Published on *METEO 300: Fundamentals of Atmospheric Science* (https://www.e-education.psu.edu/meteo300)

The atmosphere is composed of billions of billions of molecules, 10^{43} molecules to be exact. Each molecule is zooming at hundreds of meters per second but colliding with other molecules each billionth of a second, exchanging kinetic energy (½ mv^{2}), momentum (m**v**), and internal energy (rotations and vibrations). It’s impossible to calculate what all of these molecules will do, so instead we study the energy and energy changes of volumes of molecules and use this information in our forecast models. This study is called thermodynamics. Thermodynamics has some difficult concepts to master, but it is also part of your everyday experience. In this lesson, you will learn the fundamental laws of thermodynamics and see how they describe our atmosphere’s pressure and temperature structure. You will learn why the atmosphere is sometimes stable and what happens when it isn’t.

By the end of this lesson, you should be able to:

- use the fundamental gas laws - Ideal Gas Law, Dalton’s Law – to determine the relative densities of different air masses
- derive the hydrostatic equilibrium equation from force balance to show why atmospheric pressure decreases with height
- use the 1st Law of Thermodynamics and conservation of energy (i.e., adiabatic processes) to explain air parcel temperature changes
- determine stability for different dry environmental temperature profiles
- calculate buoyancy and vertical velocity with time

Please see Canvas for a list of required assignments, due dates, and submission instructions.

If you have any questions, please post them to the Course Questions discussion forum in Canvas. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.

Understanding atmospheric thermodynamics begins with the gas laws that you learned in chemistry. Because these laws are so important, we will review them again here and put them in forms that are particularly useful for atmospheric science. **You will want to memorize these laws** because they will be used again and again in many other areas of atmospheric science, including cloud physics, atmospheric structure, dynamics, radiation, boundary layer, and even forecasting.

Credit: National Scientific Balloon Facility, Palestine TX

Before you begin this lesson's reading, I would like to remind you of the discussion activity for this lesson. This week's discussion activity will ask you to take what you learn throughout the lesson to answer an atmospheric problem. You will not need to post your discussion response until you have read the whole lesson, but keep the question in mind as you read:

**This week's topic is a hypothetical question involving stability. The troposphere always has a capping temperature inversion—it's called the stratosphere. The tropopause is about 16 km high in the tropics and lowers to about 10 km at high latitudes. The stratosphere exists because solar ultraviolet light makes ozone and then a few percent of the solar radiation is absorbed by stratospheric ozone, heating the air and causing the inversion. Suppose that there was no ozone layer and hence no stratosphere caused by solar UV heating of ozone.**

**Would storms in the troposphere be different if there was no stratosphere to act like a capping inversion? And if so, how?**

You will use what you have learned in this lesson about the atmosphere's pressure structure and stability to help you to think about this problem and to formulate your answer and discussions. So, think about this question as you read through the lesson. You'll have a chance to submit your response in 2.6!

The atmosphere is a mixture of gases that can be compressed or expanded in a way that obeys the Ideal Gas Law:

$$pV=N{R}^{*}T$$

where *p* is pressure (Pa = kg m^{–1} s^{–2}), *V* is the volume (m^{3}), *N* is the number of moles, *R** is the gas constant (8.314 J K^{–1} mole^{–1}), and *T* is the temperature (K). Note also that both sides of the Ideal Gas Law equation have the dimension of energy (J = kg m^{2} s^{–2}).

Recall that a mole is 6.02 x 10^{23} molecules (Avagodro’s Number). Equation 2.1 is a form of the ideal gas law that is independent of the type of molecule or mixture of molecules. A mole is a mole no matter its type. The video below (6:17) provides a brief review of the Ideal Gas Law. Note that the notation in the video differs slightly from our notation by using *n* for *N*, *P* for *p*, and *R* for *R**.

Usually in the atmosphere we do not know the exact volume of an air parcel or air mass. To solve this problem, we can rewrite the Ideal Gas Law in a different useful form if we divide *N* by *V* and then multiply by the average mass per mole of air to get the **mass density**:

$$\rho =\frac{NM}{V}$$

where *M* is the molar mass (kg mol^{–1}). Density has SI units of kg m^{–3}. The Greek symbol *ρ* (rho) is used for density and should not be confused with the symbol for pressure, *p*.

Thus we can put density in the Ideal Gas Law:

$$p=\frac{\rho {R}^{*}T}{M}or\rho =\frac{Mp}{{R}^{*}T}$$

Density is an incredibly important quantity in meteorology. Air that is more dense than its surroundings (often called its environment) sinks, while air that is less dense than its surroundings rises. Note that density depends on temperature, pressure, and the average molar mass of the air parcel. The average molar mass depends on the atmospheric composition and is just the sum of the fraction of each type of molecule times the molar mass of each molecular constituent:

$${M}_{average}=\frac{{\displaystyle \sum _{i}{N}_{i}{M}_{i}}}{{\displaystyle \sum _{i}{N}_{i}}}=\frac{{\displaystyle \sum _{i}{N}_{i}{M}_{i}}}{N}={\displaystyle \sum _{i}\frac{{N}_{i}}{N}{M}_{i}}={\displaystyle \sum _{i}{f}_{i}{M}_{i}}$$

where the* i* subscript represents atmospheric components, *N* is the number of moles, and* M* is the molar mass. This video (3:19) shows you how to find the gas density using the Ideal Gas Law. You will note that the person uses pressure in kPa and molar mass in g/mol. Since kPa = 1000 Pa and g = 1/1000 kg, the two factors of 1000 cancel when he multiplies them together and he can get away with using these units. I recommend always converting to SI units to avoid confusion. Also, note that the symbol for density used in the video is *d*, which is different from what we have used (*ρ*, the convention in atmospheric science).

Solve the following problem on your own. After arriving at your own answer, click on the link to check your work.

Let’s calculate the density of dry air where you live. We will use the Ideal Gas Law and account for the three most abundant gases in the atmosphere: nitrogen, oxygen, and argon. *M* is the molar mass of air; *M* = 0.029 kg mol^{–1}, which is just an average accounting for the fractions of different gases:

*R** = 8.314 J K^{–1} mol^{–1}. Here, *p* = 960 hPa = 9.6 x 10^{4} Pa and *T* = 20 ^{o}C = 293 K.

Putting these values into the equation 2.3, we get that the dry air density is 1.1 kg m^{–3}.

What would the density be if the room were filled with helium and not dry air at the same pressure and temperature?

Helium density = (pressure times molar mass of helium)/(Ideal Gas Law constant in SI units times temperature in K) = 0.16 kg m^{–3}

Often in meteorology we use mass-specific gas laws so that we must specify the gas that we are talking about, usually only dry air (N_{2} + O_{2} + Ar + CO_{2} +…) or water vapor (gaseous H_{2}O). We can divide *R ^{*}* by

Thus, we will use the following form of the Ideal Gas Law **for dry air**:

$${p}_{d}={\rho}_{d}{R}_{d}T$$

where:

${R}_{d}=\frac{R*}{{M}_{dryair}}=\frac{8.314{\text{JK}}^{-1}{\text{mol}}^{-1}}{0.02897{\text{kgmol}}^{-1}}=287{\text{m}}^{2}{\text{s}}^{-2}{\text{K}}^{-1}=287{\text{Jkg}}^{-1}{\text{K}}^{-1}$ .

