Chapter 13. Maxwell 1831-1879 Boltzmann 1844-1906.
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Transcript of Chapter 13. Maxwell 1831-1879 Boltzmann 1844-1906.
Revision Task
Use the paper in front of you Summarize everything you can
remember about Gases from before summer
For example: Equations, definitions, scientists etc…
One law for all gases
2
34
1
volume V
Boyle’s law
pressure p
compress gas:pressure p increasesconstant temperature T
2
34
1
number N
Amount law
pressure p
add more molecules:pressure p increasesconstant temperature T
2
34
1
Pressure law
pressure p
heat gas:pressure p increasesconstant volume V
T/K
45.1
volume V
Charles’ law
heat gas:volume V increasesconstant pressure p
T/K
45.1
p1/V
pN
pT
VT
Combine therelationships into one
pN/V
pVNor
pVintroduceconstant k:
pVNkT
combine:
combine:
Combine unknown Nand k into measurablequantity R
Number of moleculesN not known
Nk can be measured:Nk = pV/T
For one mole, defineR = NAk
For n moles:pV = nRT
pVT
k = Boltzmann constantNA = Avogadro number(number of molecules per mole)R = molar gas constant = 8.31 J K –1 mol–1
measured from pV/T for one mole
When NA could be measured:
Avogadro number NA = 6.02 1023 particles mol–1
R = molar gas constant = NAk = 8.31 J K–1 mol–1
Boltzmann constant k = 1.38 10–23 J K –1 mol–1
combine:
4
5
0
1
2
3
105
N m–2
A stripped down view of the kinetic theory
general ideas ideas about a particle in a box
F = – pt
force equal torate of changeof momentum
p = – 2mv
t = 2cv
F = mv2
c
P = mv2
abc
P = mv2
V
P = Nmv2
V
P = Nmv2
V13
P = Nmv2
V13
particle mass, m ,speed, v, travelling2c between hits
A = a×b
box dimensions:a×b×c
V = a×b×c
P = FA
definition ofpressure
N particles
13
in each direction
randommovement
gas molecules movingat many speeds,special average, v2
Kinetic energy of gas molecules and the Boltzmann constant
compare these
Kinetic model
pV = Nmv2
Gas laws
pV = NkT
One molecule Many molecules
mv2 = kT12
32
kinetic energy permolecule = kT3
2
Nmv2 = NkT12
total kinetic energy ofmolecules = NkT3
2
32
mv2 = 3kT
total kinetic energy ofone mole ofmolecules U = RT
32
Internal kinetic energyof m olecu les o f onem ole a t T = 300 K
Boltzmann constant k
random thermalenergy of onemolecule is of orderkT
13
U = RT32
U = 3 .7 kJ m ol–1
for one moleN = NAR = NAkR = 8.31 J K–1 mol–1
Speed of a nitrogen molecule
Assume warm room temperature T = 300 K
mass of 1 mole of N2 = 28 10–3 kg mol–1
Avogadro constant NA = 6 1023 particles mol–1
Boltzmann constant k = 1.38 10–23 J K–1
kinetic energy of a molecule
from dynamics from kinetic model
mv212
kT32v2 = 3 kT
m
mass m of N2 molecule calculate speed
v = 500 m s–1 approximately
Air molecules (mostly nitrogen) at room temperature go as fast as bullets
m = mass of 1 mole of N2
Avogadro constant NA
m =
m = 4.7 10–26 kg
v2 = 3 1.4 10–23 J K–1 300 K
4.7 10–26 kg
v2 = 2.7 105 J kg–1 [same as (m s–1)2]28 10–3 kg mol–1
6 1023 mol–1
Hints for 90S1. A histogram is a bar chart where the area of a particular bar is related to
the total number in that range. As the range intervals are the same here this is not a problem.
2. Which is most common?
3. Work out the average of (100 + 100 + 100 + 100 + 100 + 150 + 150 + 150 + 150 + 150 + 150 + 150 + … + 600 + 600 + 600) / 150. This method is tedious and error-prone. Is there a quicker way?
4. Square each speed, multiply by the number of molecules with that speed, add all 11 categories up and divide by the total number of molecules. Don’t forget to take the square root of the total at the end.
10. The ideal equation is:
Density = mass / volume so eliminate the volume term.
13. You should show that the faster hydrogen and helium molecules travel at speeds comparable with the escape speed. What happens to these faster molecules? What happens to the slower ones that remain?
Transfers of energy to molecules in two ways
Hit the molecules yourself
molecules speeded up piston pushed in
work done = force distance
Let other molecules hit them
cool gas or other material hot wall
thermal transfer = mc
when both ways are used:
Engine designers arrange to transfer energy by way of heating and by way of doing work
change in internal energy U
work done W plus thermal transfer Q=
H ere the p is ton strikes m olecu les and g ives extram om entum and so extra k inetic energy.
energy transferred = work W done
H ere the m olecu les in the hot w a ll h it o ther m olecu leshard and on average g ive them extra k ine tic energy.
energy transferred = energy Q transferred thermally
U = W + Q
Looking at the kinetic energy of a single particle
Starting points
From the kinetic theory From experimental work with gases
P: pressure of the gasV: volume of the gasN: number of moleculesm: mass of one molecule
v2: mean square velocity
P: pressure of the gasV: volume of the gasn: number of molesT: temperature of gasR: universal gas constant
Note similaritiesrearrange
P = 3Nmv2
V1
PV = 3 Nmv21
PV = 2N
( 2 mv2 )13
rearrange
PV = nRT
Ek = 1
mv2
2
PV = 2N
( 2 mv2 )13
N
PV (mol–1) = 2 (Ek)3
Work with one mole
PV (mol–1) = RT
n
(Ek) = RT23rearrange
Ek = RT3
2
Ek = kT3
2
define, k, Boltzmann’s constant =RN
Boltzmann’s constant,k = 1.38 10–23 J K–1
For onemolecule
This is the energy for onemolecule, no matter whatit is, depending only ontemperature, and auniversal constant.
Something measured aboutthe Universe, turning out tobe important
.. ..