Notes States of Matter _ Gasses

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Notes States of Matter _ Gasses Powered By Docstoc
					                 Notes: States of Matter & Gas Laws
There are three common states of matter: solid, liquid and gas. Each state has some general traits which
all types of matter in this state follow.

Phases Characteristics:
          Density                 Shape                    Thermal Expansion             Compressible
 Solid     High                    fixed                       Very Little                Very Little
Liquid    Medium        takes shape of the container           Very Little                Very Little
  Gas   VERY Low        takes shape of the container             Lots                     Very Easily

What determines a substance’s state of matter? There are three factors. The first to come to mind is
probably temperature. If you heat something eventually it melts or boils. The next is not so obvious, the
pressure. For example, with a change in atmospheric pressure water boils at a lower temperature in
Denver than it does in San Francisco. Probably the most important factor is what substance we are
talking about. Water boils at a lot lower temperature than titanium.

Phase Diagram:
The phase diagram is a graphical tool used to determine what state of matter a substance will be in a given
temperature and pressure.

The phase diagram is a series of lines generated from actual experimental data. In other words, someone
took a substance and checked its state at many different combinations of temperature and pressure. They
checked the state when the substance was at 5C and at 760 mmHg, then at 5C and 761 mmHg, then at
5C and 762 mmHg and so on. As you can see it would take a while to check all the temperature/pressure
combinations there are for any given substance.

The lines on the phase diagram divide the three phases. If you cross a line, a physical process has
Generally, heating a solid will cause it to melt to a liquid (at the melting or freezing point). Continued
heating (to the boiling point) produces a gas. For most substances, at higher pressures, the freezing and
boiling points are higher.

There are two specific areas of interest, the triple point and the critical temperature. The triple point
occurs at a specific temperature and pressure at which all three phases exist in equilibrium. When
observing a substance at its triple point it appears to be boiling freezing and melting all at the same time.
Once above the critical temperature, molecules are unable to liquefy. Meaning no matter how much
pressure you apply, the molecules will remain a gas and will not reform a liquid until the temperature is
dropped. The molecules are simply moving too fast.

Phase Diagram for Water:

The same diagram for water shows the characteristic negative slope on the solid-liquid equilibrium line,
which accounts for why water floats. The manner in which water forms its crystalline structure causes it
to expand. Hence the liquid state is the most dense state. This causes ice to float.

Looking at the graph upward from the temperature axis one notices that increasing the pressure on a solid
will change it to a liquid, the opposite effect of what occurs on the first figure. When a hockey player
steps on to the ice, they apply a pressure to the ice and the ice melts. So, hockey players are not actually
skating on ice, but a thin layer of water they just melted by standing on that ice.
Units of Gasses:
   abbreviation                       name                         normal atmospheric pressure
        torr                        Torricelli                                760
      mmHg                   millimeters of mercury                           760
        psi                  pounds per square inch                           14.7
       Atm                        atmospheres                                  1
        Bar                            Bar                                    1.01
        kPa                        kilopascal                               101.325
       inHg                    inches of mercury                              29.9
      ft H2O                      feet of water                                33

Nature of Gas Pressure:
The pressure of a gas is a measure of force applied over a given area. For example, pounds per square
inch. The force being applied is due to the impact of gas particles on some object.

Right this second, as you read these notes, billions of gas particles – N2, O2, CO2, Ar, H2O – are slamming
into your eyes. Yes, your eyes and all of the rest of your body. The air is acting in the same fashion that
water does when you submerge yourself. You feel the water pressing you as you dive deeper. You feel it
on your eardrums. Just as you feel it when you come back down a mountain, say Mt. Hood.

Why do your ears hurt or feel uncomfortable when traveling up and down mountains? There is air
pressing on your eardrums, so why are your
eardrums not constantly hurting? Because your
body equalizes the pressure. Normal atmospheric
pressure is 14.7 psi. This means that the air is
pressing on everything with a pressure of 14.7 psi.
Your eardrums are no exception. Your body has
countered this by applying a pressure of 14.7 psi in
the inside of your eardrums. This results in no net
pressure and your eardrum rests happily undisturbed.
But when you drive up a mountain you get an
uncomfortable feeling, maybe even painful, why?
The pressure in your head is now greater than that
outside your head and your eardrum is pressing
outward. This stress to the drum is uncomfortable,
for some it is even incapacitating. The "popping" of
your ears is your head equalizing the pressure, which
as you travel up is some value less than 14.7 psi. As
you travel back down the mountain your drums will
be pressed in until your head can re-equalize the pressure.
Evangelista Torricelli's Barometer:
The barometer picture here was invented by Evangelista Torricelli.
This was one of the first methods used to measure atmospheric
pressure, and it is still used today. Today, most instruments named
barometers are not this large and contain no mercury, they work on
other principals of physics. But the name has been applied to all
instruments used to measure atmospheric pressure.

