physics 2 week 7 chapter 3 the kinetic theory of … 8...(a macroscopic quantity) depends on the...
TRANSCRIPT
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Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases
3.1.1. Experimental Laws and the Equation of State
3.1.2. Molecular Model of an Ideal Gas
3.2. Mean Free Path
3.3. The Boltzmann Distribution Law and
The Distribution of Molecular Speeds
3.4. The Molar Specific Heats of an Ideal Gas
3.5. The Equipartition-of-Energy Theorem
3.6. The Adiabatic Expansion of an Ideal Gas
Physics 2 – week 7
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• Number of moles in a sample:
AN
Nn N is the number of molecules
A
samplesample
mN
M
M
Mn
M is the molar mass (the mass of 1 mol)
m is the mass of one molecule
nRTpV p is the absolute pressure (Pa)
T is the temperature (in K)
11Kmol J 13.8R
• Ideal gas law:
NkTpV
1-23
A
K J1038.1N
Rk The Boltzmann constant (k)
Review
or Gas constant (R)
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f
i
V
VpdVW
:)(isochoricconstant V 1) If 0W
:(isobaric) constantp 2) If Vp)Vp(VW if
Work done by the gas:
3) If :l)(isotherma constantT
i
f
V
Vln nRTW
Review
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13. A sample of an ideal gas is taken through the cyclic process abca shown in the figure below; at point a, T=200 K. (a) How many moles of gas are in the sample? (b) the temperature of the gas at point b and point c, (c) the net energy added to the gas as heat during the cycle?
RT
pVnnRTpV
(a) Applying the equation of state:
At point a, p=2.5 kN/m2 or 2500 N/m2; V=1 m3.
(mol) 5.120031.8
12500
n
(b) 5.12 nRT
Vp
T
VpnRTpV
b
bb
a
aa
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(c) Applying the first law of thermodynamics:
WQE
For a closed cycle, E=0: WQ
W: work done by the system.
))((2
1abcb VVppW
(J) 10520.50002
1 3W
(K) 0605.12
32500
nR
VpT cc
c
Point b, p=7.5 kN/m2; V=3 m3
(K) 01805.12
37500
nR
VpT bb
b
Point c, p=2.5 kN/m2; V=3 m3
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3.1.2. Molecular Model for an Ideal Gas
In this model:
1. The molecules obey Newton's laws of motion.
2. The molecules move in all direction with equal probability.
3. There is no interactions between molecules (no collisions between molecules).
4. The molecules undergo elastic collisions with the walls.
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a. Pressure, Temperature, and RMS Speed
First, we consider a cubical box of edge length L, containing n moles of an ideal gas. A molecule of mass m and velocity v collide with the shaded wall.
Key question: What is the connection between the pressure p exerted by the gas and the speed of the molecules?
Problem: Let n moles of an ideal gas be confined in a cubical box of volume V. The walls of the box are held at temperature T.
For an elastic collision, the
particle’s momentum (=m.v) along the x axis is reserved and change with an amount:
xxxx 2mv)(mv)mv(Δp
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The average rate at which momentum is delivered to the shaded wall by this molecule:
L
mv
2L/v
2mv
Δt
Δp 2
x
x
xx
dt
pd
dt
)vd(m
dt
vdmamF
Recall: L
mvF
2
xx,1
The pressure exerted on the wall by this single molecule:
2
x,1
1L
Fp
For N molecules, the total pressure p:
2
2
Nx,
2
x,2
2
x,1
2
x
L
/Lmv.../Lmv/Lmv
L
Fp
Travel time b/w 2 walls with a speed v
Note: Pressure is the force applied perpendicular to the surface of an object
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)v...vv(L
mp 2
Nx,
2
x,2
2
x,13
The average value of the square of the x components of all the molecular speeds:
N
v...vv 2
Nx,
2
x,2
2
x,12
xv
2
x3
A vL
nmNp
gas theof massmolar the:mNM ASince
2
xvV
nMp :LV 3
For any molecule: 2
z
2
y
2
x
2 vvvv
As all molecules move in random directions: 22
x v3
1v
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2v3V
nMp
The square root of is called the root-mean-square speed: 2v
rms
2 vv
3V
nMvp
2
rms
Combining with the equation of state: nRTpV
M
3RTvrms
This relationship shows us how the pressure of the gas (a macroscopic quantity) depends on the speed of the molecules (a microscopic quantity)
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b. Translational Kinetic Energy
Consider a single molecule of an ideal gas moving around in the box (see Section a) .
