Kinetic Molecular Theory of Matter - University Courses in Electronic

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Kinetic Molecular Theory of Matter
• Heat capacity of gases and metals
• Pressure of gas
• Average speed of electrons in
semiconductors
• Electron noise in resistors
Positive metal
ion cores
Free valence
electrons forming an
electron gas
Fig. 1.7: In metallic bonding the valence electrons from the metal
atoms form a "cloud of electrons" which fills the space between the
metal ions and "glues" the ions together through the coulombic
attraction between the electron gas and positive metal ions.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Ideal gas approximation. Pressure
Square Container
Microscopic model
Area A
a
Face B
Face A
vy
Definition: P (pressure) = force per unit area
vx
Gas atoms
Pressure (P) is caused by the collisions
of molecules with walls
a
a
Fig.1.15: The gas molecules in the container are in random motion.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Our task is to show that Temperature
(T) is related to average speed of
molecules
Ideal Gas Assumptions:
1. molecules are in constant and random motions, all directions are equivalent
2. no interactions between molecules but collisions
3. all collisions are elastic
4. molecule size is negligible
5. Newtonian mechanics is valid
Connecting the pressure with the average speed of gas molecules
Step 1. Change in Momentum of a Molecule
Δp = 2mvx
N
Δp = change in momentum,
m = mass of the molecule,
vx = velocity in the x direction
Step2. Force produced by unique molecule
Δp 2mv x mv x
F=
=
=
Δt 2a / v x
a
2
Step3. Calculating pressure
2
2
2
2
mN v x
mN v x1 + v x2 + ... + v x N
Total _ force mv x1 + mv x2 + ... + mv x N
= 3
=
P=
=
2
3
a
a
a
N
V
2
Step4. Averaging speed
v x + v y + v z = v 2 = 3v x
2
2
2
2
2
2
Nmv 2 1 2
P=
= ρv
3V
3
Ν = total number of molecules
ρ = density of gas
v = velocity of molecules
Pressure ÙTemperature
Ú
Ú
Volume
Ideal Gas Equation
⎛ N ⎞
⎟⎟ RT
PV = ⎜⎜
⎝ NA ⎠
P is the pressure, i.e. force per unit area
N is the total number of molecules in volume V
R is the gas constant (8.3144 J mol-1 K-1)
NA is the Avogadro’s number (6.022 ×1023 mol-1)
Experimental law – macroscopic point of view
Gas Pressure in the Kinetic Theory
2 ⎞
⎛
m
v
1
2
⎟
P = ρ v2 = N ⎜
3
3 ⎜⎝ 2 ⎟⎠
Ideal Gas Equation
⎛ N ⎞
⎟⎟ RT
PV = ⎜⎜
⎝ NA ⎠
P = gas pressure, N = number of molecules,
m = mass of the gas molecule, v = velocity,
V = volume, ρ = density.
Mean Kinetic Energy per Atom
1 2
3
KE = mv =
kT
2
2
k =R/NA Boltzmann constant,
T = absolute temperature
0 K ≈ -273 0C
Gas constant
Boltzmann’s constant
R = 8.3144
J mol-1 K-1
k = 8.61 × 10-5
eV K-1
R = k × NA
R – macroscopic (technical, thermodynamical etc.) calculations
k – microscopic (atomic) calculations
Mole (a gram molecule) = a quantity of a substance
equal to the molecular weight of a substance
expressed in grams
Example:
Carbon has atomic mass of 12 ⇒ a mole of carbon is 12 grams
Mole contains 6.0220 × 1023 species (atoms, molecules, etc.)
NA
= 6.0220 × 1023 - an Avogadro's number
KE =
Internal Energy per Mole for a
Monatomic Gas
1 2 3
mv = kT
2
2
⎛ 1 2⎞ 3
U = N A ⎜ mv ⎟ = N A kT
⎝2
⎠ 2
U = total internal energy per mole, NA = Avogadro’s number, m =
mass of the gas molecule, k = Boltzmann constant, T = temperature
Molar Heat Capacity at Constant Volume
dU
3
3
Cm =
=
N Ak =
R
2
2
dT
Cm = specific heat per mole at constant volume (J K-1 mole-1), U =
total internal energy per mole, R = gas constant
Definition: Heat capacity is the rise of internal energy per unit temperature
Degree of freedom (DF) = the method to absorb energy
Examples: Monoatomic gas has 3 DF
Diatomic gas has 5 DF
Maxwell’s principle of equipartition of energy :
Each DF has an average energy ½ kT
Monoatomic gas has 3 DF.
