Perfect gas in a box with a piston

[Aleoa]

We have a perfect gas in a closed box. On the top of the box there is a piston.
We know that the molecules of the gas exert a pressure on the piston.
Now let us put twice as many molecules in this tank, so as to double the
density, and let them have the same speed, i.e., the same temperature.
At this point my book says :

"If we consider the true nature of the forces between the atoms, we would expect
a slight decrease in pressure because of the attraction between the atoms, and
a slight increase because of the finite volume they occupy."

I'm not able to understand what does the underlined sentence mean, can you help ?

 

[Charles Link]

They are giving a verbal description of the corrections from an ideal gas that are found in the Van der Waal's equation: $(P+\frac{a n^2}{V^2})(V-nb)=nRT$. The part of the sentence in bold letters is what the correction term $-nb$ represents. Because of this term, the pressure $P$ will be slightly larger. $\\$For the previous part of the sentence, the correction is the $+\frac{an^2}{V^2}$ term.

 

[Aleoa]

Intuitively speaking, how is connected the slight increase of pressure with the fact that the volume is finite ?

 

[Charles Link]

Intuitively speaking, how is connected the slight increase of pressure with the fact that the volume is finite ?

A given gas particle=say you put an atom in the container that has zero volume, will be moving around inside a volume that will be basically $V-nb$ because of the volume occupied by all of the other gas particles. Their finite volume essentially makes the effective volume of the container slightly smaller.

 

[Aleoa]

And this, in which way produces an increment of pressure ?

 

[Charles Link]

And this, in which way produces an increment of pressure ?

The lower volume in the equation, (i.e. $V-nb$ instead of $V$), means $P$ will need to be slightly larger in order for the product of $P+\frac{an^2}{V^2}$ and $V-nb$ to equal $nRT$.

 

[Aleoa]

If we first try to model a gas with the ideal gas law and the we use the Wan der Waal's correction, what changes it that the number of molecules are fewer in the real modeling of the problem, since every molecules has a definite volume in the real case, and cannot be considered as a point.
A part from Wan der Waal's mathematic formula, i don't understand why this cause a increment in pressure.

 

[Charles Link]

If we first try to model a gas with the ideal gas law and the we use the Wan der Waal's correction, what changes it that the number of molecules are fewer in the real modeling of the problem, since every molecules has a definite volume in the real case, and cannot be considered as a point.
A part from Wan der Waal's mathematic formula, i don't understand why this cause a increment in pressure.

In the modeling of the problem, the number of molecules is still $n N_A$, where $n$ is the number of moles and $N_A$ is Avogadro's number. If you put the same number of molecules at the same temperature into a (slightly) smaller volume, the pressure will increase.

 

[Aleoa]

I still don't understand why, physically, this happens. (A part from the math).

What i think is that, since the molecules has a certain volume, the have less space to move in the container, so they collide with each other (and with the walls) more frequently. Is this correct ?