Often scientific developments are closely related to the demands of a new technology.
Historically, the development of the science of thermodynamics, and in particular of the
thermodynamics of gases, was motivated by the desire to build better heat engines. The steam
engine, which issued in the industrial revolution, and later, the internal combustion engine, both
depend on “cycles” in which gases are alternately expanded and then compressed. The end result
of these cycles is that a portion of heat energy transferred to gas is converted into work.
KINETIC THEORY
KINETIC THEORY— UNDERPINNINGS
Models of Pressure Exerted by Molecules
So far in physics we have talked about matter as if it were continuous. We didn’t need to
invent aluminum atoms to understand how a ball rolled down the track. As early as the
fifth century B.C.E. Greek philosophers, such as Leucippus and Democritus, proposed the
idea of “atomism.” They pictured a universe in which everything is made up of tiny
“eternal” and “incorruptible” particles, separated by “a void.” Today, we think of these
particles as atoms and molecules.
In terms of everyday experience, molecules and atoms are hypothetical entities.
In just the past forty years or so, scientists have been able to “see” molecules using
electron microscopes and field ion microscopes. But long before atoms and molecules
could be “seen,” nineteenth-century scientists such as James Clerk Maxwell and Ludwig
Boltzmann in Europe and Josiah Willard Gibbs in the United States used these imaginary
microscopic entities to construct models that made the description and prediction of the
macroscopic behavior of thermodynamic systems possible. Is it possible to describe the
behavior of an ideal gas that obeys the first law of thermodynamics as a collection of
moving molecules? To answer this question, let’s observe the pressure exerted by a
hypothetical molecule undergoing elastic collisions with the walls of a two-dimensional
box. By using the laws of mechanics we can derive a mathematical expression for the
pressure exerted by the molecule as a function of the volume of the box. If we then
define temperature as being related to the average kinetic energy of the molecules in an
ideal gas, we can show that kinetic theory is compatible with the ideal gas law and the
first law of thermodynamics.
2D MOLECULAR MOTION AND PRESSURE

Consider a spherical gas atom that has velocity v  vx x  v y y and makes perfectly elastic
collisions with the walls of a two-dimensional box of length X and width Y as shown
below.
A typical screen image of a single atom moving at a speed, v, in a two-dimensional box.
Assume the atom bounces off the walls of the box with perfectly elastic collisions.
You will be making measurements on the simulated motions of an atom bouncing
in a box to derive the mathematical relationship that relates pressure of the gas to the
kinetic energy of the atom and the volume occupied by the gas. To complete this
exercise you will need:
EITHER
• Atoms in Motion Software
Load the computer simulation and run it.
Next use the hypothesis that a gas is made of a large collection or atoms behaving
like little billiard balls to explain why the ideal gas law might hold. In the next activity
you are to pretend you are looking under a giant microscope at a single spherical atom as
it bounces around in a two-dimensional box by means of elastic collisions and that you
can time its motion and measure the distances it moves as a function of time.
If the atom obeys Newton’s laws, you can calculate how the average pressure that
the atom exerts on the walls of its container is related to the volume of the box.
Activity: The Theory of 2D Molecular Motion
a. Suppose the molecule moves across a distance X completely across the box in the xdirection in a time tx. What is the equation needed to calculate its x-component of
velocity in terms of X and tx?
b. Suppose the molecule moves a distance Y completely across the box in the y-direction in
a time ty. What is the equation needed to calculate its y-component of velocity in terms
of Y and ty?
c. Suppose the box were a three-dimensional container. Write the expression for
vtotal in terms of the x, y, and z components of velocity. Hint: This is an application of
the 3-dimensional Pythagorean theorem.
d. We would like to find the average kinetic energy of each molecule. Since the
kinetic energy of a molecule is proportional to the square of its total speed, you need to
2
2
2
2
2
show that if on the average vx = vy = vz , then v total = 3vx .
e. If the molecule bounces back at the same speed in the x-direction, what is the
time, t, it takes to return to the left wall
is t = 2 tx.
f. If the collisions with the wall perpendicular to the x-direction are elastic, write an
expression for the force exerted on that wall for each collision as a function of the mass,
velocity and t for the collision. Hint F = p/t.