*M _{dry air}* is 0.02897 kg mol

Note that *p* must be in Pascals (Pa), which is 1/100th of a mb (a.k.a, hPa), and *T* must be in Kelvin (K).

We can do the same procedure **for water vapor**:

$${p}_{v}\equiv e={\rho}_{v}{R}_{v}T$$

where ${R}_{v}=\frac{R*}{{M}_{watervapor}}=\frac{8.314{\text{JK}}^{-1}{\text{mol}}^{-1}}{0.01802{\text{kgmol}}^{-1}}=461{\text{m}}^{2}{\text{s}}^{-2}{\text{K}}^{-1}=461{\text{Jkg}}^{-1}{\text{K}}^{-1}$ .

Typically *e* is used to denote the water vapor pressure, which is also called the water vapor partial pressure.

Frontispiece of John Dalton and the Rise of Modern Chemistry by Henry Roscoe. Licensed under Public Domain via Wikimedia Commons [3]

This gas law is used often in meteorology. Applied to the atmosphere, it says that the total pressure is the sum of the partial pressures for dry air and water vapor:

$$p={p}_{d}+{p}_{H2O}={p}_{d}+e$$

Imagine that we put moist air and an absorbent in a jar and screw the lid on the jar. If we keep the temperature constant as the absorbent pulls water vapor out of the air, the pressure inside the jar will drop to *p _{d}*. Always keep in mind that when we measure pressure in the atmosphere, we are measuring the

So it follows that the **density of dry air and water vapor **also add:

$$\rho ={\rho}_{d}+{\rho}_{v}$$

Solve the following problem on your own. After arriving at your own answer, click on the link to check your work.

Suppose we have two air parcels that are the same size and have the same pressure and temperature, but one is dry and the other is moist air. Which one is less dense?

We can solve this one without knowing the pressure, temperature, or volume. Let’s assume that 98% of the molecules are dry air, which means the remaining 2% are dry air in the first case and water vapor in second case. Dry air is 0.029 kg mol^{–1} and water vapor is 0.018 kg mol^{–1}, so 2% of the moist air is lighter than the 2% of dry air, and when we consider the total air, this means that *for the same temperature and pressure, moist air is always less dense than dry air*.

Suppose there are two air parcels with different temperatures and water vapor amounts but the same pressure. Which one has a lower density? We can calculate the density to determine which one is lighter, but there is another way to do this comparison. Virtual temperature, *T _{v}*, is defined as the temperature dry air must have so that its density equals that of ambient moist air. Thus, virtual temperature is a property of the ambient moist air. Because the air density depends on the amount of moisture (for the same pressure and temperature), we have a hard time determining if the air parcel is more or less dense relative to its surroundings, which may have a different temperature and amount of water vapor. It is useful to pretend that the moist parcel is a dry parcel and to account for the difference in density by determining the temperature that the dry parcel would need to have in order to have the same density as the moist air parcel.

We can define the amount of moisture in the air by a quantity called **specific humidity, q**:

$$q=\frac{{\rho}_{v}}{{\rho}_{d}+{\rho}_{v}}$$

We see that *q* is just the fraction of water vapor density relative to the total moist air density. Usually *q* is given in units of g of water vapor per kg of dry air, or g kg^{–1}.

Using the Ideal Gas Law and Dalton’s Law, we can derive the **equation for virtual temperature**:

$${T}_{v}=T\left[1+0.61q\right]$$

where *T* and *T _{v}* have units of Kelvin (not

Note that moist air always has a higher virtual temperature than dry air that has the same temperature as the moist air because, as noted above, moist air is always less dense than dry air for the same temperature and pressure. Note also that for dry air, *q* = 0 and the virtual temperature is the same as the temperature.

Solve the following problem on your own. After arriving at your own answer, click on the link to check your work.

Consider a blob of air (*T _{blob}* = 25

We will use equation (2.10). Remember to convert *T* from ^{o}C to K and *q* from g kg^{–1} to kg kg^{–1}!

$$\begin{array}{l}{T}_{vblob}=(25+273)\left[1+0.61\cdot .010\right]=299.8K=26.8{}^{o}C\\ {T}_{venv}=(26+273)\left[1+0.61\cdot .001\right]=299.2K=26.2{}^{o}C\end{array}$$

We see that the blob is less dense than its environment and so will rise. This difference of 0.6 ^{o}C may seem small, but it makes a huge difference in upward motion.

The following are some mistakes that are commonly made in the above calculations:

- not converting from
^{o}C to K:

$$\begin{array}{l}{T}_{vblob}=(25)\left[1+0.61\cdot .010\right]=25.15{}^{o}C\\ {T}_{venv}=(26)\left[1+0.61\cdot .001\right]=26.02{}^{o}C\end{array}$$

We calculate that *T _{vblob} < T_{venv}*, which is the wrong answer.

- not converting
*q*from g/kg to kg/kg:

$$\begin{array}{l}{T}_{vblob}=(25+273)\left[1+0.61\cdot 10\right]=2115K=1842{}^{o}C\\ {T}_{venv}=(26+273)\left[1+0.61\cdot 1\right]=481K=208{}^{o}C\end{array}$$

We calculate that *T _{vblob} > T_{venv}*, which is correct in this case, but the numbers are crazy! After you complete your calculations, if the numbers you get just don’t seem right—like these—then you know that you have made a mistake in the calculation. Go looking for the mistake. Don’t submit an answer that makes no sense.

Once we find *T _{v}*, we can easily find the density of a moist parcel by using equation [2.5], in which we substitute

$${\rho}_{}=\frac{{p}_{}}{{R}_{d}{T}_{v}}$$

This quiz will give you practice calculating the virtual temperature and density using the Excel workbook that you set up in the last lesson.

- Go to Canvas and find
**Practice Quiz 2-1**. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.*I strongly suggest that you enter the equations for density and for virtual temperature in your Excel worksheet and use them to do all your calculations of density and virtual temperature on both the practice quiz and the quiz.* - When you feel you are ready, take
**Quiz 2-1**. You will be allowed to take this quiz only**once**. This quiz is timed, so after you start, you will have a limited amount of time to complete it and submit it. Good luck!

The atmosphere’s vertical pressure structure plays a critical role in weather and climate. We all know that pressure decreases with height, but do you know why?

The atmosphere’s basic pressure structure is determined by the hydrostatic balance of forces [4]. To a good approximation, every air parcel is acted on by three forces that are in balance, leading to no net force. Since they are in balance for any air parcel, the air can be assumed to be static or moving at a constant velocity.

