Overview

Meteorologists devote their lives to one purpose: to forecast the weather. But many other scientists work to build our understanding of the atmosphere, which is the basis for better prediction that meteorologists can use. Atmospheric science is the foundation upon which all meteorology is based.

Just as with all other physical sciences, mathematics is the language of atmospheric science. If you had our first-year meteorology course, you learned about weather forecasting and some of the physical basis behind the forecasts. This course will introduce you to even more physical concepts of atmospheric science and some of the mathematics that describe and quantify those physical concepts.

You have learned some of the mathematics that you need for this course and will be learning more about vectors and vector calculus soon. In this lesson, we will practice some of the mathematics. You will also prepare an Excel workbook that will help you with some of this course’s assessment problems.

Learning Objectives

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

• correctly use significant figures, dimensions, and units
• solve simple problems using integral and differential calculus
• prepare and use a course Excel workbook for course calculations 

Questions?

If you have any questions, please post them to the Course Questions discussion forum. 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.

1.1 The atmosphere is …

The atmosphere is amazing, awe-inspiring, frightening, deadly, powerful, boring, strange, beautiful, and uplifting – just a few of thousands of descriptions. So much of our lives depend on the atmosphere, yet we often take it for granted.

Clouds over the Arctic Ocean at sunrise

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0 [1]

Atmospheric science attempts to describe the atmosphere with physical descriptions using words, but also with mathematics. The goal is to be able to write down mathematical equations that capture the atmosphere’s important physical properties (predictability) and to use these equations to determine the atmosphere’s evolution with time (prediction). Predicting the weather has long been a primary focus, but, increasingly, we are interested in predicting climate.

We know quite a lot about the atmosphere. It has taken decades, if not centuries, of careful observation and insightful theory that is based on solid physical and chemical laws. We have more to learn. You could help to advance the understanding of the atmosphere, but you must first understand the physical concepts and mathematics that are already well known. That is a primary purpose of this course – to give you that understanding.

What follows, below, is a series of pictures and graphical images. Each one depicts some atmospheric process that will be covered in this course. Look at these images; you will see them again, each in one of the next ten lessons. Of course, in each observation there are many processes going on simultaneously. In the last lesson, you will have the opportunity to look at an observation and attach the physical principles and the mathematics that describe several processes that are causing the phenomena that you are observing.

1.2 You won’t believe what you can do with math!

You’ve been told many times that meteorology is a math-intensive field. It is. But for this course, you already know much of the math, and what you haven’t seen, you will see in vector calculus. To get ready for the meteorology and atmospheric science in this course, you will need to refresh your ability to solve simple math problems, including solving simple problems in differential and integral calculus. At the same time, we will remind you about the importance of correctly specifying significant figures and units in your answers to the problems. The goal of this first lesson is to boost your confidence in the math you already know.

How many figures should be in my answer?

Suppose you are asked to solve the following word problem:

In the radar loop, a squall line is oriented in the north-south direction and is heading northeast at 57 km hr^-1. In the last frame of the loop, the line is 17 km west of the Penn State campus. You are out running and know that you can make it back to your apartment in 25 minutes. Will you get back to your apartment before you get soaked?

You reason that the line is moving northeast, and thus, at an angle of 45^o relative to the east. Therefore, the eastward motion of the squall line is just the velocity times the cosine of 45^o. That gives you the eastward speed. You decide to divide the distance by the eastward speed to get the amount of time before the line hits campus. You plug the numbers into your calculator and get the following result:

$\text{time}=\frac{17\text{ km}}{\text{ (}57\text{ km/h)}\cdot \cos \left(\mathrm{45º}\right)}=0.42178\text{ hours}=25.3070\text{ minutes}$

According to your calculation, you will make it back with 0.3 minutes (18 seconds) to spare. But can you really be sure that the squall line will strike in 25.3070 minutes? Maybe you should figure out how many significant figures your answer really has. To do that, you need to remember the rules:

1.3 If you thought practice makes perfect, you could be right.

Calculus is an integral part of a meteorologist’s training. The ability to solve problems with calculus differentiates meteorologists from weather readers. You should know how to perform both indefinite and definite integrals. Brush up on the derivatives for variables raised to powers, logarithms, and exponentials. We will take many derivatives with respect to time and to distance.