Fx = (2 mv)/ t
l. Substitute the expression from part a. for tx to write an expression for Fx.
mv x
Fx 
X
2
m. For simplicity, let us assume that we have a cubical box so that the length, width, and
height are all the same and hence, the volume of the box is the cube of its length. In other
3
words, V = X . Write an expression for the pressure on the wall perpendicular to the xaxis caused by the force Fx due to one atom.
mv x
P
X3
2
n. Let’s say that there are not one but N molecules in the box. What is the pressure on the
wall now?
3
o. Next, if we write the volume of our box as V = X , and recalling that v x 
can write an expression for P as a function of V.
2
v 2 total
, we
3
P N
mv 2 total
3V
p. Finally, since the average kinetic energy of a molecule is just
 E kin   21 mv 2 total , write the pressure in the box as a function of the kinetic energy
P
2 N  E kin 
3V
KINETIC ENERGY, INTERNAL ENERGY, AND TEMPERATURE
The Effect of Increasing Volume and the Number of Atoms
We have hypothesized the existence of non-interacting atoms to provide the basis for a
particle model of ideal gas behavior. We have shown that the pressure of such a gas can
be related to the average kinetic energy of each atom:
P
2 N  E kin 
3V
or
PV 
2
N  E kin 
3
Pressure increases with kinetic energy per atom and decreases with volume. This
result makes intuitive sense. The more energetic the motions of the molecules, the more
pressure we would expect them to exert on the walls. Increasing the volume of the box
decreases the frequency of collisions with the walls, since the molecules will have to
travel longer before reaching them, so increasing volume should decrease pressure if
Ekin stays the same.
Activity: Gas Law and Kinetic Theory Predictions
a. According to the ideal gas law, PV = nRT = NkBT. What should happen to the pressure of
an ideal gas as its volume increases? As the number of particles increases?
b. What do you predict will happen to the pressure in your simulated gas if you increase the
volume of its container? Explain your reasoning.
c. What do you predict will happen to the pressure in your simulated gas, if you increase the
number of atoms in the box?
Kinetic Theory and the Definition of Temperature
The model of an ideal gas we have just derived requires that:
PV 
2
N  E kin 
3
But one form of the ideal gas law is given by
PV = NkBT
where
and
N = the total number of gas molecules in the gas
-23
kB = Boltzmann’s Constant given by 1.38  10 J/K
What can we say about the average kinetic energy per molecule for an ideal gas?
You can derive a relationship between temperature and the energy of molecules that
serves as a microscopic or molecular definition of temperature.
Activity: Microscopic Definition of T
a. Use the ideal gas law and the equation relating N, P, and V to the kinetic energy of an
atom that you just developed to derive an expression relating Ekin and T. Show the steps
in your derivation.
c. Write an equation for the average velocity v based on the previous result. This
velocity is called the Root Mean Square speed vrms of the molecules.
v rms 
3RT
3kT

M
m
Kinetic Theory of Gases


The Temperature of a gas is a measure of the Average Kinetic Energy of the
molecules that make up the gas.

This relation is only valid for a monatomic molecule like Helium.
FINAL DETAILS
Mean Free Path:
Molecular Velocities:
= Normal average or mean speed. Half the molecules have speed greater than vav
vav and half are slower.
= The square root of the average of the square of the velocity of the molecules in the
vrms gas.
vp = The speed at which the largest number of molecules move.
MAXWELL - BOLTZMANN VELOCITY
DISTRIBUTION
Average Kinetic Energy of a Gas Molecule:
(FIX)
Most Probable Kinetic Energy of a Gas Molecule:
m = mass of gas molecule SI: kg
Boltzmann's Constant
k = 1.380x10-23 J/K
T = Temperature of the Gas SI: K
COMPRESSION AND EXPANSION OF GASES
DOES THE IDEAL GAS LAW TELL ALL?
