Surely the temperature of the molecules is the same throughout the water. Using $p = \rho g h$ seems to assume a constant density as well. But then how is it that the force per unit area on an object placed at the bottom of the lake will be higher than that on an object near the surface? My first thought is that the incompressible assumption is the approximation at fault, but it doesn't seem that a minute increase in the density of molecules at the bottom of the lake could account for many times more bombardments per square centimeter. What am I missing?
Answer
First of all, the temperature and pressure of a liquid are two independent intensive variables. Either of them may be lower or higher at the bottom of a lake, independently of the other. So let's focus on the pressure.
You are totally right that the incompressibility fails and this is the reason why the lake "knows" about the higher pressure: the density of atoms or molecules becomes somewhat higher. But it's still true that the liquid is "approximately incompressible" and this is exactly the reason why the changes of the pressure are so great even if the changes of the density are minuscule. In other words, $$ \frac{\partial p}{\partial \rho}$$ is a very large number or, equivalently, $$ \frac{\partial \rho}{\partial p}$$ is a very small number. That's what we mean by (approximate) incompressibility and that's why the changes of the density unavoidably linked to finite (or even huge) changes of the pressure are so tiny.
Liquids are nearly incompressible because of Pauli's exclusion principle; the electrons in the atoms are just not allowed to occupy the same state. One has to change the structure of the states but this, because of the repulsion of the charged nuclei and electrons from each other, leads to immense increases of energy. That's why the density of liquids (or, equivalently, the volume occupied by a single atom or molecule) is de facto calculable independently of the pressure.
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