E=3/2 kT
Cm=3/2 R
Diatomic gas has 5 DF.
E=5/2 kT
Cm=5/2 R
Solids have 6 DF.
E=3kT
Cm=3 R
(Dulong - Petit rule)
TRANSLATIONAL MOTION
ROTATIONAL MOTION
Ix= 0
vx
vz
x
z
Iz
y
vy
Iy
y axis out of paper
Fig. 1.16: Possible translational and rotational motions of a diatomic
molecule. Vibrational motions are neglected.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
1 2 1 2 1 2
mv x + mv y + mvz
2
2
2
Monoatomic gas:
E=
Diatomic gas:
1 2 1 2 1 2 1
1
2
E = mv x + mv y + mvz + I yω y + I zω z2
2
2
2
2
2
KE =
3
kT
2
U =
5
kT
2
Iy,z = momenta of inertia, ωy,z = angular velocities
E=
(a)
1 2 1 2 1 2
mv x + mv y + mv z +
2
2
2
+
1
1
1
K x x2 + K y y2 + K z z 2
2
2
2
y
x
Kx,y,z = spring constant,
x, y, z = extensions of springs
z
(b)
Fig. 1.17 (a) The ball-and-spring model of solids in which the springs
represent the interatomic bonds. Each ball (atom) is linked to its neighbors
by springs. Atomic vibrations in a solid involve 3 dimensions. (b) An atom
vibrating about its equilibrium position stretches and compresses its springs
to the neighbors and has both kinetic and potential energy.
6
U = kT = 3kT
2
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Dulong - Petit Rule
U = 3kT
dU
-1
-1
Cm =
= 3R = 25 J K mol
dT
R = Na × k
Cm = specific heat per mole at constant volume (J K-1 mole-1), U =
total internal energy per mole, T = temperature, R = gas constant
Relative number of molecules
per unit velocity (s/km)
What is the distribution of molecule speeds ?
2.5
v*
va v
vrms
2
1.5
1
3
KE = kT
2
298 K (25 °C)
v*
va v
vrms
1000 K (727 °C)
0.5
0
0
500
1000
1500
2000
Speed (m/s)
Fig. 1.21: Maxwell-Boltzmann distribution of molecular speeds in
nitrogen gas at two temperatures. The ordinate is dN/(Ndv),the
fractional number of molecules per unit speed interval in (km/s)-1
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
How to determine the speed of molecule ?
Relative number of molecules
per unit velocity (s/km)
What is the influence of temperature ?
2.5
v*
va v
vrms
2
1.5
1
298 K (25 °C)
v*
va v
vrms
1000 K (727 °C)
0.5
0
0
500
1000
1500
2000
Speed (m/s)
Fig. 1.21: Maxwell-Boltzmann distribution of molecular speeds in
nitrogen gas at two temperatures. The ordinate is dN/(Ndv),the
fractional number of molecules per unit speed interval in (km/s)-1
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Maxwell-Boltzmann Distribution for
Molecular Speeds
⎛ mv 2 ⎞
⎛ m ⎞3/ 2 2
⎟⎟
⎟ v exp⎜⎜ −
nv = 4π N⎜
⎝ 2π kT ⎠
⎝ 2kT ⎠
nv = the velocity density function, N = total number of molecules, m
= molecular mass, k = Boltzmann constant, T = temperature, v =
velocity
Maxwell-Boltzmann Distribution
for Molecular Speeds
⎛ mv 2 ⎞
⎛ m ⎞3/ 2 2
⎟⎟
⎟ v exp⎜⎜ −
nv = 4π N⎜
⎝ 2π kT ⎠
⎝ 2kT ⎠
E=½mv2
nv = the velocity density function, N = total number of molecules, m
= molecular mass, k = Boltzmann constant, T = temperature, v =
velocity
Maxwell-Boltzmann Distribution
for Translational Kinetic
Energies
nE =
⎛ 1 ⎞ 3 / 2 1/ 2
⎛ E⎞
N ⎜ ⎟ E exp ⎜ − ⎟
⎝ kT ⎠
π ⎝ kT ⎠
2
nE = number of atoms per unit volume per unit energy at an energy E,
N = total number of molecules per unit volume, k = Boltzmann
constant, T = temperature.
