Pressure at the base of an air column 6371 km high If you allow me some imagination, without laughing, then I can assume that I dug a very vertical hole, come on, let's say 1 m2 of section to fix the ideas, which crosses the Earth through its center, okay? (assuming the Earth is a very regular ball with a well-defined center ... and let's leave aside temperature, tectonic movements, rock density, and other technical stuff please).
This hole does not stay empty, of course, it fills with air and therefore I have a column of air, well defined, which crosses the Earth right through. From the surface to the center, this column has a height of 6371 km, the radius of the Earth.
Now let's get down to business: by what method could I calculate the "atmospheric" pressure generated by this column of air at the center of the Earth?  I wish I could say I have an idea and started some math, but unfortunately I haven't ... just wondering for now how I could start to tackle this question, even if it's wacky, in a truly scientific way: formulas, integration, differential equations, whatever, as long as it is serious. Every idea is welcomed.
 A: You'll probably need two things:

*

*The hydrostatic equilibrium equation: $$
\frac{dP}{dr} = - \rho(P,T) g(r),
$$ where $g(r)$ is the gravitational acceleration as a function of $r$, and $\rho$ is the density of the gas as a function of pressure and temperature.  This is derived through a simple force-balance equation;  see the above link.  The gravitational acceleration inside a hollow tube going through the Earth is one of those standard exercises that physics professors love to assign, so I won't give a full derivation of how it's obtained here;   but the result is that to a pretty good approximation, $g$ is linear in $r$.


*Some kind of relationship between $\rho$ and $P$.  All sorts of models can be used for this.  The simplest method (which is probably overly simplistic) is to assume a constant temperature for the gas inside the column and use the ideal gas law $PV = NkT$, with $T$ a constant.  (Note that $\rho = m N/V$, where $m$ is the molecular mass.)
More "realistically" (inasmuch as anything in this scenario is realistic) would be to find some kind of temperature profile for the interior of the Earth and assume that the gas in the tunnel at radius $r$ is in thermal equilibrium with the Earth's interior at that same radius.  Other models might be possible;  if you have any ideas, please let me know in the comments.
These two equations could then be combined to yield the differential equation
$$
\frac{dP}{dr} = - \rho(P, T(r)) g(r)
$$
which can in principle be integrated to find $P(r)$, and in particular $P(0)$.
A: Presumably you want a method that gives an approximate number, even if there are some approximations.
So here is a way.
The pressure due to a fluid, density $\rho$, height $h$, is $P=\rho g h$ where $g$ is the acceleration due to gravity, $9.81ms^{-2}$.
For this question we need to take into account the variation in $g$, $g(r) = \frac{gr}{R}$ where $R=6731$ km.  That's because the mass underneath the radius $r$ varies as $r$ cubed, but the Newtons law of gravitation is an inverse square law.
Then $$\int_0^R \rho g(r) dr = \int_0^R \rho \frac{gr}{R} dr = \frac{\rho g R^2}{2R} = \frac{\rho g R}{2}$$
If we take $\rho$ as 1.2, we get about 37.5Mpa.
This is likely to be an underestimate and an improved model would have $\rho$ increasing with depth as mentioned in Michael's answer.  You could also add atmospheric pressure of 101,000Pa, due to the height of air above the hole, but that doesn't make much difference.
A: The pressure at the base of your air column will be in the neighbourhood of 70 GPa. That's because most of the column will be filled with solid air, which has a density of ~1.1 g/mL. The mean density of the Earth is 5.51 g/mL, and the pressure at the centre of the Earth is around 350 GPa. I assume that under such high pressure that the solid air density would be >1.1 g/mL, so the actual pressure would be >70 GPa (and <350 GPa).

