Chapter 21 The Kinetic Theory of Gases
21.1 Molecular Model of an Ideal Gas
1. The number of molecules in the gas is lage, and the
average separation between them is large compared
with their dimensions.
2. The molecules obey Newton’s law of motion, but as
a whole they move randomly.
3. The molecules interact only by short-range force
during elastic collisions.
4. The molecules make elastic collisions with the walls
5. The gas under consideration is a pure substance;
that is, all molecules are identical.
2mv xi mv xi
2d
∆t =
, F∆t = 2mv xi , Fi ,on _ wall =
=
v xi
d
∆t
2
N
mv
m N
2
F = ∑ xi = ∑ v xi ,
d
d i =1
i =1
F=
∑v
2
xi
2
= N < v x > , < v 2 >= 3 < v x >
2
2
m 1
F
F
N
N1
N < v 2 > , P = = 2 = 3 m < v 2 >=
< mv 2 >
d 3
A d
V 3
3d
Molecular Interpretation of Temperature
PV = Nk B T , k B T =
T=
2 1 2
mv
3k B 2
1
1
3
< mv 2 > , < mv 2 >= kT
3
2
2
The temperature is a direct measure of average molecular
kinetic energy.
Theorem of equipartition of energy:
<
1
1
2
mv x >= kT one degree of freedom
2
2
<
1 2
3
mv >= kT three degrees of freedom
2
2
1
Total translational kinetic energy: E k = N ⋅ <
1 2
3
3
mv >= Nk B T = nRT
2
2
2
Root mean square speed: v rms = < v 2 > =
3kT
3RT
=
m
M
For hydrogen, at room temperature, v rms =
3 ⋅ 8.315 ⋅ 300
= 1.9 × 10 3 m / s
−3
2 × 10
Example: A Tank of Helium
A tank of volume 0.3 m3 contains 2 mole of helium gas at 20oC.
Assuming the
helium behaves like an ideal gas, (a) find the total internal energy of gas.
(b) What is
the rms speed of the atoms?
E=
3
nRT = 7.3 × 10 3 J , v =
2
3RT
3 ⋅ 8.31 ⋅ 293
=
= 1.35 ⋅ 10 3 m / s
−3
M
4 × 10
21.2 Molecular Specific Heat of an Ideal Gas
Energy Conservation: dE = dQ + dW
Ideal Gas: E int =
3
nRT , R = 8.31 J/mol K, PV = nRT
2
Constant volume: Q = nC v ∆T
Eint =
3
nRT
2
dW = PdV = 0 , dE = dQ , C v =
Cp
Cv
Cp-Cv Cp/Cv
He
20.8 12.5 8.33
1.67
H2
28.8 20.4 8.33
1.41
CO2 37
1 dQ 1 dE 3
J
=
= R = 12.5
n dT n dT 2
mol ⋅ K
28.5 8.5
1.31
Constant pressure: Q = nC p ∆T
dE = dQ + dW , nCv dT = nCP dT − PdV , PdV = nRdT
nCv dT = nCP dT − nRdT
C p = Cv + R
γ =
ideal gas C p =
5
R
2
C P 5R / 2
=
= 1.67
C v 3R / 2
2
Example: A cylinder contains 3.00 mol of helium gas at a temperature of 300 K.
(a) If the gas is heated at constant volume, how much energy must be transferred by
heat to the gas for its temperature to increase to 500 K?
(b) How much energy must be transferred by heat to the gas at constant pressure to
raise the temperature to 500 K?
(a) Q = nCV ∆T , CV =
3
R
2
(b) Q = nC P ∆T , CP =
5
R
2
21.3 Adiabatic Processes for an Ideal Gas
Energy Conservation: dE = dQ + dW
Ideal Gas: E int =
3
nRT , R = 8.31 J/mol K, PV = nRT
2
P1V1 = nRT1 ≠ P2V2 = nRT2
dE = dW
nC v dT = − PdV -- (1)
PV = nRT
PdV + VdP = nRdT -- (2)
R
PdV + VdP = −
PdV
Cv
 dV
C − C v dV  C P
dV dP
R dV
+
=−
=− P
=  −
+ 1
V
P
Cv V
Cv
V
 Cv
 V
C dV
dP
dV dP
dV
=− P
= −γ
,
+γ
=0
P
CV V
V
P
V
ln P + γ ln V = const , PV γ = const
since PV = nRT , TV γ −1 = const
Example: A Diesel Engine Cylinder
The fuel-air mixture in the cylinder of a diesel engine at 20.0oC is compressed from
an initial pressure of 1 atm and volume of 800 cm3 to a volume of 60 cm3.
Assuming that the mixture behave as an ideal gas with γ = 1.4 and that the
3
compression is adiabatic, find the final pressure and temperature of the mixture.
V
Pf = Pi  i
V
 f
γ
1.4


