed of the gas particles