There are **3 forces** **that determine hydrostatic balance:**

- One force is downwards (negative) onto the top of the cuboid from the pressure,
*p*, of the fluid above it. It is, from the definition of pressure [5],$${F}_{top}=-{p}_{top}A$$

- Similarly, the force on the volume element from the pressure of the fluid below pushing upwards (positive) is:
$${F}_{bottom}={p}_{bottom}A$$

- Finally, the weight [6] of the volume element causes a force downwards. If the density [7] is ρ, the volume is
*V*, which is simply the horizontal area*A*times the vertical height,*Δz*, and*g*the standard gravity [8], then:$${F}_{weight}=-\rho Vg=-\rho gA\Delta z$$

By balancing these forces, the **total force on the fluid** is:

$$\sum F={F}_{bottom}+{F}_{top}+{F}_{weight}={p}_{bottom}A-{p}_{top}A-\rho \mathrm{gA}\Delta z$$

This sum equals zero if the air's velocity is constant or zero. Dividing by *A*,

$$0={p}_{bottom}-{p}_{top}-\rho g\Delta z$$

or:

$${p}_{top}-{p}_{bottom}=-\rho g\Delta z$$

*P _{top} − P_{bottom}* is a change in pressure, and Δz is the height of the volume element – a change in the distance above the ground. By saying these changes are infinitesimally [9] small, the equation can be written in differential [10] form, where

$$dp=-\rho gdz$$

The result is the equation:

$\frac{dp}{dz}=-\rho g$

This equation is called the **Hydrostatic Equation**. See the video below (1:18) for further explanation:

Using the Ideal Gas Law, we can replace *ρ* and get the equation for dry air:

$$\frac{dp}{dz}=-g\frac{p}{{R}_{d}T}or\frac{dp}{p}=-\frac{g}{{R}_{d}T}dz=-\frac{Mg}{{R}^{*}T}dz$$

We could integrate both sides to get the altitude dependence of *p*, but we can only do that if *T* is constant with height. It is not, but it does not vary by more than about ±20%. So, doing the integral,

$$p={p}_{o}{e}^{-\raisebox{1ex}{$z$}\!\left/ \!\raisebox{-1ex}{$H$}\right.}where{p}_{o}=surfacepressureandH=\frac{{R}^{*}\overline{T}}{{M}_{air}g}$$

*H* is called a scale height because when *z = H*, we have *p* = *p _{o}*e

From the hydrostatic equation, the atmospheric pressure falls off exponentially with height, which means that about every 7 km, the atmospheric pressure is about 1/3 less. At 40 km, the pressure is only a few tenths of a percent of the surface pressure. Similarly, the concentration of molecules is only a few tenths of a percent, and since molecules scatter sunlight, you can see in the picture below that the scattering is much greater near Earth's surface than it is high in the atmosphere.

Credit: NASA

This quiz will give you practice using the hydrostatic equation to learn interesting and useful properties and quantities of the atmosphere.

*I strongly suggest that you do all your calculations on this quiz using the Excel workbook.*- There is no
**Practice Quiz 2-2**. However, you have extra time to take this quiz. - When you feel you are ready, take
**Quiz 2-2**in Canvas. You will be allowed to take this quiz only**once**. This quiz is timed, so after you start, you will have a limited amount of time to complete it and submit it. Good luck!

Weather involves heating and cooling, rising air parcels and falling rain, thunderstorms and snow, freezing and thawing. All of this weather occurs according to the three laws of Thermodynamics. The First Law of Thermodynamics tells us how to account for energy in any molecular system, including the atmosphere. As we will see, the concept of temperature is tightly tied to the concept of energy, namely thermal energy, but they are not the same because there are other forms of energy that can be exchanged with thermal energy, such as mechanical energy or electrical energy. Each air parcel contains molecules that have internal energy, which when thinking about the atmosphere, is just the kinetic energy of the molecules (associated with molecular rotations and, in some cases, vibrations) and the potential energy of the molecules (associated with the attractive and repulsive forces between the molecules). Internal energy does not consider their chemical bonds nor the nuclear energy of the nucleus because these do not change during collisions between air molecules. Doing work on an air parcel involves either expanding it by increasing its volume or contracting it. In the atmosphere, as in any system of molecules, energy is not created or destroyed, but instead, it is conserved. We just need to keep track of where the energy comes from and where it goes.

Credit: Ivana Vasilj via flickr

Let *U* be an air parcel’s internal energy, *Q* be the heating rate of that air parcel, and *W* be the rate that work is done on the air parcel. Then:

$$\frac{dU}{dt}=Q+W$$

The dimensions of energy are M L^{2} T^{–2} so the dimensions of this equation are M L^{2} T^{–3}.

To give more meaning to this energy budget equation, we need to relate *U, Q*, and *W* to variables that we can measure. Once we do that, we can put this equation to work. To do this, we resort to the Ideal Gas Law.

For processes like those that occur in the atmosphere, we can relate working,* W*, to a change in volume because work is force times distance. Imagine a cylinder with a gas in it. The cross-sectional area of the the piston is *A*. If the piston compresses the gas by moving a distance *dx*, the amount of work being done by the piston on the gas is the force (*pA*) multiplied by the distance (*dx*). *W* is then *pAdx*/*dt*. But the volume change is simply –*Adx*/*dt* and so:

$$W=-p\frac{dV}{dt}$$

Reducing a volume of gas (*dV/dt* < 0) takes energy, so working on an air parcel is positive when the volume is reduced, or *dV/dt* < 0. Thus:

$$\frac{dU}{dt}=Q-p\frac{dV}{dt}$$

The heat capacity *C* is the amount of energy needed to raise the temperature of a substance by a certain amount. Thus, $C=\frac{Q}{\frac{dT}{dt}}$
and has SI units of J/K. *C* depends on the substance itself, the mass of the substance, and the conditions under which the energy is added. We will consider two special conditions: constant volume and constant pressure.

Consider a box with rigid walls and thus constant volume: $\frac{dV}{dT}=0$ . No work is being done and only internal energy can change due to heating.

Credit: W. Brune, after Verlinde

The candle supplies energy to the box, so *Q* > 0 and *dU*/*dt* > 0. The internal energy can increase via increases in molecular kinetic and potential energy. However, for an ideal gas, the attractive and repulsive forces between the molecules (and hence the molecular potential energy) can be ignored. Thus, the molecular kinetic energy and, hence, the temperature, must increase:

$$\frac{dT}{dt}>0$$

So,

$${Q}_{\begin{array}{c}const\\ volume\end{array}}=\frac{dU}{dt}={C}_{V}\frac{dT}{dt}$$

*C _{V}*, the constant relating

Remember that ${C}_{v}\frac{dT}{dt}$ is the change in the air parcel’s internal energy.

The heat capacity, *C _{V}*, depends on the mass and the type of material. So we can write

$${C}_{V}=mass\cdot {c}_{V}$$

where *c _{V}* is called the specific heat capacity. The adjective “specific” means the amount of something per unit mass.

Some specific heat capacity values are included in the table below:

gas | c (@ 0_{V}^{o}C) J kg^{–1} K^{–1} |
---|---|

dry air | 718 |

water vapor | 1390 |

carbon dioxide | 820 |

Consider a sealed vault with an internal volume of 10 m^{3} filled with dry air (*p* = 1013 hPa; *T* = 273 K). If the vault is being heated at a constant rate from the outside at a rate of 1 kW (1,000 J s^{–1}), how long will it take for the temperature to climb by 30 ^{o}C?