Since many practical devices, from automobile engines to refrigerators, use expanding
gases to operate, it is important to understand what happens to gases when they undergo
volume changes. We know that PV = nRT for an ideal gas. Can this relationship used by
itself tell us what happens to the temperature of a gas if its volume changes? Discuss
your answer with your partner.
Activity: Ideal Gas Tells All?
a. Can the ideal gas law be used to calculate the change in temperature of a system as its
volume increases? Why or why not?
ISOTHERMAL AND ADIABATIC PROCESSES FOR AN IDEAL GAS
Imagine a piston filled with an ideal gas. What could happen to it as it is compressed?
The first law of thermodynamics tells us that

Eint = Q – W = Q – PdV
(3.1)
If we are going to describe what happens when the volume of a gas changes in more
detail, we might try using the ideal gas law in conjunction with the first law of
thermodynamics. Let’s start by considering an isothermal process in which there is no
temperature change in a gas while it is being compressed.
Activity: Isothermal Compression of a Gas
a. In an earlier section you showed that, for an ideal gas, Eint = (3/2) NkBT. Now
show that, for a fixed number of molecules of gas, N, a change in the temperature of an
ideal gas T produces a change in internal energy that is proportional to T and given by
Eint = (3/2) NkBT = (3/2)nRT.
b. In an isothermal compression, the gas is kept at a constant temperature by transferring the
appropriate amount of heat energy to its surroundings. Therefore T and Eint are both
zero in an isothermal compression. Find an expression relating Q and W and another
expression relating P and V during an isothermal compression. Hint: Use both the first
law of thermodynamics and the ideal gas law.
Another type of process that can occur during the expansion or compression of a gas is an
adiabatic change. An adiabatic process is defined as one in which a system does not
exchange heat energy with its surroundings so that Q = 0 during the process. This can be
brought about either by carefully insulating the system so that no heat energy exchange is
possible, or by carrying out the process so rapidly that heat energy does not have time to
be transferred. What happens to an ideal gas if it is compressed adiabatically? We would
like you to show on a step-by-step basis below, that for an ideal gas undergoing an
adiabatic expansion the following expression can be used to describe the relationships
between an initial volume and temperature and a final volume and temperature.
3 dT dV
3\2
3\2

 0 so that Tf Vf = Ti Vi
2 T
V
Note: The exponent of 3/2 only holds for an ideal monatomic gas. For a “real” gas the
exponent will be different.
Activity: Adiabatic Compression of a Gas
 The first step: Previously you showed that the change in the internal energy of an ideal gas
is given by Eint = (3/2)NkBT. Use the first law of thermodynamics to find a
relationship between the work done when an ideal gas is compressed adiabatically by an
amount V (with no change in pressure) and the change in the temperature of the gas, T.
(3/2) NkBT = - P V.
b. Next you can use the ideal gas law and the relationship you just derived to show that for
small temperature changes the fractional change in the temperature of an ideal gas, T /T,
can be related to the fractional change in volume, V /V, by
3 T
V

2 T
V
or
3 T V

0
2 T
V
Now we are back to proving that
3/2
Tf
3/2
Vf = Ti
T vs. V for Adiabatic Expansions
Integrate both sides of the equation given by
Vi
3 dT dV

0
2 T
V
Combine the results you just obtained above to show that, if a gas of initial temperature Ti and
volume Vi is compressed adiabatically to a final volume Vf with final temperature Tf, then
Tf
3/2
3/2
Vf = Ti
Vi
Note: Strictly speaking, this result only holds for ideal monatomic
gases composed of single molecules such as helium. A similar
equation holds for other gases at modest densities.
THE FIRE SYRINGE AND THE RAPID COMPRESSION OF AIR
A device known as a fire syringe allows a rapid compression of
air in a small glass tube that is inside a safety tube of Plexiglas. If
pushed very hard, the piston in the glass tube can be forced
almost down to the end of the straight-walled section of the
tube. If this is done rapidly, the compression can be nearly
adiabatic. Air is not a monatomic gas, but the formulas derived
above work well enough. As you can tell from the equation you
derived in the last activity, the air in the fire syringe should
increase in temperature as its volume decreases. Examine a fire
syringe and make some reasonable assumptions about the initial
and final volumes of air in the chamber. You can then calculate
the approximate final temperature of the compressed air.
