3.1.1 Atmospheric Chemical Composition

The Earth’s atmosphere is the gas layer that surrounds the Earth. It is composed mostly of molecular nitrogen (N₂) and molecular oxygen (O₂) in proportions of 78 % and 21 %, respectively. Argon is present at <1 %. Carbon dioxide (CO₂) is present at about 400 ppm (parts per million), i.e., about 0.04 %. Water vapor is on average about 1 %, but it displays great spatio-temporal variability.

3.1.2 Atmospheric Pressure

The atmosphere is a gas that can be considered to be ideal at the pressures observed in the Earth’s atmosphere. Therefore, the ideal gas law applies:

where P is atmospheric pressure (atm), V is the air volume of interest (m³), T is temperature (K), n is the number of moles of air in the volume of interest, and R is the ideal gas law constant (R = 8.206 × 10⁻⁵ atm m³ mole⁻¹ K⁻¹; if P is expressed in pascal (Pa), then R = 8.314 J mol⁻¹ K⁻¹).

Atmospheric pressure results from the gravitational force of the planet Earth on the atmospheric gases. It decreases with height. The hydrostatic equation can be written as follows (where z = 0 at the Earth’s surface):

where z is altitude (m), P is pressure (here in Pa, i.e., kg m⁻¹ s⁻²), ρ_a is the density of the air (kg m⁻³), and g is the gravitational constant (9.81 m s⁻²). For an incompressible fluid: ρ_a(z) = constant; therefore, the relationship between pressure and altitude is linear. For example, pressure increases almost linearly with depth in the oceans (with small variations due to changes in temperature and salt-content and a slight compressibility at very high pressure). The atmosphere is a gas and, therefore, a compressible fluid, implying that its density increases with pressure. The density of the air may be calculated according to the ideal gas law as follows:

ρa(z)=MMairnV=MMair P(z)R T(z)(3.3)

where MM_air is the average molar mass of the air, i.e., a weighted average of the molar masses of the various constituents of the air (0.029 kg mole⁻¹ for dry air). Therefore, the dry air density is 1.184 kg m⁻³ at 1 atm and 25 °C; it is 1.225 kg m⁻³ at 1 atm and 15 °C. Replacing ρ_a(z) in the hydrostatic equation leads to:

dP(z)P(z)=−MMair g dzR  T(z)(3.4)

Thus:

P(z) =P(0) exp(−∫0zMMair g dzR T(z))(3.5)

This equation may be written as a function of a characteristic height, h_a(z):

ha(z)=R T(z)MMair g(3.6)

P(z)=P(0) exp(−∫0zdzha(z))(3.7)

This characteristic height varies as a function of altitude since temperature varies as a function of altitude. However, the variation of atmospheric temperature (expressed here in K) as a function of altitude is significantly less than that of pressure. If one assumes that temperature is constant, atmospheric pressure may then be written as a decreasing exponential function:

P(z)=P(0) exp(−zha)(3.8)

where h_a ≈ 8 km for a temperature of 273 K and R = 8.314 J mol⁻¹ K⁻¹. Thus, atmospheric pressure is about 0.88 atm at an altitude of 1 km and 0.29 atm at an altitude of 10 km. Figure 3.1 depicts this variation of atmospheric pressure with altitude.

Figure 3.1. Vertical profile of atmospheric pressure. Atmospheric pressure is shown as a function of altitude with a logarithmic scale (left figure) and a linear scale (right figure). The tropopause and stratopause altitudes are approximate and given only as qualitative information.

The assumption of a quasi-constant temperature is in part justified by the fact that absolute temperature does not vary as much as pressure if one assumes that atmospheric processes are adiabatic (see Section 3.1.3). For example, the relationship between temperature and pressure results in a change of only 3 % in temperature for a change of 10 % in pressure.

The concentration of air molecules may be calculated according to the ideal gas law. Therefore, for an atmosphere with a pressure of 1 atm and a temperature of 25 °C (i.e., 298 K), the molar concentration is as follows:

nV=PRT=18.206×10-5×298=40.9 moles m−3(3.9)

The air density (ρ_a = MM_air n / V) is proportional to its molar concentration; it increases proportionately with pressure and inversely proportionately with temperature (K). Therefore, warm air is less dense than cold air.

The unit of pressure “atmosphere” (atm) corresponds to the average atmospheric pressure observed at sea level at the latitude of Paris, France (≈ 49 °N). It varies at the Earth’s surface depending on altitude and temperature. For a given location, atmospheric pressure varies slightly, by a few % (see the discussion of high and low pressure systems as part of the general circulation description in Section 3.2.3). On average, it is 0.977 atm at the Earth’s surface because of altitude. Atmospheric pressure may be expressed with different units. For a reference pressure of 1 atm: P = 1 atm = 1.013 × 10⁵ Pa = 1,013 mbar (millibar) = 1,013 hPa (hectopascal) = 760 Torr (or mm of mercury).

The total mass of the Earth’s atmosphere, M_A, may be calculated according to the definition of pressure being the force exerted by the atmosphere (i.e., the mass of the atmosphere multiplied by the acceleration of gravity) per unit surface area. The Earth’s surface area is 5.1 × 10⁸ km², i.e., 5.1 × 10¹⁴ m². Therefore, the corresponding pressure expressed in pascal (i.e., N m⁻² or kg m⁻¹ s⁻²) is:

P=MA×9.815.1×1014 =0.977 atm=9.9×104 Pa(3.10)

Thus: M_A = 5.15 × 10¹⁸ kg.

3.1.3 Vertical Structure of the Atmosphere

The temperature vertical profile may be derived from the first principle of thermodynamics and from the ideal gas law, with the assumption that the atmosphere is adiabatic. The first principle of thermodynamics may be written as follows:

where the first term, dU, represents the change in the internal energy of the system (expressed here per mass, J kg⁻¹), dQ (J kg⁻¹) represents the amount of heat added to the system, and dW (J kg⁻¹) represents the amount of work done by the system (or on the system if dW <0). For an adiabatic process, dQ = 0. Furthermore, for a parcel of air moving vertically, the expression for an ideal gas leads to:

where c_v is the specific heat capacity of the air at constant volume (J kg⁻¹ K⁻¹):

where V’ is the specific volume (i.e., the volume of an air parcel of 1 kg), which is equivalent to the inverse of the air density (V’ = V / (n MM_air)), where MM_air is expressed here in kg mole⁻¹. Therefore:

It is more convenient to use only T and P as state variables and V’ may be replaced as a function of T and P. The ideal gas law may be written as a function of the specific volume:

PV‘=RMMair T(3.15)

By derivation:

dV‘=RMMair(dTP−T dPP2)(3.16)

Thus:

 cv dT= R T dPMMair P−RMMairdT(3.17)

To obtain the vertical profile of temperature, one changes dP to dz according to the hydrostatic relationship (see Equation 3.2):

 cv dT+RMMair dT=− R T ρa g dzMMair P(3.18)