Download Lecture 20 - Eunil Won

PHYS151
Lecture 20
Ch 20 The Kinetic Theory of Gases
Eunil Won
Korea University
Fundamentals of Physics by Eunil Won, Korea University
Avogadro’s 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:
NA = 6.02 × 1023 mol−1
NA is called Avogadro’s number after Italian scientist Amadeo Avogadro
The number of moles n contained in a sample of
any substance is :
N
n=
NA
N: the number of molecules
NA: Avogadro’s number
molar mass M = the mass of 1 mol
Msam
Msam
=
n=
M
mNA
M = mNA
Fundamentals of Physics by Eunil Won, Korea University
Ideal Gases
pV = nRT
When the density of a given gas is low, we found that:
(Ideal gas law)
Ideal gas law holds for any single gas or for any
mixture of different gases
p : pressure
V : volume
n : the number of moles
R : gas constant ( 8.31 J/mol K)
T : temperature
Boltzmann constant k:
R
8.31 J/mol K
−23
k=
=
=
1.38
×
10
J/K
−1
23
NA
6.02 × 10 mol
Since nR = Nk,
pV = N kT
(2nd expression for the ideal gas law)
Fundamentals of Physics by Eunil Won, Korea University
Ideal Gases
Work Done by an Ideal Gas at Constant Temperature
: we allow the gas to change its volume while we keep
the temperature constant (isothermal expansion)
1
1
p = nRT = (a constant)
V
V
work done by the gas:
W =
!
W =
!
Vf
p dV
Vi
Vf
Vi
nRT
V
dV = nRT [ln V ]Vfi
V
Vf
W = nRT ln V
Vi
(isothermal expansion)
Work Done by at Constant Volume and at Constant Pressure
W =0
W = p(Vf − Vi ) = p∆V
Fundamentals of Physics by Eunil Won, Korea University
(constant-volume process)
(constant-pressure process)
Pressure, Temperature, and RMS Speed
Let n moles of an ideal gas be confined in a cubical box
of volume V
change in the particle’s momentum (along the x axis):
∆px = (−mvx ) − mvx = −2mvx
: the momentum delivered to the wall is +2mvx
: the average rate of the momentum (or force) delivered to the shaded wall by the single molecule is
2mvx
∆px
mvx2
=
=
∆t
2L/vx
L
: the pressure on the wall becomes then
p
2
2
2
Fx
mvx1
/L + mvx2
/L + ... + mvxN
/L
=
=
2
2
L
L
!m"
2
2
2
)
=
(v
+
v
+
...
+
v
x1
x2
xN
3
L
N : number of molecules in the box
Fundamentals of Physics by Eunil Won, Korea University
Pressure, Temperature, and RMS Speed
N = nNA and
N
!
2
vxi
= N (vx2 )avg
i
(v2x)avg : average value of the square of the x components
of all the molecular speeds
nmNA 2
(vx )avg
p=
3
L
2
v =
vx2
+
vy2
+
vz2
root-mean square speed :
vx2
1 2
= v
3
!
(v 2 )avg = vrms
finally we get:
2
nM vrms
p=
3V
vrms =
Fundamentals of Physics by Eunil Won, Korea University
so
!
3RT
M
or
nM (vx2 )avg
p=
V
nM (v 2 )avg
p=
3V
Translational Kinetic Energy
We again consider a single molecule of an ideal gas in a box:
1
1
1
2
2
2
Kavg = ( mv )avg = m(v )avg = mvrms
2
2
2
!
3RT
Since we have
vrms =
M
average translational kinetic
energy becomes
using M/m = NA
using k = R/NA
Kavg
3RT
1
= ( m)
2
M
Kavg
3RT
=
2NA
Kavg
3
= kT
2
: At a given temperature T, all ideal gas molecules have the same average translational kinetic
energy, namely 3/2 kT
Fundamentals of Physics by Eunil Won, Korea University
Mean Free Path
The figure left shows the path of a typical molecule as it moves,
changing both speed and direction as it collides with other
molecules
mean free path λ: the average distance traversed by a
molecule between collisions
1
λ= √
2πd2 N/V
d : diameter of a molecule (we assume molecules are spheres)
N/V : the number of molecules per unit volume
a bit quantitative analysis:
λ
∼
=
length of path during ∆t
v∆t
=
number of collisions in ∆t
πd2 v∆tN/V
1
πd2 N/V
we are missing
√
2 and that’s because
vrel =
√
2vavg
(Two v’s in the above equation are different: vavg in the numerator and vrel
Fundamentals of Physics by Eunil Won, Korea University
in the denominator)
The Distribution of Molecular Speeds
What is the distribution of molecular speeds in a gas at a given temperature?