Simple Integrals and Derivatives That are Frequently Used to Describe the Behavior of Atmospheric Phenomena

1. 

$$\begin{array}{l}\frac{da}{dt}=-ka\\ \frac{da}{a}=-kdt\\ {\displaystyle \underset{a_o}{\overset{a_1}{\int }}\frac{da}{a}}=-{\displaystyle \underset{t_o}{\overset{t_1}{\int }}kdt}\\ \ln (a_1)-\ln (a_0)=-k\left(t_1-t_0\right)\\ \ln (a_1/a_0)=-k\left(t_1-t_0\right)\\ a_1/a_0=e^{\left(-k\left(t_1-t_0\right)\right)}=\exp \left(-k\left(t_1-t_0\right)\right)\\ a_1=a_0{e}^{\left(-k\left(t_1-t_0\right)\right)}=a_0\exp \left(-k\left(t_1-t_0\right)\right)\end{array}$$

2.  $p=p_o{e}^{\left(-z/H\right)};{\displaystyle \underset{0}{\overset{\infty }{\int }}pdz=\text{? (Do the definite integral.)}}$

$${{\displaystyle \underset{0}{\overset{\infty }{\int }}pdz=-H{p}_o{e}^{-z/H}|}}_0^{\infty }=-H{p}_o\left(0-1\right)=p_{o}H$$

3.  $p=p_0{e}^{\left(-\frac{z}{H}\right)};\frac{1}{p}\frac{dp}{dz}=?$

$$\frac{dp}{dz}=-\frac{1}{H}{p}_0{e}^{\frac{-z}{H}}=-\frac{1}{H}p;\frac{1}{p}\frac{dp}{dz}=-\frac{1}{H}$$

4. $\begin{array}{l}\frac{d\ln (ax)}{dt}=?\frac{d\ln (ax)}{dt}=\frac{1}{ax}\frac{d(ax)}{dt}=\frac{1}{ax}\frac{adx}{dt}=\frac{1}{x}u\text{ , where u = velocity}\\ \end{array}$

5. $d(\cos (x))=?d(\cos (x))=-\sin (x)dx$

You have the power.

Often in meteorology and atmospheric science you will need to manipulate equations that have variables raised to powers. Sometimes, you will need to multiply variables at different powers together and then rearrange your answer to simplify it and make it more useful. In addition, it is very likely that you will need to invert an expression to solve for a variable. The following rules should remind you about powers of variables.

Are you ready to give it a try? Solve the following problem on your own. After arriving at your own answer, click on the link to check your work. Here we go:

1.4 Are you ready to get with the programming?

Meteorologists and atmospheric scientists spend much of their time thinking deep thoughts about the atmosphere, the weather, and weather forecasts. But to really figure out what is happening, they all have to dig into data, solve simple relationships they uncover, and develop new ways to look at the data. Much of this work is now done by programming a computer. Many of you haven’t done any computer programming yet, and for those of you who have, congratulations – put it to good use in this class. For those who are programming novices, we can introduce you to a few of the concepts of programming by getting you to use Excel or another similar spreadsheet program.

To help you learn and retain the concepts and skills that you will learn in this course, you will solve many word problems and simple math problems. For several activities, we give you the opportunity to practice solving particular types of problems enough times until you gain confidence that you can solve those same types of problems on a quiz. That means that you will be solving some types of problems several times and only the numbers for the variables will change. The simplest way for you to solve these problems is to program a spreadsheet to do that repetitive math for you.

Spreadsheet Screenshot 1

Credit: Image developed by W. Brune © Penn State with Excel [7]  2019 and licensed under CC BY-NC-SA 4.0 [8]

Click for a text description of the spreadsheet screenshot.

Screenshot shows an Excel Spreadsheet

A text box says "put activity number in row 1" and an arrow points to cell A1.

A second text box says "put variable names in row 2" with an arrow pointing to cells A2 and B2

A third text box says "start numbers for variables in row 3" with an arrow pointing to cells A3 and B3

A final text box says "calculations follow variable numbers" with an arrow pointing to C3.