2rms
22 mv2
1v
2
1
2
1K mmv
M/m
3RT
2
1
M
3RT
2
1K
m
A2N
3RTK
The Boltzmann constant k: AN
Rk
kT2
3K
22z
2y
2x v
3
1vvv kT
2
1vm
2
1vm
2
1vm
2
1 2z
2y
2x
Kdoes not depend on the mass of the molecule
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24. At 273 K and 1.0 x 10-2 atm, the density of a gas is 1.24 x 10-5 g/cm3. (a) Find vrms for the gas molecules. (b) Find the molar mass of the gas and (c) identify the gas (hint: see Table 19-1).
)1(3
M
RTvrms
Root-mean-square speed:
)2(n
VM
V
nM
V
M gas
(1) and (2):
p
V
nRTvrms
33
3235 kg/m1024.1g/cm1024.1
Pa1001.1atm100.1 32 p
m/s494rmsv
(a)
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)2(n
VM
Equation of state:
)3(nRTpV
p
RT
n
VM
g/mol28kg/mol028.0 M
From Table 19.1, the gas is nitrogen (N2)
(b)
(c)
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Homework: 18, 20, 23, 25, 27 (p. 531-532)
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Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases
3.1.1. Experimental Laws and the Equation of State
3.1.2. Molecular Model of an Ideal Gas
3.2. Mean Free Path
3.3. The Boltzmann Distribution Law and
The Distribution of Molecular Speeds
3.4. The Molar Specific Heats of an Ideal Gas
3.5. The Equipartition-of-Energy Theorem
3.6. The Adiabatic Expansion of an Ideal Gas
Physics 2 – week 8
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3.2. Mean Free Path
3.2.1 Concept
•A molecule traveling through a gas changes both speed and direction as it elastically collides with other molecules in its path.
•Between collisions, the molecules moves in a straight line at constant speed.
•The mean free path is the average
distance traversed by a molecule between
collisions.
V
N
1
density
1λ
V
N is the number of molecules per unit volume or the density of molecules
molecules ofnumber theis N
gas theof volume theis V where
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To count the number of collisions: We further consider that this single molecule has an equivalent radius of d and all the other molecules are points (see cartoons next slides for an equivalent problem).
Our goal: Estimate of of a single molecule.
Assumptions: + Our molecule is traveling with a constant speed v and all the other molecules are at rest. + All molecules are spheres of diameter d a collision occurs as the centers of 2 molecules come within a distance d.
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= 1 collision
d
= 1 collision . d
Equivalent problem
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1st collision
2nd collision
3rd collision
Equivalent problem
d
2d
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The number of collisions = the number molecules lie in a cylinder of length vt and cross-sectional area d2
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If all the molecules are moving:
V
Nd 22
1
Using the equation of state: pV = NkT
pd
kT22
The average time between collisions (the mean free time):
vpd
kT
vt
22
The average time between collisions (the mean free time):
The frequency of collisions:
tf
1
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3.3. The Boltzmann Distribution Law and the Distribution of Molecular Speeds
The Boltzmann distribution law: if the energy is associated with some state or condition of a system is then the frequency with which that state or condition occurs, or the probability of its occurrence is proportional to:
kTe /
constantBoltzmann the:k
Many of the most familiar laws of physical chemistry are special cases of the Boltzmann distribution law:
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P(v)dv is the fraction of molecules with speeds in the infinitesimal range (v,v+dv).
