1
Ideal Gasses and Equipartition.
1. For any ideal gasses we have the equation of state
PV
=
Nk
B
T
(1)
Here
P
is the pressure,
V
is the volume (for a vixed particle particle number
N
),
T
is the temperature
in Kelvin and
k
B
is the Boltzmann constant
k
B
= 1
.
38
×
10

23
J
◦
K
(2)
2. A microscopic analysis shows that the pressure is related to the average translational kinetic energy of
the molecules (see below)
PV
=
2
3
N
×
1
2
mv
2
(3)
3. Usually instead of expressing the total number of particles with
N
, we express it in moles
n
, which
counts particles in units of Avagadros number 6
.
02
×
10
23
N
=
nN
A
(4)
Then we have the following facts about moles and Avagadros number:
(a) Then the ideal gas constant is deﬁned as
R
≡
N
A
k
B
R
= 8
.
3
J
◦
K
(5)
So that the ideal gas equation of state is
PV
=
nRT
(6)
(b) The mass of a
n
moles of a substance is
M
=
n
M
(7)
where
M
is the molar mass, i.e. the mass of the one avagadros number of the molecule in question.
(c) The molar mass is easy to estimate. The weight of an avagadros number of protons is approxi
mately 1 gram, i.e.
M
= 1
g
. The mass of a nuetron is approximately the mass of of a proton
m
n
≈
m
p
. The mass of an electron is neglible
m
e
/m
p
≈
1
/
2000. Thus oxygen which which has 8
protons and 8 neutrons and 8 electrons the molar mass is approximately
M
= 16
g
.
4. The energy per particle of a general (i.e. nonideal) gas is a function (which can be measured by
measuring speciﬁc heats) of temperature and density
U//N
=
f
(
N/V,T
). But the energy of an ideal
gas (where there is no interaction between the molecules) is a function of temperature only
U/N
=
f
(
T
)
(8)
5. The equipartition theorem states:
The average energy of each degree of freedom is
1
2
k
B
T
(a) The degreees of freedom of a monoatomic are simply the motion in
x,y,z
1
2
mv
2
x
=
1
2
mv
2
y
=
1
2
mv
2
z
=
1
2
k
B
T
(9)
1
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View Full Document(b) The degrees of freedom for a diatomic gas are simply motion in
x,y,z
and rotation around the
two axes
x,y
1
2
mv
2
x
=
1
2
mv
2
y
=
1
2
mv
2
z
=
1
2
Iω
2
x
=
1
2
Iω
2
y
=
1
2
k
B
T
(10)
6. From the equipartition theorem we conclude that for a monoatomic or a diatomic gas we hve
1
2
mv
2
=
3
2
k
B
T
(11)
i.e. the rootmean square velocity is
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 Fall '10
 teaney
 Physics, Thermodynamics, Energy, Qin

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