Let’s do a simple example. Suppose we have several boxes, some with different shapes and sizes, and we want to calculate the volume of the boxes and find the total volume. I have put in the names of the variables (with units!) and then the numbers for the length, width, and height of each box type and the total number of each box. To calculate the volume of each box, click on E3 and put an “= a3*b3*c3” in the equation line. Hit enter and it will do the calculation and put the answer in E3. A small square will appear in the lower right corner of E3. Click on this square with the mouse and pull down over the next three rows. Excel will automatically do the calculations for those rows. To calculate the total volume, go to F3 and enter “=d3*e3,” and hit “enter.” Grab the small box and pull down to get the total volume of each type of box. To get the total volume, click on F7, click on “Formulas,” and then “AutoSum,” and finally “Sum.” Excel will show you which cells it intends to sum. You can change this by adjusting the edges of the box it shows. Finally, pay attention to significant digits in your calculated volumes. You can adjust the number of decimal places by highlighting a cell or group of cells and then by clicking on the appropriate icon on the Excel tool bar. Note that an incorrect number of significant figures is displayed in some of the answers--see if you can figure out where.

Spreadsheet Screenshot 2

Credit: Image developed by W. Brune © Penn State with Excel [7] 2019 and licensed under CC BY-NC-SA 4.0 [8]

Click for a text description of the spreadsheet example part 1.

length (m)	width (m)	height (m)	number of boxes	box volume (m3)	total volume (m3)
0.72	0.52	0.52	3	=A3*B3*C3
0.52	0.52	0.52	4
1.2	0.85	1.01	1
0.62	0.62	0.95	2

Spreadsheet Screenshot 3

Credit: Image developed by W. Brune © Penn State with Excel [7] 2019 and licensed under CC BY-NC-SA 4.0 [8]

Click for a text description of the spreadsheet example part 2.

length (m)	width (m)	height (m)	number of boxes	box volume (m3)	total volume (m3)
0.72	0.52	0.52	3	0.19	0.58
0.52	0.52	0.52	4	0.14	0.56
1.2	0.85	1.01	1	1.03	1.03
0.62	0.62	0.95	2	0.37	0.73

Hopefully this example is a refresher for most of you. For those who are totally unfamiliar with Excel, please click on the question mark in the upper right of the screen and type in the box “creating your first workbook.” You can also visit Microsoft's help page for additional step-by-step instructions for how to Use Excel as Your Calculator [9]. The best way to learn, after the introduction, is by doing. The Keynote Support website [10] also lists helpful summaries of instructions.

Summary

There is a very good reason that you are taking this class and I am teaching it – all of us are fascinated by the weather, awed by the atmosphere’s power, and passionate about learning more about it. Quite honestly, I can’t imagine a more rewarding career than the one that you are embarking upon or the one that I have. Nothing could be more rewarding than saving lives by making the atmosphere more predictable or by making the perfect prediction. Nothing.

But, do you know what? The best forecasters are the ones who can not only read weather maps, but who also know physically what the atmosphere is doing. The best forecasters know how to translate the physics into mathematics so that hand-waving can be turned into usable numbers. This course will start to make all of these connections between observations and physical cause-and-effect and help us find numerical solutions to questions.

For those of you who are in related disciplines, this course will give you a solid basic understanding of the atmosphere that you can apply in your studies and career, whether it be civil engineering, mechanical engineering, environmental engineering, chemistry, hydrology, or many other fields. 

We have now reviewed some important concepts like significant figures and dimensions and units. You will continue to gain confidence in using the differential and integral calculus that you already know. As you go through the course, I want you to look back at the pictures of the atmosphere and imagine which equations are governing the processes that are causing your observations.

Reminder - Complete all of the Lesson 1 tasks!

You have reached the end of Lesson 1! Make sure that you have completed all of the tasks in Canvas.

Overview

The atmosphere is composed of billions of billions of molecules, 10⁴³ 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 mv²), momentum (mv), 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.

Learning Objectives

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

Questions?

If you have any questions, please post them to the Course Questions discussion forum. 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.

2.1 Gas Laws

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.

A constant pressure balloon stays aloft for weeks at an altitude of 100,000 ft so that the instruments in the attached gondola can make long-term measurements.

Credit: Credit: Super Pressure Balloon. National Aeronautics and Space Administration [11] (NASA) (Public Domain)

Ideal Gas Law

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

where p is pressure (Pa = kg m^–1 s^–2), V is the volume (m³), 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 m² s^–2).

Recall that a mole is 6.02 x 10²³ molecules (Avogadro’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:

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:

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 mole fraction of each type of molecule times the molar mass of each molecular constituent:

where the i subscript represents atmospheric components, N is the number of moles, M is the molar mass, and f is the mole fraction. 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.

Dry Air

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 (N₂ + O₂ + Ar + CO₂ +…) or water vapor (gaseous H₂O). We can divide R^* by M_i to get a mass-specific gas constant, such as R_d = R*/M_{dry air}.

