Why is air more dense closer to the Earth's surface? - particle's view

[Richard R Richard]

Hello, many correct explanations have been written, but also some that are not so, I hope what I comment will serve you.
There are no molecules that can descend in free fall to the surface, so analyzing density as a function of speed, as expressed in the first message, is an error, a collision-free trip is statistically very unlikely due to the large number of molecules that found in the trajectory, the collision is inevitable, every atmospheric model starts from this assumption. (A simple cross section analysis will tell you that they can travel a short distance without interacting with another molecule)
We can imagine a simplified model of the atmosphere, as a series of "almost spherical" layers that surround the surface of the Earth "almost spherical (geoid)". Well in each layer, as already expressed previously, the own weight of the layer increases the pressure of the immediately lower layer. Air is a compressible fluid, the pressure increases at a "almost constant" temperature between layers of "almost constant" volume when the radial difference tends to zero, ends up increasing the density of the gas. (This is applied by applying the gas law ideals)
Therefore there is more pressure due to the weight of the upper layers, simply using Archimedes' hydrostatic principle, convective currents only occur if the lower density is less than the upper one, so the hotter lower layers are confined, given the ascent of a layer, tends to expand in the 3 dimensions not only the radial. As a result, the contact surface between layers increases and so does the thermal exchange, quickly neutralizing any attempt at abrupt ascent… (Volcanic eruptions, atomic fungi, only rise to a certain height). Friction (collisions) and gravity end up prevailing over the average kinetic energy of a group of particles. As the atmosphere is a large thermodynamic system, in which the composition of the layers varies depending on the molecular size, since in the lower layers, there are more greenhouse gases, $CO_2$, $H_2O$ something of $CH_4$, which are responsible for absorbing infrared shocks from the photons emitted by the sun and increasing their own kinetic energy, but by sharing the volume with $O_2$ and $N_2$ through collisions elastic, they transfer that kinetic energy to these molecules, increasing the temperature of the layers closest to the Earth's surface.
As the largest and heaviest molecules cannot break through easily, they do not reach high altitudes, but if $O_2$ and $N_2$ do, and not just up to 100km but well beyond, the ISS, must correct periodically its altitude due to atmospheric friction, and travels more than 400 km high.
These molecules/atoms, in the plasma state, are also part of the pressure supported by the lower layers. At these heights they are scarce, because only a very low percentage of molecules can by collisions in relatively close layers take speeds close to those of escape. This is seen in the Maxwell-Boltzman velocity distribution, the probability density of high velocities is really small.
I prefer not to comment on the technical aspects of the documents ...just a snapshot of the fundamental ideas extracted from the atmospheric models

 

[Dale]

many correct explanations have been written, but also some that are not so, I hope what I comment will serve you.
There are no molecules that can descend in free fall to the surface, so analyzing density as a function of speed, as expressed in the first message, is an error

I wouldn’t say that it is incorrect. It would apply for very thin atmospheres. So it wouldn’t be valid on Earth, but it could be valid on other astronomical bodies.

And in any case it is a foundational assumption of the kinetic theory of ideal gasses, so even if it did not characterize any actual atmosphere it is an interesting theoretical problem to derive the correct density distribution for an ideal gas.

That said, I do like your approach of analyzing it more classically as layers of a continuum.

 

[snorkack]

Now imagine gas which is so non-Maxwellian in velocity distribution that speed of particles increases downwards causing the density to fall. Such as gas poured out at a certain height.
The density will indeed decrease downwards... but note that pressure still increases downwards as the molecules speed up. The density decreases because temperature increases faster than pressure.

 

[sophiecentaur]

The density decreases because temperature increases faster than

Not on our Earth, it doesn't. Why does everyone ignore the supply of Energy? Look at the distant large moons to see the effect of (low) heat supply.

 

[snorkack]

Not on our Earth, it doesn't. Why does everyone ignore the supply of Energy? Look at the distant large moons to see the effect of (low) heat supply.

The situation of a gas poured out from height is a special experiment. And it is not stable anyway. Molecules may fall with equal speed at first, but they will undergo oblique collisions causing them to acquire different energies.

Situation where air gets denser upwards is commonly seen on Earth. Look on distant roads. They reflect light because air is less dense near ground.
But this situation is unstable, which is why only a thin layer of air above ground is less dense than the air above.

 

[sophiecentaur]

Situation where air gets denser upwards is commonly seen on Earth. Look on distant roads.

Yes - of course but that is all about 'weather', which is basically due to instability and energy transfer. It's the opposite extreme to the attempted 'simplest' model discussed in this thread. I really don't think there is a valid model that is representative enough of a real atmosphere in sunlight. If you take even a perfectly insulated vertical tube, then you could expect something like the temperature lapse rate over the first few thousand metres of altitude.

 

[Richard R Richard]

The small layer in thickness relative to the thickness of the atmosphere where the lower density is less than the upper one, is located in contact with the Earth's surface, generally the surface is at a higher temperature than the air, and by conduction it transfers thermal energy That is why we see effects such as the loss of horizon, mirage or, on the contrary, the "fata morgana" where the refractive index changes with the density and humidity composition of the air (explaining these effects is not the subject of the thread).
All these situations are due to a thermodynamic imbalance between atmospheric layers, the speed of energy transfer by convection is slower than that of conduction because air is a good thermal insulator, thermal energy expands the air, lowers its density and does work raising the upper layers of the atmosphere a bit, always maintaining a relatively stable pressure balance. If there are very pronounced density differences, the effects of convective currents occur, such as those used by birds, gliders and hang gliders, even more pronounced is the effect of dust devils and tornadoes as extreme.
But returning to the topic of the thread, the relationship of the thermal gradient with respect to the density and pressure is not a constant curve, in the absence of pronounced temperature gradients, the normal thing is that the density increases closer to the surface in relation to the increase of pressure justified by equilibrium between fluid push and gravity "the weight of the atmosphere", in an environment with a very low temperature gradient, 1 degree every 150 m of altitude, if I remember correctly, correct me if it were not , for the layers Lower parts of the atmosphere each layer of 150 has 1.8% less air than the layer immediately above it. This small gradient (or almost constant temperature) is where, by the ideal gas law, we see a direct relationship between density and pressure, so that closer to the surface there is more pressure and density than away from it.
$$P \cong P_{atm} (1-0.018) ^ {(\frac {h} {150m})}$$
It is a fairly accurate empirical formula for low altitudes.
Where 1.8% comes out of $$\dfrac {\Delta P} {P {atm}} = \dfrac {g \rho_{air} 150m} {101325Pa}$$

At 15000 m we have only 16% of the weight of the atmosphere above. For Mount Everest 8800 m I get 34% and from tables I see 32% quite accurate because I assume that the 1.8% drop between layer and layer is constant when it doesn't have to be ...
The books talk about what
$$\dfrac {dP} {dh} = \dfrac {-mgh} {kT}$$
then
$$P (h) = P_o\, e ^{\frac {-mgh} {kT}}$$

Where the linear relationship established between density and pressure is clear.