Would the relationship of pressure and volume of gas change if the gas type is changed?

Question

Question. There is an experiment investigating how a gas's occupying volume changes when its exerting pressure alters. As a result, a graph of pressure versus volume has been plotted, which is half of a reciprocal graph (like the bottom half of the capital letter C). Then the question comes: how would the graph change if the gas type changes from air molecules (say nitrogen gas) to hydrogen gas?

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My Attempt. My thinking is that nitrogen has a larger atomic mass than hydrogen, so the latter would occupy less space. Consequently, I argue that the new graph would be relatively shifted to the left, i.e. a same pressure would now corresponds to a less volume.

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Comment. The given answer in book is "no change" which confused me. Thanks for any help in advance!

Answer 1

The general answer is yes, the nature of the gas matter. The practical answer is; it depends. In your conditions, the effect might well be negligible.  

As joseph established, if you are under the conditions where you can call your gas an ideal gas, the nature of it matters not. The P-V relation is simply given by, well, you guessed it, the ideal gas law.  

However, I would argue that assuming the gas ideal is already assuming the answer. The model of an ideal gas corresponds to a gas of non-interacting particles. Much of the difference between two gases is encoded in this very thing that we neglect for an ideal gas. A PbI2 gas will have pretty different Van Der Walls interactions form an O2 gas. Therefore, neglecting this would definitely lead to the conclusion "gases behave the same because we pretty much assumed it in the first place"  

The question then arises: what happens in a more sophisticated model? Well, luckily, taking into account the interactions and deriving a more precise form for the P-V relation is actually rather easy with what we call a virial expansion. Then, if we fix the temperature constant in our problem (isothermal PV curve), the relation takes the general form: $$P=\frac{A}{\bar{V}} + \frac{B}{\bar{V}^2} + \frac{C}{\bar{V}^3} + …$$ And that goes on for as long as you want (tho in practice you don’t need more than  2-3 terms). $\bar{V}$ is the molar volume that joseph introduced to you, and the coefficents $A,B,C$ … depend on the nature of your gas, because they are derived from the inter-molecular interactions (that we neglected in the ideal gas law). So in general, the nature of the gas does impact the P-V curve.

Answer 2

Provided that the number of ideal$^1$ gas particles is the same, the pressure-volume relationship is the same for different ideal gases.

Avogadro's law states that $$\frac{V}{n}=k$$ where $n$ is the number of mole and $k$ is a constant, meaning $$\frac{V_1}{n_1}=\frac{V_2}{n_2}$$

This tells us that even if the number of moles of a gas increases/decreases, then the volume of the gas will also increase/decrease proportionally.

Thus, Avogadro's law, under the conditions of the same temperature and pressure, states that equal volumes of different gases will contain an equal number of particles. That is, it is independent of the molar mass of the gas. Therefore, one would expect that under these conditions, the pressure-volume relation for different gases would be identical, provided there are identical numbers of particles.

In fact, independent of the type of gas, we can calculate the volume of one mole of a gas. This is called the molar volume, and at standard temperature and pressure, the molar volume is $$V_M=\frac{V}{n} = \frac{RT}{P} \approx 22.4 \ \text{litre} \ \text{mole}^{-1}$$ and $R$ is the gas constant.

$^1$ This answer comes with the understanding that we are dealing with ideal gases, so we ignore inter-molecular interactions (Van der Waals interactions). The question is based on why changing the type of gas results in identical, or "no change" as OP states, PV-dependence. This automatically implies we are talking about ideal gases, and therefore answers would be based on ideal gas behavior.