Physics 2A Chapter 13
Transcript of Physics 2A Chapter 13
• Temperature ~ Average KE of each particle
• Particles have different speeds
• Gas Particles are in constant RANDOM motion
• Average KE of each particle is: 3/2 kT
• Pressure is due to momentum transfer
Speed ‘Distribution’ at
CONSTANT Temperature
is given by the
Maxwell Boltzmann
Speed Distribution
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A single molecule follows a
zig-zag path through a gas
as it collides with other
molecules.
The average distance
between the collisions is
called the mean free path:
(N/V) is the number density of the gas in m−3.
r is the the radius of the molecules when modeled as
hard spheres; for many common gases r ≈ 10−10 m.
Mean Free Path
Slide 18-20
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The temperature of a rigid container of oxygen gas
(O2) is lowered from 300C to 0C. As a result, the
mean free path of oxygen molecules
A. Increases.
B. Is unchanged.
C. Decreases.
QuickCheck 18.1
Slide 18-21
© 2013 Pearson Education, Inc.
The temperature of a rigid container of oxygen gas
(O2) is lowered from 300C to 0C. As a result, the
mean free path of oxygen molecules
A. Increases.
B. Is unchanged.
C. Decreases.
QuickCheck 18.1
Slide 18-22
λ depends only on N/V, not T.
Pressure and Kinetic Energy
• Assume a container is a cube with edges d.
• Look at the motion of the molecule in terms of its velocity components and momentum and the average force
• Pressure is proportional to the number of molecules per unit volume (N/V) and to the average translational kinetic energy of the molecules.
• This equation also relates the macroscopic quantity of pressure with a microscopic quantity of the average value of the square of the molecular speed
• One way to increase the pressure is to increase the number of molecules per unit volume
• The pressure can also be increased by increasing the speed (kinetic energy) of the molecules
___22 1
3 2o
NP m v
V
Molecular Interpretation of
Temperature• We can take the pressure as it relates to the kinetic
energy and compare it to the pressure from the
equation of state for an ideal gas
• Temperature is a direct measure of the average
molecular kinetic energy
___2 B2 1
3 2
N nRT Nk TP mv
V V V
___2
B
1 3
2 2om v k T
2 3rms
kTv v
m Root-mean-square speed:
The Kelvin Temperature of
an ideal gas is a measure of
the average translational
kinetic energy per particle:
k =1.38 x 10-23 J/K Boltzmann’s Constant
___2
tot trans B
1 3 3 3
2 2 2 2K N mv Nk T nRT PV
Total Kinetic Energy
• The total kinetic energy is just N times the kinetic
energy of each molecule
• If we have a gas with only translational energy, this is
the internal energy of the gas
• This tells us that the internal energy of an ideal gas
depends only on the temperature
___2
tot trans B
1 3 3
2 2 2K N mv Nk T nRT
Kinetic Theory Problem
A 5.00-L vessel contains nitrogen gas at
27.0C and 3.00 atm. Find (a) the total
translational kinetic energy of the gas
molecules and (b) the average kinetic energy
per molecule.
Kinetic Theory ProblemCalculate the RMS speed of an oxygen molecule
in the air if the temperature is 5.00 °C.
The mass of an oxygen molecule is 32.00 u
(k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)
3rms
kTv
m What is m?
m is the mass of one
oxygen molecule in kg.
What is u?
How do we get the mass in kg?
Kinetic Theory ProblemCalculate the RMS speed of an oxygen molecule
in the air if the temperature is 5.00 °C.
The mass of an oxygen molecule is 32.00 u
(k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)
3rms
kTv
m
23
27
3(1.38 10 / )278
(32 )(1.66 10 / )
x J K K
u x kg u
466 /m s
What is m?m is the mass of one
oxygen molecule.
Is this fast? YES!Speed of
sound:
343m/s!
Distribution of Molecular Speeds• The observed speed distribution of gas
molecules in thermal equilibrium is
shown at right
• NV is called the Maxwell-Boltzmann
speed distribution function
• mo is the mass of a gas molecule, kB is
Boltzmann’s constant and T is the
absolute temperature
Ludwig Boltzmann
1844 – 1906
Austrian physicist
Contributed to
Kinetic Theory of Gases
Electromagnetism
Thermodynamics
Pioneer in statistical mechanics
2
3 / 2
/ 22
B
42
Bmv k ToV
mN N v e
k T
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A rigid container holds both hydrogen gas (H2)
and nitrogen gas (N2) at 100C. Which statement
describes their rms speeds?
A. vrms of H2 < vrms of N2.
B. vrms of H2 = vrms of N2.
C. vrms of H2 > vrms of N2.
QuickCheck 18.3
Slide 18-34
© 2013 Pearson Education, Inc.
A rigid container holds both hydrogen gas (H2)
and nitrogen gas (N2) at 100C. Which statement
describes their rms speeds?
A. vrms of H2 < vrms of N2.
B. vrms of H2 = vrms of N2.
C. vrms of H2 > vrms of N2.
QuickCheck 18.3
Slide 18-35
More Kinetic Theory Problems
A gas molecule with a molecular mass of 32.0 u has a speed of 325 m/s. What is the temperature of the gas molecule?
A) 72.0 K B) 136 K C) 305 K D) 459 K
E) A temperature cannot be assigned to a single molecule.
