Modeling the Atmosphere, pt. 2

The previous Modeling the Atmosphere post derived an expression for describing the change in pressure dP with respect to change in altitude dz in terms of density \rho and acceleration due to gravity g:

\frac{dP}{dz} = -\rho g

This equation is short to write, but it is not particularly useful. Before applying it to real-life situations, it must be transformed into the barometric equation.

First, recall that density \rho represents mass per unit volume. In Modeling the Atmosphere, pt. 1, \rho was used to replace the expression \frac{M}{V}, where M represents the mass of a thin slice S of air with volume V. Now,

\frac{dP}{dz} = -\frac{Mg}{V}

Air is not actually composed of slices, so it would be more convenient to represent the mass M in terms of particles. By volume, Earth’s atmosphere is composed of about 78% N_2, 21% O_2, and 1% Ar. Approximating Earth air as an ideal gas and using some information from the periodic table, the mass of one mole (6.022 \times 10^{23}) of air particles can be calculated:

nitrogen oxygen argon

0.78(2 \cdot 14.02 \frac{\text{g}}{\text{mol}}) + 0.21(2 \cdot16.00 \frac{\text{g}}{\text{mol}}) + 0.01(39.95 \frac{\text{g}}{\text{mol}}) \approx 29.98 \frac{\text{g}}{\text{mol}} \approx 0.029 \frac{\text{kg}}{\text{mol}}

So on average, 6.022 \times 10^{23} air particles (a.k.a. one mole of air particles) masses approximately 0.029 \text{ kg}. Using this knowledge, the mass M of one slice S of air may be re-expressed as 0.029 \frac{\text{kg}}{\text{mol}} \cdot N \text{mol}, where N represents the number of moles of air particles in the slice S of atmosphere. The same process holds for any other collection of particles that can be modeled as an ideal gas. For a general gas, each mole might mass m \text{ kg} instead of 0.029 \text{ kg}. Generally:

\frac{dP}{dz} = \frac{-mNg}{V}

and for Earth’s atmosphere in particular,

\frac{dP}{dz} = \frac{-0.029 N g}{V}

Now that the mass M of the slice S of atmosphere has been re-expressed as mN, let’s re-express the volume V of the atmosphere slice using the ideal gas law:

PV = NkT

Solving for volume yields:

V = \frac{NkT}{P}


\frac{dP}{dz} = \frac{-mNg}{\frac{NkT}{P}}

and simplify:

\frac{dP}{dz} = \frac{-mg}{kT} P

At this point, we’ve almost transformed the starting equation into the barometric equation. To review, m represents the mass per mole of the ideal gas (for Earth’s atmosphere, m \approx 0.029 \frac{\text{kg}}{\text{mol}} ), g \approx 9.81 \frac{\text{m}}{\text{s}^2} is the acceleration due to gravity, k \approx 1.38 \times 10^{-23} \frac{\text{J}}{\text{K}} is the Boltzmann constant, and T is the temperature in Kelvin (assumed to be constant – regardless of height – in this idealized atmosphere).

The next step requires calculus. First, “multiply both sides” by the differential dz:

dP = \frac{-mg}{kT} P dz

and divide both sides by P:

\frac{1}{P} dP = \frac{-mg}{kT} dz

Using the classic technique for solving differential equations, integrate both sides:

\int \frac{1}{P} dP = \int \frac{-mg}{kT} dz

\ln{P} = \frac{-mg}{kT} z + c

Exponentiate both sides:

P(z) = e^{\frac{-mg}{kT}z+c} = e^c e^{\frac{-mg}{kT}z}

Using the boundary condition when z = 0, it’s clear that the constant e^c corresponds to the pressure at zero altitude: P_0 = e^c. Thus,

P(z) = P_0 e^{\frac{-mg}{kT}z}

Finally, we have derived the barometric equation. Using this equation, the pressure P of an idealized atmosphere can be determined as a function of altitude z. Interestingly, this derivation is not the only method for obtaining the barometric equation. Check back for an alternate derivation and an application to satellite testing with weather balloons.

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