Lecture Notes

Of the three states of matter - solid, liquid, and gas - the gaseous state is the simplest to understand because the molecules behave as if they were isolated from one another. Let's see if we can figure out why this is the case.

Consider one mole of water. The densities of solid water (ice) and that of liquid water are very similar - approximately 1.0 g/mL. The density of gaseous water at room temperature and pressure is about 8 x 10-4 g/mL.

What is the average volume of a water molecule in the condensed states (solid and liquid) and the gaseous state?

In the condensed states, one mole of water occupies about 18 cm3. Thus, the average volume of a molecule is:

18 cm3/ 6 x 1023 = 3 x 10-23 cm3

If we think of this volume as a tiny cube, we can get the width of the cube by taking the cube root:

(3 x 10-23 cm3)0.33 = 3 x 10-8 cm

This is about the size of a small molecule and it means that in the condensed states, the molecules are essentially arranged very intimately; that is, there is little empty space between them.

Atomic distances are measured either in Angstroms (1 Å = 1 x 10-10 m, 1 Å = 1 x 10-8 cm), in nanometers (1 nm = 1 x 10-9 nm) or in picometers (1 pm = 1 x 10-12 m; 1 pm = 1 x 10-10 cm.). Thus, 1 Å = 0.1 nm = 100 pm. The diameter of an atom like oxygen is about 1.5 Å.

In the gaseous state the volume of a mole of water is:

18 g / 8 x 10-4 g/mL = 2.2 x 104 cm3

and the average volume of a molecule is:

(2.2 x 104 cm3) / 6 x 1023 = 4 x 10-20 cm3

and the approximate width of the cube is:

(4 x 10-20 cm3)0.33 = 3 x 10-7 cm

When you compare the condensed states with the gas state you can see that in the gas state there is about 1000 times more "effective" volume for molecules.

Problem One
Based on these calculations, are the molecules in the gas phase essentially stacked one on top of the other; that is, closely packed?

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No, the 'effective' volume is much larger than the average volume of an isolated molecule and therefore the molecules are separated from one another (on average) by a considerable distance.

This much larger distance between the molecules in the gas phase has an important consequence: gases can easily change their volume. We have all observed the change in volume of an automobile tire or balloon as the temperature changes. This dependence of the volume of a gas on temperature and pressure is in fact the basis for some important laws: Boyle's Law, on the effect of pressure on volume, Charles's Law, on the effect of temperature on volume, Gay-Lusac's Law on the effect of temperature on pressure, and Dalton's Law on partial pressures. We will examine the interrelationship of pressure, volume and temperature by examining the behavior of a gas confined to a cylinder.


Before we begin our discussion we must be clear about the pressure of a gas. Suppose that we have a gas confined to a sealed vessel. The gas exerts a pressure from within the vessel that is a result of the impact of the gas molecules on the walls of the vessel. The greater the velocity of the molecule the greater the impact, or force, that it will exert when it hits the walls of the vessel. Although molecules move about within their container with different velocities, we can change the average velocity by increasing their temperature (the energy available to the molecule increases). An increase in temperature also increases the number of times a particular molecule will hit the walls of the box.

Perhaps the following analogy will be helpful. Let's suppose that a group of people are trapped in an elevator because the door is stuck. After awhile, one desperate man begins to pound on the door in an effort to break it down. The force exerted by this one man is insufficient, however, to open the door. Meanwhile, another person panics and begins to pound of the back of the elevator, which, of course, is totally ineffective in opening the door. A third person joins the first in pounding on the door, and finally, a fourth person of considerable strength adds his efforts to the door. As they begin to panic they all hit the door with greater force (the velocity of the their movements increases). Finally, the combined forces of the three people on the door are sufficient to cause it to collapse.

If we consider the elevator door a unit of area, we notice that the pressure on that unit increases as the number of people hitting the door increases. The pressure on that unit also increases with the velocity of the impact made by the person hitting the door. Thus, pressure is force per unit area.

pressure = force / area

The pressure exerted by the atmosphere is usually measured by a device called a barometer, which is constructed by filling a glass tube with mercury or some other liquid and then inverting it in a pan or beaker (Figure 1). When the tube is inverted some of the mercury will run out into the pan. The flow of mercury out of the pan is stopped, however, when the pressure of the atmosphere on the mercury in the pan is equal to the pressure exerted by the column of mercury. At standard atmospheric pressure the height of the column of mercury will be 760 mm. Of course atmospheric pressure depends upon a number of factors, most important of which is the elevation of the point at which the pressure is measured.

mercury barometer

Figure 1. A Mercury Barometer

Problem Two
Which of the following describes how you would make up a 1.0 M NaCl solution in water?