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

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^–1, which is the average of the molar masses of the gases in a dry atmosphere computed to four significant figures.

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).

Water Vapor

We can do the same procedure for water vapor:

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.

Dalton’s Law

John Dalton

Credit: Dalton John desk [14] by H. Roscoe, W.H. Worthington, and J. Allen from Wikimedia [15] Public Domain [16] 

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:

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 total pressure, which includes the partial pressures of dry air and water vapor.

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

Virtual Temperature

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:

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:

where T and T_v have units of Kelvin (not ^oC and certainly not ^oF!) and q must be unitless (e.g., kg kg^–1).

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.

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

• not converting from ^oC 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 T_v for T. Thus,

$${\rho }_{}=\frac{p_{}}{R_d{T}_v}$$

2.2 The Atmosphere’s Pressure Structure: Hydrostatic Equilibrium

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?

Air parcel at rest with three forces in balance

The atmosphere’s basic pressure structure is determined by the hydrostatic balance of forces [17]. 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:

1. 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 [18],

   $$F_{top}=-p_{top}A$$

   [2.11]
2. Similarly, the force on the volume element from the pressure of the fluid below pushing upwards (positive) is:

   $$F_{bottom}=p_{bottom}A$$

   [2.12]
3. Finally, the weight [19] of the volume element causes a force downwards. If the density [20] is ρ, the volume is V, which is simply the horizontal area A times the vertical height, Δz, and g the standard gravity [21], then: 

   $$F_{weight}=-\rho Vg=-\rho gA\Delta z$$

   [2.13]

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

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

or:

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 [22] small, the equation can be written in differential [23] form, where dp is top pressure minus bottom pressure just as dz is top altitude minus bottom altitude.

The result is the equation:

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:

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,

$$p= p_o{e}^{-\raisebox{1ex}{$z$}\!\left/ \!\raisebox{-1ex}{$H$}\right.} where p_o=surface pressure and H= \frac{R^{*}\overline{T}}{M_{air}g}$$

H is called a scale height because when z = H, we have p = p_oe^–1. If we use an average T of 250 K, with M_air = 0.029 kg mol^–1, then H = 7.3 km. The pressure at this height is about 360 hPa, close to the 300 mb surface that you have seen on the weather maps. Of course the forces are not always in hydrostatic balance and the pressure depends on temperature, thus the pressure changes from one location to another on a constant height surface.

From equation 2.20, the atmospheric pressure falls off exponentially with height at a rate given by the scale height. Thus, for every 7 km increase in altitude, the pressure drops by about 2/3. 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.

Scattered light near Earth's surface.

Credit: NASA

2.3 First Law of Thermodynamics

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.

Floating molecules

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:

The dimensions of energy are M L² T^–2 so the dimensions of this equation are M L² 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 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:

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:

Heat Capacity

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.

Heat Capacity at Constant Volume

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.

Heating a box with rigid walls.

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:

So,

C_V, the constant relating Q to temperature change, is called the heat capacity at constant volume. Heat capacity has units of J K^-1.

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 as:

where c_V is called the specific heat capacity. The adjective “specific” means the amount of something per unit mass. The greater the heat capacity, the smaller the temperature change for a given amount of heating.

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

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

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:

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

Heat Capacity Constant Pressure

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.

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:

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

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 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:

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

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

where C_p is the heat capacity at constant pressure and c_p is the specific heat capacity at constant pressure.

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 c_v? You will see the derivation of the relationship, but I will just present the results:

• 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: $c_{pv}=c_{Vv}+R_v$

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.

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

2.4 The higher the temperature, the thicker the layer

A surprising way to relate the distance between two pressure surfaces to the temperature of the layer between them.

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.

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:

Integrate both sides:

or

where T is the average temperature of the layer between p₁ and p₂. So, the thickness is actually a measure of the average temperature in the layer.

2.5 Adiabatic Processes: The Path of Least Resistance

Adiabatic Process

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:

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.

Changes in volume and temperature as an air parcel ascends and the pressure decreases.

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.

Derivation of the Poisson Relations

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 (Equation 2.41) and then assume an adiabatic process (q = 0):

Divide both sides by T:

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

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

Divide by c_p:

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

We can rewrite R_d/c_p as a new term denoted by the Greek letter gamma, γ

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

Potential Temperature

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:

or

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).

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.