The 1^{st }Law can be rewritten as:

$$Q=\frac{dU}{dt}+p\frac{dV}{dt}={C}_{V}\frac{dT}{dt}+p\frac{dV}{dt}=mass\cdot {c}_{V}\frac{dT}{dt}+p\frac{dV}{dt}$$

However, *dV/dt* = 0 because the vault’s volume isn’t changing. So, we can use the equation, rearrange it and integrate it:

$$Q=mass\cdot {c}_{V}\frac{dT}{dt}$$

How do we find the mass of the air inside the vault? Use the Ideal Gas Law to find the number of moles and then multiply by the mass per mole!

$$mass={M}_{dryair}\cdot n={M}_{dryair}\cdot \frac{pV}{{R}^{*}T}=0.029\cdot \frac{1.013x{10}^{5}\cdot 10}{8.314\cdot 273}=12.9kg$$

$$\Delta t=\frac{mass\cdot {c}_{V}\cdot \Delta T}{Q}=\frac{12.9\cdot 718\cdot 30}{{10}^{3}}=278\mathrm{s}\text{\hspace{0.17em}}(~5\mathrm{min})$$

Often we do not have a well-defined volume, but instead just an air mass. We can easily measure the air mass’s pressure and temperature, but we cannot easily measure its volume. Often we can figure out the heating rate per volume (or mass) of air. Thus:

$$\frac{Q}{mass}\equiv q=\frac{mass\cdot {c}_{V}\frac{dT}{dt}}{mass}={c}_{V}\frac{dT}{dt};\text{\hspace{1em}}q={c}_{V}\frac{dT}{dt}$$

where *q* is the specific heating rate (SI units: J kg^{–1} s^{–1}).

The atmosphere is not a sealed box and when air is heated it can expand. We can no longer ignore the volume change. On the other hand, as the volume changes, any pressure changes are rapidly damped out, causing the pressure in an air parcel to be roughly constant even as the temperature and volume change. This constant-pressure process is called isobaric.

$$Q=\frac{dU}{dt}+p\frac{dV}{dt}$$

Now the change in the internal energy could be due to changes in temperature *or *changes in volume. It turns out that *internal energy does not change with changes in volume. It only changes due to changes in temperature*. But we already know how changes in internal energy are related to changes in temperature from the example of heating the closed box. That is, the internal energy changes are related by the heat capacity constant volume, *C _{v}*. Thus:

$$Q={C}_{V}\frac{dT}{dt}+p\frac{dV}{dt}$$

Note that when volume is constant, we get the expression of heating a constant volume.

Suppose we pop the lid off the box and now the air parcel is open to the rest of the atmosphere. What happens when we heat the air parcel? How much does the temperature rise?

It’s hard to say because it is possible that the air parcel’s volume can change in addition to the temperature rise. So we might suspect that, for a fixed heating rate *Q*, the temperature rise in the open box will be less than the temperature rise in the sealed box where the volume is constant because the volume can change as well as the temperature.

Enthalpy (*H*) is an energy quantity that accounts not only for internal energy but also the energy associated with working. It is a useful way to take into consideration *both ways that energy can change* in a collection of molecules – by *internal energy changes* and by *volume changes that result in work being done*.

$$enthalpy\equiv H=U+pV$$

Enthalpy is the total energy of the air parcel including effects of volume changes. We can do some algebra and use the **Chain Rule** to write the First Law of Thermodynamics in terms of the enthalpy:

$$Q=\frac{dU}{dt}+p\frac{dV}{dt}=\frac{dU}{dt}+\frac{d\left(pV\right)}{dt}-V\frac{dp}{dt}=\frac{d\left(U+pV\right)}{dt}-V\frac{dp}{dt}=\frac{dH}{dt}-V\frac{dp}{dt}$$

If the pressure is constant, which is true for many air parcel processes, then *dp/dt* = 0 and:

$$Q=\frac{dH}{dt}$$

- In a constant volume process, heating changes only the internal energy,
*U*. - In a constant pressure process, heating changes enthalpy,
*H*(both internal energy and working).

In analogy with constant volume process, for a constant pressure process, we can write:

$${Q}_{\begin{array}{c}constant\\ pressure\end{array}}=\frac{dH}{dt}={C}_{p}\frac{dT}{dt}$$

where *C _{p}* is the heat capacity at constant pressure and

Note that *c _{p}* takes into account the energy required to increase the volume as well as to increase the internal energy and thus temperature.

What is the difference between *c _{p}* and

- by mole:${c}_{(p,m)}={c}_{(V,m)}+{R}^{*}$
- by mass for dry air: ${c}_{pd}={c}_{Vd}+{R}_{d}$
- by mass for water vapor: ${c}_{pv}={c}_{Vv}+{R}_{v}$

gas | c_{V} (@ 0^{o}C) J kg^{–1} K^{–1} |
c_{p} (@ 0^{o}C) J kg^{–1} K^{–1} |
---|---|---|

dry air | 718 | 1005 |

water vapor | 1390 | 1858 |

Since* c _{p} > c_{v}*, the temperature change at constant pressure will be less than the temperature change at constant volume because some of the energy goes to increasing the volume as well as to increasing the temperature.

$Q={C}_{V}\frac{dT}{dt}+p\frac{dV}{dt}$

$Q={C}_{p}\frac{dT}{dt}-V\frac{dp}{dt}$

$Q=\frac{dU}{dt}+p\frac{dV}{dt}$

$Q=\frac{dH}{dt}-V\frac{dp}{dt}$

and *c _{p} = c_{V} + R, C_{p} = c_{p}m = c_{p} ρV, C_{v} = c_{v}m = c_{v} ρV; α = V/M* (specific volume)

We can look at specific quantities, where we divide variables by mass.

$q={c}_{V}\frac{dT}{dt}+p\frac{d\alpha}{dt}$

$q={c}_{p}\frac{dT}{dt}-\alpha \frac{dp}{dt}$

$q=\frac{du}{dt}+p\frac{d\alpha}{dt}$

**You can figure out which form to use by following three steps:**

- Define the system. (i.e., what is the air parcel and what are its characteristics?)
- Determine the process(es) (i.e., constant pressure, constant volume, heating, cooling?). Choose the form of the equation by making a term with a conserved quantity go away (i.e.,
*dp/dt*= 0 or*dV/dt*= 0) because then you have a simpler equation to deal with. - Look at which variables you have and then choose the equation that has those variables.

Consider the atmospheric surface layer that is 100 m deep and has an average density of 1.2 kg m^{–3}. The early morning sun heats the surface, which heats the air with a heating rate of *F* = 50 W m^{–2}. How fast does the temperature in the layer increase? Why is this increase important?

- What is the system? Air layer. Since we know the heating per unit area, work the problem per unit area.
- What is the process? Constant pressure and heating by the sun.
- Which variables do we have? $Q={C}_{p}\frac{dT}{dt}$

$$Q=FA=50\text{}W\text{}{m}^{-2}A$$

$${C}_{p}={c}_{p}\rho V={c}_{p}\rho \text{}\Delta z\text{}A$$

$$Q=FA={c}_{p}\rho \text{}\Delta z\text{}A\frac{dT}{dt}$$

$$\frac{dT}{dt}=\frac{FA}{{c}_{p}\text{}\rho \text{}\Delta z\text{}A}=\frac{F}{{c}_{p}\rho \text{}\Delta z}=\frac{50}{1005\text{1.2100}}=4.2x{10}^{-4}K\text{}{s}^{-1}=1.5\text{}K\text{}h{r}^{-1}$$

This temperature increase is important because it is one of the most important factors in determining whether convection will occur later in the day. We will talk more about instability soon.

