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As stated above, in the theory of ideal gases, do we care about chemical species or can all be treated as the same i.e. $\text{N}_2 = \text{O}_2$?

My initial thought is no, since for the same system conditions, $PV=nRT$ outputs the same value irregardless of chemical species.

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  • $\begingroup$ Within the approximation of ideal gas, you are right. $\endgroup$ – Deep Sep 27 '16 at 5:01
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No, we don't care about chemical species, because an ideal gas is composed by point particles which do not interact except when they collide.

Basically, we are completely disregarding the structure of the atoms/molecules and their interaction potential, setting it to $0$ (a pretty heavy approximation!), so of course we won't be able to distinguish between different chemical species. In fact, as you correctly remarked, the ideal gas equation doesn't care at all about chemical species.

The difference starts to manifest itself when we consider more realistic models, such as the van der Waals gas. In the vdW approximation the molecules are approximated as hard spheres, not as point particles, so they occupy a certain amount of volume, called the excluded volume. Furthermore, we introduce an average short-ranged attractive potential between the particles, which results in an overall decrease in pressure, because it tries to keep the particles which are near the walls of the container close to each other.

The resulting equation of state is

$$ \left(p + \frac {n^2 a} {V^2}\right) (V-nb) = nRT $$

where $a$ and $b$ are fitting parameters which are different for different species (check it out).

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Dalton's law of partial pressures has the total pressure as the sum of the partial pressures exerted by the individual gases in the mixture.

If you mix oxygen and nitrogen the number of oxygen and the number of nitrogen molecules does not change but what might change is the average kinetic energy of the molecules if before addition the temperatures of the gases were different.

On mixing, and hence allowing the molecules to collide with one another, the average kinetic energies of the two types of molecules become the same so that, after the equilibrium state has been reached, on average there is no net transfer of kinetic energy between the colliding molecules. The gases are at the same temperature and each gas behaves as though the other one was not there.

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  • $\begingroup$ I think you may have misinterpreted/misread the OP's question. He is asking whether an ideal gas of $N_2$ particles is the same as an ideal gas of $O_2$ particles or not, he is not asking about mixing. $\endgroup$ – valerio Sep 27 '16 at 10:19

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