The variation of pressure with altitude -- the "exponential atmosphere".

(Click here to learn more about the variation of temperature with altitude.)

When the LVMC group summited Telescope Peak, Mike pulled out his super Nike barometer, and reported a pressure of about 666 millibars (I would guess that number has special significance in Death Valley). A sea-level standard atmosphere is 1.01325 bars, so at 11,049’, we were experiencing 0.666/1.01325 = 0.657 atmospheres, thus less than 2/3 the oxygen density we would get at sea level.

I mentioned the "exponential atmosphere model" for estimating the pressure change with altitude, and people looked at me like I had a penguin on my head. So here is the scientific rationale for one bounding limit on the variation of pressure with altitude.

First, I’ll start with the "exponential atmosphere" model beloved by 2^{nd}-year engineering students who take thermodynamics. I’ll try to be consistent and use the MKS units system.

The first relation, from elementary mechanics, is:

dp = -r g dh |
(1) |

where p is pressure, r is the density of the fluid (= m/V, where m is mass and V is volume), h is the height above sea level in meters, and g is the acceleration due to gravity (9.8 m/s^{2}). Air is a fluid, like water, and equation (1) just tells us that the deeper we are in a fluid, the more pressure we feel – any deep-sea diver should agree. But unlike water, air is quite compressible, and its density changes a lot as the weight of the air above causes compression.

We also have the trusty ideal gas equation, which is pretty good at "low" pressures (like the surface of the earth):

pV = nRT |
(2) |

where T is the temperature (in Kelvin), R is the universal gas constant (~8314 J/(degree Kelvin kilogram-mol)), and n is the number of kilogram-mols. When (2) is recast in density, we obtain

(W p)/(RT) = r |
(3) |

where W is the average molecular weight of the gas. The W of air is ~0.8(28 for N_{2}) + 0.2(32 for O_{2}) = 28.8 kg/kg-mol (roughly). Now we substitute (3) into (1), and rearrange to obtain:

dp/p = -(W g/RT) dh |
(4) |

…let’s initially assume a constant temperature for the first-order calculation, picking 25C = 298.15 Kelvins, in between the temperatures we experienced that day. We’ll define A = W g/RT = (28.8*9.8)/(8314*298.15) = 1.1386 e-4 (i.e. 0.00011386).

Integrating (4), we obtain:

*ln*(p/p_{0}) = -A(h – h_{0}); and h_{0} is defined as 0, and p_{0 }as 1 atm (atmosphere), so

p/p |
(5) |

…hence the term "exponential atmosphere".

Thus at 11,049’ ~ 3368 m, equation (5) gives

p/p_{0} = *exp*(-.00011386*3368) = *exp*(-0.38348) = 0.6815 (fraction of an atmosphere) = 690.5 millibars*.

That’s a bit higher than Mike’s value, but we made a lot of approximations – for one, we assumed a constant temperature, when in reality, the temperature varies with elevation (hence pressure). It is relatively easy to substitute a temperature that is a linear function of elevation – we could use the known temperature-height variation on that day (it was about 105F at Stovepipe Wells, near sea level). Or we could use the familiar "adiabat" … but I have to get back to work now, so we’ll leave that for another day.

Note that the mountaintop pressure is also affected by weather systems – that’s one reason we calibrate aneroid altimeters. Finally, you may note that when pressure is reported for high-altitude places like Denver, Santa Fe, the North Rim, etc., the numbers tend to be suspiciously like those at LA, San Francisco, New York City – i.e. they are close to one atmosphere. That’s because standard weather reports normalize to the pressure expected at that altitude.

And -- I know this is hard to believe -- Mike's Nike barometer may not have been completely accurate. Most "less expensive" portable altimeters are not temperature-compensated.

*Note that the absolute value of x in exp(x) is <0.8 for any point in North America, so an approximation for the exp function usually suffices. In fact, the truncated Taylor series

exp(x) ~ 1 + x + x^{2}/2

is only 3% in error with an x corresponding to the height of Mount Whitney. Hence the "exponential atmosphere" is really close to a mild parabola, in any place where the assumed constant temperature is even mildly accurate.

Click here to learn more about the variation of temperature with altitude.

Epilogue: Non-Isothermal

Here it is, 4.5 years later, and I've finally decided to add the calculation for the non-isothermal pressure variation. From the link above, we know that the adiabatic atmosphere gives us a temperature variation like:

T = T |
(6) |

…where T_{0}^{ }is the temperature at h=0 (say ~Stovepipe Wells).

Let's go back to equation (4), and substitute in this relation for T:

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

and for ease of integration, we define u = T_{0} + b
h, so du = b
dh. Thus

dp/p = -(W g/[b R]) du/u . |
(8) |

Now from the discussion of the adiabat, we know b is actually -Wg/(C+R), so

dp/p = ((C+R)/R) du/u . |
(9) |

Remember that the C above is the heat capacity at constant volume. Now we have an interesting relation that you'll find in thermodynamics books; for a diatomic gas (like the N_{2 }and O_{2}^{ }that make up most of air), C ~ 2.5R at modest temperatures (like those in the atmosphere), so we get

dp/p = 3.5 du/u . |
(10) |

So by making the derivation "more complicated" and adding a temperature variation, we have simplified the equation! Now we integrate and do minor algebra to get:

ln(p/p |
(11) |

Where the "0" subscript signifies h=0, and T_{0} is the temperature at h=0. Now we have to back-substitute

p/p |
(12) |

From the discussion of the adiabat, we know b = -0.0097277 degrees (Kelvin or Celsius) per meter (actually, use of that b is very slightly inconsistent, but the difference is trivial here).

Let's take T_{0} as the temperature of Stovepipe Wells (very close to sea level), which reached 105F ~ 40.55C ~ 313.7K that afternoon, and again take 3368 m as the elevation of Telescope Peak. That gives:

p/p_{0}= (1 + b
h/T_{0})^{3.5} = (1 - 0.0097277*3368/313.7)^{ 3.5} = 0.6797 (fraction of the atmospheric pressure at h=0), which is very close to the previous 0.6815. However, now we are taking the reference T as 313.7K, not 298.15, so we must correct the to that temperature; so we predict:

p = 0.6797*298.15/313.7 = 0.6460 atmospheres, or 0.6546 bars,

which is within error of the barometer measurement of 0.666 bars on top of the peak. Considering the assumptions made in the calculations, the constant temperature and adiabatic calculation are very close.