Potential temperature (solid lines, K) as a function of latitude and altitude. Note that the decrease in potential temperature with height is small in the troposphere and large in the stratosphere.

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

Dry Adiabatic Lapse Rate

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:

Take the derivative w.r.t. z:

But we also know from the hydrostatic equation that:

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

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.

2.6 Stability and Buoyancy

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.

Stability

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?

Examples of instability (left) and stability (right).

Credit: W. Brune, after H. 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.

Determining instability (left) and stability (right) of an air parcel by moving it on the DALR slope and seeing if the parcel’s temperature is greater or less than the environmental temperature.

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.

An atmospheric temperature profile (red) and the movement of individual air parcels (numbered circles) along dry adiabats (dashed lines).

Credit: W. Brune

Buoyancy

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).

Forces on an air parcel that has a different density from its environment.

Credit: W. Brune

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:

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:

If B > 0, then the parcel accelerates upwards; if B < 0, then the parcel accelerates downwards.

We look at the instability at each point in the environmental temperature profile and can determine Γ_env for each point.

Thus,

so that:

If Γ_env < Γ_d, the parcel accelerates downward for positive Δz (statically stable environment).

If Γ_env > Γ_d, the parcel accelerates upward for positive Δz (statically unstable environment).

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

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:

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.

Layer stability for different profiles of 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.

Summary and Final Tasks

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.

Reminder - Complete all of the Lesson 2 tasks!

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.

Overview

The atmosphere’s most abundant chemicals are molecular nitrogen (N₂), molecular oxygen (O₂), and Argon (Ar). These are all only in the gas phase. Water vapor, the next most abundant, can exist as vapor, liquid, or solid. The phase changes of water have a major role in weather and in climate. In the atmosphere, water is always trying to achieve a balance between evaporation and condensation while never really succeeding. In this lesson, you will discover the conditions under which the phases of water are in balance and will see that they depend on only two quantities—the amount of water and the temperature. Equilibrium conditions, often called saturation, are expressed mathematically by the Clausius–Clapeyron Equation. We will see that phase changes of water create weather, including severe weather, and that we can use the 1^st Law of Thermodynamics to do many calculations involving situations where there are phase and temperature changes. Combining the Clausius–Clapeyron equation with the equations of thermodynamics, we can construct a diagram called the skew-T. The skew-T is useful in helping us understand both the atmosphere’s temperature structure and the location and behavior of clouds.

Learning Objectives

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

• differentiate among the different ways that moisture can be expressed and choose the correct one for finding an answer
• explain the meaning of the lines and spaces on a water vapor phase diagram
• calculate relative humidity using the Clausius–Clapeyron Equation
• solve energy problems related to temperature and phase changes
• demonstrate proficiency with using the skew-T to find the lifting condensation level (LCL), potential temperature, relative humidity, wetbulb temperature, dry and moist adiabats, and equivalent potential temperature

Questions?

If you have any questions, please post them to the Course Questions discussion forum. 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.

3.1 Ways to Specify Water Vapor

Student using a sling psychrometer to measure the dewpoint temperature, which is one way to specify atmospheric water vapor.

Credit: W. Brune

Up to now, we have dealt with water vapor only as the specific humidity in order to determine the virtual temperature. But there are many ways for us to quantify the amount of water vapor in the atmosphere. The most common are specific humidity, water vapor mixing ratio, relative humidity, and dewpoint temperature.

Specific humidity (q) is the density of water vapor (mass per unit volume) divided by the density of all air, including the water vapor:

We have already seen that specific humidity is used to calculate virtual temperature. Specific humidity is unitless, but often we put it in g kg^–1.

Water vapor mixing ratio (w) is the density of water vapor divided by the density of dry air without the water vapor:

Water vapor mixing ratio is widely used to calculate the amount of water vapor. It is also the quantity used on the skew-T diagram, which we will discuss later in this lesson. Water vapor mixing ratio is unitless, but often we put it in g kg^–1.

Since ρ_d = ρ – ρ_v we can rearrange the equations to get the relationship between w (water vapor mixing ratio) and q (specific humidity):

The water vapor mixing ratio, w, is typically at most about 40 g kg^–1 or 0.04 kg kg^–1, so even for this much water vapor, q = 0.040/(1 + 0.040) = 0.038 or 38 g kg^–1.

Thus, water vapor mixing ratio and specific humidity are the same to within a few percent. But specific humidity is less than the water vapor mixing ratio if the humidity is more than zero.