The apparatus is set up by taking a tube filled with Hg, placing your
thumb over the top, inverting the tube and submerging the open end
in another container, a dish, also filled with Hg. The Hg in the tube
falls, leaving a vacuum at the top of the tube. But not all of the Hg
drains out of the tube. What force is holding this dense liquid up
against gravity’s pull? Air pressure. Normal atmospheric pressure
will hold a column of Hg 760 mm tall. This is one of our units listed
above for atmospheric pressure; it is equivalent to the Torr,
obviously named after Torricelli.

Gas Laws:
There are several laws used to describe the behavior of gasses. What follows are descriptions of the most
important. There about 8 laws we will discuss. These laws we will be discussing pressure, temperature,
volume and moles or numbers of gas particles.

When doing these problems, keep in mind that there are only 4 parameters to consider: P, T, V, and n.
For many of these problems you will be keeping two of the four parameters steady as you modify one of
these four and calculate that modification of the fourth parameter.

Boyle's Law:
This law defines the relationship between pressure and volume if temperature and amount of gas is held
constant. If the volume of a container is increased, the pressure decreases. If the volume of a container is
decreased, the pressure increases. The law is described by the following equation:

                                                P1V1 = P2V2

Example: A sample of gas is in a 2.00 L contain at a pressure of 740.0 mmHg. What is the new pressure
of the sample if the container’s volume is reduced to 1.25 L?

Answer: This problem is solved by inserting values into the given equation:

                                  (740.0 mmHg) (2.00 L) =(X) (1.25 L)

Solving for X will give you a new pressure of 1184 mmHg.
Charles's Law:
This law defines the relationship between volume and temperature if pressure and amount of particles are
held constant. If the temperature of a gas is increased, the volume of the gas will increase. If the
temperature of a gas is decreased, the volume of the gas will decrease. This is a direct relationship. One
goes down, so does the other.

Picture the gas particles flying around inside a balloon. If you were to put the balloon in the freezer, the
gas particles would slow down, therefore they would not hit the balloon walls as hard and the balloon
would shrink in size.

                                          V1 V2
                                               or V1T2 = V2T1
                                          T1 T2

Example: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard

Answer: Convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our
equation like this:

                                               2.85L   V2
                                               298K 273K

Solving for the new volume gives a value of 2.61 liters. The volume has decreased as the temperature has

Gay-Lussac's Law:
This law characterizes the relationship between pressure and temperature when volume and amount are
held constant. If the temperature of a container is increased, the pressure increases. If the temperature of
a container is decreased, the pressure decreases. This is another example of a direct relationship. One
goes up, so does the other.

Think about this law in this manner, if the gas particles are moving faster, as happens when the
temperature of a gas is increased, the force of the impact will increase. Therefore, increasing the
temperature will increase the pressure exerted by a gas.

Example: 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature
(in Celsius) to change the pressure to standard pressure?

Answer: Change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values
into the equation and get:

The answer is 311.3 K, but the question asks for Celsius, so you subtract 273 to get the final answer of
38.3°C, but then you knew that. Right?
Avogadro's Law:
This law gives the relationship between volume and number of gas particles when pressure and
temperature are held constant. Remember the number is measured in moles. If the amount of gas in a
container is increased, the volume increases. If the amount of gas in a container is decreased, the volume
decreases. Another direct relationship.

The volume of a container holding a gas will increase with increasing numbers of gas particles because
there are more particles impacting the wall of the container.

Example: A 5.00 L sample of a gas is known to contain 0.965 mol. If the amount of gas in this container
is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

                                               V1n2 = V2n1
                                  (5.00 L) (1.80 mol) = (x) (0.965 mol)

Combined Gas Law:
To derive the Combined Gas Law, follow these steps:
Step 1: Write Boyle's Law
                                            P1V1 = P2V2

Step 2: Multiply by Charles’s Law
V1T2 = V2T1

                                          P1V12 / T1 = P2V22 / T2

Step 3: Multiply by Gay-Lussac's Law
P1T2 = P2T1
                                        P12V12 / T12 = P22V22 / T22

Step 4: Take the square root to get the combined gas law:

                                           P1V1 / T1 = P2V2 / T2

As a side note:
If we include Avogadro’s Law the following equation is generated:

                                        P1V1 / n1T1 = P2V2 / n2T2
A 2.00 L sample of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP?

