-2
$\begingroup$

Is there any reason as to why we assume the volume of ideal gas to be negligible in comparison to the container? What is the underlying reason to assume that the ideal gas molecules are point molecules? Why can we not assume them to be non-point particles?

$\endgroup$
8
  • 2
    $\begingroup$ If you assume point particles and no inter-particle forces, as simplifying approximations, then you have an easy path to this useful approximation: pV = nRT, i.e., the ideal gas law. Of course, you can assume non-negligible particle sizes and assume that inter-particle forces exist, so then the equation of state will necessarily be more complicated. $\endgroup$
    – Ed V
    Commented Jan 26, 2021 at 17:02
  • 2
    $\begingroup$ If it weren't negligible, we would expect different behavior of molecules of distinct size. However, the ideal gas law is not dependent on particle size. You might try to derive a gas law for uniform volume particles rather than zero-volume particles but then you would expect that chosen uniform volume to somehow appear in your alternative gas law (for example, the volume of the gas could not be smaller than the total volume of particles). In the ideal gas law, the volume of the gas approaches zero as the temperature approaches zero (with fixed n and p). $\endgroup$
    – Karsten
    Commented Jan 26, 2021 at 17:28
  • 2
    $\begingroup$ The experiment shows that boiling water at 100°C produces a vapor which is about $1000$ times more cumbersome that the liquid water from which it is produced. As in this process the molecules have not been chemically modified, it means that in the vapor, the volume occupied by the molecules is negligible, as if they were point particles. $\endgroup$
    – Maurice
    Commented Jan 26, 2021 at 17:28
  • 1
    $\begingroup$ I'd use voluminous instead of cumbersome, but Maurice makes a good point. Chemistry typically uses 3 or 4 significant figures at best. The equation PV=nRT isn't going to yield 15 digits of precision. // Van der Waals equation makes some rudimentary corrections to the ideal gas equation. $\endgroup$
    – MaxW
    Commented Jan 26, 2021 at 17:59
  • $\begingroup$ Saving the best for last [ ;-) ], see equation of state in wikipedia! Loads of increasingly complicated equations of state! Imagine being a beginning chemistry student and having all that dropped on you! Hence the practical reality that beginning chemistry courses are heavily simplified (and often seriously over-simplified) so as to be manageable and so that there are survivors! $\endgroup$
    – Ed V
    Commented Jan 26, 2021 at 18:34

1 Answer 1

3
$\begingroup$

Because it gives simpler-to-derive laws which are often very good approximations

Clearly real gases do not always follow the ideal gas laws. They mostly liquefy under some conditions, for example, and, under those conditions they are clearly not ideal.

But in practice gas laws are used for things far away from those non-ideal regions. When we are applying the laws we are usually applying them to gases where the deviations from ideality are small. At normal lab conditions (1 atm pressure, room temperature) most of the gases you will ever deal with will be as close to ideal as matters within measurement error. Given that you want the simplest law possible and that law is easily derived by making the assumptions that particles have no size and don't stick together.

More complicated gas laws can and have been developed (van der Waals equation was a 19th century development of a simple extended law that captures some features about liquefaction and non-ideality under cold or high pressure conditions). But the deviations of the results from the simplest ideal gas equation are small under normal conditions (van der Waals calculations for carbon dioxide differ from ideal results by 0.5% under normal lab conditions). This is such a small amount that more complex equations are not worth bothering with unless you know you are using very low temperatures or high pressures.

And, when you are in one of those regions, you will find that the gas equations get very complicated very quickly. It takes a lot of extra empirical data and mathematical complexity to handle those cases and under normal conditions this just isn't worth the effort. Why waste effort when you don't gain any practical benefit?

$\endgroup$

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