Here is a video (1:30) explanation of the above problem:

Consider the atmospheric surface layer that is 100 m deep and has an average density of 1.2 kg m^{–3}. It is night and dark and the land in contact with the air is cooling at 50 W m^{–2}. If the temperature at the start of the night was 25 ^{o}C, what is the temperature 8 hours later?

- What is the system? Air layer. Since we know the cooling per unit area, work the problem per unit area.
- What is the process? Constant pressure and cooling by the land radiating energy to space and the air cooling by being in contact with the land.
- Which variables do we have? $Q={C}_{p}\frac{dT}{dt}$

$$Q=FA=-50\text{}W\text{}{m}^{-2}A$$

$${C}_{p}={c}_{p}\rho V={c}_{p}\rho \text{}\Delta z\text{}A$$

$$Q=FA={c}_{p}\rho \text{}\Delta z\text{}A\frac{dT}{dt}$$

$$\frac{dT}{dt}=\frac{FA}{{c}_{p}\text{}\rho \text{}\Delta z\text{}A}=\frac{F}{{c}_{p}\rho \text{}\Delta z}=\frac{-50}{1005\text{1.2100}}=-4.2x{10}^{-4}K\text{}{s}^{-1}=-1.5\text{}K\text{}h{r}^{-1}$$$${\mathrm{}}^{}$$

Since the cooling continues for 8 hours, the total amount of cooling is –1.5 K/hr x 8 hr = 12 K or 12 ^{o}C. Thus, the temperature 8 hours later will be 13 ^{o}C.

This cooling near the surface creates a layer of cold air near the surface with a layer of warmer air above it. The layering of warm air over colder air creates an temperature inversion, which suppresses convection and lock pollutants into the air layer near Earth's surface.

Consider a column of air between two pressure surfaces. If the mass in the column is conserved, then the column with the greater average temperature will be less dense and occupy more volume and thus be higher in altitude. But the pressure at the top of the column is related to the weight of the air above the column, which is constant, and so the upper pressure surface is higher in altitude. If the temperature of the column is lower, then the pressure surface at the top of the column will be lower in altitude.

We can look at this behavior from the point-of-view of hydrostatic equilibrium.

$\frac{\text{d}p}{dz}=-\rho g=-\frac{pg}{{R}_{d}T}$[11]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaqGKbGaamiCaaWdaeaapeGaamizaiaadQhaaaGaeyypa0JaeyOeI0IaeqyWdiNaam4zaiabg2da9iabgkHiTmaalaaapaqaa8qacaWGWbGaam4zaaWdaeaapeGaamOua8aadaWgaaWcbaWdbiaadsgaa8aabeaak8qacaWGubaaaaaa@42C0@

If the temperature is greater, then the change in *p* with height is less, which means that any given pressure surface is going to be higher.

The difference between any two pressure surfaces is called the **thickness**.

We can show that the thickness depends only on temperature:

$dz=-\frac{dp}{p}\frac{{R}_{d}}{g}T$[11]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaadsgacaWG6bGaeyypa0JaaiiOamaalaaapaqaa8qacaWGKbGaamiCaaWdaeaapeGaamiCaaaadaWcaaWdaeaapeGaamOua8aadaWgaaWcbaWdbiaadsgaa8aabeaaaOqaa8qacaWGNbaaaiaadsfaaaa@3E3B@

Integrate both sides:

$\underset{{z}_{1}}{\overset{{z}_{2}}{{\displaystyle \int}}}dz=-\underset{{p}_{1}}{\overset{{p}_{2}}{{\displaystyle \int}}}\frac{dp}{p}\frac{{R}_{d}}{g}T$[11]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaawahabeWcpaqaa8qacaWG6bWdamaaBaaameaapeGaaGymaaWdaeqaaaWcbaWdbiaadQhapaWaaSbaaWqaa8qacaaIYaaapaqabaaaneaapeGaey4kIipaaOGaamizaiaadQhacqGH9aqpcaGGGcWaaybCaeqal8aabaWdbiaadchapaWaaSbaaWqaa8qacaaIXaaapaqabaaaleaapeGaamiCa8aadaWgaaadbaWdbiaaikdaa8aabeaaa0qaa8qacqGHRiI8aaGcdaWcaaWdaeaapeGaamizaiaadchaa8aabaWdbiaadchaaaWaaSaaa8aabaWdbiaadkfapaWaaSbaaSqaa8qacaWGKbaapaqabaaakeaapeGaam4zaaaacaWGubaaaa@4C07@

or

${z}_{2}-{z}_{1}=\frac{{R}_{d}}{g}\text{ln}\left(\frac{{p}_{1}}{{p}_{2}}\right)\overline{T}$[11]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaadQhapaWaaSbaaSqaa8qacaaIYaaapaqabaGcpeGaeyOeI0IaamOEa8aadaWgaaWcbaWdbiaaigdaa8aabeaak8qacqGH9aqpcaGGGcWaaSaaa8aabaWdbiaadkfapaWaaSbaaSqaa8qacaWGKbaapaqabaaakeaapeGaam4zaaaacaqGSbGaaeOBamaabmaapaqaa8qadaWcaaWdaeaapeGaamiCa8aadaWgaaWcbaWdbiaaigdaa8aabeaaaOqaa8qacaWGWbWdamaaBaaaleaapeGaaGOmaaWdaeqaaaaaaOWdbiaawIcacaGLPaaaceWGubWdayaaraaaaa@46A3@

where *T* is the average temperature of the layer between *p _{1}* and

As some of you already know, you can use the thickness between different pressure surfaces to estimate the type of precipitation that will fall - snow, rain, or a mixture. You can check out these resources for some more information and example problems:

Suppose that the 500 mb surface is at 560 dam (decameters, 10s of meters) and the 1000 mb surface is at 0 dam. What is the average temperature of the layer between 1000 mb and 500 mb?

Rearrange equation 2.4 to get an expression in terms of the average temperature and then put all the numbers into the equation to find the average temperature of the layer. Make sure that all the units are correct.

$$\overline{\text{T}}=\frac{10*\left({z}_{2}-{z}_{1}\right)}{\left(\raisebox{1ex}{$287$}\!\left/ \!\raisebox{-1ex}{$9.8$}\right.\right)\mathrm{ln}\left(\raisebox{1ex}{${p}_{1}$}\!\left/ \!\raisebox{-1ex}{${p}_{2}$}\right.\right)}=\frac{10*\left(560-0\right)}{\left(\raisebox{1ex}{$287$}\!\left/ \!\raisebox{-1ex}{$9.8$}\right.\right)\mathrm{ln}\left(\raisebox{1ex}{$1000$}\!\left/ \!\raisebox{-1ex}{$500$}\right.\right)}=276\text{\hspace{0.17em}}K$$So far, we have covered **constant volume (isochoric)** and **constant pressure (isobaric)** processes. There is a third process that is very important in the atmosphere—the **adiabatic process**. Adiabatic means *no energy exchange* between the air parcel and its environment: *Q *= 0. **Note: adiabatic is not the same as isothermal.**

Consider the Ideal Gas Law:

$pV=n{R}^{*}T$

If an air parcel rises, the pressure changes, but how does the temperature change? Note that the volume can change as well as the pressure and temperature, and thus, if we specify a pressure change, we cannot find the temperature change unless we know how the volume changed. Without some other equation, we cannot say how much the temperature will rise for a pressure change.