You have to recognize that five values are given in the problem and the sixth is the only unknown. Also,
remember to change the Celsius temperatures to Kelvin.

When problems like this are solved it is very helpful to write out all the variables in the equation as shown

Next fill in the data given in the problem. Here is the right-hand side filled in with the STP values:

You can be pretty sure that the term "STP" will appear in these types of problems. I recommend you
memorize these standard conditions.

Here are all the given values:

Insert the values in their proper places in the combined gas law equation:

                                            P1V1 / T1 = P2V2 / T2

and solve for x.

PV = nRT: The Ideal Gas Law:
The derivation of this law is a lot of math. So, I will just give you the equation and examples of how to
use it.
                                             PV = nRT
The Numerical Value for R:
R's value can be determined many ways. This is just one way:

We will assume we have 1.000 mol of a gas at STP. The volume of this amount of gas under the
conditions of STP is known to a high degree of precision. We will use the value of 22.414 L.
By the way, 22.414 L at STP has a name. It is called molar volume. It is the volume of ANY ideal gas at
standard temperature and pressure. As far as you are concerned, all the gasses we discuss will behave as
ideal gasses. So, if you have a sample of gas containing 6.022 x 1023 gas particle, this sample of gas will
have a volume of 22.414 L, or about 5 gallons. Think about an object that is 5 gallons. I picture an office
water cooler bottle, or a one of those LARGE buckets of paint.

Let's plug our numbers into the equation:

                           (1.000 atm) (22.414 L) = (1.000 mol) (R) (273.15 K)

Notice how atmospheres were used as well as the exact value for standard temperature.

Solving for R gives 0.08206 L atm / mol K, when rounded to four significant figures. This is usually
enough. Remember the value. You'll need it for problem solving.

Notice the weird unit on R. Say out loud "liter atmospheres per mole Kelvin."

This is not the only value of R that can exist. It depends on which units you select. Those of you that take
more chemistry or physics will most likely meet up with 8.3145 Joules per mole Kelvin, but that's for
another course. We will only use the 0.08206 value in gas-related problems.

Example: A sample of gas with a mass of 2.1025 grams is found to occupy a volume of 2.850 L at
22.0°C at a pressure of 740.0 mmHg. How many moles of the gas are present?

Notice that the units for pressure MUST be in atm., so the 740.0 mm Hg must be converted first.

740.0 mm Hg ÷ 760.0 mm Hg/atm = 0.9737 atm

However, the unrounded-off value should be used in the calculation just below.

Now, plug into the equation:

(0.9737 atm) (2.850 L) = (n) (0.08206 L atm / mol K) (295.0 K)

and solve for n = 0.115 mol

Example: Using the problem above, what is the molar mass of the gas?

This is a very common use of this law and the odds are very good you will see this type of question on a

The key is to remember the units on molar mass: grams per mole. We know from the problem statement
that 2.1025 grams of the gas is involved and we also know how many moles that is.

We know that from doing the calculation above and getting 0.1146 mol.
So all we have to do is divide the grams of gas by how many moles it is:

2.1025 g ÷ 0.1146 mol = 18.34 g/mol

With a molar mass of 18.34 g per mol can you make an educated guess as to what gas this might be?
Let's go over those steps for using the Ideal Gas Law to calculate the molar mass of the gas:
1. You have to know the grams of gas involved. Usually the problem will just give you the value, but
    not always. You might have to calculate it.
2. You are going to have to calculate the moles of gas. Use PV = nRT and solve for n. Make sure to use
    L, atm and K.
3. Divide grams by moles and there's your answer.

Dalton's Law of Partial Pressures:
For any pure gas (let's use helium) PV = nRT holds true. Therefore, P is directly proportional to n if V
and T remain constant. As n goes up, so would P. Or the reverse.

Suppose you were to double the moles of helium gas present. What would happen?

Answer: the gas pressure doubles.

However, suppose the new quantity of gas added was a DIFFERENT gas. Suppose that, instead of
helium, you added neon.

What would happen to the pressure?

Answer: the pressure doubles, same as before.

Dalton's Law immediately follows from this example since each gas is causing 50% of the pressure.
Summing their two pressures gives the total pressure.

