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?
Fluid Statics – How Pressure Varies from Top to Bottom of a Lake
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Related Solutions
The pressure on each side of the pipe is given by $P = \rho g h$ where $\rho$ is the density, $g$ is the acceleration due to gravity and $h$ is the depth of the pipe.
From the Euler equations of motion in 1D and steady state, we have:
$$ \frac{1}{2}\frac{d u^2}{d x} = - \frac{1}{\rho} \frac{dP}{dx}$$
If we make some more assumptions, namely that the pipe is full of seawater (which makes sense in the steady state because the water will flow from the sea to the lake) and that the pressure at the fresh water end is constant (so we ignore dilution/mixing, the sea water just instantly drops under the fresh), and we integrate from $x = 0$ at the sea end to $x = L$ at the fresh end:
$$ u^2 |_0^L = - 2\frac{1}{\rho} P|_0^L$$
and taking the velocity to be zero at $x = 0$ gives:
$$ u(L)^2 = -2\frac{1}{\rho_s}(P(L)-P(0))$$ $$ u(L)^2 = -2\frac{1}{\rho_s}(\rho_f g h - \rho_s g h)$$
Taking $\rho_s = 1020 \text{kg}/\text{m}^3$ and $\rho_f = 1000 \text{kg}/\text{m}^3$ with $g = 9.8 \text{m}/\text{s}^2$ yields:
$$ u(L)^2 = 0.0961h $$
or
$$ u(L) \approx 0.31h^{1/2}$$
Obviously this makes some pretty big assumptions. No viscosity, which is probably not that bad of an assumption unless your pipe is really deep, and the pressure on the fresh water end is constant implying the salt water just "disappears" by dropping very quickly out of the pipe under the fresh water.
Why does the length of the pipe matter
It doesn't actually. You'll notice $L$ doesn't appear anywhere in the expression. The velocity at the end of a mile long pipe or a 1 inch long pipe is the same and given by that expression.
What is the significance of $h^{1/2}$
Again, there really isn't any significance. The units of pressure/density are $\text{m}^2/\text{s}^2$ which is what RHS of $ u(L)^2 = 0.0961h $ is. So the units on the 0.0961 are $\text{m}/\text{s}^2$ and $h$ is $\text{m}$. So when you take the square root of both sides to get into $\text{m}/\text{s}$, you end up with the $h^{1/2}$.
So the significance is really just that there is a non-linear relationship between velocity and the depth of the pipe. If you put your pipe four times deeper, you'll only get twice the velocity.
How do slight changes in these properties result in a large change in pressure, microscopically?
Slight change of volume is not so easy to accomplish for solids - it takes a great force to achieve it. Considerable external force applied by different body (wall) needs to be maintained. The pressure is a measure of this force per unit area and since the force is great, also the pressure is great.
One way to explain this is the following. In solids, the atoms are so close to each other that the inter-atomic repulsion forces become close to comparable to intra-atomic forces between charged particles that constitute them.
The electric force between particles gets stronger at small distances, since the Coulomb law says it changes with distance of particles $r$ as $1/r^2$. To change the volume of a solid appreciably, the force per area associated with one atom would have to be comparable to the Coulomb electric force of the nucleus on the electron. Its value for atoms is so immense that it is unattainable by common standards. It would be also hard to find a container that would keep its shape and resist expansion under such great forces. Hence changes in volume by forces of common magnitude are commonly negligible.
Best 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.