Credit: W. Brune, after Lamb and Verlinde

However, we can use the First Law of Thermodynamics to relate changes in temperature to changes in pressure and volume for adiabatic processes.

I do not expect you to be able to do this derivation, but you should go through it to make sure that you understand all the steps as a way to continue to improve your math skills. Start with the following specific form of the 1^{st} Law for dry air:

$q={c}_{p}\frac{dT}{dt}-\alpha \frac{dp}{dt}$

$q=0={c}_{p}\frac{dT}{dt}-\alpha \frac{dp}{dt}$

Divide both sides by *T* and note that $This\; equation\; is\; not\; rendering\; properly\; due\; to\; an\; incompatible\; browser.\; See\; Technical\; Requirements\; in\; the\; Orientation\; for\; a\; list\; of\; compatible\; browsers.$ (α is called the specific volume):

$\frac{{c}_{p}}{T}\frac{dT}{dt}-\frac{{R}_{d}T}{pT}\frac{dp}{dt}=0=\frac{{c}_{p}}{T}\frac{dT}{dt}-\frac{{R}_{d}}{p}\frac{dp}{dt}={c}_{p}\frac{dln\left(T\right)}{dt}-{R}_{d}\frac{dln\left(p\right)}{dt}$

where we figured out the two terms were just derivatives of the natural log of *T* and *p*.

$\frac{d}{dt}({c}_{p}ln\left(T\right)-{R}_{d}ln\left(p\right))=0$

But *d/dt *= 0 just means that the value is constant:

$({c}_{p}ln\left(T\right)-{R}_{d}ln\left(p\right))=constant$

Divide by *c _{p}*:

$ln\left(T\right)-\frac{{R}_{d}}{{c}_{p}}ln\left(p\right)=ln\left(T\right)+ln\left({p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}\right)=ln\left(T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}\right)=constant$

If the natural log of a variable is constant then the variable itself must be constant:

$T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}=constant$

We can rewrite *R _{d}/c_{p}* as a new term denoted by the Greek letter gamma,

$\gamma \equiv \frac{{c}_{p}}{{c}_{v}}=\frac{{c}_{v}+{R}_{d}}{{c}_{v}}=1.4$

$\frac{{R}_{d}}{{c}_{p}}=\frac{\gamma -1}{\gamma}=0.286$

We can use the Ideal Gas Law to get relations among *p*, *V*, and* T*, called the Poisson’s Relations:

$T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}=T{p}^{\left(\raisebox{1ex}{$\left(1-\gamma \right)$}\!\left/ \!\raisebox{-1ex}{$\gamma $}\right.\right)}=constant$

${\alpha}^{\gamma}p=constant$

$T{\alpha}^{\gamma -1}=constant$

The Poisson Relation that we use the most is the relation of pressure and temperature because these are two variables that we can measure easily without having to define a volume of air:

$\frac{T}{\theta}={\left(\frac{p}{{p}_{o}}\right)}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}$

or

$\theta =T{\left(\frac{{p}_{o}}{p}\right)}^{\raisebox{1ex}{${R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.}=T{\left(\frac{1000}{p}\right)}^{0.286}$

We call *θ* the potential temperature, which is the temperature that an air parcel would have if the air is brought to a pressure of *p _{o}* = 1000 hPa. Potential temperature is one of the most important thermodynamic quantities in meteorology.

Adiabatic processes are common in the atmosphere, especially the dry atmosphere. Also, adiabatic processes are often the same as isentropic processes (no change in entropy).

Air coming over the Laurel Highlands descends from about 700 m (*p* ~ 932 hPa) to 300 m (*p* ~ 977 hPa) at State College. Assume that the temperature in the Laurel Highlands is 20 ^{o}C. What is the temperature in State College?

We can find the temperature in State College due only to adiabatic changes by the following equation:

$$\begin{array}{l}{T}_{7}{}_{00}{p}_{7}{{}_{00}}^{-0.286}={T}_{3}{}_{00}{p}_{3}{{}_{00}}^{-0.286}\\ {T}_{300=}{T}_{700}{\left(\frac{{p}_{300}}{{p}_{700}}\right)}^{0.286}=\left(273+20\right){\left(\frac{977}{932}\right)}^{0.286}=297K=24C\end{array}$$[11]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaaaaaaWdbiaadsfapaWaaSbaaSqaa8qacaaI3aaapaqabaGcpeWaaSbaaSqaaiaaicdacaaIWaaabeaakiaadchapaWaaSbaaSqaa8qacaaI3aaapaqabaGcdaWgaaWcbaGaaGimaiaaicdaaeqaaOWaaWbaaSqabeaacqGHsislcaaIWaGaaiOlaiaaikdacaaI4aGaaGOnaaaakiabg2da98qacaWGubWdamaaBaaaleaapeGaaG4maaWdaeqaaOWdbmaaBaaaleaacaaIWaGaaGimaaqabaGcpaGaamiCamaaBaaaleaapeGaaG4maaWdaeqaaOWdbmaaBaaaleaacaaIWaGaaGimaaqabaGcpaWaaWbaaSqabeaapeGaeyOeI0IaaGimaiaac6cacaaIYaGaaGioaiaaiAdaaaaak8aabaGaamivamaaBaaaleaacaaIZaGaaGimaiaaicdacqGH9aqpaeqaaOGaamivamaaBaaaleaacaaI3aGaaGimaiaaicdaaeqaaOWaaeWaaeaadaWcaaqaaiaadchadaWgaaWcbaGaaG4maiaaicdacaaIWaaabeaaaOqaaiaadchadaWgaaWcbaGaaG4naiaaicdacaaIWaaabeaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaicdacaGGUaGaaGOmaiaaiIdacaaI2aaaaOGaeyypa0ZaaeWaaeaacaaIYaGaaG4naiaaiodacqGHRaWkcaaIYaGaaGimaaGaayjkaiaawMcaamaabmaabaWaaSaaaeaacaaI5aGaaG4naiaaiEdaaeaacaaI5aGaaG4maiaaikdaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIWaGaaiOlaiaaikdacaaI4aGaaGOnaaaakiabg2da9iaaikdacaaI5aGaaG4naiaadUeacqGH9aqpcaaIYaGaaGinaiaadoeaaaaa@745A@

This temperature change is 4 ^{o}C, or 7 ^{o}F just from adiabatic compression.

We can plot adiabatic **(isentropic)** surfaces in the atmosphere. An air parcel needs no energy to move along an adiabatic surface. Also, it takes energy for an air parcel to move from potential surface to another potential energy surface.

Credit: W. Brune, after Andews, Holton, and Leovy

Suppose an air parcel has *p* = 300 hPa and *T* = 230 K. How much heating per unit volume of dry air would be needed to increase the potential temperature by 10 K?

The heating raises the temperature, and the amount of heating required depends on the heat capacity, constant pressure, which depends on the mass of air, or the density times the volume. Let's do the calculation for a volume of air; that way we can use the density.

First we need to find the temperature rise that is the same as a potential temperature rise of 10 K at a pressure of 300 hPa.

