Why does pressure behave exponentially in the atmosphere?

Question

I wonder why the pressure is exponentially within the atmosphere but (almost?) linear in the sea?

Answer 1

Let us consider the differences between the sea and the atmosphere. The atmosphere is a layer of gas, while the sea is a layer of (nearly)-incompressible fluid. We can use the properties of these to deduce the pressure variation with depth.

In both situations, we need to balance the forces from air (water) pressure and gravity.

Consider the air first. Because air is (approximately) an ideal gas, the density is given by $\mu p / RT$ where $\mu$ is the average molar mass of the air. The force from air pressure on a thin "slab" of air at height $z$ with thickness $\Delta z$ and area $A$ is $( p(z) - p(z + \Delta z) )A$, where up is considered positive. The force of gravity is equal to the density times the volume, or $ - \left( \frac{\mu p g}{RT} \right) (A \Delta z) $. The force balance reads

$$ (p(z + \Delta z) - p(z))A = - \left( \frac{\mu p g}{RT} \right) (A \Delta z) $$ $$ \frac{dp}{dz} = \frac{ p(z + \Delta z) - p(z) }{\Delta z} = -\frac{\mu p g}{RT} $$

in the limit as $\Delta z \to 0$. Because $dp / dz$ is proportional to $p$, we have an exponential dependence, $p \propto \exp(-\frac{ \mu g z }{ RT })$.

Now, consider the water. Because water is an incompressible fluid, the density is constant throughout, let it be $\rho$. Then, the force balance reads, at depth $z$,

$$ (p(z) - p(z + \Delta z))A = - (\rho g) (A \Delta z) . $$

Note the difference in signs: by convention, we let $z$ be the depth beneath the surface of the water, as opposed to the height above the bottom of the sea.

Thus, we can write

$$ \frac{dp}{dz} = \frac{ p(z + \Delta z) - p(z) }{\Delta z} = \rho g . $$

Since the derivative is constant, we can write $p = p_0 + \rho g z$, which is a linear gradient.

Summary. The difference in the types of pressure gradients arises due to the fact that density is (nearly) constant in the sea, whereas density is linearly proportional to pressure in the air.

Answer 2

Gasses are compressible; so as the weight from the layers above applies pressure on the gas; the gas will have a tendency to compress itself as well.

This means that as you get a thicker atmosphere; the weight doesn't just increase because there is more fluid above it; it also changes because the more fluid above it; the greater the density of the fluid, as it gets compressed more by the weight. This all relates to the ideal gas law.

Water is practically incompressible at moderate pressures, so it doesn't behave this way and approximately only has the linear effect of hydrostatic pressure.

Answer 3

Nitrogen and oxygen molecules are not bound to each other and move independently. Their height distribution is given by the Boltzmann factor:

$$P(E) \propto \exp\frac{mgh}{kT},$$ where $mgh$ is the gravitational potential energy of an air molecule and is $kT$ the thermal energy at the temperature $T\approx 260$ kelvin. This exponential distribution is in reasonable agreement with the barometric formula and the scaling height of the atmosphere.

Water molecules stick together. Gravitation acts on the condensed liquid as a whole. The density is high and almost constant because the compressibility of the liquid is low. With pressure as the weight of the material above, this gives a linear pressure increase.