temperature ~ average ke of each particle gas particles …lwillia2/41/41ch2122.pdf · •...
TRANSCRIPT
• 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
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
B
2 1
3 2
NP mv nRT Nk T
V
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
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.
Hot Question Suppose you apply a flame to 1 liter of water for a certain
time and its temperature rises by 10 degrees C. If you apply
the same flame for the same time to 2 liters of water, by how
much will its temperature rise?
a) 1 degree b) 5 degrees c) 10 degrees d) zero degrees
Ludwig Boltzmann or Dean Gooch?
• 1844 – 1906
• Austrian physicist
• Contributed to
– Kinetic Theory of Gases
– Electromagnetism
– Thermodynamics
• Pioneer in statistical
mechanics
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
2
3 / 2
/ 22
B
42
Bmv k ToV
mN N v e
k T
Molecular Speeds and
Collisions
Speed Summary
• Root mean square speed
• The average speed is somewhat lower than the rms speed
• The most probable speed, vmp is the speed at which the distribution curve reaches a peak
• vrms > vavg > vmp
B Brms
31.73
o o
k T k Tv
m m
B Bmp
21.41
k T k Tv
m m
B B
avg
81.60
o o
k T k Tv
m m
Some Example vrms Values
At a given temperature, lighter molecules move faster, on
the average, than heavier molecules
Speed Distribution • The peak shifts to the right
as T increases – This shows that the average
speed increases with increasing temperature
• The asymmetric shape occurs because the lowest possible speed is 0 and the highest is infinity
23/ 2 1/ 2 rmskT KE mv
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
Kinetic Theory Problem Calculate 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 Problem Calculate 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!
A cylinder contains a mixture of helium
and argon gas in equilibrium at 150°C.
(a) What is the average kinetic energy
for each type of gas molecule?
(b) What is the root-mean-square speed
of each type of molecule?
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
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
• With complex molecules, other contributions to
internal energy must be taken into account
• One possible energy is the translational motion of
the center of mass
• Rotational motion about the various axes also
contributes
• There is kinetic energy and potential energy
associated with the vibrations
Monatomic and Diatomic Gases The 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 Heat
• We define specific heats for two processes
that frequently occur:
– Changes with constant pressure
– Changes with constant volume
• Using the number of moles, n, we can
define molar specific heats for these
processes
• Molar specific heats:
– Q = nCV DT for constant-volume processes
– Q = nCP DT for constant-pressure processes
Ideal Monatomic Gas
• Therefore, DEint = 3/2 nRT
DE is a function of T only
• In general, the internal energy of an ideal
gas is a function of T only
– The exact relationship depends on the type of
gas
• At constant volume, Q = DEint = nCV DT
– This applies to all ideal gases, not just
monatomic ones
Ratio of Molar Specific Heats
• We can also define the ratio of molar specific heats
• Theoretical values of CV , CP , and g are in excellent agreement for monatomic gases
• But they are in serious disagreement with the values for more complex molecules
– Not surprising since the analysis was for monatomic gases
5 / 21.67
3 / 2P
V
C R
C Rg
Agreement with Experiment • 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
Sample Values of Molar Specific
Heats
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
Molar Specific Heats of Other
Materials
• The internal energy of more complex gases
must include contributions from the
rotational and vibrational motions of the
molecules
• In the cases of solids and liquids heated at
constant pressure, very little work is done,
since the thermal expansion is small, and CP
and CV are approximately equal
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
Adiabatic Process
• The PV diagram shows
an adiabatic expansion
of an ideal gas
• The temperature of the
gas decreases
– Tf < Ti in this process
• For this process
Pi Vig = Pf Vf
g and
Ti Vig-1 = Tf Vf
g-1
A 2.00-mol sample of a diatomic ideal
gas expands slowly and adiabatically
from a pressure of 5.00 atm and a
volume of 12.0 L to a final volume of
30.0 L.
(a) What is the final pressure of the gas?
(b) What are the initial and final
temperatures?
(c) Find Q, W, and DEint.
