METR125: Physical Meteorology: Lecture: Atmospheric Thermodynamics (1) Prof. Menglin S. Jin San Jose State University, Meteorology Acknowledgements: modified from Prof Peter Lynch’s online notes

Atmospheric Thermodynamics Thermodynamics plays an important role in our quantitative understanding of atmospheric phenomena, ranging from the smallest cloud microphysical processes to the general circulation of the atmosphere. The purpose of this section of the course is to introduce some fundamental ideas and relationships in thermodynamics and to apply them to a number of simple, but important, atmospheric situations. The course is based closely on the text of Wallace & Hobbs and G&Y

Outline WH 3.1 1 The Gas Laws WH3.2, not review in this class? 2 The Hydrostatic Equation WH3.2, not review in this class? 3 The First Law of Thermodynamics WH3.3 4 Adiabatic Processes WH3.4 5 Water Vapor in Air WH3.5 6 Static Stability WH3.6 7 The Second Law of Thermodynamics WH3.7

The Kinetic Theory of Gases The atmosphere is a gaseous envelope surrounding the Earth. The basic source of its motion is incoming solar radiation, which drives the general circulation. To begin to understand atmospheric dynamics, we must first understand the way in which a gas behaves, especially when heat is added or removed. Thus, we begin by studying thermodynamics and its application in simple atmospheric contexts.

The Kinetic Theory of Gases Fundamentally, a gas is an agglomeration of molecules. We might consider the dynamics of each molecule, and the interactions between the molecules, and deduce the properties of the gas from direct dynamical analysis. However, considering the enormous number of molecules in, say, a kilogram of gas, and the complexity of the inter-molecular interactions, such an analysis is utterly impractical.

The Kinetic Theory of Gases We resort therefore to a statistical approach, and consider the average behavior of the gas. This is the approach called the kinetic theory of gases. The laws governing the bulk behavior are at the heart of thermodynamics. We will not consider the kinetic theory explicitly, but will take the thermodynamic principles as our starting point.

The Gas Laws The pressure, volume, and temperature of any material are related by an equation of state, the ideal gas equation. For most purposes we may assume that atmospheric gases obey the ideal gas equation exactly.

The Gas Laws The pressure, volume, and temperature of any material are related by an equation of state, the ideal gas equation. For most purposes we may assume that atmospheric gases obey the ideal gas equation exactly. The ideal gas equation may be written pV = mRT Where the variables have the following meanings: p = pressure (Pa) V = volume (m3) m = mass (kg) T = temperature (K) R = gas constant (JK−1 kg−1)

The value of R depends on the particular gas. Again, the gas law is: pV = mRT The value of R depends on the particular gas. For dry air, its value is R = 287 JK−1 kg−1. Exercise: Check the dimensions of R.

The value of R depends on the particular gas. Again, the gas law is: pV = mRT The value of R depends on the particular gas. For dry air, its value is R = 287 JK−1 kg−1. Class Exercise: Check the dimensions of R. Since the density is ρ= m/V , we may write p = ρRT . Defining the specific volume, the volume of a unit mass of gas, as α = 1/ρ, we can write pα = RT .

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Class Practice At an altitude of 5600 m above sea level, where the standard sir pressure is 500 millibars and the standard air density is 0.69 kg/m3, calculate the standard air temperature

Class Practice At an altitude of 5600 m above sea level, where the standard sir pressure is 500 millibars and the standard air density is 0.69 kg/m3, calculate the standard air temperature P=ρRT 50000b = 0.69kg/m3 x 287 JK-1Kg-1 x T T = 50000bar/(0.69kg/m3 x 287 JK-1Kg-1 ) = 252.48 K

Class Participation We know that averaged global surafce temperature is 15°C. If the average air density at sea level is 1.226 kg/m3, what would be the average sea level pressure? P= ρRT = 1.226 kg/m3 x 287J-1K-1Kg-1 x (15+273.15) K = 1013 mb