# Chapter 19 The Kinetic Theory of Gases

```Chapter 19
The Kinetic Theory of Gases
In this chapter we will introduce the kinetic theory of
gases, which relates the motion of the constituent atoms to
the volume, pressure, and temperature of the gas. The
following topics will be covered:
Ideal gas law
Internal energy of an ideal gas
Distribution of speeds among the atoms in a gas
Specific heat under constant volume
Specific heat under constant volume
Adiabatic expansion of an ideal gas
(19–1)
n
M sam M sam
N


M
mN A N A
A &quot;mole&quot; of any substance is defined as the quantity contained in a mass equal to
the molar mass of the substance. The mole of any substance
contains the same number of atoms (or molecules). This is known as
&quot;Avogadro's number &quot;: N A  6.02 1023 atoms/mole.
The number n of moles in a sample of mass M sam of a substance is given by the ratio
M sam
N
n
. Here M is the molar mass of the substance. The number n 
.
M
NA
Here N is the number of atoms in the mass M sam .
The mass M sam  mN A . Here m is the mass of each molecule.
(19–2)
pV  nRT
pV  NkT
Ideal Gases, Ideal Gas Law
It was found experimentally that if 1 mole of any gas is placed in containers that
have the same volume V and are kept at the same temperature T , approximately all
have the same pressure p. The small differences in pressure disappear if lower gas
densities are used. Further experiments showed that all low-density gases obey the
equation pV  nRT . Here R  8.31 K/mol  K and is known as the &quot;gas constant.&quot;
The equation itself is known as the &quot;ideal gas law.&quot; The constant R can be expressed
as R  kN A . Here k is called the Boltzmann constant and is equal to 1.38 10-23 J/K.
If we substitute R as well as n 
N
in the ideal gas law we get the equivalent form:
NA
pV  NkT . Here N is the number of molecules in the gas.
The behavior of all real gases approaches that of an ideal gas at low enough densities.
Low densities means that the gas molecules are far enough apart that they do not
interact with one another, but only with the walls of the gas container.
(19–3)
Example 1:
Work Done by an Ideal Gas at Constant Temperature
Consider the gas shown in the figure. It is held at a
constant temperature T and undergoes an isothermal
expansion from volume Vi to volume V f . The process
follows the red line on the lower figure. The work
W done by the ideal gas is given by the equation
Vf
W
 pdV .
From the ideal gas law we have that
Vi
Vf
Vf
nRT
nRT
dV
V
p
W  
dV  nRT 
 nRT  ln V V f ;
i
V
V
V
Vi
Vi
W  nRT ln
Vf
Vi
.
For expansion we have : V f  Vi  ln
Vf
Vi
For compression we have : V f  Vi  ln
 0  W  0.
Vf
 0  W  0.
Vi
(19–4)
Example 2:
Example 3:
Example 4:
Work Done by an Ideal Gas at Constant Volume
p
Consider process a  f . During this process
the volume of the ideal gas is kept constant.
Thus the work W done by the gas is
Vi
Vf
W   pdV  0.
Work Done by an Ideal Gas at Constant Pressure
Consider process i  a. During this process
the pressure is kept constant at p and the volume
changes from Vi to V f . The work W done by
Vf
the gas is W 
Vf
 pdV  p  dV  p V
Vi
f
 Vi .
Vi
(19–5)
Example 5:
Example 6:
Ideal Gas Pressure, Temperature, and RMS Speed
Consider the molecule of mass m moving inside a
container of dimensions L  L  L as shown in the
figure. We will follow the motion of the molecule
along the x-axis. The molecule bounces off the walls
2L
with time interval t 
between collisions.
vx
px mvx   mvx  2mvx
mvx 2
The ratio



 Fx . Here px is the momentum
t
t
2 L / vx
L
transfer to the wall. The force exerted by one molecule Fx 
px
. Thus the pressure
t
p exerted by all the molecules on the wall is given by
2
Fx mvx21 / L  mvx22 / L  ...  mvxN
/L m 2
2
2
p 2 

v

v

...

v

xN  .
 3  x1 x 2
L
L2
L
 
The root mean square (RMS) value for vx is defined as
v 
2
x avg
v


2
x1
2
 vx22  ...  vxN

N
2
  vx21  vx22  ...  vxN
  N  vx2 
avg
.
(19–6)
2
nMvrms
p
3V
vrms 
3RT
M
Nm 2
v
.
3  x avg
L
For each molecule the speed v 2  vx2  v y2  vz2 .
Thus the gas pressure p 
The average values of the squares for each
v2
component are equal. Thus: v  .
3
2
x
2
nMvrms
Thus p 
. This equation tells us how the gas pressure depends on the speed
3V
of the gas molecules. If we solve this equation for vrms we get vrms 
vrms
3Vp
,
nM
3nRT
3RT


.
nM
M
(19–7)
Example 7:
The molar mass of argon is 39.95 g/mol.
vrms
3  8.31J/mol  K  313K 
3RT


 442 m/s.