We can make a crude estimate of the air pressure in the upper part of the column by using the scale height of the atmosphere.
In this post I'll assume that the air column is magically isolated from the high temperature inside the Earth, and that the Earth is an ideal sphere of uniform density.
At high temperature and low to moderate pressure, air behaves like an ideal gas, so its density is proportional to the pressure. If we rewrite the ideal gas equation in terms of density $\rho$, and the specific gas constant $R_s$, we get
$$\rho = \frac{P}{R_sT}$$
where $P$ is the pressure, and $T$ is the absolute temperature.
Pressure is defined as force per unit area. The pressure at height $z$ in a column of air of cross-section area $A$ is determined by the weight of all the air above $z$. Consider the small section of the column at $z$ with height $\Delta z$ and density $\rho$.
The volume of that section is $A\Delta z$, so its weight is $Ag\rho\Delta z$, where $g$ is the acceleration due to gravity. So the increment in pressure, $\Delta P$, caused by that section is
$$\Delta P = \frac{Ag\rho\Delta z}A = g\rho\Delta z$$
The pressure increases as $z$ decreases, so we need a minus sign in there.
$$\frac{\Delta P}{\Delta z} = -g\rho$$
Taking limits,
$$\frac{dP}{dz} = -g\rho$$
Plugging in our previous expression for $\rho$,
$$\frac{dP}{dz} = -\frac{Pg}{R_sT}$$
Rearranging,
$$\frac{dP}{P} = -\frac{g}{R_sT}dz$$
Let $H = R_sT/g$, and temporarily assume that $g$ is constant, which is reasonable if the range of $z$ is small relative to the radius of the Earth. Then
$$\frac{dP}{P} = -\frac{1}{H}dz$$
Integrating,
$$\ln(P) = -z/H + C$$
Let $P_0$ be the pressure at $z=0$. Then $C = \ln(P_0)$, so
$$\ln(P) = \ln(P_0) -z/H$$
that is,
$$P = P_0\exp(-z/H)$$
$H$ is known as the scale height. If we go up by $H$, the pressure is divided by $e$ (~2.71828), and if we go down by $H$, the pressure is multiplied by $e$.
Using the standard value $R_s = 287.058\, J/(kg\cdot K)$ of dry air, the standard gravitational acceleration $g=9.80665\, m/s^2$, and $T = 290\, K = 16.85°C = 62.33°F$, we get
$$H = 8.4888\, km$$
So if we descend in our air column to a depth of 84.888 km, the pressure is $P=P_0\exp(10)\approx22026.5P_0$. Setting $P_0=101325\,Pa$, which is the standard atmospheric pressure, we get $P\approx 2.23\,GPa$. And if we went down a further 84.888 km, the pressure would again be multiplied by 22026.5 to 49,159 GPa! Air is not an ideal gas at those pressures, so those numbers aren't reliable.
Let's what happens at the more modest depth of 60 km.
Inside a sphere of uniform density, the gravitational acceleration is proportional to the distance from the centre, thanks to the Shell theorem, so the true scale height is slightly higher than what we calculated above. The radius of the Earth is ~6371 km, so at a depth of 60 km $g$ is reduced by a factor of $6311/6371\approx 0.99$, which doesn't make much difference.
Using that reduced $g$, we get $H=8.5695\,km$, so at 60 km
$$P = 111.29\, MPa$$
and
$$\rho = 1.337\, g/mL$$
which is greater than the density of water (1 g/mL), and the density of solid air. So even at this depth, the numbers aren't very reliable.

There's a parameter called the compressibility factor which measures the deviation of the volume of a real gas relative to what the ideal gas law predicts. But it's not easy to find measured or predicted values for it at extreme pressures. Wikipedia has a table and a couple of graphs for air, but they only go up to 500 Bar (50 MPa). NIST has data up to 80 MPa for oxygen, and around 1000 MPa for nitrogen. Here are their graphs for $T = 290 K$.
oxygen

nitrogen


Eric W. Lemmon, Mark O. McLinden and Daniel G. Friend, "Thermophysical Properties of Fluid Systems" in NIST Chemistry WebBook, NIST Standard Reference Database Number 69, Eds. P.J. Linstrom and W.G. Mallard, National Institute of Standards and Technology, Gaithersburg MD, 20899, https://doi.org/10.18434/T4D303, (retrieved June 15, 2021).

FWIW, owlapps has some interesting info on solid nitrogen, mostly taken from Wikipedia.