Written as an equation, it looks like this:
                                               PHe + PNe = Ptotal

Dalton's Law of Partial Pressures: Each gas in a mixture creates pressure as if the other gases were not
present. The total pressure is the sum of the pressures created by the gases in the mixture. Ptotal = P1 + P2
+ P3 + .... + Pn

Where n is the total number of gases in the mixture.

The only necessity is that the two gases do not interact in some chemical fashion, such as reacting with
each other.

The pressure each gas exerts in mixture is called its partial pressure.

A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three
gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container?

PT = PO2 + PCO2 + PHe

PT = 2 atm + 3 atm + 4 atm = 9 atm
Graham's Law:
Consider samples of two different gases at the same Kelvin temperature.

Since temperature is proportional to the kinetic energy of the gas molecules, the kinetic energy (KE) of
the two gas samples is also the same.

In equation form, we can write: KE1 = KE2

Since KE = (1/2) mv2, (m = mass and v = velocity) we can write the following equation:

m1v12 = m2v22

Note that the value of one-half cancels out.

The equation above can be rearranged algebraically into the following:

The square root of (m1 / m2) = v2 / v1

You may wish to assure yourself of the correctness of this rearrangement.

Another way you may see this written is:

                                               Rate1    MM 2
                                               Rate 2   MM1

                                                Rate1 of gas 1

                                                Rate2 of gas 2

                                         MM1 molar mass of gas 1

                                         MM2 molar mass of gas 2

This last equation is the modern way of stating Graham's law.

This is a good way to determine the ratio of the speeds of the gasses.

This will tell you which gas will make it through a small hole quicker, i.e. the small holes you find in a
balloon. For example, which gas will leave the balloon quicker?

Intermolecular Forces:
Intermolecular forces are the forces of attractions that exist between molecules in a compound. These
cause the compound to exist in a certain state of matter, solid, liquid or gas, and affect the melting and
boiling points of compounds as well as the solubilities of one substance in another.

The melting point of a compound is the temperature at which a compound turns from a solid to a liquid or
a liquid to a solid.
The boiling point of a compound is the temperature at which a compound turns from a liquid to a gas or a
gas to a liquid. This temperature is a true measure of the forces of attractions between molecules as
molecules separate from one another when they turn from a liquid to a gas.

The stronger the attractions between particles, the more difficult it will be to separate the particles. When
substances melt, the particles are still close to one another but the forces of attraction that held the
particles rigidly together in the solid state have been sufficiently overcome to allow the particles to move.
When substances boil, the particles are completely separated from one another and the attractions between
molecules are completely overcome and the particles float away as a gas. The energy required to cause
substances to melt and to boil, and thus disrupt the forces of attraction, comes from the environment
surrounding the material. If you place a piece of ice in your hand, the ice will melt more quickly than if it
is placed on a cold counter top. The energy required to melt the ice comes from your hand, your hand
gets colder and the ice gets warmer.

Look at the table of melting points and boiling points for the halogens, shown below.

Melting Points and Boiling Points of Similar Substances with Increasing Formula Weights
SUBSTANCE                 FW (g/mole)                mp (oC)                   bp (oC)
F2                        38                         -220                      -188
Cl2                       71                         -100.98                   -34.6
Br2                       160                        -7.2                      58.78
I2                        254                        113.5                     184.35

As the size of the halogens increases, the melting and boiling points increase. The energy required to
move and separate the molecules from one another increases as the size of the molecules increases. More
massive molecules have more inertia, which must first be overcome before the molecules can be

If it takes more energy to separate the molecules, the attractions between molecules must be greater. The
types of intermolecular forces responsible for the increase in melting points and boiling points of these
non-polar covalent compounds are called dispersion forces also named London forces.

Now look at the table below:

Melting Points and Boiling Points of Substances with Similar Formula Weights
SUBSTANCE                 FW (g/mole)                mp (oC)                       bp (oC)
F2                        38                         -220                          -188
NO                        30                         -164                          -152
CH3OH                     32                         -94                           65
Ca                        40                         893                           1484
NaF                       42                         993                           1695

All the substances in this table have similar formula masses, thus they have similar dispersion forces. If
the only attractions between substances have to do with size, then they should have similar melting points
and boiling points. They do not. Let’s look more closely at the nature of the substance to see if we can
relate the structure of the material with its properties.
Fluorine and Nitrogen Monoxide:
Fluorine and nitrogen monoxide are similar in size and therefore have similar dispersion forces. Fluorine
is a non-polar covalent molecule while nitrogen monoxide is a polar covalent molecule; it has an overall
dipole. Since nitrogen monoxide has the higher melting point and boiling point, it must have the stronger
intermolecular forces. Given the same size, polar covalent molecules must have stronger forces of
attraction than non-polar covalent molecules. These forces of attractions are called dipole-dipole forces.