Here is one example of global specific humidity.

Specific humidity (q) for December 2014. The color scale goes from 0 at blue to 30 g kg^–1 for pink. Green is about 10 g kg^–1 and red is about 20 g kg^–1.

Credit: IRI [26]

The greatest absolute specific humidity is in the tropics with maximum values approaching 30 g kg^–1. The smallest values are at the high latitudes and are close to zero. Why is specific humidity distributed over the globe in this way?

Relative humidity (RH) is another measure of water vapor in the atmosphere, although we must be careful when using it because a low relative humidity may not mean a low water vapor mixing ratio (i.e., at high temperatures) and a high relative humidity might still be quite dry air (i.e., at low temperatures).

According to the World Meteorological Organization (WMO) definition,

where w_s is the saturation mixing ratio (the mixing ratio at which RH = 100%). w and w_s can both have units of g kg^–1 or kg kg^–1, as long as they are consistent. Relative humidity is usually expressed as a percent. Thus, when w = w_s, RH = 1 = 100%. In most problems involving RH, it is important to keep in mind conversions between decimal fractions and percent.

A more physically based definition of the relative amount of moisture in the air is the saturation ratio, S:

where e is the vapor pressure and e_s is the saturation vapor pressure. The saturation ratio is used extensively in cloud physics (Lesson 5). To see how RH and S are related, start with the Ideal Gas Law and then do some algebra:

where ε = 0.622 is just the molar mass of water (18.02 kg mol^–1) divided by the mass of dry air (28.97 kg mol^–1). e and e_s are typically less than 7% of p, and since e is usually 20%–80% of e_s, the difference between the two definitions is usually less than a few percent.

Note that at saturation, you can replace w with w_s and e with e_s in the equation that relates w to e.

Some processes depend upon the absolute amount of water vapor, which is given by the specific humidity, water vapor mixing ratio, and water vapor pressure, and other processes depend on the relative humidity. For example, the density of a moist air parcel depends on the absolute amount of water vapor. So does the absorption and emission of infrared atmospheric radiation. On the other hand, cloud formation depends on the relative humidity, although the cloud might be kind of wimpy if the absolute humidity is small.

One of the most common indicators of absolute humidity is the dewpoint temperature. We will postpone the discussion of it until after we learn about the relationship between temperature and saturation vapor pressure, e_s.

3.2 Condensation and Evaporation

What is vapor pressure? Because of the Ideal Gas Law (Equation 2.1), we can think of vapor pressure e (SI units = hPa or Pa) as being related to the concentration of water vapor molecules in the atmosphere,

where n is the number of moles per unit volume (n = N/V).

What makes liquid water different from ice or water vapor? It is actually the weak bonds between water molecules that are called hydrogen bonds. These bonds are 20 times weaker than the bonds between hydrogen and oxygen in the same molecule and can be broken by collisions with other molecules if they are traveling fast enough and have enough kinetic energy to break the bonds. So the differences between vapor, liquid, and ice are related to the number of hydrogen bonds. In vapor, there are essentially no hydrogen bonds between molecules. In ice, each water molecule is hydrogen bonded to four other water molecules. And in liquid, only some of those hydrogen bonds are made and they are constantly changing as the water molecules and clusters of water molecules bump into and slide past each other.

Water vapor condensed into liquid drops on a spider web overnight. As the sun rises and the air heats up, there will be net evaporation and the beads will shrink and disappear during the day.

Credit: Spider web with dew drops by devra [27] from Flickr [28] is licensed under CC BY 2.0 [29].

Think about a liquid water surface on a molecular scale. What is happening all the time is that some water molecules in the gas phase are hitting the surface and sticking (i.e., making hydrogen bonds), while at the same time other water molecules are breaking free from the hydrogen bonds that tie them to other molecules in the liquid and are becoming water vapor. The water vapor surface is like a Starbucks, but even busier. We can easily calculate the flux of molecules that are hitting the surface using simple physical principles, although it is harder to calculate the number that are leaving the liquid. Both are happening all the time, although usually the amount of condensation and evaporation aren't the same, so that we usually have net evaporation or net condensation.

In equilibrium, the flux of molecules leaving the surface exactly balances the flux of molecules that are hitting the surface. This condition is called equilibrium, or saturation. We can show that:

Thus, when S = 1, e = e_s, RH is approximately 100%, and w is approximately w_s. Condensation and evaporation are in balance. These two processes are going on all the time, but sometimes there can be more evaporation than condensation, or more condensation than evaporation, or evaporation equaling condensation. However, water is always trying to come into equilibrium.

