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 part 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 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 it gains or loses energy. Thus, we begin by studying thermodynamics and its application in simple atmospheric contexts.

The Kinetic Theory of Gases

• Fundamentally, a gas is an collection 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.

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.

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

p = pressure [hPa]

V = volume [m3]

m = mass [kg]

T = temperature [K]

R = gas constant [J K−1 kg−1]

The value of R depends on the particular gas. For dry air, it is 287 J K−1 kg−1.

Gas Laws

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

Gas Laws

Ideal gas Law:

𝑝𝑉 = 𝑚𝑅𝑇 or 𝑉 =𝑚𝑅𝑇

𝑝

For a fixed volume of gas at constant temperature, mRT is constant so volume is inversely proportional to pressure:

𝑉 ∝ 1 𝑝 𝐁𝐨𝐲𝐥𝐞

′𝐬 𝐋𝐚𝐰

For a fixed mass of gas at constant pressure, 𝑚𝑅 𝑝 is constant so volume is inversely proportional to temperature:

𝑉 ∝ 𝑇 𝐂𝐡𝐚𝐫𝐥𝐞′𝐬 𝐋𝐚𝐰

Gas Laws

Avogadro’s law: equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

For a given mass of an ideal gas, the volume and amount (number of molecules, or moles) of the gas are directly proportional if the temperature and pressure are constant.

𝑉

𝑛=𝑅𝑇

𝑝

where n is the number of moles of an ideal gas.

In other words, gases with the same number of molecules occupy the same volume at a given p and T.

Something wrong here?

Gas Laws

Avogadro’s Number: the number of molecules in a mole of any gas is a universal constant called Avogadro’s Number, NA, and is equal to 6.022× 1023

For a gas of molecular weight M, with mass m, the number n

of kilomoles is 𝑛 =𝑚

𝑀 so

𝑝𝑉 = 𝑛𝑀𝑅𝑇

Since equal volumes of different gases have the same number of molecules, MR must be the same for any gas.

We call MR the universal gas constant, R*: 𝑅∗ = 𝑀𝑅 = 8.3145J K−1 mol−1

and 𝑝𝑉 = 𝑛𝑅∗𝑇

Gas Laws

• For a single molecule, another universal constant is Boltzmann’s constant, k:

𝑘 =𝑅∗

NA

For n0 molecules per unit volume: 𝑝 = 𝑛0𝑘𝑇

Virtual Temperature

The mean molecular weight, Md, of dry air is about 29. The mean molecular weight, Mv, of water vapor is about 18.

Thus, the mean molecular weight, Mm, of moist air is between the two: Mv < Mm < Md . The gas constant R for water vapor must be larger than that for dry air.

Rd =𝑅∗

Md

= 287 J K−1 kg−1

and

Rv =𝑅∗

Mv

= 461 J K−1 kg−1

The ratio is defined:

ε =RdRv= 0.622

Virtual Temperature

For moist air the mean molecular weight and therefore, the gas constant depend on the amount of water vapor.

It is inconvenient to use a gas constant that varies in this way; it is simpler to use Rd and a modified, or virtual temperature, Tv, in the ideal gas equation.

Mixing Ratio

Consider a fixed volume, V, of moist air at temperature T and pressure p which contains a mass md of dry air mv of water vapor. The total mass is 𝑚 = 𝑚𝑑 +𝑚𝑣and the mass mixing ratio, w, is:

𝑤 =𝑚𝑣𝑚𝑑

Mixing ratio is usually defined as grams of vapor to kg of dry air. Typical values are a few g/kg at mid-latitudes 20 g/kg in the tropics.

Mixing Ratio and Vapor Pressure

Consider: 𝑝𝑑𝑉 = 𝑛𝑑𝑅

∗𝑇 𝑒𝑉 = 𝑛𝑣𝑅

∗𝑇 𝑝𝑉 = 𝑛𝑅∗𝑇

where pd, e, and p are the pressures due to dry air, water vapor, and dry plus vapor, respectively.

𝑒

𝑝=𝑛𝑣𝑛=

𝑛𝑣𝑛𝑣 + 𝑛𝑑

=𝑚𝑣 𝑀𝑣

𝑚𝑣 𝑀𝑣 +𝑚𝑑 𝑀𝑑 =

𝑤

𝑤 + 휀

Virtual Temperature The density is

𝜌 =𝑚𝑑 +𝑚𝑣𝑉

= 𝜌𝑑 + 𝜌𝑣

𝑝𝑑 = 𝑅𝑑𝜌𝑑𝑇 𝑒 = 𝑅𝑣𝜌𝑣𝑇

Dalton’s Law of partial pressure: 𝑝 = 𝑝𝑑 + 𝑒 so

𝜌 = 𝜌𝑑 + 𝜌𝑣 =𝑝𝑑𝑅𝑑𝑇

+𝑒

𝑅𝑣𝑇 =

𝑝

𝑅𝑑𝑇 1 −

𝑒

𝑝1 − 휀

Gas law: 𝑝 = 𝑅𝑑𝜌𝑇𝑣

where virtual temperature, 𝑇𝑣, is:

𝑇𝑣 =𝑇

1 −𝑒𝑝1 − 휀

Virtual Temperature

𝑇𝑣 =𝑇

1 −𝑒𝑝1 − 휀

Total density and pressure are related by the ideal gas law with the gas constant the same as that for dry air.

