Chapter 19
The Kinetic Theory of Gases
From the macro-world to the micro-world
Key contents:
Ideal gases
Pressure, temperature, and the RMS speed
Molar specific heats
Adiabatic expansion of ideal gases
19.2 Avogadro’s Number
Italian scientist Amedeo Avogadro (1776-1856) suggested that
all gases occupy the same volume under the condition of the
same temperature, the same pressure, and the same number of
atoms or molecules. => So, what matters is the ‘number’ .
One mole is the number of atoms in a 12 g sample of carbon-12.
The number of atoms or molecules in a mole is called
Avogadro’s Number, NA.
If n is the number of moles contained in a sample of any
substance, N is the number of molecules, Msam is the mass of
the sample, m is the molecular mass, and M is the molar mass,
then
19.3: Ideal Gases
The equation of state of a dilute gas is found to be
Here p is the pressure, n is the number of moles of gas present,
and T is its temperature in kelvins.
R is the gas constant that has the same value for all gases.
Or equivalently,
Here, k is the Boltzmann constant, and N the number of molecules.
(# The ideal gas law can be derived from the Maxwell distribution; see slides below.)
19.3: Ideal Gases; Work Done by an Ideal Gas
Example, Ideal Gas Processes
Example, Work done by an Ideal Gas
19.4: Pressure, Temperature, and RMS Speed
The momentum delivered to the wall is +2mvx
Considering
, we have
Defining
, we have
Comparing that with p =
nRT
, we have
V
3kT
=
m
The temperature has a direct connection to the RMS speed squared.
Translational Kinetic Energy
19.4: RMS Speed
Example:
19.7: The Distribution of Molecular Speeds
Maxwell’s law of speed distribution is:
The quantity P(v) is a probability distribution function:
For any speed v, the product P(v) dv is the fraction of molecules with speeds in the
interval dv centered on speed v.
Fig. 19-8 (a) The Maxwell speed distribution for
oxygen molecules at T =300 K. The three
characteristic speeds are marked.
kT
= 1.41
m
kT
= 1.59
m
= 1.73
kT
m
ò
¥
2 -ax 2
xe
dx =
p
16a 3
¥ 3 -ax 2
1
ò 0 x e dx = 2a 2
¥ 2 -ax 2
9p
ò 0 x e dx = 64a 5
0
Example, Speed Distribution in a Gas:
Example, Different Speeds
19.8: Molar Specific Heat of Ideal Gases: Internal Energy
The internal energy Eint of an ideal gas is a function of the gas
temperature only; it does not depend on any other variable.
For a monatomic ideal gas, only translational kinetic energy is
involved.
19.8: Molar Specific Heat at Constant Volume
where CV is a constant called the molar specific
heat at constant volume.
But,
Therefore,
With the volume held constant, the gas cannot
expand and thus cannot do any work.
Therefore,
# When a confined ideal gas undergoes temperature change DT, the resulting change in
its internal energy is
A change in the internal energy Eint of a confined ideal gas depends on only the
change in the temperature, not on what type of process produces the change.
19.8: Molar Specific Heat at Constant Pressure
DEint = nCV DT
Example, Monatomic Gas:
Molar specific heats at 1 atm, 300K
g=CP/CV
CV (J/mol/K)
CP-CV (J/mol/K)
monatomic
1.5R=12.5
R=8.3
He
12.5
8.3
1.67
Ar
12.5
8.3
1.67
diatomic
2.5R=20.8
H2
20.4
8.4
1.41
N2
20.8
8.3
1.40
O2
21.0
8.4
1.40
Cl2
25.2
8.8
1.35
polyatomic
3.0R=24.9
CO2
28.5
8.5
1.30
H2O(100°C)
27.0
8.4
1.31
19.9: Degrees of Freedom and Molar Specific Heats
Every kind of molecule has a certain number f of
degrees of freedom, which are independent ways in
which the molecule can store energy.
Each such degree of freedom has associated with it —
on average — an energy of ½ kT per molecule (or ½
RT per mole). This is equipartition of energy.
Recall that
DEint = nCV DT
g=CP/CV
CV (J/mol/K)
CP-CV (J/mol/K)
monatomic
1.5R=12.5
R=8.3
He
12.5
8.3
1.67
Ar
12.5
8.3
1.67
diatomic
2.5R=20.8
H2
20.4
8.4
1.41
N2
20.8
8.3
1.40
O2
21.0
8.4
1.40
Cl2
25.2
8.8
1.35
polyatomic
3.0R=24.9
CO2
28.5
8.5
1.30
H2O(100°C)
27.0
8.4
1.31
Example, Diatomic Gas:
19.10: A Hint of Quantum Theory
A crystalline solid has 6 degrees of
freedom for oscillations in the lattice.
These degrees of freedom are frozen
(hidden) at low temperatures.
# Oscillations are excited with 2 degrees of freedom
(kinetic and potential energy) for each dimension.
# Hidden degrees of freedom; minimum amount of energy
# Quantum Mechanics is needed.
19.11: The Adiabatic Expansion of an Ideal Gas
with Q=0 and dEint=nCVdT , we get:
From the ideal gas law,
and since CP-CV = R,
we get:
With g = CP/CV, and integrating, we get:
Finally we obtain:
19.11: The Adiabatic Expansion of an Ideal Gas
19.11: The Adiabatic Expansion of an Ideal Gas, Free Expansion
A free expansion of a gas is an adiabatic process with no work or change in
internal energy. Thus, a free expansion differs from the adiabatic process
described earlier, in which work is done and the internal energy changes.
In a free expansion, a gas is in equilibrium only at its initial and final points;
thus, we can plot only those points, but not the expansion itself, on a p-V
diagram.
Since ΔEint =0, the temperature of the final state must be that of the initial
state. Thus, the initial and final points on a p-V diagram must be on the same
isotherm, and we have
Also, if the gas is ideal,
Example, Adiabatic Expansion:
Four Gas Processes for an Ideal Gas
Homework:
Problems 13, 24, 38, 52, 59