Say there is a sealed cylinder of air that has a height $\mathrm{h}$ and area $\mathrm{A}$ on the ends. The initial temperature throughout the column is $T_0$ and has a uniform initial density $\rho_0$. If the bottom of the cylinder is at sea level, what is the temperature at the top of the cylinder when the system reaches equilibrium?

As we know, hot air rises and cold air sinks. So it stands to reason that the bottom will be cooler then the top in a very tall air column. However is there a formula?

  • $\begingroup$ Does the cylinder allow heat exchange between the air inside the cylinder and the atmosphere? If yes, what is the temperature of the atmosphere outside? $\endgroup$ Jul 20, 2016 at 10:47
  • $\begingroup$ No heat exchange. Assume cylinder has perfect insulation. $\endgroup$
    – cspirou
    Jul 21, 2016 at 11:32
  • 1
    $\begingroup$ You are looking for the adiabatic lapse rate. There is a good discussion at physics.stackexchange.com/q/433840 $\endgroup$ Oct 29, 2018 at 11:20

5 Answers 5


CAVEAT - I am giving a possible calculation, but I believe the answer may be off by a factor 2x (compared to the lapse rate observed in the atmosphere). I am leaving it here for you to ponder. Perhaps it can inspire you to find the correct solution yourself. Or perhaps the difference is due to the fact that this calculation doesn't assume convection - so that the adiabatic expansion terms in the derivation of lapse rate don't apply.

If the column of air is sealed we should probably assume there is not much air flow. In the steady state, we can use conservation of energy to solve this.

First - assume that the sum of (mean) kinetic and potential energy of the molecules is constant, independent of height. We know that at a given temperature the kinetic energy of a light diatomic gas (like most of the components of air) is

$$KE = \frac52 k_B T$$

Where the factor 5 comes from 5 degrees of freedom (3 translation, 2 rotation). Now the potential energy for a molecule of mass $m$ is $PE = m~g~h$. For air we will use an "average" mass of 28.8 amu (20% oxygen, 80% nitrogen; ignoring CO2, water, argon, ...). If the sum of $KE+PE$ is constant, then

$$m~g~h + \frac52 k_B T = \rm{C}$$

This means that there will be a linear change in temperature with height:

$$T(h) = T(0) - \frac{2mgh}{5 k_B}$$

Putting in numbers, we get

$$T(h) = T(0) - 0.014 h$$

which results in a temperature change of 1 K for every 70 meters. In reality, the slope in the atmosphere (according to this NASA page) is about 0.00649 K/m

That's about a factor 2x off from my calculation. I don't know what simplifying assumption I am making (or whether there is simply an arithmetic error in my work).

  • $\begingroup$ In equilibrium, should not temperatures be the same throughout? $\endgroup$
    – user137289
    Oct 29, 2018 at 9:05
  • $\begingroup$ @Pieter not necessarily. Equilibrium means there is no net energy needed to go from one state to the other. For air to rise, energy is needed; when air cools down, energy is made available. These two cancel exactly so you end with with an equilibrium where temperatures are not constant. $\endgroup$
    – Floris
    Oct 29, 2018 at 12:17
  • $\begingroup$ Of course there will be a transient if one could instantly switch on gravity in such a column, adiabatic compression at the bottom of the cylinder, adiabatic expansion at the top, leading to a temperature difference. But after equilibrium, I agree with the answer by @Kalliope that temperatures must be the same. Otherwise this would be a nice perpetual motion machine. $\endgroup$
    – user137289
    Oct 29, 2018 at 12:24
  • $\begingroup$ @Floris I also get 0.014 K/m for the calculation. FYI, from Britannica: "the adiabatic lapse rate of temperature, which equals about 1 °C per 100 metres (about 2 °F per 300 feet) for dry air and 0.5 °C per 100 metres (about 1 °F per 300 feet) for saturated air, in which condensation (with liberation of latent heat)" $\endgroup$
    – Roger Wood
    Jan 19, 2021 at 4:43

To know whether a column of fluid is in equilibrium, you take a fluid element anywhere in the column, displace it by a small amount (compared to column height) in any direction, and see what forces come to act on it; if the forces are such that they push (or pull) the fluid element back to its initial position, then the fluid column may be said to be in equilibrium. Under the action of gravity, we check for displacement along vertical direction. Forces that act on displaced fluid parcel are then due to fluid parcel's density being different from that of its surroundings. Therefore we need to know how fluid parcel's density changes when it is displaced. For that we need to assume some thermodynamic process that fluid element goes through when it is displaced.