The virtual temperature is the temperature that dry air must have in order to have the same density as moist air at the same pressure. Therefore,𝑇𝑣 ≥ 𝑇.

The Hydrostatic Equation

Balance of vertical forces in an atmosphere in hydrostatic balance

The Hydrostatic Equation

• Air pressure at any height in the atmosphere is due to the force per unit area exerted by the weight of all of the air lying above that height. Consequently, atmospheric pressure decreases with increasing height above the ground.

• The net upward force acting on a thin horizontal slab of air, due to the decrease in atmospheric pressure with height, is generally very closely in balance with the downward force due to gravitational attraction that acts on the slab.

• If the net upward pressure force on the slab is equal to the downward force of gravity on the slab, the atmosphere is said to be in hydrostatic balance.

The Hydrostatic Equation

For an atmosphere in hydrostatic equilibrium, the balance of forces in the vertical require that

−𝛿𝑝 = 𝑔𝜌𝛿𝑧

In the limit as 𝛿𝑧 → 0, 𝑑𝑝

𝑑𝑧= −𝑔𝜌 or 𝑔𝑑𝑧 = −𝛼𝑑𝑝

This is the hydrostatic equation. Integrating:

− 𝑑𝑝 = 𝑝 𝑧 = 𝑔𝜌 𝑑𝑧∞

𝑧

𝑝(∞)

𝑝(𝑧)

The pressure at height z is equal to the weight of the air in the vertical column of unit cross sectional area lying above that level.

Geopotential

The geopotential, , at any point in the atmosphere is defined as the work that must be done against the gravitational field to raise a mass of 1 kg from sea level to that point. is thus the gravitational potential energy per unit mass, with units

J kg-1.

The work required to raise 1 kg from z to z + dz is gdz. Therefore, 𝑑 = 𝑔𝑑𝑧 = −𝛼𝑑𝑝. Integrating, we get

𝑧 = 𝑔𝑑𝑧𝑧

0

Define geopotential height:

𝑍 = 𝑧

𝑔0=1

𝑔0 𝑔𝑑𝑧𝑧

0

𝑔0is mean gravitational acceleration at surface, ~ 9.81 m s-1

Geopotential Height

Geopotential height is used as a vertical coordinate in many applications in which energetics is important. Z is close to z in the lower atmosphere.

Hypsometric Equation

It is often inconvenient to work explicitly with density, , a quantity that is seldom measured. It can be eliminated in the hydrostatic equation using the gas law:

𝑑𝑝

𝑑𝑧= −

𝑝𝑔

𝑅𝑇= −

𝑝𝑔

𝑅𝑑𝑇𝑣

𝑑 = 𝑔𝑑𝑧 = 𝑅𝑇𝑑𝑝

𝑝= −𝑅𝑑𝑇𝑣

𝑑𝑝

𝑝

Integrating:

𝑑𝑧2

𝑧1

= 2 −1 = −𝑅𝑑 𝑇𝑣𝑑𝑝

𝑝

𝑝2

𝑝1

and

𝑍2 − 𝑍1 = −𝑅𝑑𝑔0 𝑇𝑣

𝑑𝑝

𝑝

𝑝2

𝑝1

Hypsometric Equation

If 𝑇𝑣 constant: 𝑍2 − 𝑍1 = −𝑅𝑑𝑇𝑣𝑔0

𝑑𝑝

𝑝

𝑝2

𝑝1

= −𝐻 ln𝑝2𝑝1

𝑝2 = 𝑝1𝑒−𝑍2−𝑍1𝐻

What is H? e-folding depth, or scale height:

𝐻 =𝑅𝑑𝑇𝑣𝑔0

= 29.3𝑇𝑣

Instead, can define mean virtual temperature, 𝑇𝑣:

𝑇𝑣 = 𝑇𝑣 𝑑ln 𝑝𝑝2𝑝1

𝑑ln 𝑝𝑝2𝑝1

= 𝑇𝑣 𝑑ln 𝑝𝑝2𝑝1

ln𝑝2𝑝1

and

𝑍2 − 𝑍1 = −𝐻 ln𝑝2𝑝1

𝑤ℎ𝑒𝑟𝑒 𝐻 =𝑅𝑑 𝑇𝑣𝑔0

For a mean 𝑇𝑣 of 255 K, 𝐻 ≈ 7.5 km

Reduction of Pressure to Sea Level

In mountainous regions (like here!) the difference in surface pressure from one station to another may be due to differences in elevation. To isolate that part of the pressure field that is due to passage of weather systems is it necessary to reduce pressure to a common reference level, usually sea level. Using subscript g for ground and 0 for sea level:

𝑍𝑔 − 𝑍0 = 𝑍𝑔 = −𝐻 ln𝑝𝑔

𝑝0 so

𝑝0 = 𝑝𝑔𝑒𝑍𝑔𝐻

Homework

• Read sections 3.1-3.4 in text

• Problems to be assigned …