Climate Physics Chapter 2: Thermodynamics

How can we use thermodynamics to understand the structure of our atmosphere?

From weather balloon data, we know that pressure in the atmosphere decreases monotonically with height. We use pressure as a fundamental vertical coordinate because it is easily measurable, and it is a fundamental thermodynamic variable that tells us about the state of gases in the atmosphere. Because height is linearly related to the log of pressure, we also often use \(-\ln{\frac{p}{p_{o}}}\), where \(p_{o}\) is the largest relevant pressure, as a height-like coordinate that increases with height.
Temperature decreases with height until a critical height, called the tropopause, above which it increases with height. The atmosphere below the tropopause is called the troposphere, and the portion immediately above is the stratosphere. This is a typical pattern that we observe in many planetary atmospheres.

Balloon Sounding of the atmosphere over the tropical Pacific, taken during the CEPEX field campaign on March 12, 1993.

Dry Thermodynamics of an Ideal Gas

The Ideal Gas Equation

In thermodynamics, we are concerned with three main variables:
Temperature (\(T\)): Proportional to the average kinetic energy per molecule. We say two objects have the same temperature when they are in thermodynamic equilibrium, where heat flows from hotter areas to cooler areas no longer occur.
Pressure(\(p\)): Force per unit area exerted on a surface in a direction perpendicular to the surface.
Density(\(\rho\)): mass divided by volume.
The ideal gas law relates these three variables:

\begin{equation} p=k_{B}nT=R\rho T\\ \end{equation}

In the first equality, \(k_{B}\) is Boltzmann’s constant, with a value of \(1.3806\textrm{ kg}\left(\frac{\textrm{m}}{\textrm{s}}\right)^{2}\textrm{K}^{-1}\), and \(n\) is the number of molecules per unit volume. In the second equality, we instead write the equation in terms of mass density \(\rho\) and \(R\), the gas constant, defined as \(R=\frac{R^{*}}{M}\), where the universal gas constant \(R^{*}=8314.5\left(\frac{\textrm{m}}{\textrm{s}}\right)^{2}\textrm{K}^{-1}\) and the molecular weight \(M\) is an integer telling the number of protons and neutrons in a molecule.

Example Problem: The Ideal Gas Law

A bicycle tire with mass \(0.1\) kg when empty and a volume of \(3\) liters is pumped up with Earth air to a pressure of 4 bars. Its temperature remains at the ambient air temperature of \(290\) K. What is the mass of the tire after it has been pumped up?
We have \(R_{dryair}=287\frac{\textrm{m/s}^{2}}{\textrm{K}}\).

\begin{equation} m=\frac{pV}{RT}=\frac{4\cdot 10^{5}\textrm{Pa}\cdot 0.003\textrm{m}^{3}}{287\frac{\textrm{m/s}^{2}}{\textrm{K}}\cdot 290\textrm{K}}=0.015\textrm{kg}\\ \end{equation}