Why does temperature get lower as I climb a mountain? (Adiabatic cooling)

**(**click here to learn about the variation of pressure with altitude**)**

Generally, as we gain elevation, the air gets cooler – at least, relative to the nearby air at lower elevations, at the same time. You may have heard the rule "5 degrees F/1000 feet for dry air, 3 degrees F/1000 feet for moist air". But what is the origin of this rule? Why should air get cooler as elevation increases?

Let’s look at the scientific basis for this rule, and see why the rule-of-thumb is often inaccurate.

First, an implicit assumption is that air moves up and down elevation, as needed, to keep the atmosphere "gravitationally stable" relative to the – g r dh rule we learned about before. For example, if air at low elevation (in a valley) gets heated by the sun-baked ground, and thus becomes less dense, the atmosphere will overturn given the slightest chance (say a light wind). The rearrangement will generally put denser air near the ground, and the less-dense air will rise to higher elevations. But the pressure is also lower at higher elevations, so the air expands; and when it expands, it cools. (This "cooling with expansion" is the principle behind most refrigerators and air conditioners. You compress the gas outside your house and it warms up and gives off heat; then you bring the gas inside the house, expand it, and the gas cools down, along with the insides of your house.)

But wasn’t the air warmer to start with, and that’s why it rose? Ah, there are competitive forces at work, and now we rely on thermodynamics to do the bookkeeping for us. Here’s where the awkward term "adiabatic" comes in.

An adiabatic process is one that doesn’t involve transfer of heat from a system. "Heat" is just energy flow in response to a temperature difference. No process is completely adiabatic, but some processes are darn close. "System" is a somewhat slippery term, but here it will refer to a big parcel of air – meters, maybe kilometers across. Some parcels go up, some go down, to maintain the gravitational balance. (The actual size of the parcel (in mols) cancels out in the calculation.) A process can be nearly adiabatic when there is a really good thermal insulator between regions with different temperatures; or when the rate of heat transfer is really slow, compared to the rate at which the fluid moves. The latter condition often holds for the atmosphere. For example, a parcel of air that is just a 10 meters across requires a thermal diffusion time of ~5 million seconds, or about 58 days, to reach thermal equilibrium with its surroundings (** by conduction**). That parcel of air could have moved to the mountaintop and beyond in that time.

Another way a system can lose energy is by expanding – i.e. the volume becomes greater, against the surrounding pressure, so the system does "PV work". Conversely, the system gains energy if work is done ON it – e.g., if it is compressed against an outside pressure.

The First Law of Thermodynamics sums up these competing energy sources, as follows:

D U = q – w |
(1) |

where q is the heat added to the system, w is the work done BY the system, and D U is the change in the total internal energy of the system – in this case, the system is a parcel of gas. In differential calculus, the rule becomes:

dU = dq – pdV |
(2) |

where p is pressure, and V is the volume of the gas parcel. In an adiabatic process, dq=0 by definition, so

dU = -pdV. |
(3) |

If the gas is an ideal gas (a very good assumption, at low pressures), then it turns out – and this is not an obvious fact – that the internal energy is a function of temperature alone, so

nC dT = -pdV |
(4) |

in an adiabatic process, where T is the absolute temperature in Kelvin, n is the number of mols gas in our system, and C is the heat capacity (technically, C is Cv, the heat capacity at constant volume, but let’s avoid that potential can of worms, which would take another 2 pages of explanation!).

Now we add two other equations into the mix; the familiar

dp = -r g dh |
(5) |

where r is the density, g is the acceleration due to gravity, and h is the elevation above sea level; and

pV = nRT |
(6) |

which is an explicit statement of the Ideal Gas Law, where n is the number of mols in the system, and R is the universal gas constant. From this point on, the derivation is just algebraic slogging; I’ll give my path, which is by no means the best or fastest way to get the answer.

Remember that density is mass/volume, so r = n W/V where W is the average molecular weight of the gas (about 28.8 for air). Combining equations (5) and (6) with the definition of density, we get:

dp = -p W g dh /(RT) |
(7) |

Now we play some games with the Ideal Gas Law, differentials and the Chain Rule:

d(pV) = d(nRT)

pdV + Vdp = nR dT |
(8) |

(since n and R are constants in our system).

Combining the (8) above with (4), we get

n(C + R) dT = V dp |
(9) |

Substitute (7) into the right side of (9), and get

(C + R) dT = -(PV/(nRT)) W g dh |
(10) |

…but (PV/(nRT)) = 1 by the Ideal Gas Law, so we finally get:

dT/dh = -W g /( C + R) |
(11) |

Equation (11) is rather remarkable – it tells us that the rate of drop in temperature is constant with altitude, if the atmosphere rearranges itself to maintain adiabatic conditions. The C for a 20-80% mix of O2 and N2 is about 20.7 kJ/(K * kg-mol) (W.J. Moore, Physical Chemistry 4^{th} edition, table 4.4). (The C of the gas doesn’t vary much with temperature, in the range we consider.) Thus the value of the constant is:

-W g /( C + R) =

-(28.8 kg/kg-mol)*(9.8 m/s^{2})/(20700 + 8314) = -0.0097277 degrees (Kelvin or Celsius) per meter,

or roughly -1 degree Celsius per 100 meters of elevation gain, or -5.34 degrees Fahrenheit per 1000 feet. That’s the adiabatic rule for "dry air" that we all know and love. The rate for wet air calculates to be lower, because when wet air (100% humidity) rises and gets colder, some of the water turns from vapor to liquid and gives up latent heat, counterbalancing the effects of expansion somewhat.

Now, some flies in the ointment. The movement of dry air gets little chance to exchange heat by conduction, as long as the flow is laminar, because the thermal diffusion coefficient of gas is pretty low; so far, so good. But there are other ways to exchange heat, beside conduction; the sun heats up the rocks on the mountain, and they in turn can radiate heat. In addition, flow may be turbulent, so warm parcels of air (near the hot rocks of the mountain) swirl and mix with the cooler rising air. And of course, there are local weather patterns that can screw with the adiabat either way.

The adiabat is usually the ** maximum** gradient one can expect. In extreme cases, the cold air just sits in low places, undisturbed by wind, with warmer air above -- and

If we totally believed the adiabat, when it was 105 F down at Stovepipe Wells, it should have been about 105 – (5.34)*11 ~46 F at the top of Telescope Peak. Actually, it was more like 60 F on top.

**(**click here to learn about the variation of pressure with altitude**)**