I have a question about deriving the equation of kinetic theory of ideal gas

$$PV=\frac{1}{3} Nmc_r^2$$

where $N$ is number of atoms, $c_r$ is root mean square of atom speed and $m$ is mass of one atom.

In deriving the equation, many text books consider the total rate of change of momentum within a certain time and calculate the pressure. However, in actual case the atoms collide on the wall of container separately at different position at different time, how can these changes of momentum be considered as a whole?


1 Answer 1


In kinetic theory we always try to derive equations that relate macroscopically averaged properties of the system. This means we only consider quantities that are averaged over distances much greater than the average distance $d$ between the atoms, and times much longer than the average time $\tau$ between collisions. This is because these averages are all we can measure, given that our measuring devices (thermometers, pressure gauges etc.) are also macroscopically large.

If you could actually build a "pressure gauge" that measured the force acting over an area $d^2$, with the ability to respond to changes over times shorter than $\tau$, then your gauge would register a wildly fluctuating value. At this point the very concept of the pressure stops being useful, because you can resolve what the individual atoms are doing. But normally your pressure gauge measures the force over an area that is huge compared to $d^2$, and responds much more slowly than $\tau$. Therefore the effect of these fluctuations cancels out and you only see the average. So the derivations you are looking at are actually only working out the average force over some area and time that are much larger than $d$ and $\tau$.

  • $\begingroup$ You answer helps me to understand better, but I still have a point needs to clarify. In kinetic theory, is it a necessary assumption that huge amount of atoms are present in the container of gas? Imagine in the case there is a few number of atoms, the container wall is just collided occasionally and it doesn't seem make sense to consider the average pressure for a relatively large area of container wall. $\endgroup$
    – Kelvin S
    Aug 2, 2014 at 3:54
  • 1
    $\begingroup$ Exactly right, you need there to be a macroscopically large number of atoms in the gas. For small numbers of atoms the fluctuations of extensive quantities will be very large compared to the average, so that using the average does not give a very good description. For a very large number of atoms, the size of fluctuations (of extensive quantities) tends to scale like $\sqrt{N}$, while the averages scale with $N$, so that the relative size of fluctuations is very small for large $N$. $\endgroup$ Aug 3, 2014 at 13:32

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