$d\theta =dT{\left(\frac{1000}{p}\right)}^{0.286}\text{}dT=10{\left(\frac{300}{1000}\right)}^{0.286}=7.1K$[11]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaapeGaamizaiabeI7aXjabg2da9iaadsgacaWGubWaaeWaa8aabaWdbmaalaaapaqaa8qacaaIXaGaaGimaiaaicdacaaIWaaapaqaa8qacaWGWbaaaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaIWaGaaiOlaiaaikdacaaI4aGaaGOnaaaak8aacaqGGaGaaeiiaiaabccacaqGGaGaamizaiaadsfacqGH9aqpcaaIXaGaaGima8qadaqadaWdaeaapeWaaSaaa8aabaWdbiaaiodacaaIWaGaaGimaaWdaeaapeGaaGymaiaaicdacaaIWaGaaGimaaaaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGaaGimaiaac6cacaaIYaGaaGioaiaaiAdaaaGcpaGaeyypa0JaaG4naiaac6cacaaIXaGaam4saaaa@5831@

Then we need to find the density so that we can calculate the heat capacity:

$$\rho =\frac{p}{{R}_{d}T}=\frac{{\mathrm{3x10}}^{4}}{287\cdot 230}=0.45\text{\hspace{0.17em}}kg\text{\hspace{0.17em}}{m}^{-3}$$

Now we can put it all together:

$$\frac{Q\Delta t}{V}=\frac{\rho \cdot V\cdot {c}_{p}\cdot \Delta T}{V}\Rightarrow \frac{Q\Delta t}{V}=\rho \cdot {c}_{p}\cdot \Delta T=0.45\cdot 1005\cdot 7.1=3.2x{10}^{3}\text{\hspace{0.17em}}J\text{\hspace{0.17em}}{m}^{-3}$$

Note that we were asked to provide the total heating per unit volume, which is just the heating rate times time divided by the unit volume. So the quantity on the left is what we want. Is this heating large? Yes! So it takes a lot of heating or cooling the raise or lower an air parcel potential temperature just 10 K.

The temperature change with change in pressure (and thus change in altitude) is a major reason for weather. For dry air, the main effect is buoyancy. So because the pressure change generally follows the hydrostatic equation, the *change in height translates into a change in pressure which translates into a change in temperature due to adiabatic expansion*. Note that as the air parcel rises, its pressure quickly adjusts to the pressure of the surrounding air. Thus we can determine the dry adiabatic lapse rate by starting with the Poisson relation between pressure and temperature:

$$T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}=constant$$

Take the derivative w.r.t. *z*:

$$\frac{dT}{dz}{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}+T\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)-1}\frac{dp}{dz}=\frac{dT}{dz}+T\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{-1}\frac{dp}{dz}=0$$

But we also know from the hydrostatic equation that:

$$\frac{dp}{dz}=-\rho g$$

Substituting –*ρg* for *dp*/*dz* into the equation and rearranging the terms:

$$\frac{dT}{dz}=T\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{-1}\rho g=-T\left(\raisebox{1ex}{${R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{-1}g\left(\frac{p}{{R}_{d}T}\right)$$

$$ -\frac{d T}{d z} \equiv \Gamma_{d}=\frac{g}{c_{p}}=\frac{9.8 \frac{m}{s}}{1005 J k g^{-1} K^{-1}}=9.8 K k m^{-1} $$

*Γ _{d}* is called the dry adiabatic lapse rate. Note that the temperature decreases with height, but the dry adiabatic lapse rate is defined as being positive.

Air coming over the Laurel Highlands descends from about 700 m to 300 m at State College. Assume that the temperature in the Laurel Highlands is 20 ^{o}C. What is the temperature in State College?

We can find the temperature in State College due only to adiabatic changes by using the dry adiabatic lapse rate multiplied by the height change:

$$dT=-{\Gamma}_{d}(300-700)=9.8Kk{m}^{-1}\cdot 0.4={4}^{\circ}C$$

This temperature change is 4.0 ^{o}C, or 7 ^{o}F just from adiabatic compression. This answer is very similar to the answer we obtained using the change in potential temperature.

This quiz is practice calculating energy balance using the First Law of Thermodynamics. Some of the problems will involve potential temperature.

- Go to Canvas and find
**Practice Quiz 2-3**. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 2-3**. You will be allowed to take this quiz only**once**. This quiz is timed, so after you start, you will have a limited amount of time to complete it and submit it. Good luck!

We know that an air parcel will rise relative to the surrounding air at the same pressure if the air parcel’s density is less than that of the surrounding air. The difference in density can be calculated using the virtual temperature, which takes into account the differences in specific humidity in the air parcel and the surrounding air as well as the temperature differences.

In equilibrium, the sum of forces are in balance and the air parcel will not move. The question is, what happens to the parcel if there is a slight perturbation in its vertical position?

Credit: W. Brune, after Verlinde

For the figure on the left, if the ball is displaced a tiny bit to the left or the right, it will be pulled by gravity and will continue to roll down the slope. That position is unstable. For the figure on the right, if the ball is displaced a little bit, it will be higher than the central position and gravity will pull it back down. It may rock back and forth a little bit, but eventually it will settle down into its original position.

To assess instability of air parcels in the atmosphere, we need to find out if moving the air parcel a small amount up or down causes the parcel to continue to rise or to fall (instability) or if the air parcel returns to its original position (stability).

Now look at some atmospheric temperature profiles. **Important: a dry air parcel that is pushed from its equilibrium position always moves along the dry adiabatic lapse rate (DALR) line.**

Credit: W. Brune

Note that we can also show that if the air parcel is pushed down, it will keep going if the atmospheric (environmental) profile looks like the one on the left and will return to the original position if it looks like the one on the right.

Credit: W. Brune

Use the image above to determine the following:

Is the air parcel stable or unstable at each of the points, 1-5?

The red line is the atmospheric temperature profile; the dashed lines are the dry adiabatic lapse rate lines (-9.8 K/km). Consider points 1-5. When an air parcel is pushed up the DALR and its temperature is greater than the atmospheric temperature at the new level, it is warmer and thus less dense. It will continue to rise. When an air parcel is pushed down the DALR and its temperature is less than the atmospheric temperature at that new level, it is colder and thus more dense. It will continue to fall. Both cases are unstable. However, if when an air parcel is pushed up the DALR and its temperature is less than the atmospheric temperature at that new level, it is colder and thus more dense. It will sink back down to its original position and is stable.

Using this thinking, air parcels at points 1, 2, and 5 are stable and at points 3 and 4 are unstable.

We can calculate the acceleration an unstable air parcel will have and, from this, can determine the parcel’s velocity at some later point in time. This acceleration is called buoyancy (B).

Credit: W. Brune

$\sum F={F}_{bottom}+{F}_{top}+{F}_{weight}={p}_{bottom}A-{p}_{top}A-{\rho}^{\prime}gAh$

Let’s look at the forces on an air parcel again, like we did to derive the hydrostatic equilibrium. But this time, let’s assume that the parcel has a different density than the surrounding air. We will designate quantities associated with the air parcel with an apostrophe (’); environmental parameters will have no superscript.