For a short fluid column, we may safely assume that displacement process is isobaric. Further if the fluid is single phase, single component, then and only then does density of fluid become function of its temperature alone, and so you can say "hot air rises and cold air sinks". This is the situation we mostly encounter in day-to-day life.

However you may easily set up a stable "cold on top, warm at bottom" fluid configuration by using say, water on top of brine , or oil on top of water, with bottom fluid heated to an appropriate temperature. My point is, just because the system is stable does not imply that bottom fluid is colder than top fluid (and vice versa). You must always compare density.

If the fluid column is very tall, like in atmosphere, pressure decreases with height, and assumption of isobaric displacement process would not be proper. Usually reversible-adiabatic (viz. isentropic) displacement process is assumed, and this is justified by saying that air being poor conductor of heat, in the time the displacement of a fluid element takes place, it would lose negligible amount of heat. If an air parcel is displaced upward then due to decrease in pressure, its temperature and density decrease (recall, $p/\rho ^\gamma=$constant). The question is whether the density has decreased to such an extent that it is less than atmospheric density at that level; if it has then the column is unstable, otherwise stable. Same calculation ought to be repeated for downward displacement too (just to be sure), but the logic is the same.

For more details read up any book on atmospheric thermodynamics.


I would expect an isotherm systen since diffusion, heat conduction and radiation should eliminate any differences in the adiabatic system after some time.

The reason the atmosphere of earth has a temperature gradient is just that the earth is not adiabatic, but has a heat source at the ground (heated by the sun) as well as some absortion in the atmosphere (probably as a function of the pressure (or height)).

  • $\begingroup$ Indeed, the temperature gradient in the troposphere is mostly due to absorption/emission of infrared by CO2 etc, the greenhouse effect. $\endgroup$
    – user137289
    Oct 29, 2018 at 9:00

A more simple answer would be Boyles law:


since volume remains constant a first approximation would be $T2=P_2T_1/P_1$ and using $P_2=P_1-\text{density}\cdot\text{height}\cdot\text{gravity}$. A more complete answer would need the density to be corrected for temperature though out the column


An isolated system with Hamiltonian H has distribution function in thermal equilibrium, has an exponential distribution in phase space over the energy with parameter $k T$ as

$$Pr[x\in (x,x+dx) \wedge p \in (p,p+ dp)] = \frac{1}{Z}\ e^{- \frac{H(p,x)}{k T}} dp \wedge dx$$

where Z is the total integral, called the partition function.

The constancy of the temperature at constant volume results from the equlibirium by maximizing the entropy and thereby bringing any thermal transport currents to zero.

For a real gas, the Hamiltonian is the sum of kinetic energy, pair potentials and external potential energy. Neclecting the pair potential, that produces the real gas corrections to an ideal gas, yields

$$Pr[x\in (x+dx) \wedge p \in (p + dp)]= \frac{1}{Z} \ e^{- \frac{p^2}{2m k T}} \ dp \wedge e^{-\frac{V(x)}{k T}} \ dx$$

So the probability in space and momentum space factorize, that is, the two random variables p and x are statistically independent. The momentum p is always Maxwell normal distributed with variance $m k T$. In a constant gravitional field the position is distributed exponentially with parameter $m g h/(k T)$, called barometric hight formula.

The non-constant temperature of the atmosphere below 5 km in equilibrium is goverened by vapor thermal transport from 12C sea temperature to 0C ice temperatur at 5 km, in the higher atmosphere mainly by the different photon temperatures at the different frequencies and their interaction with O_2/N_2. Here temperature always characterizes momentum variances of the distribution in k-space, not the energy, that is their randomness.

The abrupt changes of temperature in large hights are the effect of parts of the external radiation, that have large cross sections with athmospheric molecules and therefore are compeletely absorped before reaching the vapor dominated sphere.


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