So we know that the amount of water in vapor phase determines the condensation rate and thus e. So what determines e_s? We will see next that e_s depends on only one variable: temperature!

3.3 Phase Diagram for Water Vapor: Clausius–Clapeyron Equation

The Clausius–Clapeyron Equation

We can derive the equation for e_s using two concepts you may have heard of and will learn about later: entropy and Gibbs free energy, which we will not go into here. Instead, we will quote the result, which is called the Clausius–Clapeyron Equation,

where l_v is the enthalpy of vaporization (often called the latent heat of vaporization, about 2.5 x 10⁶ J kg^–1), R_v is the gas constant for water vapor (461.5 J kg^–1 K^–1), and T is the absolute temperature. The enthalpy of vaporization (i.e., latent heat of vaporization) is just the amount of energy required to evaporate a certain mass of liquid water.

What is the physical meaning? The right-hand side of [3.9] is always positive, which means that the saturation vapor pressure always increases with temperature (i.e., de_s/dT > 0). This positive slope makes sense because we know that as water temperature goes up, evaporation is faster (because water molecules have more energy and thus a greater chance to break the bonds that hold them to other water molecules in a liquid or in ice). At saturation, condensation equals evaporation, and since evaporation is greater, condensation must be greater as well. Much of the higher condensation comes from having more water vapor molecules hitting the liquid surface, which according to the Ideal Gas Law, means that the water vapor pressure is higher.

The temperature sensitivity of e_s is quite high. Plugging in the appropriate values to the right side of [3.9] yields 0.07 K^–1, which means that the saturation vapor pressure increases by 7% for every 1 K increase in temperature. This high sensitivity has profound implications for weather and climate.

Separating variables (e_s and T) in [3.9] and integrating, assuming that l_v is a constant with temperature (it is not quite constant!), yields:

Generally T_o is taken to be 273 K and e_so is then 6.11 hPa.

What does the plot of this equation look like?

The equilibrium vapor pressure above a liquid water surface as calculated by the Clausius–Clapeyron Equation. ***The y-axis label is incorrect; it should say "equilibrium vapor pressure (hPa)".*** The line for liquid water can be extended below 273 K, the freezing point, because water can remain liquid at those low temperatures and become a “supercooled” liquid.

What happens between vapor and ice? The same methods can be applied and the same basic equations are obtained, except with a different constant:

where e_si is the saturation vapor pressure for the ice vapor equilibrium and l_s is the enthalpy of sublimation (direct exchange between solid water and vapor = 2.834 x 10⁶ J kg^–1).

Equations for e_s and e_si that account for variations with temperature of l_v and l_s, respectively, can be found in Bohren and Albrecht (Atmospheric Thermodynamics, Oxford University Press, New York, 1998, ISBN 0-19-509904-4):

Dewpoint Temperature as a Measure of Water Vapor

Simply put, the dewpoint temperature is the temperature at which the atmosphere’s water vapor would be saturated. It is always less than or equal to the actual temperature. Mathematically,

which means that the water vapor pressure at some temperature T (not multiplied by T) equals the water vapor saturation pressure at the dewpoint temperature, T_d. So we see that because w_s depends only on T_d at a given pressure, T_d is a good method for designating the absolute amount of water vapor.

The Phase Diagram for Water

We can draw the phase diagram for water. There are three equilibrium lines that meet at the triple point, where all three phases exist (e_s = 6.1 hPa; T = 273.14 K). Along the line for e_s, vapor and liquid are in equilibrium, and evaporation balances condensation. Along the line for e_si, vapor and ice are in equilibrium and sublimation equals deposition. Along the line for e_sm, liquid and ice are in equilibrium and melting balances fusion.

Phase diagram for water for most water pressures and temperatures that are relevant to the atmosphere. e_s is the saturation pressure (i.e., equilibrium pressure) between liquid and vapor and e_si  is the saturation pressure between vapor and ice. The dashed line is a continuation of e_s into the ice phase and represents supercooled water. All three phases meet at the triple point. This diagram is not to scale.

Credit: W. Brune

Is it possible to have water in just one phase? Yes!

The simplest case is when all the water is vapor, which occurs when the water vapor pressure is low enough and the temperature (and thus saturation vapor pressure) is high enough that all the water in the system is evaporated and in the vapor phase.