Nitrogen Monoxide and Methanol:
Nitrogen monoxide and methanol are similar in size and thus have similar dispersion forces. Nitrogen
monoxide and methanol are polar covalent molecules and thus have dipole-dipole forces. Since methanol
has the higher melting point and boiling point, it must have the stronger intermolecular forces. The
difference in these molecules is the presence of a certain extremely polar bond present in methanol that is
not present in nitrogen monoxide. This is the hydrogen bond or H-bond, formed between the oxygen and

Oxygen is more electronegative than hydrogen and pulls the electrons to the oxygen, away from the
hydrogen. The oxygen is now holding a partial negative charge. The hydrogen is left with very little
electron density, and since hydrogen has no core electrons, a large partial positive charge develops.

The lack of electrons around the hydrogen leaves the nucleus relatively bare. As the nucleus is positively
charged it is attracted to the lone pairs of electrons on the oxygen.

This interaction of a non-bonding pair with a hydrogen attached to an electronegative element such as
oxygen is called a hydrogen bond. Other elements that may hydrogen bond include nitrogen and the

Calcium and Sodium Fluoride:
A large jump in melting points and boiling points is observed when we turn from covalent compounds to
metals and ionic compounds. Both metals and ionic compounds involve the interaction of particles with
full charges.

      Metals: Metal ions interact with the sea of electrons that surround them. This attraction must be
       very strong as the melting point and boiling point of calcium is much higher than the covalent
       compounds which share a similar formula mass.

      Ionic Compounds: Substances which bear full charges, anions and cations, are attracted very
       strongly as evidenced by the melting point and boiling point of sodium fluoride. Full charges are
       much more difficult to separate then partial charges.

The types of interactions responsible for the extremely high melting and boiling points of metals and ionic
compound are called electrostatic forces and are the strongest of all the intermolecular forces.
Intermolecular Forces Overview:
     Ionic Compounds and Metals:
Electrostatic forces - these forces occur between charged species and are responsible for the extremely
high melting and boiling points of ionic compounds and metals.

     Covalent Compounds:
Dispersion forces - all molecules have the capability to form London forces. These are solely dependent
on the surface area and the polarizability of the surface of the molecule. These are the only types of forces
that non-polar covalent molecules can form. They result from the movement of the electrons in the
molecule which generates temporary positive and negative regions in the molecule.

Dipole-dipole forces - only polar covalent molecules have the ability to form dipole-dipole attractions
between molecules. Polar covalent molecules act as little magnets; they have positive ends and negative
ends which attract each other.

Hydrogen bonding - these occur between polar covalent molecules that possess a hydrogen bonded to an
extremely electronegative element, specifically - N, O and the halogens.

Vapor Pressure/Evaporation:
Vapor pressure is the pressure of the vapor over a liquid (and some solids, Vick’s Vapor Rub) at

Now, what does that definition mean? I'm going to go through some explanation steps that, hopefully,
give you a correct idea of vapor pressure.

1. Imagine an air tank, like the one below, of several liters in size. It has rigid walls and is totally empty
   of all substances.

2. Now, I inject some liquid into the tank, but I do not fill the tank with liquid. What will happen to the
3. That's right. Some, maybe all, of the liquid will evaporate into gas, filling the empty space. Now if all
   the liquid evaporates, we just have a tank of gas. That's not what we want. So let's suppose that only
   some of the liquid evaporated and that there are both the liquid state and the gas state present in the

The gas that is above the liquid is called its vapor and it creates a pressure called vapor pressure.
What I mean is, suppose you attached a pressure gauge to the tank, would a gas pressure be recorded?
The answer is YES!

However, here is a key point. The vapor must be in contact with the liquid at all times. Remove the liquid
and you just have a box of gas; you do not have vapor or vapor pressure.

Let's emphasize this point!!!! For vapor pressure to exist, the vapor (gas phase) MUST be in physical
contact with the liquid (or solid) it came from. You CAN'T have vapor pressure without two phases being
present and in contact!!!!!!

How is vapor pressure created? Another way to put it - how do molecules of the liquid become molecules
of gas?

