# Why does R. Feynman states that in a low-density gaseous enviornment, the pressure exerted by the molecules is proportional to density?

I was reading Dr. Feynman's Physics Lectures Vol 1 and in the first chapter about the atomic theory, he gives the example of a piston-type container where he explains the proportionality of force and area when density remains constant. But then he proposes to double the density of the gas in the container (let it be steam for example), and says

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. Then, to a close approximation, the number of collisions will be doubled, and since each will be just as “energetic” as before, the pressure is proportional to the density. 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. Nevertheless, to an excellent approximation, if the density is low enough that there are not many atoms, the pressure is proportional to the density.

My problem is the last part of the statement 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. Nevertheless, to an excellent approximation, if the density is low enough that there are not many atoms, the pressure is proportional to the density.

Why is it so? I have spent hours thinking about why is there decreases in pressure if the attraction is supposed to make atoms "more compact" and therefore denser, should we not appreciate an increase in pressure? Does the repulsion of molecules play an important role in exerting force against a surface?

What would happen if the density is not low, would it remain constant? In fact, how low is low enough?

• Increasing the attractive interaction doesn't make the gas more dense, because a gas will always occupy the entire volume of its container. Commented Mar 20, 2023 at 22:20

The first point is summarized by the ideal gas equation, $$PV=nRT,$$ where the pressure $$P$$ is proportional to the density $$n$$. The argument regarding the attraction between the molecules is one of the two modifications that lead to the van der Waals equation $$\left(P+\frac{n^2a}{V^2}\right)(V-nb)=nRT$$ where $$\frac{n^2a}{V^2}$$ models the reduction of pressure due to the attractive forces between the molecules, whereas $$nb$$ is the excluded volume occupied by the molecules (the short-range repulsion between the molecules.)
Next, we introduce a (not necessarily pairwise) attractive force between the particles. Van der Waals assumed that, notwithstanding the existence of this force, the density of the fluid is homogeneous; furthermore, he assumed that the range of the attractive force is so small that the great majority of the particles do not feel that the container is of finite size.[citation needed] Given the homogeneity of the fluid, the bulk of the particles do not experience a net force pulling them to the right or to the left. This is different for the particles in surface layers directly adjacent to the walls. They feel a net force from the bulk particles pulling them into the container, because this force is not compensated by particles on the side where the wall is (another assumption here is that there is no interaction between walls and particles, which is not true, as can be seen from the phenomenon of droplet formation; most types of liquid show adhesion). This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density. On considering one mole of gas, the number of particles will be N_A $$C=\frac{N_A}{V_m}$$ The number of particles in the surface layers is, again by assuming homogeneity, also proportional to the density. In total, the force on the walls is decreased by a factor proportional to the square of the density, and the pressure (force per unit surface) is decreased by $$a'C^2=a'\left(\frac{N_A}{V_m}\right)^2=\frac{an^2}{V_m^2}$$