Why does pressure due to van der Waals' forces increase as the square of density?

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In van der Waals' equation for real gases, the adaptation to account for intermolecular attractions in real gas is [itex]a\frac{n^2}{V^2}[/itex]. This implies that the pressure due to the VDW forces on the container is proportional to the square of the density, [itex]\rho^2 = \frac{n^2}{V^2}[/itex].

When I do calculations following from [itex]P_{VDW}=\frac{\pi \rho^2 \lambda}{H^3}[/itex], I end up with a cubic dependence.

Is there a qualitative way to imagine why the pressure is proportional to the square of the molecular density, [itex]\rho[/itex]?
 

TeethWhitener

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Is there a qualitative way to imagine why the pressure is proportional to the square of the molecular density, ##\rho##?
Precisely speaking, it's the deviation of pressure from ideality that's proportional to the square of the number density. Pressure is given as force per unit area (we can think of this as the force being applied to the wall of a container with unit area). The force comes from the collisions of individual gas molecules with the container wall. So clearly the total pressure will depend on the number density of those particles: ##P \propto \rho##. Van der Waals's insight was that the deviation from ideality also depends on the density, so that total pressure will look like: ##P \propto \rho(1-\rho)##. The dependence of the deviation on density comes from the fact that, at the (idealized, non-interacting) wall of the container, a particle feels an attractive short-range force from the other particles in the gas, but not from the wall, so that there's a net force away from the wall that depends on how close the particle at the wall is to other particles. In other words, that net force depends on the number density of the particles. Since that force is density-dependent, and the number of particles at the wall is density-dependent, the deviation from ideality is quadratic in density.
 

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