As you ascend a mountain, for example, Pike's Peak, the distance from the center of the earth increases and the pull of gravity decreases. Thus, the higher we go, the fewer molecules of N2 and O2 are prevented from leaving the atmosphere. At heights of 10000 feet some people experience altitude sickness due to lack of oxygen and at heights above 18000 feet bottles of oxygen are necessary to supplement the oxygen in the atmosphere.

The standard atmospheric pressure (called one atmosphere) is 760 mm of mercury, which is also referred to as 760 torr (1 torr = 1 mm Hg). The SI unit for pressure, the pascal (Pa), is one newton per square meter. One atmosphere is equal to 101,325 Pa. Most commonly, we will refer to pressure in atmospheres or torr.

A newton is a unit of force. If you have had a course in physics you know that force is the product of mass and acceleration. When the mass is expressed in kilograms and the acceleration is in meters per second per second, the force will come out in newtons.


We are now ready to examine Boyle's Law:

For any given mass of gas, the volume varies inversely with the pressure, provided the temperature is held constant.

Notice several important things about this law: First, it tells us that we must keep two variables constant: the mass of the gas and the temperature. This is true for all of the gas laws. Two of the four variables (number of moles, temperature, pressure, and volume) must be kept constant while we examine the relationship between the other two. Second, it uses the term "varies inversely", which we must be sure we understand.

Problem Three
If the volume varies inversely with the pressure (we could also say that the volume was inversely proportional to the pressure), what would happen to the volume as we increase the pressure?


This means that if we double the pressure the volume will decrease by a factor of 2.


Thus, if two variables are directly proportional, an increase in one leads to an increase in the other by the same proportion. Lets suppose that x and y are directly proportional. If the value of x is 2 when the value of y is 6 and we then increase x to 6, the value of y will increase to 18. The value of x increased by 200% and the value of y increased by 200%.

y = 3x

The constant 3 is called the proportionality constant. Notice, however, that no matter what the value of the constant, a doubling of x leads to a doubling of y, and so on.

When variables are inversely proportional, the equation becomes:

y = k/x or, yx = k, k is the proportionality constant

The inverse relationship of Boyle's Law can be illustrated by Figure 2, where a gas is shown confined within a cylinder.

inverse relationship of Boyle's Law

Figure 2. The inverse relationship between P and V

The cylinder contains a piston that can be used to change the pressure resting on the gas. On the left, the pressure is simply designated as P; in the middle the pressure is doubled, and on the right the pressure is increased three-fold relative to the original pressure. The volume of the gas decreases by a factor of two and three as the pressure is doubled and tripled, respectively.

Problem Four
A given mass of gas at a given temperature has a volume of 10 L and a pressure of 100 kPa (1 kPa = 103 Pa). What volume will it have if the pressure is changed to 150 kPa?

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The pressure has increased by a factor of 150 kPa / 100 kPa = 1.50 and the volume must decrease by the same factor. Thus, the volume will fall from 10 L to 10 L / 1.50 = 6.7 L. Notice, that this is equivalent to multiplying 10 L by (100 kPa / 150 kPa) :
(10 L x 100 kPa) / 150 kPa = 6.7 L
It is always important to think about your answer for a minute. Is it reasonable to have a new volume that is smaller than the old? Yes, because an increase in pressure should produce a smaller volume. Or, in other words, you are reaffirming the inverse relationship between volume and pressure.

Problem Five
Determine the pressure required to change the volume of the gas from 10 L to 5.0 L.

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The volume has decreased by a factor of 10 L / 5.0 L = 2.0. Hence, the pressure must increase by the same factor:
(100 kPa x 10 L) / 5.0 L = 200 kPa


We are now ready to take a look at the second Law related to volume. Charles's Law states that:

For a given mass of gas, the volume varies directly with the absolute temperature, provided the pressure remains constant.

Again, we notice that two variables (mass and pressure) must remain constant. Now, however, the relationship between volume and absolute tempeature is a direct proportion. Interestingly, Charles found that it was absolute temperature - that is, temperature expressed in the Kelvin scale, not the Celsius or Fahrenheit scales - that was involved in the relationship. This was a result of his observation that a given volume of gas expands by 1/273 of its volume at 0°C for every degree Celsius that the temperature is raised.

Problem Six
Suppose that you have 273 mL of a gas at 0°C and raise the temperature 10°C. What would be the new volume?

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273 mL + (273 mL x (1/273)/(1°C) x 10 °C) = 283 mL

Problem Seven
Suppose that you have 273 mL of a gas at 0°C and lower the temperature by 273 °C. What would be the new volume?

