0.1 The kinetic molecular theory  (Page 4/7)

Based on our observations and deductions, we take as the postulates of our model:

• A gas consists of individual particles in constant and random motion.
• The individual particles have negligible volume.
• The individual particles do not attract or repel one another in any way.
• The pressure of the gas is due entirely to the force of the collisions of the gas particles with the walls of thecontainer.

This model is the Kinetic Molecular Theory of Gases . We now look to see where this model leads.

Derivation of boyle's law from the kinetic molecular theory

To calculate the pressure generated by a gas of $N$ particles contained in a volume $V$ , we must calculate the force $F$ generated per area $A$ by collisions against the walls. To do so, we begin by determining the number of collisions of particles withthe walls. The number of collisions we observe depends on how long we wait. Let's measure the pressure for a period of time $\Delta (t)$ and calculate how many collisions occur in that time period. For a particle to collide with the wall within the time $\Delta (t)$ , it must start close enough to the wall to impact it in that period of time. If the particle is travelling with speed $v$ , then the particle must be within a distance $v\Delta (t)$ of the wall to hit it. Also, if we are measuring the force exerted on the area $A$ , the particle must hit that area to contribute to our pressure measurement.

For simplicity, we can view the situation pictorially here . We assume that the particles are moving perpendicularly to the walls. (This is clearly not true. However,very importantly, this assumption is only made to simplify the mathematics of our derivation. It is not necessary to make thisassumption, and the result is not affected by the assumption.) In order for a particle to hit the area $A$ marked on the wall, it must lie within the cylinder shown, which is of length $v\Delta (t)$ and cross-sectional area $A$ . The volume of this cylinder is $Av\Delta (t)$ , so the number of particles contained in the cylinder is $\times ((Av\Delta (t))(), \frac{N}{V})$ .

Not all of these particles collide with the wall during $\Delta (t)$ , though, since most of them are not traveling in the correct direction. There are six directions for aparticle to go, corresponding to plus or minus direction in x, y, or z. Therefore, on average, the fraction of particles moving inthe correct direction should be $\frac{1}{6}$ , assuming as we have that the motions are all random. Therefore, the numberof particles which impact the wall in time $\Delta (t)$ is $\times ((Av\Delta (t))(), \frac{N}{6V})$ .

The force generated by these collisions is calculated from Newton’s equation, $F=ma$ , where $a$ is the acceleration due to the collisions. Consider first a singleparticle moving directly perpendicular to a wall with velocity $v$ as in . We note that, when the particle collides with the wall, the wall does not move, so the collision must generally conservethe energy of the particle. Then the particle’s velocity after the collision must be $-v$ , since it is now travelling in the opposite direction. Thus, the change in velocity of the particle inthis one collision is $2v$ . Multiplying by the number of collisions in $\Delta (t)$ and dividing by the time $\Delta (t)$ , we find that the total acceleration (change in velocity per time) is $\frac{2ANv^2}{6V}$ , and the force imparted on the wall due collisions is found bymultiplying by the mass of the particles: