Thermodynamic Derivation of the Heat Insulation Coefficient
Formula
Given that $m$ is the mass of a gas molecule, $h$ is the height, and $T$ is the temperature, the following equation holds:
$$ \dfrac{dT}{dh} = - {{ \gamma -1} \over { \gamma }} \dfrac{ mg }{k_{B}} $$
Here, $\gamma = \dfrac{C_{p}}{C_{V}}$ is the ratio of the isobaric heat capacity to the isochoric heat capacity.
Explanation
As is well known, the higher the altitude, the lower the temperature, and this ratio is represented mathematically. Of course, this is derived using thermodynamics alone without considering various other factors such as humidity. At this time, it is assumed that the gas molecules only have a difference in altitude and do not exchange heat with the outside in an adiabatic process. It is helpful to imagine that, in the atmosphere, warm air goes up and cold air goes down, rather than mixing when winds meet.
Derivation
Part 1. $\dfrac{T}{p} dp = - \dfrac{mg}{k_{B} T} dh$
Considering the atmosphere has a thickness of $dh$ and a density of $\rho$, the pressure applied is $p$, the following holds:
$$ dp = - \rho g dh $$
$$ pV = N k_{B} T $$
The density is $\rho = Nm$ when the mass is $m$ and there are $N$ molecules, and from the ideal gas equation, $N = \dfrac{p}{k_{B} T}$, the following holds:
$$ dp = - {{p} \over {k_{B} T}} m g dh $$
Further simplification yields the following:
$$ \dfrac{T}{p} dp = - {{mg} \over {k_{B}}} dh $$
Part 2. $\dfrac{T}{p} dp = \dfrac{ \gamma }{ \gamma -1} dT$
Adiabatic expansion of an ideal gas
$p V^{\gamma}$ is a constant.
$p V^{\gamma}$ is a constant, and from the ideal gas equation, since $V^{\gamma} \propto ( p^{-1} T )^{\gamma}$, the following equation is constant:
$$ p V^{\gamma} = p ( p^{-1} T )^{\gamma} = p^{1- \gamma} T^{\gamma} = C $$
Taking the logarithm of both sides of the above equation yields:
$$ (1- \gamma) \ln p + \gamma \ln T = \ln C $$
Taking the total differential gives:
$$ (1 - \gamma ) {{1} \over {p}} dp + \gamma {{1} \over {T}} dT = 0 $$
Simplification yields the following:
$$ \dfrac{T}{p} dp = \dfrac{ \gamma }{ \gamma -1} dT $$
Part 3.
Combining the results of Part 1. and Part 2. yields the following:
$$ -\dfrac{mg}{k_{B} T} dh = \dfrac{ \gamma }{ \gamma -1} dT $$
Simplification yields the following:
$$ \dfrac{dT}{dh} = - \dfrac{ \gamma -1}{ \gamma } \dfrac{ mg }{k_{B}} $$
■