Why 1 mole of H2 occupied the same volume occupied by 1 mole of O2?

I'm asking a question about the volume occupied by gasses in standard temperature and pressure. My textbook said that a mole of any gas occupies 22.4 L at standard temperature and pressure. But $$\ce{O2}$$ is greater than $$\ce{H2}$$ in size $O$ and $$H$$ atoms compared"> I asked my teacher why the volume occupied by 1 mole of $$\ce{H_2}$$ is equal to the volume occupied by $$\ce{O_2}$$ at standard temperature and pressure, but he seemed not to know the answer, and I remain confused. Please answer in simple words.

• The basic assumption is ideal gas behavior. // PS -- STP changed in 1982. An ideal gas has a volume of 22.7 liters at STP. – MaxW Mar 25 at 18:31
• Think of tennis and football balls in a very big hangar to see that the frequency of collision is basically the same, unless you pack the hangar with a lot more balls. What you refer to is the proper volume, which is not the volume of a gas. Reread what your book say about ideal gas, if the level of your class includes it. Else use my pictorial example. – Alchimista Mar 26 at 8:37

On the same page, one can find a value of 68 nm ($$6.8 \cdot 10^{-8}$$ m) for the mean free path at room temperature and a pressure of about 1 atmosphere. This is to be compared with the bond lengths of $$\ce{H2}$$ and $$\ce{O2}$$, which are about 74 pm ($$0.74 \cdot 10^{-10}$$ m) and 121 pm ($$1.21 \cdot 10^{-10}$$ m) respectively. Even if we generously assume and effective molecular diameter of three times the bond length, we can conclude that molecular size has little impact given that there are two orders of magnitude difference between size and mean free path.