0
$\begingroup$

The ideal gas law has in it the number of particles N, in PV=NKT. Now, if N=1, does the ideal gas law still apply?

$\endgroup$
2
  • $\begingroup$ The criterion is that the mean free path is small compared to the physical dimensions of the container. $\endgroup$ Commented Oct 23 at 22:23
  • $\begingroup$ It depends on what you are looking for. But surely, in the case of a ideal gas, you can take one particle in a box. If you average every quantity over a large period you will find exactly the ideal gas law. Note that it would not be the case for an interacting system where you would have finite size correction. For a system with a large number of particle, you would not have to average over a long period too $\endgroup$
    – Syrocco
    Commented Oct 23 at 22:29

1 Answer 1

5
$\begingroup$

It technically only applies for $N=\infty$, and only there. That being said, we do indeed apply it for small N.

The ideal gas law is an an asymptotic approximation. There is no cutoff where it starts to apply, it simply gives more and more accurate results as N increases.

Whether the ideal gas law applies is not just dependent on N, but what you intend to model and what you seek to derive from it. The more precise you need your results to be, the higher N has to be before the statistical likelihoods behind it become sufficiently strong that you can forget about the statistics and claim things like "pressure" are simply a number.

In practice, N is always large enough, until one is consciously trying to experiment on small numbers of gas molecules. On that day, you will understand the requirements on your modeling, and will have enough knowledge to formulate whether or not the ideal gas law meets your needs or not.

An example of where it starts to fall apart is in space or an artificial vacuum. Once the density gets low enough and the chamber small enough, may need to worry about the rarified gas behaviors which stray from the ideal gas law. Or its entirely possible that the thing you need to derive from your model is low enough precision that the ideal gas law is still sufficient. For some purposes, we even get away with claiming that the pressure in space is 0, because we get a good enough answer with that assumption (even if its technically not accurate)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.