Each molecule in the liquid has energy, but not the same amount. The energy is distributed according to
the Maxwell-Boltzmann distribution. Even if you don't know what that is, the point is that some
molecules have a fairly large amount of energy compared to the average. Those are the ones we are
interested in.

We are ESPECIALLY interested if one of these high energy molecules happens to be sitting right at the
surface of the water. Now, all the molecules are in motion because of their energy, but none have
sufficient energy to break the mutual attractive force molecules have for each other. Suppose that our
surface molecule was moving up away from the surface AND had enough energy to break away from the
attractive forces of the molecules around it.

Where would that molecule go? It would continue to move away from the liquid surface AND IT
BECOMES A MOLECULE OF GAS. This is great because we are now making some vapor pressure. It
happens to another molecule and another and another.

But wait! The vapor pressure stops going up and winds up staying at some fixed value. What's going on?
Here's the answer, I hope you can handle it!

As more and more molecules LEAVE the surface, what do some start to do? That's right, some RETURN
to the surface and resume their former life as a liquid molecule. Soon the number of molecules in the
vapor phase is constant because the rate of returning equals the rate of leaving and so the pressure stays

This image is my attempt to summarize this process. I'll put it here without comment:
Factors that will affect vapor pressure:
1. Temperature of the substance
2. The specific substance you are studying

1. If you  the temperature you ↑ the kinetic energy of the gas particles  ↑ speed at which these
   particles are moving, then the particle has a better chance of breaking free of the liquid.
2. Different substances have different intensities of self-attraction. The name for these attraction forces
   are intermolecular forces.

Heating Diagram:

This diagram depicts the curve of temperature verses time of a given substance. Starting with a solid that
is melted into a liquid that is boiled to a vapor. As heat is added this is the shape of the curve generated.
Flat lines are due to intermolecular forces being broken, solid ice turning to liquid water, liquid water
being turned into steam. Once boiling begins the temperature remains fixed at the boiling point until all
of the liquid is converted to gas. Thus, at the boiling point liquid and gas coexist in equilibrium.
Similarly, solid and liquid coexist in equilibrium at the melting or boiling point.

The angled lines are where the temperature is rising in a given phase. So, flat lines indicate a phase
change is occurring and angled lines indicate no phase change.
The first angled line is the rising of temperature of a solid. Once the melting temperature has been
reached, the line becomes flat and a phase change begins. The energy needed for this to occur is called
the Heat of Fusion and is defined as that amount of heat needed to melt a specific mass of a specific
substance. If this curve were a cooling curve the line would be called the Heat of Solidification, defined
as that amount of heat given off a specific mass of a specific substance when it freezes. Yes, freezing is
exothermic. It usually gets warmer right when it begins to snow.

The second angled line is the rising of temperature of a liquid. Once the boiling or vaporization has been
reached, the line again becomes flat, indicating a phase change is occurring. The energy needed for this
to occur, the Heat of Vaporization, is that amount of heat needed to vaporize a specific mass of a specific
substance. If this curve were a cooling curve the line would be called the Heat of Condensation, defined
as that amount of heat given off a specific mass of a specific substance when it condenses. Yes,
condensing is exothermic. It usually gets warmer right when it begins to rain.

A Detailed Heating Curve for Water:

The following equation is used to calculate the amount of heat needed during a phase change of specific
substance. The phase change can be melting or freezing or it can be vaporization or condensation. The
amount of heat needed for melting is the same as that given off in freezing. Likewise, the amount of heat
needed for vaporization is the same as that given off in condensation.
The equation will take the form of one of the following:

                       Q = m • Hvap       heat = mass  heat of vaporization
                       Q = m • Hcond      heat = mass  heat of condensation
                       Q = m • Hfus       heat = mass  heat of fusion
                       Q = m • Hsol       heat = mass  heat of solidification

Would it take more heat to melt 10 grams of solid water, ice or vaporize 10 grams of liquid water? Look
at the lengths of the flat lines on the above plot.
It takes more heat to boil something than it does to melt it. Think about water on the stovetop and an ice
cube on the counter. Plus, look at the Heat of Vaporization & Heat of Fusion; the Heat of Fusion is 80
cal/gram and the Heat of Vaporization is 540 cal/g.

How much heat would it take to melt 25 grams of solid water?
Q = 25 grams  80 cal/gram = 2000 cal

How much heat would it take to vaporize 25 grams of liquid water?
Q = 25 grams  540 cal/gram = 13500 cal

Longer Problem: Start from ice go to steam.

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