Cyclic Processes
• A cyclic process is one that starts and ends in the same state
– On a PV diagram, a cyclic process appears as a closed curve
• DEint = 0, Q = -W
• In a cyclic process, the net work done on the system per cycle equals the area enclosed by the path representing the process on a PV diagram
intE Q WD
Isothermal Process • At right is a PV diagram of an isothermal
expansion
• The curve is a hyperbola
• The curve is called an isotherm
• The curve of the PV diagram
indicates PV = constant
– The equation of a hyperbola
• Because it is an ideal gas and the
process is quasi-static,
PV = nRT and
f f f
i i i
V V V
V V V
nRT dVW P dV dV nRT
V V
ln i
f
VW nRT
V
Isothermal Process
• An isothermal process is one that occurs at
a constant temperature
• Since there is no change in temperature,
DEint = 0
• Therefore, Q = - W
• Any energy that enters the system by heat
must leave the system by work
intE Q WD f
i
V
VW P dV PV nRT
Isobaric Processes
• An isobaric process is one that occurs at a
constant pressure
• The values of the heat and the work are
generally both nonzero
• The work done is W = P (Vf – Vi) where P
is the constant pressure
intE Q WD f
i
V
VW P dV PV nRT
Isovolumetric Processes
• An isovolumetric process is one in which there is
no change in the volume
• Since the volume does not change, W = 0
• From the first law, DEint = Q
• If energy is added by heat to a system kept at
constant volume, all of the transferred energy
remains in the system as an increase in its internal
energy
intE Q WD f
i
V
VW P dV PV nRT
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
Thermo Processes • Adiabatic
– No heat exchanged
– Q = 0 and DEint = W
• Isobaric
– Constant pressure
– W = P (Vf – Vi) and DEint = Q + W
• Isovolumetric
– Constant Volume
– W = 0 and DEint = Q
• Isothermal
– Constant temperature
DEint = 0 and Q = -W
intE Q WD
ln i
f
VW nRT
V
A gas is taken through the cyclic process as shown.
(a) Find the net energy transferred to the system by heat during one complete cycle. (b) What If? If the cycle is reversed—that is, the process follows the path ACBA—what is the net energy input per cycle by heat?
intE Q WD f
i
V
VW P dV
A sample of an ideal gas goes through the process as shown. From A to B, the process is adiabatic; from B to C, it is isobaric with 100 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kJ of energy leaving the system by heat. Determine the difference in internal energy E(B) – E(A).
intE Q WD f
i
V
VW P dV PV nRT
Important Concepts
Heat Engine
• A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work
• A heat engine carries some working substance through a cyclical process
• The working substance absorbs energy by heat from a high temperature energy reservoir (Qh)
• Work is done by the engine (Weng)
• Energy is expelled as heat to a lower temperature reservoir (Qc)
DEint = 0 for the entire cycle
eng
h
We
Q
DEint = 0 for the entire cycle
Thermal Efficiency of a Heat
Engine
• Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature
• We can think of the efficiency as the ratio of what you gain to what you give
eng1h c c
h h h
W Q Q Qe
Q Q Q
DEint = 0 for the entire cycle
Rank in order, from largest to smallest, the work Wout
performed by these four heat engines.
A. Wb > Wa > Wc > Wd
B. Wb > Wa > Wb > Wc
C. Wb > Wa > Wb = Wc
D. Wd > Wa = Wb > Wc
E. Wd > Wa > Wb > Wc
eng1h c c
h h h
W Q Q Qe
Q Q Q
Rank in order, from largest to smallest, the work Wout
performed by these four heat engines.
A. Wb > Wa > Wc > Wd
B. Wb > Wa > Wb > Wc
C. Wb > Wa > Wb = Wc
D. Wd > Wa = Wb > Wc
E. Wd > Wa > Wb > Wc
eng1h c c
h h h
W Q Q Qe
Q Q Q
Perfect Heat Engine
• No energy is expelled to the cold reservoir
• It takes in some amount of energy and does an equal amount of work
• e = 100%
• It is an impossible engine
Could this heat
engine be built?
A. Yes.
B. No.
C. It’s impossible to tell without knowing
what kind of cycle it uses.
e
H C
H
W Q Q
W
Q
and 1c c c
c
h h h
Q T Te
Q T T
Could this heat
engine be built?
A. Yes.
B. No.
C. It’s impossible to tell without knowing
what kind of cycle it uses.
e
H C
H
W Q Q
W
Q
and 1c c c
c
h h h
Q T Te
Q T T
Analyze this engine to determine (a) the net work done per cycle, (b) the engine’s thermal efficiency and (c) the engine’s power output if it runs at 600 rpm. Assume the gas is monatomic and follows the ideal-gas process above.