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273 mL - (273 x (1/273)/°C x 273) = 0
This calculation shows that at a temperature of -273 °C, according to Charles's observations, the volume of a gas should be zero. Of course, this could not in fact occur because as the temperature decreases the molecules will acquire slower velocities and at the boiling point they will coalesce into the liquid state and then upon further cooling the velocities will slow even further until, at the freezing point, the solid state is achieved. Nevertheless, this temperature of -273 °C is termed the absolute zero of temperature and represents the starting point of the absolute or Kelvin temperature scale.

Problem Eight
Give the relationship in mathematical form between degrees Kelvin (indicated by K) and degrees Celsius.

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K = °C + 273

We can convert the proportionality of Charles's Law:

V µ T

into an equation, using k as a proportionality constant:

V = kT

If we are mathematically inclined we might be tempted to use this relationship to develop a little equation that allows us to calculate a new volume from an old volume as the temperature changes. We can rearrange V = kT into:

VT = k

which tells us that for a given amount of gas at a constant pressure VT will be a constant. Thus, if we change the temperature or the volume, the ratio will be the same. This means that if we start with V1 and T1, the ratio:

V1 / T1

will be the same as some new V2 and T2. Thus:

V1 / T1 = V2 / T2

However, this equation is not necessary to solve problems such as the following and we encourage you to simply use the fact that volume is directly proportional to absolute temperature.

Problem Nine
A gas has a volume of 100 mL at a temperature of -50 °C. What will the volume be at 0 °C?

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Remembering that we need absolute temperature we first convert to K :
-50 °C = 223 K, 0 °C = 273 K
Because volume is directly proportional to T, the new volume will be:
100 mL x (273/223) = 122 mL
Notice that the increase in temperature means that the new volume must be larger, not smaller, than the original volume.

Problem Ten
What temperature is necessary to decrease the volume of 100 mL of a gas at 0 °C to 50 mL?

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If the volume decreases by a factor of 50100 , the absolute temperature must decrease by the same factor. Thus, the new temperature will be:
(273 K x 50)/100 = 137 K


From the two relationships already developed - Boyle's and Charles's Laws - you can guess at the relationship between the pressure and the absolute temperature of a gas.

Problem Eleven
If the volume and mass of a gas are held constant, how will a two-fold increase in absolute temperature affect the volume?

As the temperature is increased the energy of the molecules and their velocities increase, they impact the walls of the container with greater force and more molecules impact the walls per unit area. Consequently, the pressure must increase.

As the temperature is increased the energy of the molecules and their velocities increase, they impact the walls of the container with greater force and more molecules impact the walls per unit area. Consequently, the pressure must increase.

If we allowed the volume to expand as the temperature increased, the pressure could remain constant. But, in fact, the volume remains constant.

It is this direct relationship between pressure and absolute temperature that accounts for the mishaps that occasionally occur in the laboratory when a stoppered vessel is heated. The temperature increase produces an increase in the pressure of the air inside the flask. This increase in pressure may be sufficient to blow the stopper out of the flask or even to break the flask.

This relationship is known as Gay-Lussac's Law:

At constant volume, the pressure and absolute temperature of a gas are directly proportional.

Problem Twelve
If the volume is held constant and the temperature of a gas at a pressure of 2.2 atmospheres is increased from 10 °C to 100 °C, what will the new pressure of the gas be?

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2.2 atm x (373 K / 283 K) = 2.9 atm

It is sometimes necessary to know how a change in two of three variables will affect the third. For example, suppose that we want to know how a change in both the temperature and pressure will affect the volume of a gas. This kind of problem is easily solved by considering the effect of one variable on the volume followed by the effect of the second. Suppose that we have 10 L of a gas at 0 °C and 1 atmosphere (these conditions, that is, 0 °C and 1 atmosphere pressure are referred to as standard temperature and pressure (STP)) and we want to know the volume at 25 °C and 0.7 atmospheres. We certainly know how to deal with the change in temperature and we know how to deal with the change in pressure. So we first find the volume that we get if we changed from 0 °C (273 K) to 25 °C (298 K):

10 L x 298 K / 273 K = 10.9 L

Next, we take this intermediate volume and subject it to a change in pressure from 1 atmosphere to 0.7 atmospheres (remember, this is an inverse relationship):

10.9 L x 1 atm / 0.7 atm = 15.6 L

Thus, the combined change of T and P has produced a volume of 15.6 L.