Finally, you can attempt to ignite a tiny piece of tissue paper
with the fire syringe.
Activity: The Fire Syringe—Fahrenheit 451
a. Before approximating the final temperature of air in the
syringe, estimate the following quantities:
Initial length of air column Li =
Final length of air column Lf =
Inner radius of the tube R =
Initial volume Vi =
Final volume Vf =
____________________
____________________
____________________
3
____________________
cm
3
____________________
Initial temperature Ti =
cm
____________________
K
cm
cm
cm
Fig. 3.2. A fire syringe
that allows a rapid
compression of
trapped air to ignite a
piece of paper.
b. Calculate the final temperature in the cylinder in Kelvin.
Final Temperature Tf =
____________________
K
c. Compare this to the “flash point” or burning temperature of paper, which is 451°F.*
What do you expect to happen to the tissue paper in the fire syringe when the plunger is
pushed down rapidly?
d. Put on safety gloves and carry out the fire syringe experiment by rapidly and forcefully
depressing the plunger. What happens?
State Variables
A gas is described by several quantities. These quantities are called state variables--they
are like the quantities listed on a person's driver's license (brown hair, blue eyes, height,
weight) that describe the status or state of the person. The state variables for a gas
describe the status of the gas.
 the pressure p in the gas in kPa (1 kPa = 103 N/m2)
 the volume V of space occupied by the gas in dm3 [1dm3 = (10-3) m3]
 the absolute temperature T of the gas in kelvin (K)
 the number N of atoms or molecules in the gas or the number of moles n of atoms or
molecules in the gas
WORK - W, HEAT - Q, and INTERNAL ENERGY - U
WORK W:
Useful Energy Transferred across the System's Boundaries, capable of producing
Macroscopic-Mechanical Motion of a the system's Center-of-Mass.
W = Work done by (or on) one system on another system
*
The paper flashpoint of 451 degrees Fahrenheit is well known to readers of Ray Bradbury’s famous
science fiction novel, Fahrenheit 451, about book burning.
ENERGY FLOW
OUT:
W>0
System Does External Work
Sys --> Work
INTO:
W<0
Work Done on the System
Work --> Sys
Work done by a Gas :
Constant Volume Process
Constant Pressure Process
Constant Temperature Process
Adiabatic Process Q = 0
HEAT Q:
Energy Transfer across the System's Boundaries that cannot produce MacroscopicMechanical Motion of the system's Center-of-Mass. Energy Transfer at the Molecular
Level .
Q = Microscopic Energy flow into (or out of) the System
ENERGY FLOW
INTO: Q > 0 System Absorbs Heat Heat --> Sys
OUT: Q < 0 System Releases Heat Sys --> Heat
Some common types of Heat Lost:
Solids or Liquids
Gas- Constant Pressure Process
Gas - Constant Volume Process
Gas - Constant Temperature Process
Gas - Adiabatic Process
INTERNAL ENERGY U
Energy Stored in a System at the Molecular Level.
The System's Thermal Energy -the Kinetic Energy of the atoms due to their random
motion relative to the Center of Mass plus the binding energy (Potential Energy) that
holds the atoms together.
U = Microscopic Energy contained in the System
FIRST LAW of THERMODYNAMICS:
Any Change in the Internal Energy of a System U is due to either the Heat Flow into/outof the System or due to Work Done by/on the System provided the system's center-ofmass energy does not change.



It is important to observe that the  in U is absolutely necessary because both
work W and heat Q represent a transfer of energy whereas the internal energy U
is a quantity of energy that a system contains.
Another way to state his difference is that U is a state variable were as Q and W
are not state variables. What this means is that if you take the system from one
state to another state by two different processes, U will be the same independent
of the path taken but not Q or W.
The quantities Qin, Qout, Win, and Wout are all taken to be positive quantities
whereas Q and W can be either positive or negative.
ISOBARIC Pressure is Constant (P = 0)
Example:
Gas Heated in a Cylinder fitted with a movable frictionless piston.