 = 1 ⋅  800  = 37.6atm , T f = Ti  Vi

V
 60 

 f




γ −1
 800 
= 293.15

 60 
0.4
= 826 K
21.4 The Equipartition of Energy
The theorem of equipartition of energy: at
equilibrium , each degree of freedom contributes,
on the average,
1
k B T of energy per molecule
2
Monatomic gas: three degrees of freedom
more complex molecules, the vibrational and
rotational motions contribute to the internal energy
Diatomic gas:
1
including rotational energy: E = 5 Nk B T
2
CV =
5
7
R , CV = R
2
2
1
including vibrational energy: E = 7 Nk B T
2
CV =
7
9
R , CV = R
2
2
Agrees with equipartition theorem at high
temperature
classical limit, Boltzmann
statistics
A Hint of Energy Quantization:
classical statistics
or quantum statistics
energy level splitting:
∆Etranslation < ∆Erotation < ∆Evibration
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21.5 Distribution of Molecular Speeds
Distribution functions (number of occurrence times): f i =
∑f
∑n
= 1 since
i
i
ni
N
value si
=N
i
The average value will be sav = ∑ si f i .
i
The average of the square of the value will be (s 2 )av = s 2 = ∑ si2 f i .
i
The root mean square value will be srms =
s2 .
The standard deviation will be σ 2 = (si − sav ) = s 2 − sav2
2
Change to the scheme of continuous distribution:
f (x )
ni
N
fi =
∑f
i
∫ s(x ) f (x )dx
∫ fdx = 1 ; ∑ s f
=1
i i
i
i
(s )
2
av
∫ [s(x )] f (x )dx
= s 2 = ∑ si2 f i
2
i
Maxwell-Boltzmann distribution function:
m 3/ 2 2 −
Number of atoms with speed v: N v = 4πN (
) v e
2πkT
v rms = 1.73
v mp =
kT
m
2kT
kT
= 1.4
m
m
8 kT
π m
v=
∞
I = ∫ e − Av dv , −
2
0
∞
∞
2
d
I = ∫ v 2 e − Av dv
dA
0
2
2
1 π
1
π
I = ∫ e − A( x + y ) dxdy =
, I=
,
4 −∞
4A
2 A
2
mv 2 / 2
kT
∞
∫v e
2
0
− Av 2
=
π
4
A
−
3
2
5
∞
∫ N v dv = 4πN (
0
∞
m 3 / 2 2 − 2 kT v2
m 3 / 2 π 2kT 2
) ∫v e
= 4πN (
)
(
) =N
2πkT
2πkT
4
m
0
3
m
∞
∫ vN dv
∞
v
v=
0
N
∞
3 − Av
∫ v e dv = −∫
=?,
2
0
0
∞
2
2
v2
1
1
d (e − Av ) = ∫ e − Av vdv =
2A
A0
2 A2
∞
m 3 / 2 3 − Av 2
m 3 / 2 1 2kT 2
8 kT
v = 4π (
) ∫v e
= 4π (
)
(
) =
2πkT
2πkT
2 m
π m
0
Example: A System of Nine Particles
Nine particles have speeds of 5, 8, 12, 12, 12, 14, 14, 17, and 20 m/s.
(a) Find the
average speed.
v = 12.7 m / s
(b) What is the rms speed?
vrms = < v 2 > = 13.3m / s
(c) What is the most probable speed of the particles?
12m/s
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