If the forces are not in balance, then we need to keep the acceleration that we set to zero in the hydrostatic equilibrium case. We can also divide both sides of the equation by the mass of the air parcel:

$\frac{\sum F}{{\rho}^{\prime}V}=\frac{-\frac{{p}_{top}-{p}_{bottom}}{\Delta z}-{\rho}^{\prime}g}{{\rho}^{\prime}}\to \frac{-\left(-\rho g\right)-{\rho}^{\prime}g}{{\rho}^{\prime}}$

$a\equiv B=\frac{\left(\rho -{\rho}^{\prime}\right)g}{{\rho}^{\prime}}$

where we have used the hydrostatic equilibrium of the environment to replace the expression for the pressure change as a function of height with the density times the acceleration due to gravity.

We can then use the Ideal Gas Law to replace densities with virtual temperatures because the pressure of the parcel and its surrounding air is the same:

$B=\frac{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${T}_{v}$}\right.-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${T}_{v}^{\text{'}}$}\right.\right)g}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${T}_{v}^{\text{'}}$}\right.}=\frac{\left({T}_{v}^{\text{'}}-{T}_{v}\right)}{{T}_{v}}g$

If *B*>0, then the parcel rises; if *B*<0, then the parcel descends.

We look at the instability at each point in the environmental temperature profile and can determine* Γ _{env}* for each point.

Thus,

${T}_{env}={T}_{0}+{\frac{\partial T}{\partial z}|}_{env}\Delta z={T}_{0}-{\Gamma}_{env}\Delta z;{T}_{parcel}={T}_{0}+{\frac{\partial T}{\partial z}|}_{parcel}\Delta z={T}_{0}-{\Gamma}_{d}\Delta z;$

so that:

$B=\frac{\left({T}_{0}-{\Gamma}_{d}\Delta z-{T}_{0}+{\Gamma}_{env}\Delta z\right)}{{T}_{v}}g=\frac{\left({\Gamma}_{env}-{\Gamma}_{d}\right)}{{T}_{v}}g\Delta z$

If *Γ _{env} < Γ_{d}*, the parcel accelerates downward for positive

If *Γ _{env} > Γ_{d}*, the parcel accelerates upward for positive

We can put this idea of buoyancy in terms of potential temperature.

$\theta =T{\left(\frac{{p}_{o}}{p}\right)}^{\raisebox{1ex}{${R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.}$

We want to find *dθ/dz*. Taking the log of both sides of the equation and replacing a *dp/dz* term with *–gρ*, we are able to find the following expression for buoyancy in terms of potential temperature:

$B=-g\Delta z\frac{1}{\theta}\frac{d\theta}{dz}$

Remember that no matter what the environmental temperature or potential temperature profiles, a change in height of an air parcel will result in a temperature that changes along the dry adiabat and a potential temperature that does not change at all. As you can see below, the stability of a layer depends on the change in environmental potential temperature with height. Air parcels try to move vertically with constant potential temperature.

Credit: W. Brune

Parcels will move to an altitude (and air density) for which B = 0. However, if they still have a velocity when they reach that altitude, they will overshoot, experience a negative acceleration, and then descend, overshooting the neutral level again. In this way, the air parcel will oscillate until its oscillation is finally damped out by friction and dissipation of the air parcel. Note that in the neutral section of vertical profile where potential temperature does not change, it is not possible to determine if an air parcel will be stable or unstable. For instance, if the air parcel in the neutral region is given a small upward push, it will continue to rise until it reaches a stable region.

This quiz provides practice determining stability or instability of an air parcel and in calculating the buoyancy of air parcels.

- Go to Canvas and find
**Practice Quiz 2-4**. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 2-4**. You will be allowed to take this quiz only**once**. Good luck!

**(3 discussion points)**

This week's discussion topic is a hypothetical question involving stability. The troposphere always has a capping temperature inversion - it's called the stratosphere. The tropopause is about 16 km high in the tropics and lowers to about 10 km at high latitudes. The stratosphere exists because solar ultraviolet light makes ozone and then a few percent of the solar radiation is absorbed by stratospheric ozone, heating the air and causing the inversion. Suppose that there was no ozone layer and hence no stratosphere caused by solar UV heating of ozone.

Would storms in the troposphere be different if there were no stratosphere to act like a capping inversion? And if so, how?

Use what you have learned in this lesson about the atmosphere's pressure structure and stability to help you to think about this problem and to formulate your answer and discussions. It's OK to be wrong, as long as you have some solid reasoning to back up your ideas. My goal is to get you all to communicate with each other and think hard about atmospheric science.

- You can access the
**Storms in the Troposphere Discussion Forum**in Canvas. - Post a response that answers the question above in a thoughtful manner that draws upon course material and outside sources.
- Keep the conversation going!
**Comment on at least one other person's post**. Your comment should include follow-up questions and/or analysis that might offer further evidence or reveal flaws.

This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:

Evaluation | Explanation | Available Points |
---|---|---|

Not Completed | Student did not complete the assignment by the due date. | 0 |

Student completed the activity with adequate thoroughness. | Posting answers the discussion question in a thoughtful manner, including some integration of course material. | 1 |

Student completed the activity with additional attention to defending his/her position. | Posting thoroughly answers the discussion question and is backed up by references to course content as well as outside sources. | 2 |

Student completed a well-defended presentation of his/her position, and provided thoughtful analysis of at least one other student’s post. | In addition to a well-crafted and defended post, the student has also engaged in thoughtful analysis/commentary on at least one other student’s post as well. | 3 |

We started with the Ideal Gas Law and Dalton’s Law to develop an understanding of the atmosphere’s behavior and atmospheric density, a fundamental driving force for vertical motion in the atmosphere. We saw that composition matters, and developed a new quantity, virtual temperature, that allows us to compare the densities of different moist air parcels. Using these laws along with Newton’s laws, we were able to derive a fundamental property of the atmosphere: the hydrostatic equation, which states that the change in pressure with altitude is proportional to the negative of the density times gravity. We applied a new constraint – the First Law of Thermodynamics, which states that energy is conserved and saw that combining it with the gas laws enabled us to calculate temperature changes and to derive an important atmospheric quantity – the potential temperature. With these concepts, we were able to determine the stability (and instability) of an air parcel. We could also determine the buoyancy of an air parcel, which allows us to calculate the acceleration and thus the velocity of an air parcel after it has accelerated for a while.

You have reached the end of Lesson 2! Be sure to complete the Activities within three days of the end of the lesson. I strongly encourage you to do them as you are going through the lesson.

**Links**

[1] https://www.youtube.com/channel/UCj3EXpr5v35g3peVWnVLoew

[2] https://www.youtube.com/channel/UCK4ojE1uJYPabSP0ABkM6EA

[3] https://commons.wikimedia.org/wiki/File:Dalton_John_desk.jpg#/media/File:Dalton_John_desk.jpg

[4] http://en.wikipedia.org/wiki/Hydrostatic_equilibrium

[5] http://en.wikipedia.org/wiki/Pressure

[6] http://en.wikipedia.org/wiki/Weight

[7] http://en.wikipedia.org/wiki/Density

[8] http://en.wikipedia.org/wiki/Standard_gravity

[9] http://en.wikipedia.org/wiki/Infinitesimal

[10] http://en.wikipedia.org/wiki/Differential_equation

[11] mailto:MathType@MTEF

[12] http://www.wxonline.info/topics/thickness_pcp.html#q1

[13] http://weatherfaqs.org.uk/node/152