Maxwell’s speed distribution law:
the probability is given by
P (v) = 4π
!
"3/2
M
2πRT
Since P(v) is the probability:
2 −M v 2 /2RT
v e
!
∞
P (v) dv = 1
0
By integrating v and v2, we get (without showing actual
calculation)
"
! ∞
vavg =
vP (v) dv =
0
(v 2 )avg =
!
∞
v 2 P (v) dv =
0
Also the most probable speed
vP is:
Fundamentals of Physics by Eunil Won, Korea University
vP =
!
8RT
πM
3RT
M
2RT
M
The Molar Specific Heats of an Ideal Gas
Internal Energy Eint
Let us assume that
1) our ideal gas is a monatomic gas: has individual atoms
ex) helium, neon, or argon
2) the internal energy is simply sum of the translational kinetic energies
Eint = (nNA )Kavg
since k = R/NA
Eint
3
= (nNA )( kT )
2
3
= nRT
2
: the internal energy Eint of an ideal gas is a function of the temperature
only; it does not depend on any other variable
Fundamentals of Physics by Eunil Won, Korea University
The Molar Specific Heats of an Ideal Gas
Molar Specific Heat at Constant Volume
n moles of an ideal gas at pressure p and temperature T, confined to
a cylinder of fixed volume V
We add a small amount of energy (Q) then we find
Q = nCV ∆T
CV: molar specific heat at constant volume
From the 1st law of thermodynamics (ΔEint=Q - W)
∆Eint = nCV ∆T − W
since ΔV= 0, W must be 0 :
the change in internal energy:
∆Eint
CV =
n∆T
∆Eint
3
= nR∆T
2
3
CV = R = 12.5 J/mol K
2
Fundamentals of Physics by Eunil Won, Korea University
the internal energy of any ideal gas is then:
Eint = nCV T
the resulting change in its internal energy :
∆Eint = nCV ∆T
The Molar Specific Heats of an Ideal Gas
Molar Specific Heat at Constant Pressure
n moles of an ideal gas at pressure p and temperature T, confined to
a cylinder of constant pressure p
Q = nCP ∆T
CP > CV: because energy must now be
supplied to raise T and for gas to do work
W = p∆V = nR∆T
(constant pressure process)
comparing with the previous results, we get:
CV = CP − R
or
CP = CV + R
Fundamentals of Physics by Eunil Won, Korea University
Degrees of Freedom and Molar Specific Heats
Note that:
3
CV = R
2
Every kind of molecule has a certain number 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 1/2 kT per molecule
there are three axes to rotate, so
associated energy :
3
!
1
kT
2
"
generalizing this idea, if we let f to be the number of degrees of freedom:
! "
f
CV =
R = 4.16 f J/mol K
2
Fundamentals of Physics by Eunil Won, Korea University
A Hint of Quantum Theory
Quantum theory shows that there are degrees of freedom due to rotations and
oscillations
Fundamentals of Physics by Eunil Won, Korea University
The Adiabatic Expansion of an Ideal Gas
For adiabatic process (Q=0) of an ideal gas:
pV γ = a constant
since pV = nRT,
!
Fundamentals of Physics by Eunil Won, Korea University
nRT
V
"
V
γ
= a constant
Summary
The number of atoms (or molecules) in a mole:
Degree of freedom:
NA = 6.02 × 1023 mol−1
Ideal gas law:
pV = nRT
Mean free path:
1
λ= √
2πd2 N/V
! "
f
CV =
R = 4.16 f J/mol K
2
For adiabatic process (Q=0) of an ideal gas:
pV γ = a constant
Fundamentals of Physics by Eunil Won, Korea University