Help understanding the pressure's effect on equilibrium and Le Chatelier's principle So let's say we have a closed system in the following reaction:

$\ce{N2O4 -&gt; 2 NO2}$

The only way to increase the system's pressure without adding any mole of gas and keeping temperature constant is reducing the system's volume. Once we increase the pressure by reducing the total volume, formation of $\ce{N2O4}$ is observed.

In many text books, the explanation for this is that 2 moles of $\ce{NO2}$ occupy a bigger volume than 1 mole of N2O4, thus the equilibrium is restablished when more $\ce{N2O4}$ is formed. What doesn't make sense for me is that both the substances are contained in the same volume, so how 1 mole of $\ce{N2O4}$ would occupy less volume than 2 moles of $\ce{NO2}$? It only makes sense to think about partial pressures, not partial volumes, right?

I'm trying to use entropy and energy microstates to solve this problem, because it seems more fundamental than this mantra in the general chemistry textbooks. From quantum mechanics, I know (can't explain, just know; need further studying on this subject) that translational energy states become closer when a gas expands in a bigger volume, so I'm assuming that in the reaction above, with a bigger pressure (and thus a smaller volume) these energy states will be more spaced between each other. With energy states more far apart, why are there more possible energy microstates when $\ce{N2O4}$ is formed?

I can't get past this! Why does this implies in the formation of the substances with less gaseous coeficients? How are the energy states of reactants and products and their changes on pressure related with that?

Edit: to make myself clearer, I'll use this example from this textbook I'm reading: 



Chem1 Virtual Textbook, Stephen Lower, link

In the endothermic reaction H–H (g) ⟶ 2 H (g), you only get hydrogen atoms at very high temperatures, because you'll have a higher distribution of populated states on the 2 H side, thus you favor the products at higher temperatures for endothermic reactions. I can completely understand the effects of themperature in Le Chatelier's principle, and I'm trying to understand an analogous thought for changes in pressure in the microscopic point of view.
What the textbook means when it is saying one occupies more volume than the other is that the reactants (in this case) will occupy less volume than the product under standard conditions (~ atmospheric pressure). Therefore you have to use more force (pressure) to confine the products to the given volume than you need to do so for the reactants.
Therefore there is less enforcement on the reactants than there is on the products. Therefore it is energetically more favorable to be in the reactants-form.  

In order to understand this a little better you can have a look at the ideal gas law:
$$
p = \frac{nRT}{V}
$$
As you can clearly see the pressure is linearly dependent on the amount of particles in the system. Given that during the reaction (in the way you stated it in your question) this amount increases (doubles if the reaction did run through completely). Thus the pressure increases as well. However a higher pressure is connected to a higher energy state of the particles in the "box" as they repulse one another.
Why don't the reactants stay as they are then as this is the state that produce the least pressure? Because by forming the products, energy is released ($\Delta_rG &lt; 0$). Therefore the reaction can take place until the increase in pressure results in a "raise of energy" that is equal to the energy released during the reaction. Any further reaction would be energetically unfavorable.
And this point is the one that can be influenced by the volume. Let's say that the reaction can take place until the pressure reaches $1.5p_0$ and $p_0 = \frac{n_0RT}{V_0}$ (and for simplicity let's assume a isothermal reaction). This means that 50% of the the reactants can react until that limit is reached. If we increase the volume, this point is reached only after having a higher particle count than with a lower volume which is why the equilibrium shifts to the side with more gaseous particles if the volume is increased and vice versa.