Temperature ~ Average KE of all particles
Molecular Interpretation of
Temperature
• Simplifying the equation relating
temperature and kinetic energy gives
• This can be applied to each direction,
– with similar expressions for vy and vz
___2
B
1 3
2 2om v k T
___2
B
1 1
2 2xmv k T
Equipartition of Energy• Each translational degree of freedom contributes an
equal amount to the energy of the gas
– In general, a degree of freedom refers to an independent means by which a molecule can possess energy
• Each degree of freedom contributes ½kBT to the
energy of a system, where possible degrees of
freedom are those associated with translation,
rotation and vibration of molecules
Monatomic and Diatomic GasesThe thermal energy of a monatomic gas of N atoms is
A diatomic gas has more thermal energy than a monatomic
gas at the same temperature because the molecules have
rotational as well as translational kinetic energy.
Molar Specific HeatsIsobaric requires MORE HEAT than Isochoric for the
same change in Temperature!!!!
The total change in thermal
energy for ANY PROCESS,
due to work and heat, is:
This applies to all ideal gases, not
just monatomic ones! WOW!
CP and CV Note that for all ideal gases:
whereR = 8.31 J/mol K is the universal gas constant.
Slide 17-80
Important Concepts
Agreement with Experiment: Diatomic Hydrogen
acts like a monatomic gas at low temperature!
•Molar specific heat is a function of
temperature.
•At low temperatures, a diatomic gas
acts like a monatomic gas.
– CV = 3/2 R
•At about room temperature, the value
increases to CV = 5/2 R.
– This is consistent with adding
rotational energy but not
vibrational energy.
•At high temperatures, the value
increases to CV = 7/2 R.
– This includes vibrational
energy as well as rotational
and translational.
Quantization of Energy.
•To explain the results of the various molar
specific heats, we must use some quantum
mechanics.
– Classical mechanics is not sufficient
•This energy level diagram shows the rotational
and vibrational states of a diatomic molecule.
•The lowest allowed state is the ground state.
•The vibrational states are separated by larger
energy gaps than are rotational states.
•At low temperatures, the energy gained during
collisions is generally not enough to raise it to the
first excited state of either rotation or vibration.
Section 21.4
In a constant-volume process, 209 J of
energy is transferred by heat to 1.00 mol
of an ideal monatomic gas initially at
300 K. Find (a) the increase in internal
energy of the gas, (b) the work done on
it, and (c) its final temperature
What additional kind of energy makes CV larger
for a diatomic than for a monatomic gas?
A. Charismatic energy.
B. Translational energy.
C. Heat energy.
D. Rotational energy.
E. Solar energy.
Reading Question 18.2
Slide 18-12
What additional kind of energy makes CV larger
for a diatomic than for a monatomic gas?
A. Charismatic energy.
B. Translational energy.
C. Heat energy.
D. Rotational energy.
E. Solar energy.
Reading Question 18.2
Slide 18-13
An adiabatic process is one for which:
where:
Adiabats are steeper than
hyperbolic isotherms because
only work is being done to
change the Temperature. The
temperature falls during an
adiabatic expansion, and rises
during an adiabatic
compression.
Adiabatic Processes
Slide 17-88
Adiabatic Processes: Q=0
Ti Vig-1 = Tf Vf
g-1
A 4.00-L sample of a nitrogen gas confined to a
cylinder, is carried through a closed cycle. The
gas is initially at 1.00 atm and at 300 K. First, its
pressure is tripled under constant volume. Then,
it expands adiabatically to its original pressure.
Finally, the gas is compressed isobarically to its
original volume. (a) Draw a PV diagram of this
cycle. (b) Find the number of moles of the gas. (c)
Find the volumes and temperatures at the end of
each process (d) Find the Work and heat for each
process. (e) What was the net work done on the
gas for this cycle?
Adiabatic Processes for an
Ideal Gas• An adiabatic process is one in which no energy is
transferred by heat between a system and its surroundings (think styrofoam cup)
• Assume an ideal gas is in an equilibrium state and so PV = nRT is valid
• The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV g = constant
g = CP / CV is assumed to be constant
All three variables in the ideal gas law (P, V, T ) can change during an adiabatic process
Special Case: Adiabatic Free
Expansion
• This is an example of adiabatic free expansion
• The process is adiabatic because it takes place in an insulated container
• Because the gas expands into a vacuum, it does not apply a force on a piston and W = 0
• Since Q = 0 and W = 0, DEint = 0 and the initial and final states are the same and no change in temperature is expected. – No change in temperature is expected
The second law of thermodynamics says that
A. The entropy of an isolated system never
decreases.
B. Heat never flows spontaneously from
cold to hot.
C. The total thermal energy of an isolated
system is constant.
D. Both A and B.
E. Both A and C.
Reading Question 18.3
Slide 18-15
A. Both microscopic and macroscopic processes
are reversible.
B. Both microscopic and macroscopic processes
are irreversible.
C. Microscopic processes are reversible and
macroscopic processes are irreversible.
D. Microscopic processes are irreversible and
macroscopic processes are reversible.
In general,
Reading Question 18.4
Slide 18-16
Systems A and B are both monatomic gases. At this instant,
A. TA > TB.
B. TA = TB.
C. TA < TB.
D. There’s not enough information to compare their
temperatures.
QuickCheck 18.6
Slide 18-51
Systems A and B are both monatomic gases. At this instant,
A. TA > TB.
B. TA = TB.
C. TA < TB.
D. There’s not enough information to compare their
temperatures.
QuickCheck 18.6
Slide 18-52
A has the larger average energy per atom.