The partial pressure is the pressure one gas in a mixture would exert if it were the only gas present in the volume under discussion. There's a simple formula for it:

P_{i} = x_{i} P |
[1] |

where P_{i} is the partial pressure (in pascals in the SI), x_{i} is the volume fraction or mole fraction of the gas in the mixture under discussion, and P is the total pressure exerted by all the gases.

Let's find the partial pressure of carbon dioxide in a sample of dry air at sea-level. Sea-level pressure averages 101,325 Pa. As of this year (2008), carbon dioxide makes up 0.000385 of that (385 parts per million) by volume. The partial pressure of carbon dioxide should then be P_{i} = 0.000385 x 101325 = 39.0 Pa.

Now let's verify that this is the pressure that would be exerted if the carbon dioxide alone occupied that volume. The ideal gas law is:

P V = n R T | [2] |

where P is pressure, V volume, n number of particles, R the universal gas constant, and T the temperature.

According to the US Standard Atmosphere, T at sea-level averages 288.15 K. Let's work with one cubic meter of air, V = 1 m. The latest NIST CODATA figure for the universal gas constant is 8,314.472 Joules per Kelvin per kilomole (J K^{-1} kmol^{-1}). We begin with dry air: P is 101,325 Pa. The number of molecules n must therefore be, by equation [2], n = 0.042292 kilomoles.

We know the mole fraction (or kilomole fraction, or volume fraction) of CO_{2} is 0.000385 this year. Plugging V = 1.0 m, n = 0.000385 x 0.042292 = 0.000016282 kmol, R = 8314.472 J K^{-1} kmol^{-1} and T = 288.15, the pressure should be 39.0 Pa. We're back where we started.

For the atmosphere as a whole, we can write:

P = (M / A) g | [3] |

where M is the mass of the atmosphere, A the surface area of the Earth, and g the Earth's gravity. Solving for M, and knowing that P = 101,325 Pa, A = 5.1007 x 10^{14} m^{2}, and g = 9.80665 m s^{-2}, the mass M of Earth's atmosphere comes out as 5.2767 x 10^{18} kilograms.

The figure is a bit high. Because land sticks out above sea level, some of the volume of air that should be there is rock and water instead (and people and buildings, etc., etc.). Recent estimates of the actual mass of Earth's atmosphere cluster around 5.136 x 10^{18} kg. But for purposes of illustration here I will stick to the high figure, 5.277 x 10^{18} kg.

Now here's the paradox. If we try using this formula to find the pressure exerted by one gas due to gravity, we get a different answer than the partial pressure equation gives.

Again I will use the example of carbon dioxide. CO_{2}'s volume fraction is 0.000385, as discussed earlier. But that is not its *mass* fraction, since a molecule of carbon dioxide is heavier than an average molecule of air. CO_{2}'s molecular weight of 44.0096 AMU, compared to a figure for normally wet air of around 28.92, means its mass fraction in Earth's atmosphere is m_{i} = 0.000586. From the known total mass of Earth's atmosphere, there must then be 3.00 x 10^{15} kg of carbon dioxide floating around up there.

But plug that mass figure into equation [3], and we get a pressure of 57.7 pascals, not 39.0. We must therefore conclude that partial pressure is not the same as gravitationally induced pressure.

*Note added 6/05/2008:* This page originally included the following text:

I would be grateful to anyone who could explain the discrepancy to me. Please email me here and include the word "Paradox" in the subject line. Thanks.

Please note that I have received adequate replies and the above is no longer necessary. Thanks to everyone who wrote in.

Thanks to Martin Vermeer and Simon Dalley, I think I now understand the paradox. The conclusion above -- "partial pressure is not the same as gravitationally induced pressure" -- was correct. CO_{2} in air does not exert the same pressure as it would if it were the sole atmospheric constituent!

Note that in the example above, I solved for the pressure 385 ppmv of CO_{2} would have in one cubic meter of air at sea level. If CO_{2} were the only gas present, 0.000016282 kilomoles of it would not occupy one cubic meter at sea level! Volume would be lower, and thus pressure higher.

The difference can be understood through the concept of scale height. Gases in an atmosphere under gravity lose pressure approximately exponentially with height. The increase in altitude needed for pressure to fall by a factor of *e* (about 2.718) is the "scale height:"

H = R T / (MW g) | [4] |

where H is the height in meters, R the universal gas constant (see above), MW the mean molecular weight of the gas in question, and g gravity.

For the U.S. Standard Atmosphere sea-level temperature of T = 288.15 K, and the mean molecular weight of wet air being 28.92 (for technical reasons I should really use the figure for dry air here, but bear with me), and g = 9.80665 m s^{-2} at sea level, H comes out as 8,448 meters. But with MW = 44.0096 AMU, the figure is h = 5,551 meters!

Carbon dioxide, at least in the troposphere and stratosphere which make up 99% of the mass of the atmosphere, is a well-mixed gas. Mixed with air, it has the air scale height. As sole constituent of an atmosphere, it would have the CO_{2} scale height -- less "volume" (if we can speak of the volume of an infinite space), and consequently, higher pressure. The air mixed with it "buoys up" the carbon dioxide, so to speak, and it exerts a different pressure when mixed with air.

Moral: Do not use equation [3] on one constituent of a mixed-gas atmosphere! It works only for the atmosphere as a whole. Partial pressure is not the same as gravitationally-induced pressure. And, when constructing semigray models of greenhouse warming in planetary atmospheres, it's partial pressure that appears to be related to optical thickness (usually as τ = k P^{1/2}, where k is a konstant). It's a useful concept.

Page created: | 05/17/2008 |

Last modified: | 02/13/2011 |

Author: | BPL |