5.1: Classical Model of Ideal Gas Pressure

Introduction

Classical physics can be used to derive the form of the ideal gas law,  which can be combined with the root-mean-square average speed from the Maxwell-Boltzmann distribution of speeds (derived classically or via statistical mechanics) to recover the ideal gas law. The discussion of collisions with the walls to produce a pressure provides a place to begin understanding how collisions are involved in the properties of gases.

The assumptions are those of the kinetic molecular theory of gases:

1. Gas particles obey Newton’s laws of motion and travel in straight lines unless they collide with other particles or the walls of the container.
2. the particle trajectories and velocities are random.
3. Gas particles are very small compared to the averages of the distances between them.
4. Molecular collisions are perfectly elastic so that kinetic energy is conserved.

Ideal gas law from the kinetic molecular model of gases

Ideal gases have no interactions between the particles, hence, the particles do not exert forces on each other. However, particles do experience a force when they collide with the walls of the container. Let us assume that each collision with a wall is elastic. Let us assume that the gas is in a cubic box of length \(a\) and that two of the walls are located at \(x = 0\) and at \(x = a\). Thus, a particle moving along the \(x\) direction will eventually collide with one of these walls and will exert a force on the wall when it strikes it, which we will denote as \(F_x\). Since every action has an equal and opposite reaction, the wall exerts a force \(-F_x\) on the particle.

According to Newton’s second law, the force \(-F_x\) on the particle in this direction gives rise to an acceleration via

\[-F_x = ma_x = m\dfrac{\Delta v_x}{\Delta t} \label{2.1} \]

Here, \(t\) represents the time interval between collisions with the same wall of the box. In an elastic collision, all that happens to the velocity is that it changes sign. Thus, if \(v_x\) is the velocity in the \(x\) direction before the collision, then \(-v_x\) is the velocity after, and \(\Delta v_x = -v_x - v_x = -2v_x\), so that

\[-F_x = -2m\dfrac{v_x}{\Delta t} \label{2.2} \]

Since the particles have no forces acting upon them, except for when they collide iwht the wall container, the particles move at constant speed. Thus, a collision between a particle and, say, the wall at \(x = 0\) will not change the particle’s speed. Before it strikes this wall again, it will proceed to the wall at \(x = a\) first, bounce off that wall, and then return to the wall at \(x = 0\). The total distance in the \(x\) direction traversed is \(2a\), and since the speed in the \(x\) direction is always \(v_x\), the interval \(\Delta t = \dfrac{2a}{v_x}\). Consequently, the force is:

\[-F_x = -\dfrac{mv_x^2}{a} \label{2.3} \]

Thus, the force that the particle exerts on the wall is:

\[F_x = \dfrac{mv_x^2}{a} \label{2.4} \]

The mechanical definition of pressure is the average force over area:

\[P = \dfrac{\langle F \rangle}{A} \label{2.5} \]

where \(\langle F \rangle\) is the average force exerted by all \(N\) particles on a wall of the box of area \(A\). Here \(A = a^2\). If we use the wall at \(x = 0\) we have been considering, then

\[P = \dfrac{N \langle F_x \rangle}{a^2} \label{2.6} \]

because we have \(N\) particles hitting the wall. Hence:

\[P = \dfrac{N m \langle v_x^2 \rangle}{a^3} = \dfrac{N m \langle v_x^2 \rangle}{V}, \label{2.7} \]

where we have recognized that \(a^3 = V\). Equation \(\ref{2.7}\) shows that with this simple physical model we predict that the pressure depends on the density of particles (N/V) and the average kinetic energy (\(1/2 m \langle v_x^2 \rangle\)).

To go any further towards the ideal gas law, we need an expression for \(\langle v_x^2 \rangle\). From our study of the Maxwell-Boltzmann distribution, we know:

\[\langle v^2 \rangle = \dfrac{3k_B T}{m}, \label{2.8} \]

where \(v\) is the speed in 3 dimensions. From the definition of vector length \(v^2 = v_x^2 + v_y^2 + v_z^2 \implies \langle v^2 \rangle= \langle v_x^2 \rangle+\langle v_y^2 \rangle+ \langle v_z^2 \rangle\). In addition we know that all three dimensions have the same velocity distribution. Thus, \(\langle v^2 \rangle= 3\langle v_x^2 \rangle = 3\langle v_y^2 \rangle = 3 \langle v_z^2 \rangle \implies \langle v_x^2 \rangle = \langle v^2 \rangle/3 \). Substituting equation \(\ref{2.8}\) divided by 3 into equation \(\ref{2.7}\) for \( \langle v_x^2 \rangle\) yields:

\[P = \dfrac{N k_B T}{V} = \dfrac{n R T}{V} \label{2.9} \]

which is the ideal gas law. 

Contributors and Attributions

• Mark Tuckerman (New York University)

• Jerry LaRue (Chapman University)

• Jonathan Gutow (UW Oshkosh)