The pressure the atmosphere and the pressure due to the weight of the piston
remains constant as the gas heats up and expands.
First Law Implications: U = Q - W
Unlike some of the other processes below neither the heat Q , the work W, or
the change in internal energy U are necessarily zero in a constant pressure
process.
For an ideal gas, constant pressure work is easily determined, W =  PdV =
PV
Part of the heat that flows into the system causes the temperature to rise, Q =
n cp T = m Cp T, the rest goes into work.
ISOTHERMAL Temperature is Constant (T = 0)
Example1:
Boiling of water in the open air.
In general most isobaric phase changes are isothermal.
In this example the system does work as the steam-produced
pushes against
the atmosphere as it expands. Neither the heat Q , the work W, or
the change in internal energy U are zero. In this case Q = mLv
since the water changes phase.
Example 2:
In general for an Ideal gas U is only a function of the temperature
so that U is always equal to zero for an isothermal process.Since
U = 0 then W = Q from the First Law.
What has to happen for this process to be isothermal is that the gas
in a cylinder is compressed slowly enough that heat flows out of
the gas at the same rate at which is being done on the gas.
The ideal gas law can be used to determine the work done W = PV
ln(Vf/Vo) which is also the equation for Q.
Note that P1V1 = P2V2 = nRT, the ideal gas law for an isothernal
process.
ISOCHORIC (Isovolumetric) Volume is constant (V = 0)
Example:
Heating of a Gas in a Rigid, Closed container.
In this case no work is done on the gas because W =  PdV =  P 0 = 0.
As a result the FirstLaw implies that the change in internal energy must
equal any heat flowing into or out of the system, U = Q = n cv T = m Cv
T.
Note that V1 = V2 = nRT1/P1 = nrT2/P2, the ideal gas law for constant
volume process.
ADIABATIC No heat flows into or out of the system (Q = 0)
Example
Compression of a Gas in an Insulated Cylinder.
In this case any change in the internal energy of the gas is due to work
done on it or by it, U = W.
Normally if U changes the temperature of a system will change. Any
temperature rise or fall is due to the work done or by the gas alone and not
due to heat flowing into or out of the system since Q = 0.
If a process is carried out fast enough the heat flow will be small and the
process can be approximate as being adiabatic. This happen because heat
flow is in general a slow process.
Observe that we did not say that Q is constant because it not a state
variable. Q represent an energy transfer not the heat energy of the system.
In addition to the ideal gas law PV = NkT, the quantity PV is constant for
an ideal gas where  = cP/cV, the ratio of molar specific heats.
For an ideal gas the work W = (P1V1 - P2V2)/( - 1)
State Variables and Ideal Gas Law
Question 1: Isobaric Processes
An isobaric process is one in which the pressure p and the number N of atoms remain
constant as the gas volume V and temperature T change. Use the ideal gas law to select
below the graph line that best represents the relation between volume and temperature for
an isobaric process. Your correct selection will open and run an isobaric simulation.
Question 2: Isochoric Processes
An isochoric process is one which the volume V and the number N of atoms remain
constant as the gas pressure p and temperature T change. Use the ideal gas law to select
below the graph line that best represents the relation between pressure and temperature
for an isochoric process. Your correct selection will open and run an isochoric
simulation.
Question 3: Isothermal Processes
An isothermal process is one in which the temperature T and the number N of atoms
remain constant as the gas pressure p and volume V change. Use the ideal gas law to
select below the graph line that best represents the relation between pressure and volume
for an isothermal process. Your correct selection will open and run an isothermal
simulation.
Question 4: Isobaric Process
Adjust the pressure to 100kPa (about atmospheric pressure). The process starts with 1.0
moles of gas at temperature 500K and volume 41.6dm3. Determine the volume of the
container when the temperature is reduced to 301.8K, a little above room temperature.
Question 5: Isochoric Process
The process starts with 1.0 mole of gas at temperature 100K and pressure 42kPa.
Determine the pressure in the gas when the temperature is increased to 300K, about room
temperature.