This result could also have been obtained by multiplying the original volume of 10 L by both factors simultaneously:

10 L x (298 K / 273 K) x (1 atm / 0.7 atm) = 15.6 L

Problem Thirteen
0.10 L of a gas has a pressure of 700 torr and a temperature of 50 °C. If the termperature is decreased to 10 °C, what pressure is necessary to increase the volume to 0.20 L?

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First, we determine the volume at 10 °C.
0.10 L x (283 / 323) = 0.088 L
Next, we use x as the pressure necessary to bring the volume to 0.20 L.
0.088 L x (x torr / 700 torr) = 0.20 L
x = 1.6 x 103 torr


We have now examined three of the four variables that determine the condition of a gas. The one variable remaining is the number of moles of gas. The relationship between volume and number of moles of gas was determined primarily by Amedeo Avogadro, who determined that equal volumes of gases contain equal number of molecules at a given temperature and pressure.

Problem Fourteen
At standard temperature and pressure (STP), 1.0 liter of nitrogen gas contains 0.045 moles of N2 molecules. How many O2 molecules are there in 2.0 liters of oxygen gas at STP?

Avogardro found that equal volumes contain equal numbers of moles. Thus, one liter of any gas at STP contains 0.045 moles of molecules.

The nature of the gas, according to Avogadro, is unimportant. If one liter of one gas contains 0.045 moles, then two liters of any other gas at the same conditions contains twice as many moles (0.090 moles).

The nature of the gas, according to Avogadro, is unimportant. If one liter of one gas contains 0.045 moles, then two liters of any other gas at the same conditions contains twice as many moles (0.090 moles).

Clearly, Avogadro's Law implies that at a given temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas.

V µ n

Or, in equation form, using k as a proportionality constant:

V = kn

As a result of much experimental work, we now know that one mole of a gas at STP occupies 22.4 L. This volume is referred to as the molar gas volume at standard temperature and pressure. This experimentally determined fact allows us to solve any type of gas problem. The drill questions below exemplify a few of the various types of problems that can be tackled with a combination of the molar gas volume at STP and the laws that we have previously discussed.

Problem Fifteen
What volume will 1.0 g of H2 occupy at STP?

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1.0 g of H2 is 1.0 g 2.0 g/mole = 0.50 mole
It will occupy 0.50 mole x 22.4 L/mole = 11 L

Problem Sixteen
What volume will 2.8 g of N2 occupy at 350 torr and 100 °C?

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The number of moles = 2.8 g / 28 g/mole = 0.10 mole
At STP, 0.10 mole x 22.4 L/mole = 2.24 L
At 350 torr and 100 °C, 2.24 L x (760 torr / 350 torr) x (373 K / 273 K) = 6.65 L

Problem Seventeen
What is the density of acetylene (C2H2) at STP?

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The molecular weight of C2H2 = 26 g/mole.
Density is mass per volume and one mole occupies 22.4 L at STP.
Therefore, (26 g/mole) / (22.4 L/mole) = 1.16 g/L


Let's summarize how the three variables - n (number of moles), P, and T - affect the volume:

V = kn
V = k'T
V = k"/P

We can combine these three relationships into one by combining the three proportionality constants:

V = kk'k"nT/P

or if we now call the product of the constants R and multiply both sides of the equation by P, we get:

PV = RnT

This relationship, usually called the ideal gas equation, is extremely useful and can be used to solve most gas problems. The proportionality constant R can be obtained from the molar gas volume.

Problem Eighteen
Determine R from the fact that one mole of a gas at STP occupies 22.4 L.

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PV = RnT, or:
solving for R, R = PV / nT
For one mole at 273 K and 1 atmosphere,
R = (1 atm x 22.4 L) / (1 mole x 273 K)
= 0.0821 L-atm/K-mole

This value of R (0.0821 L-atm/K-mole) requires that the following units be used:

volume = L
pressure = atm
temperature = K

If you prefer to work in torr, you can easily derive R in terms of torr.

Problem Nineteen
Calculate the value of R in units of L-torr/K-mole.

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R = (760 torr x 22.4 L)/ (1 mole x 273 K)
= 62.4 L-torr/K-mole

Now take a look at the following examples of the application of the ideal gas law.

Problem Twenty
A cylinder of N2 has a volume of 20 L, a pressure of 2000 psi (pounds per square inch), and a temperature of 25 °C.
Determine the mass of N2 in the cylinder. [14.7 lb/in2 is one atmosphere.]

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We can solve PV= nRT for n; n = PV/RT.
The pressure is 2000 psi / 14.7 psi/atm = 136 atm, and the temperature is 273 + 25 °C = 298 K. Thus:
n = PV / RT
= (136 atm x 20 L) / (0.0821 L-atm/K-mole x 298 K)
= 1.1 x 102 moles N2
The mass of 1.1 x 102 moles of N2 is:
1.1 x 102 moles x 28 g/mole = 3.1 x 103 g

Problem Twenty One
Calculate the pressure in a 20 L cyclinder at 0 °C filled with 320 g O2.