1)(0
dvvP
The fraction of molecules with speeds from v1 to v2:
2
1)(frac
v
vdvvP
3.3.1. The distribution of molecular speeds (or the Maxwell speed distribution law):
Let M be the molar mass of the gas, v be the molecular speed, and P(v) be the speed distribution function:
)1(2
4)( 2/22
2/3
kTMvevRT
MvP
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Average, RMS, and Most Probable Speeds
)2()(0
dvvvPvThe average speed:
from (1) & (2): M
RTv
8
0
22 )( dvvPvv
M
RTv
32
The root-mean-square speed:
M
RTvvrms
32
The most probable speed is the speed at which P(v) is maximum:
0)(
dv
vdP
M
RTvP
2
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3.3.2. The barometric distribution law:
This law gives the number density (h), i.e. number of molecules per unit volume, of an ideal gas of uniform temperature T as a function of height h in the field of the Earth’s gravity.
kThhmgehh
/)0(
0 )()(
where h0 is an arbitrary fixed reference height; m is the mass of a molecule.
Abell 1982 nasa.gov
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Homework: 28, 32, 33, 40 (Page. 531-532)
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Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases
3.1.1. Experimental Laws and the Equation of State
3.1.2. Molecular Model of an Ideal Gas
3.2. Mean Free Path
3.3. The Boltzmann Distribution Law and
The Distribution of Molecular Speeds
3.4. The Molar Specific Heats of an Ideal Gas
3.5. The Equipartition-of-Energy Theorem
3.6. The Adiabatic Expansion of an Ideal Gas
Physics 2 – week 9
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3.4. The Molar Specific Heats of an Ideal Gas
Let’s consider our ideal gas of n moles that is a monatomic gas, which has individual atoms, e.g. helium, argon, neon. For a single atom, the average translational KE:
kTK2
3
The internal energy Eint of the gas (no rotational KE for monatomic gases):
nRTnkTKEN
2
3N
2
3A
1int
Recall molar specific heat: TCnQ
a. Molar specific heat at constant volume:
• Consider n moles of an ideal gas at state i: p, T, and fixed V state f: p+p, T+T
TnCQ V
CV is a constant and called the molar specific heat at constant volume.
Þ DEint =3
2nRDT
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TnRWTnCWQE V 2
3int
11
V K mol J5.122
3C 0 Since RW
Note: For diatomic and polyatomic gases, their CV is greater than that of monatomic gases.
TnCE V int
So, the change in internal energy can be calculated by:
TnRE 2
3int
or
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b. Molar specific heat at constant pressure:
TnCQ p
WQE int
Cp is the molar specific heat at constant pressure.
TnRVpW
TnRTnCTnR p 2
3
RRRCp2
5
2
3
RCC Vp
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Checkpoint 4 (p. 522): The figure here shows 5 paths traversed by a gas on a p-V diagram. Rank the paths according to the change in internal energy of the gas, greatest first.
TnRE 2
3int
123 TTT
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Example: (Problem 8, page 530) Suppose 1.8 mol of an ideal gas is taken from a volume of 3.0 m3 to a volume of 1.5 m3 via an isothermal compression at 300C. (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?
(a) We have:
An isothermal process: T=constant
Work done by the gas for isotherm:
(b) Q<0: heat transferred from the gas
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3.5. The Equipartition-of-Energy Theorem
Every kind of molecule has a certain number f of degrees of freedom. For each degree of freedom in which a molecule can store energy, the average internal energy is per molecule. kT
2
1
Molecule Example
Degrees of freedom
Translational Rotational Total (f)
Monatomic He
Diatomic O2
Polyatomic CH4
3 0 3
3 2 5
3 3 6
RCC
Rf
C
Vp
V
2
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Six degrees of freedom Technical Aspects of robotics
wac.nsw.edu.au
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5 degrees of freedom of a diatomic molecule
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3.6. The Adiabatic Expansion of an Ideal Gas
What is an adiabatic process?: a process for which Q = 0
constantpV
Vp/CC where
nRTpV
constant1 TV
Proof of the equations above, see p. 526-527 (homework)
Equation of state:
Free expansions:
0 :Recall WQ
fi TTE 0int
ffii VpVp
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Homework: 42, 44, 46, 54, 56, 78 (p. 533-535)