Boltzmann Energy
Distribution
⎛ E⎞
nE
= C exp⎜ − ⎟
⎝ kT ⎠
N
nE = number of atoms per unit volume per unit energy at an energy E, N = total
number of atoms per unit volume in the system, C = a constant that depends on
the specific system, k = Boltzmann constant, T = temperature
Number of atoms per unit energy, nE
Where lies the average kinetic energy ?
Average KE at T1.
T1
Average KE at T2
T2 > T1
EA
Energy, E
Fig. 1.22: Energy distribution of gas molecules at two different
temperatures. The number of molecules that have energies greater
than EA is the shaded area. This area depends strongly on the
temperature as exp(-EA/kT).
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
What is “thermal equilibrium”?
E=
1 2 1 2 1 2
mv x + mv y + mv z +
2
2
2
+
1
1
1
K x x2 + K y y2 + K z z 2
2
2
2
SOLID
GAS
M
KE=3/2 kT
1 2 1 2 1 2
mv x + mv y + mvz
2
2
2
1
1
+ I yω y2 + I zω z2
2
2
E=
V
v
m
Gas Atom
KE=3/2 kT
Fig. 1.23: Solid in equilibrium in air. During collisions between the
gas and solid atoms, kinetic energy is exchanged.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
What is “thermal equilibrium”?
SOLID
GAS
KE=3/2 kT
M
V
KE=3/2 kT
v
m
Gas Atom
Heat = amount of energy transferred from kinetic energy of the
atoms in solid to the kinetic energy of gas molecules
In equilibrium heat transfer = 0
Mechanical “noise”
Compression
Equilibrium
Extension
Δx
m Δx = 0
Compression
Mean displacement
Δx = 0
Instantaneous potential
energy
Δx < 0
PE(t)=½K(Δx)2
m
Mean potential energy
½K(Δx)2=½kT
Δx> 0
Extension
m
t
Fig.1.24: Fluctuations of a mass attached to a spring due
to random bombardment by air molecules
(Δx )rms
kT
=
K
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
K = spring constant, T = temperature, (Δx)rms = rms
value of the fluctuations of the mass about its
equilibrium position.
Electrical noise
A
B
B
A
v=0V
v = -3 μV
Voltage, v(t)
B
A
Time
v = +5 μV
Fig.1.25: Random motion of conduction electrons in a conductor results
in electrical noise.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Average value = 0
Average power = ?
Root Mean Square Noise Voltage Across a Resistance
1. Instantaneous energy
Current
v(t)
Electron
Flow
R
E(t)=½ C V(t)2
2. Mean energy
C
Time
1
1
E = CV (t ) 2 = kT
2
2
3. RMS voltage
kT
V (t ) 2 =
C
Electron
Flow
Current
4. Bandwidth
Fig.1.26: Charging and discharging of a capacitor by a conductor due
to the random thermal motions of the conduction electrons.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Vrms = [4kTRB]1/2
B=
1
2πRC
Johnson resistor noise equation
R = resistance, B = bandwidth, Vrms = root mean square
noise voltage, k = Boltzmann constant, T = temperature
Root Mean Square Noise Voltage
Across a Resistance
E(t)=½ C V(t)2
Instantaneous energy
1
1
E = CV (t ) 2 = kT
2
2
Mean energy and equipartition principle
V (t ) 2 =
B=
kT
C
1
2πRC
RMS voltage
Bandwidth
Vrms = [4kTRB]1/2
Johnson resistor noise equation
R = resistance, B = bandwidth, Vrms = root mean square noise voltage,
k = Boltzmann constant, T = temperature

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