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 <span class=$O$ 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.

| improve this question | | | | |
  • 2
    $\begingroup$ The basic assumption is ideal gas behavior. // PS -- STP changed in 1982. An ideal gas has a volume of 22.7 liters at STP. $\endgroup$ – MaxW Mar 25 '19 at 18:31
  • $\begingroup$ 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. $\endgroup$ – Alchimista Mar 26 '19 at 8:37

Let's take a look at the mean free path in a gas, which is defined by Wikipedia as

the average distance travelled by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions)

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.

This is true as long as we deal with a gas that is reasonably close to ideal. In reality, the size does matter indirectly. Larger and heavier molecules typically have more polarizable electron clouds, leading to stronger intermolecular forces. This is the reason why at low densities, gases with otherwise rather different boiling points (that are largely dependent on the strength of the intermolecular forces) still behave rather similar.

| improve this answer | | | | |
  • 3
    $\begingroup$ I down voted this answer. The OP is obviously an absolute novice. I think you need to first explain the basic notions of ideal gas behavior. $\endgroup$ – MaxW Mar 25 '19 at 20:37
  • $\begingroup$ I'd restructure that last paragraph since "This is the reason..." doesn't follow from the preceding sentence (it follows from the first sentence). Also, nonideality is probably unimportant except for charged particles, polarization being important only when you get to much higher densities compared to a gas at STP (maybe a dipole moment will have some effect). $\endgroup$ – Buck Thorn Mar 25 '19 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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