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The ideal gas law states:

One mole of an ideal gas will occupy a volume of $\pu{22.4 liters}$ at STP.

Does this mean that both

  • $\pu{1 mole}$ of $\ce{O}$ would occupy $\pu{22.4 L}$ (or if this doesn't usually occur in nature, say \pu{1 mole }of $\ce{He}$ or any another monoatomic gas)
  • $\pu{1 mole}$ of $\ce{O2}$ would occupy $\pu{22.4 L}$

I am assuming the answer is yes based on:

  • reading the ideal gas law definition, it wouldn't make sense if it was true only for $\ce{He}$ but not for $\ce{O}$.
  • in class when we are finding the volume of e.g. $\pu{100 grams}$ of $\ce{Cl2}$, it was always $\frac{\pu{100 grams}}{(2\times\pu{35 grams} \text{[molar mass of $\ce{Cl2}$]}} \times\pu{22.4L}$ and not $\pu{1\times35 grams}$

But why? Wouldn't $\ce{O2}$ contain twice the amount of $\ce{O}$ matter and hence take up twice the volume?

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    $\begingroup$ The ideal gas laws assume that the volume occupied by the gas itself is negligible, zero essentially. So twice zero is still zero. As you realize this isn't really true. Van der Waals equation for gases tries to correct for this. It is a bit better than PV=nRT but still not great. There are other models that are even better. see en.wikipedia.org/wiki/Real_gas $\endgroup$
    – MaxW
    Commented Sep 21, 2016 at 0:15
  • $\begingroup$ "it was always 100 grams / (TWO . 35 grams [molar mass of Cl2])" - 35 grams is the molar mass of Cl. And you are talking about Cl2, not Cl. 35 grams * 2 is the molar mass of Cl2. $\endgroup$
    – Stack Exchange Supports Israel
    Commented Sep 21, 2016 at 5:20
  • $\begingroup$ Think of it this way: while the container filled with O₂ does contain twice the matter, because the systems are held at STP, each O₂ molecule must have the same momentum as each molecule of O. So, even though the O₂ weighs twice as much, the O molecules are moving twice as fast so they both exert the same pressure on their environment (and therefore take up the same volume). $\endgroup$
    – sig_seg_v
    Commented Sep 21, 2016 at 8:26
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    $\begingroup$ Strictly speaking, the volume at STP is now 22.7 L. It went up recently (1980's!) when the standard pressure was reduced from one atmosphere to one bar. $\endgroup$
    – Oscar Lanzi
    Commented Feb 19, 2017 at 9:24

2 Answers 2

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Does this mean that both

  • 1 mole of $\ce O$ would occupy $22.4~\mathrm L$ (or if this doesn't usually occur in nature, say 1 mole of $\ce{He}$ or another monoatomic gas)
  • 1 mole of $\ce{O2}$ would occupy $22.4~\mathrm L$

Yes, it means exactly that. And you're right, a stable gas of $\ce O$ atoms is a pretty exotic thing, so $\ce{He}$ is a much better example.

Wouldn't $\ce{O2}$ contain twice the amount of $\ce O$ matter

Yes and no. The $\ce{O2}$ gas would indeed contain twice the mass of the $\ce O$ gas, but it would contain the same number of particles ("molecular entities") as the $\ce O$ gas.

and hence take up twice the volume?

Nope. There is a lot of empty space in gases. A lot. So much, in fact, that molecules have to get pretty darn large in the gas phase$^\dagger$ before they start to get "cramped" and "insist" on taking up more space (viz., occupying a larger volume).

To give you an idea of just how much room there is, consider what's called the van der Waals volume of $\ce{O2}$. This is an estimate of the total volume actually occupied by one molecule of $\ce{O2}$, and has a value of about $0.053~\mathrm{nm}^3$ (that's 'cubic nanometers'). For one mole of $\ce{O2}$ molecules, that works out to just under $32~\mathrm{cm^3\over mol}$ of volume actually occupied by the oxygen molecules themselves.

However, as you note, that mole of $\ce{O2}$ gas actually expands to fill a full $22.4~\mathrm L$ of space at STP. So, that space is actually only about ${32~\mathrm{cm^3}\over 22.4~\mathrm{L}} = 0.14\%$ full of $\ce{O2}$.

So, even though one $\ce{O2}$ molecule does take up more space than one $\ce O$ atom, compared to the amount of space available, they're both really, really tiny. And, thus, the fact that the two $\ce O$'s in $\ce{O2}$ are bound so closely to one another means that they both behave essentially identically in the pressure they exert when they fill up a container of a given size.

$^\dagger$ At ambient temperature and pressure, anyways.

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  • $\begingroup$ RE: and hence take up twice the volume? // This is a bit ambiguous. I took it to mean that the Van der Waals correction for $\ce{O2}$ would be twice as big as for the correct for $\ce{O}$ which would be fairly accurate. $\endgroup$
    – MaxW
    Commented Sep 21, 2016 at 7:52
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    $\begingroup$ Isn't volume at stp is 22.7 L and at ntp is 22.4 L ? $\endgroup$
    – user31607
    Commented Sep 21, 2016 at 9:49
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    $\begingroup$ @MaxW Agreed, both about the ambiguity and about the $\sim\!2\times$ vdW volume of $\ce{O2}$ vs $\ce O$. The whole topic is rife with ambiguity, actually, especially for a new learner. $\endgroup$
    – hBy2Py
    Commented Sep 21, 2016 at 13:35
  • $\begingroup$ @A---B Very possibly. There was no such thing as NTP when I was learning chemistry 20 years ago ... it may have been introduced in the interim. $\endgroup$
    – hBy2Py
    Commented Sep 21, 2016 at 13:36
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Yes. Because STP is the standard temperature which is 0 degree celcius or 273 K nad standard pressure which is 760 mm Hg. So, if there's no given of volume, you can use the standard volume which is 22.4 L. No matter what kind of element is that, if we say STP, the number of volume at standard is 22.4 L

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