Let’s think about what it would take to have all the water in the liquid phase. Suppose we have a vertical cylinder closed on one end and a sealed piston at the other end. The whole cylinder is immersed in a constant-temperature water bath so that we can hold the cylinder and its contents at a fixed temperature (i.e., isothermal). Initially we fill the cylinder with liquid water and have a small volume of pure water vapor at the top. If we set the bath temperature to, say, 280 K and let the system sit for a while, the vapor will become saturated, which is on the e_s line. For isothermal compression, in which energy is removed from the system by the bath in order to keep the temperature constant, a push on the piston will slightly raise the vapor pressure above e_s and there will be net condensation until equilibrium is obtained again. If we continue to slowly push in the piston, eventually all the cylinder’s volume will be filled with liquid water and the cylinder will contain only one phase: liquid. If we continue to push the piston and the bath keeps the temperature constant, then the water pressure will increase.

In the atmosphere, ice or liquid almost always has a surface that is exposed to the atmosphere and thus there is the possibility that water can sublimate or evaporate into this large volume. Note that the presence or absence of dry air has little effect on the condensation and evaporation of water, so it is not the presence of air that is important, but instead, it is the large volume for water vapor that is important.

Conditions can exist in the atmosphere for which the water pressure and temperature are in the liquid or sometimes solid part of the phase diagram. But these conditions are unstable and there will be condensation or deposition until the condensation and evaporation or sublimation and deposition come into equilibrium, just as in the case of the piston above. Thus, more water will go into the liquid or ice phase so that the water vapor pressure drops down to the saturation value. When the water pressure increases at a given temperature to put the system into the liquid region of the water phase diagram, the water vapor is said to be supersaturated. This condition will not last long, but it is essential in cloud formation, as we will see in the lesson on cloud physics.

Note also that the equilibrium line for ice and vapor lies below the equilibrium line for supercooled liquid and vapor for every temperature. Thus e_si < e_s for every temperature below 0 ^oC because l_s > l_v in the Clausius–Clapeyron Equation. This small difference between e_si and e_s can be very important in clouds, as we will also see in the lesson on cloud physics.

3.4 Solving Energy Problems Involving Phase Changes and Temperature Changes

When a cloud drop evaporates, the energy to evaporate it must come from somewhere because energy is conserved according to the 1^st Law of Thermodynamics. It can come from some external source, such as the sun, from chemical reactions, or from the air, which loses some energy and thus cools. Thus, temperature changes and phase changes are related, although we can think of phase changes as occurring at a constant temperature. The energy associated with phase changes drives much of our weather, especially our severe weather, such as hurricanes and deep convection. We can quantify the temperature changes that result from phase changes if we have a little information on the mass of the air and the mass and phases of the water.

In the previous lesson, we said that all changes of internal energy were associated with a temperature change. But the phase changes of water represent another way to change the energy of a system that contains the phase-shifter water. So often we need to consider both temperature change and phase change when we are trying to figure out what happens with heating or cooling.

For atmospheric processes, we saw that we must use the specific heat at constant pressure to figure out what the temperature change is when an air mass is heated or cooled. Thus the heating equals the temperature change times the specific heat capacity at constant pressure times the mass of the air. For dry air, we designate the specific heat at constant pressure as c_pd. For water vapor, we designate the specific heat at constant pressure as c_pv. So for example, the energy required to change temperature for a dry air parcel is c_pd m ΔT = c_pd ρV ΔT, where c_pd is the specific heat capacity for dry air at constant pressure. If we have moist air, then we need to know the mass of dry air and the mass of water vapor, calculate the heat capacity of each of them, and then add those heat capacities together.

For liquids and solids, the specific heat at constant volume and the specific heat at constant pressure are about the same, so we have only one for liquid water (c_w) and one for ice (c_i).

For phase changes, there is no temperature change. Phase changes occur at a constant temperature. So to figure out the energy that must be added or removed to cause a phase change, we only need to know what the phase change is (melting/freezing, sublimating/depositing, evaporating/condensing) and the mass of water that is changing phase. So, for example, the energy needed to melt ice is l_f m_ice.

Icicles melting. The energy for the phase change from ice to liquid water comes from the air, which must be warmer than freezing.

Credit: double drops [30] by liz west from Flickr [28] is licensed under CC BY 2.0 [29].

The following tables provide numbers and summarize all the possible processes involving dry air and water in its three forms.