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We can solve PV = nRT for P:
P = nRT/V
n = 320 g / 32 g/mole = 10 moles
P = (10 moles)(0.0821 L-atm/K-mole)(273 K) / 20 L = 11 atm

One of the most important applications of the ideal gas equation is the determination of the molecular weight of a gas. Usually the procedure is to introduce a small amount of a gas into a previously-weighed bulb of known volume at a particular temperature and pressure. The bulb including the gas is then reweighed to obtain the weight of gas inside the bulb. Sometimes this weight is then expressed as density. Let's take a specific example. Suppose that an unknown gas is introduced into a one liter bulb that weighs 25.553 g. The temperature of the bulb is 25°C and the pressure of the gas is 720 torr. The bulb is closed and now weighs 26.553 g. Thus, the mass of the gas is 26.553 - 25.553 = 1.000 g. Now let's use PV = nRT to calculate the number of moles.

n = PV/RT
= ((720/760) atm) (1.00 L) / ((0.0821 L-atm/K-mole )(298 K))
= 0.0387 moles

Hence the number of moles in the bulb is 0.0387 moles and weighs 1.000 g. The molecular weight is:

1.000 g / 0.0387 moles = 25.8 g/mole

The density of the gas is:

d = m/V = 1.000 g/1.000 L = 1.000 g/L

and the problem could also be expressed in terms of the density.

Problem Twenty Two
A gas has a density of 0.850 g/L at STP. What is the molecular weight of the gas?

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We can do this in several ways. First, we calculate the number of moles of gas in one liter:
n = PV/RT
= (1 atm) (1.000 L) /((0.0821 L-atm/K-mole) (273 K))
= 0.0446 moles
This number of moles has a mass of 0.850 g and the molecular weight is:
0.850 g / 0.0446 moles = 19.1 g/mole
We could also simply recognize that at STP one mole occupies 22.4 L, and that therefore 1.000 L is:
1.000 L / 22.4 L/mole = 0.0446 mole
This number of moles has a mass of 0.850 g and the molecular weight is:
0.850 g / 0.0446 moles = 19.1 g/mole
Finally, we could use the density to rewrite PV = nRT. Notice that the number of moles is mass/M (the molecular weight). Thus, we can write:
PV = (mass/M)RT
P = ((mass/M)(RT))/ V
Since mass/V = d, this becomes P = (d/M)RT,
which we can rearrange in terms of M:
M = (d/P)RT
M = (d/P)RT = ((0.850 g/L)(0.0821 L-atm/K-mole)(273 K))/1 atm
= 19.1 g/mole.

Problem Twenty Three
Calculate the molecular weight of a gas with a density of 1.28 g/L at 40 °C and 520 torr.

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48.1 g/mole


In some cases we must work with a mixture of gases and it is then necessary to know the contribution of each gas to the total pressure of the mixture. Dalton's law states that:

In a mixture of gases, each gas exerts the same pressure as it would if it were in the container alone. The sum of these individual pressures, called the partial pressures, is the total pressure of the mixture.

Suppose that we have 0.5 mole N2 and 0.5 mole of O2 in a one liter flask at a total pressure of 1 atmosphere. Dalton's Law and common sense tells us that if just the 0.5 mole N2 were present, the pressure would only be 0.5 atmosphere. This is necessarily the case because the pressure is a result of collisions of molecules with the walls. The greater the number of collisions, the greater the pressure. If we decrease the number of moles of molecules by half, we must decrease the pressure by the same amount. Hence, we say that the partial pressure of N2 is 0.5 atm.

Dalton's Law is useful when a gas is collected over water. Because the water has a significant vapor pressure, the water molecules in the gas phase will contribute to the total pressure observed. If the pressure of the "wet" gas is observed to be 320 torr at 25 °C (where the vapor pressure of water is 25 torr) the partial pressure of the gas is 320 - 25 = 295 torr.

Problem Twenty Four
Determine the partial pressure of each gas in a mixture of 0.1 mole of H2 , 0.3 mole of N2 , and 0.6 mole O2 at a total pressure of 1 atmosphere.

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The total number of moles of gases is 0.1 + 0.3 + 0.6 = 1.0. The fraction of moles due to the H2 is 0.11.0 = 0.1 and its partial pressure is 0.1 x 1 atm = 0.1 atm. Likewise, for the other gases, 0.3 atm for N2 , 0.6 atm for O2.