3
M
39.95  10 kg/mol
Example 8:
K avg 
Translational Kinetic Energy
mv 2
The kinetic energy of a gas molecule K 
.
2
2
 mv 2 
mvrms
Its average kinetic energy K avg  
.
 
2
 2 avg
Thus K avg 
3kT
2
m 3RT 3RT

.
2 M
2NA
We finally get:
K avg 
3kT
2
At a given temperature T all ideal gas molecules, no matter what their mass,
have the same average translational kinetic energy. When we measure the
temperature of a gas, we are also measuring the average translational
kinetic energy of its molecules.
(19–8)
Example 9:
K avg
3
 (1.38  1023 J/K) (1600 K) = 3.31  1020 J .
2
3nRT
Eint 
2
Internal Energy of an Ideal Gas
Consider a monatomic gas such as He, Ar, or Kr. In this case the internal energy
Eint of the gas is the sum of the translational kinetic energies of the constituent
atoms. The average translational kinetic energy of a single atom is given by the
equation K avg 
3kT
.
2
A gas sample of n moles contains N  nN A atoms.
The internal energy of the gas Eint  NK avg 
nN A 3kT 3nRT

.
2
2
The equation above expresses the following important result:
The internal energy Eint of an ideal gas is a function of gas temperature
only; it does not depend on any other parameter.
(19–11)
CV 
3R
2
Eint  nCV T
Molar Specific Heat CV at Constant Volume
Consider n moles of an ideal gas at pressure p and
temperature T . The gas volume is fixed at V .
These parameters define the initial state of the gas.
A small heat quantity Q is added from the reservoir
that changes the temperature to T  T and the
pressure to p  p and brings the system to its
final state. The heat Q  nCV T . The constant
CV is called the molar specific heat at constant
volume. From the first law of thermodynamics
we have: Q  Eint  W and W  pV  0.
Eint
Thus Q  Eint  nCV T  CV =
.
nT
3nRT
3nRT
3R
Eint 
 Eint 
 CV 
2
2
2
We can write the internal energy of the
gas in the following form: Eint  nCV T
(19–12)
 E  nC T .
int
V
C p  CV  R
Molar Specific Heat C p at Constant Pressure
We assume that we add a heat amount Q to
the gas and change its temperature from T
to T  T and its volume from V to V  V
while keeping the pressure constant at p.
The heat Q  nC p T . The constant C p is called
molar specific heat at constant pressure. The
first law of thermodynamics gives: Q  W  Eint
 nC p T  pV  nCV T .
Using the law of ideal gases pV  nRT we get:
pV  nRT  nC p T  nRT  nC p T .
Thus: C p  CV  R.
(19–13)
Checkpoint 4:
The figure here shows five paths traversed by a gas on a
p-V diagram. Rank the paths according to the change in
internal energy of the gas, Greatest First.
Example 10:

pV
i i  p f Vf
TV
i i
 1

 Tf V f
Adiabatic Expansion of an Ideal Gas
 1
Consider the ideal gas in fig. a. The container is well
insulated. When the gas expands, no heat is transferred
to or from the gas. This process is called adiabatic.
Such a process is indicated on the p - V diagram of fig. b
by the red line. The gas starts at an initial pressure pi and
initial volume Vi . The corresponding final parameters
are p f and V f . The process is described by the equation


piVi  p f V f . Here the constant  
Cp
CV
.
Using the ideal gas law we can get the equation
TV
i i
 1
 Tf V f
 1
Vi  1
 T f  Ti  1
Vf
If V f  Vi we have adiabatic expansion and T f  Ti .
If V f  Vi we have adiabatic compression and T f  Ti .
(19–15)
Ti  T f
piVi  p f V f
Free Expansion
In a free expansion, a gas of initial volume Vi and initial
pressure pi is allowed to expand in an empty container
so that the final volume is V f and the final pressure p f .
In a free expansion Q  0 because the gas container is insulated. Furthermore,
since the expansion takes place in vacuum the net work W  0.
The first law of thermodynamics predicts that Eint  0.
Since the gas is assumed to be ideal there is no change in temperature:
Ti  T f . Using the law of ideal gases we get the following equation,
which connects the initial with the final state of the gas:
piVi  p f V f .
(19–16)
Checkpoint 5:
Rank paths 1,2 and 3 in Fig according to the energy transfer to
the gas as heat Greatest First.
Example 11:
Example 18:
Example 19:
Example 20:
Example 21:
Example 22:
Example 23:
Example 24:
Example 25:
Example 26:
Example 27:
Example 28:
Example 29:
Example 30:
Example 31:
Example 32:
Example 33:
Example 34:
Example 35:
Example 36:
Example 37:
Example 38:
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