Relationship between pressure, temperature and density in meterology

Question

I have been learning meteorology a bit on my own, and one thing that never gets answered clearly is the relationship of pressure, temperature, and density.

When I looked it up, the density of air at various temperatures varies quite a bit - if I recall, from something like 1.1 to 1.5 kg/m³.

My confusion comes from trying to understand weather formation. A typical text says that, for instance, heating of the earth leads to thermals that rise due to their higher temperature and lower density, leading altitude increasing and to temperature dropping at the dry adiabatic lapse rate, then the moist adiabatic lapse rate, etc. But it is always done using "parcel" arguments.

From basic thermodynamics, heating could lead to a combination of temperature increase, pressure increase, and physical expansion (if there is no fixed container).

Specifically, the ideal gas law gives:

$$PV=nRT $$ or $$P=\frac{n}{V}T$$ (ignoring the constant of R) or $$P=\rho T$$where $\rho$ equals density.

Am I wrong to assume that the simplifying assumption is that, to a first order approximation, the density can't really change much about an area since a parcel of air could expand, but the whole atmosphere locally can't (since the 'parcels' all abut)? In other words, ground heating causes temperature to rise, and thus presumably pressure, but to a negligible degree decreases density?

This would seem to imply higher pressure from heating actually leads to a net kinematic effect of driving the column of air upwards (as molecular pressure against the ground increases), where it can expand against lower pressure air above due to the fact pressure falls quickly with altitude - but there really isn't expansion or a decrease in density near the surface itself.

The typical "parcel lifted by a helicopter" argument fails to explain how it really all holds together and actually happens, since there is no 'container' for the gas in reality.

Another way to see this is in this example. Suppose I have a hot tub in Antarctica in winter (I have one in Idaho in winter which is close enough). The ambient temperature may be -30C or about 240K, and when I open the cover there is water at about 30C or 300K. I can see thermals climb; the air, perhaps, is heated to 270K (being generous). But the new air that gets sucked in and heated over the surface to 270K and then rises would immediately seem to be at higher pressure than ambient air since $$P=\rho T$$ I could even believe the kinematically more energetic air could cause an expanding 'plume' radially a bit as it rises, since the higher P and T slightly pushes out on less energetic ambient air. But that just means $\Delta P$ is slighlty reduced to take into account the expanded volume of the plume; $\rho$ has dropped slightly, but nowhere near how fast it will drop if it gains altitude and has truly lower pressure about it.

In reality, it seems:

1. locally the (synoptic) pressure is close to fixed - and varies only a few percent globally at sea level;
2. the gradient of pressure change is in the $z$ direction;
3. local heating increases T and thus P, except to the degree that increased P allows it to form an expanding plume (which is a few percent, so second order);
4. Really, the more energetic (kinetic) molecules have no where to go but up, what is really causing any updraft.

Answer 1

'Locally' and 'synoptically' are conflicting adjectives. Synoptic means overall, or large horizontal scale- think thousands of kilometers.

You are right on some accounts. Synoptically, pressure doesn't change much, depending on your qualitative definition of 'much.' On a percent scale or day-to-day experience, absolute temperature fluctuates more than pressure. But the gap of those fluctuations result in changes of density, leading to changes in buoyancy.

The vertical momentum equation :$$\frac{\partial w}{\partial t}+(\vec{v}\cdot\nabla_H)w+w\frac{\partial w}{\partial z}=-\frac{1}{\rho}\frac{\partial P}{\partial z}- g + \nabla^2w$$ when applied to synoptic scales can be treated as approximately hydrostatic: $$\frac{\partial P}{\partial z}=-\rho g$$. From this equation, as well as the ideal gas law, the hypsometric equation may be derived.

Although the vertical gradient of pressure is large, gravity is also large, which explains why the atmosphere is not lost to space and vertical motions aren't usually large in horizontal scale.

Using said hypsometric equation, it can be shown that changes in vertically-averaged temperature lead to synoptic scale changes in surface pressure.