Question 6: Isothermal Process
. Set the thermometer to 300K, about room temperature. The process starts with 1.0 mole
of gas at volume 40dm3 and pressure 62kPa. Determine the pressure of the gas as the
volume decreases to 20dm3 and then to 10dm3, as the simulation runs.
Please skip question 7 for now.
Elevated Tank Problem
A 50.0 m 3 closed tank is initially filled with air at 20.0 0C and 100 kPa. A pump fills the
tank 3/4 full of water in such a way that the temperature of the air and water in the tank
remain constant at 20.0 0C. Assume that the height of the center of mass of the water
when the tank is 3/4 full is 40.0 m above the level of a lake from which the tank is filled.
(A) What is the final pressure of the air in the tank when the tank is 3/4 full ?
(B) How much heat is lost by the air in the tank while the tank is filled to 3/4 full?
(C) What is the minimum amount of work that the pump must do to fill the tank 3/4
full?
Sketch and Process:
Physical Process:
Water is pumped into an elevated storage. Air above the water in the tank is compressed
isothermally.
Relevant Physics:
There are two different systems which we must consider and to which we must apply the
first law of thermodynamics.
Air in the tank:
The temperature of the air in the tank remains constant while it is compressed from the
volume of tank to 1/4 the volume of the tank. For an isothermal process the internal
energy of the of the gas is zero, so that by the first law, any work done on the gas must be
equal to the heat lost by the gas.
This will be a negative value since it represents work done on the air, not by the air.
Equivalently, it also represents the heat lost.
Water in tank when it is 3/4 full:
Since the center of mass of the water changes, we must use the more general form of the
first law, the conservation of energy.
If the water is pumped into the tank slow enough, then any heat lost by the air in the tank
will exit the tank altogether. Some heat will go into the water but we assume that this heat
flows out so that the temperature of the water also remains constant and Qw = 0.
If we assume that both the temperature and density of the water remain constant, then
change in the water's internal energy will be zero, U = 0. Thus the change in the water's
mechanical energy must be equal to the negative of the net work done by the water.
The change in mechanical energy of the water due to the rise in elevation of its center of
mass is,
The work Ww consists of three different sources of work.
(a) The work done on the water by the pump. Since this represents work done on
the system, this work must be a negative number. If Wp is the positive work done
by the pump, then -Wp is work done on the water.
(b) The work that the water does on the air in the tank. By conservation of energy
this must also be equal to the positive work done on the air to compress the air
isothermally. If Wgas is the work done on the gas which is negative, then the work
done by water is -Wgas.
(c) The pressure of the air outside the tank acting on the surface of the water will
do some work on the water. This is a constant pressure process as the volume of
the water in the lake is reduced. Again this work must be negative since it is done
on the water.
The first law becomes
Thus the work done by the pump is equal to the work needed to raise the water plus the
positive work need to compress the air less the work done by the outside air pressure.
(A) Find the final pressure of the gas P2.
Applying the ideal gas law,
B) Find heat lost by the gas Qgas.
Since the gas is compressed isothermally, the heat loss by the gas must equal the work
done by the gas.
(C) Find the work done by the pump, Wp.
The work done by a pump with no frictional losses must be equal to the work needed to
raise the water's center-of-mass plus the positive work needed to compress the air in the
tank less the work done by the external air pressure helping to push the water into the
tank.
The work needed to elevate the water is equal to the water potential energy when it is in
the tank. The mass of the water can be found since we know its volume and density.
The work needed to isothermally compress the air in the tank is the same as we found in
part B.
The external air pressure pushes down on the surface of the lake at a constant pressure.
As the volume of water in the lake decreases the work done by the external air pressure is
a constant pressure process.
Thus the minimum work needed to be done by the pump is,
We can see that most of the work needed is due to raising the water to an elevated
position. If the tank were open to the air rather than closed, then the work needed would
be just that of the work done against gravity to elevate the water. Here the work on the
gas in the tank would be the work needed to push the air in the tank out of the tank at a
constant pressure. This would be equal to the work done by the external air pressure, i.e.
the last two terms cancel each other out.