Gasoline Engine
• In a gasoline engine, six processes occur
during each cycle
• For a given cycle, the piston moves up and
down twice
• This represents a four-stroke cycle
• The processes in the cycle can be
approximated by the Otto cycle
The Conventional Gasoline
Engine
Gasoline Engine – Intake Stroke
• During the intake stroke,
the piston moves
downward
• A gaseous mixture of air
and fuel is drawn into the
cylinder
• Energy enters the system
as potential energy in the
fuel
Gasoline Engine – Compression
Stroke
• The piston moves upward
• The air-fuel mixture is
compressed adiabatically
• The temperature increases
• The work done on the gas is
positive and equal to the
negative area under the curve
Gasoline Engine – Spark
• Combustion occurs when the
spark plug fires
• This is not one of the strokes
of the engine
• It occurs very quickly while
the piston is at its highest
position
• Conversion from potential
energy of the fuel to internal
energy
Gasoline Engine – Power Stroke
• In the power stroke, the gas expands adiabatically
• This causes a temperature drop
• Work is done by the gas
• The work is equal to the area under the curve
Gasoline Engine – Valve Opens
An exhaust valve opens as the piston reaches
its bottom position
• The pressure drops suddenly
• The volume is approximately constant
– So no work is done
• Energy begins to be expelled from the
interior of the cylinder
Gasoline Engine – Exhaust
Stroke
• In the exhaust stroke, the
piston moves upward
while the exhaust valve
remains open
• Residual gases are
expelled to the
atmosphere
• The volume decreases
Otto Cycle
The Otto cycle
approximates the
processes occurring in an
internal combustion
engine
Otto Cycle Efficiency
• If the air-fuel mixture is assumed to be an
ideal gas, then the efficiency of the Otto cycle
is
g is the ratio of the molar specific heats
• V1 / V2 is called the compression ratio
1
1 2
11e
V Vg
Otto Cycle Efficiency, cont
• Typical values:
– Compression ratio of 8
g = 1.4
– e = 56%
• Efficiencies of real engines are 15% to 20%
– Mainly due to friction, energy transfer by conduction, incomplete combustion of the air-fuel mixture
g
D34. The compression ratio of an Otto cycle, as shown in Figure 22.13, is VA/VB = 8.00. At the beginning A of the compression process, 500 cm3 of gas is at 100 kPa and 20.0C. At the beginning of the adiabatic expansion the temperature is TC = 750C. Model the working fluid as an ideal gas with Eint = nCVT = 2.50nRT and
= 1.40. (a) Fill in this table to follow the states of the gas:
T (K) P (kPa) V (cm3) Eint
A 293 100 500
B
C 1 023
D
A
(b) Fill in this table to follow the processes:
Q (input) W(output) Eint
AB
BC
CD
DA
ABCDA
Heat Pumps and Refrigerators
• Heat engines can run in reverse
– This is not a natural direction of energy transfer
– Must put some energy into a device to do this
– Devices that do this are called heat pumps or
refrigerators
• Examples
– A refrigerator is a common type of heat pump
– An air conditioner is another example of a heat pump
Coefficient of Performance
• The effectiveness of a heat pump is
described by a number called the
coefficient of performance (COP)
• In heating mode, the COP is the ratio of the
heat transferred in to the work required
energy transferred at high tempCOP =
work done by heat pump
hQ
W
COP, Heating Mode
• COP is similar to efficiency
• Qh is typically higher than W
– Values of COP are generally greater than 1
– It is possible for them to be less than 1
• We would like the COP to be as high as
possible
COP, Cooling Mode
• In cooling mode, you “gain” energy from a
cold temperature reservoir
• A good refrigerator should have a high COP
– Typical values are 5 or 6
COP cQ
W
Carnot Engine – Carnot Cycle
A heat engine operating in an ideal, reversible cycle (now called a
Carnot cycle) between two reservoirs is the most efficient engine
possible. This sets an upper limit on the efficiencies of all other engines
Carnot Cycle, PV Diagram
• The work done by the
engine is shown by the
area enclosed by the
curve, Weng
• The net work is equal
to |Qh| – |Qc|
DEint = 0 for the entire
cycle
Efficiency of a Carnot Engine
• Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs
• Temperatures must be in Kelvins
• All Carnot engines operating between the same two temperatures will have the same efficiency
and 1c c c
c
h h h
Q T Te
Q T T
Carnot’s Theorem
• No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs
– All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle
– The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc.
Notes About Carnot Efficiency
• Efficiency is 0 if Th = Tc
• Efficiency is 100% only if Tc = 0 K
– Such reservoirs are not available
– Efficiency is always less than 100%
• The efficiency increases as Tc is lowered and as Th
is raised
• In most practical cases, Tc is near room
temperature, 300 K
– So generally Th is raised to increase efficiency
Carnot Cycle in Reverse
• Theoretically, a Carnot-cycle heat engine
can run in reverse
• This would constitute the most effective
heat pump available
• This would determine the maximum
possible COPs for a given combination of
hot and cold reservoirs
Carnot Heat Pump COPs
• In heating mode:
• In cooling mode:
C
h h
h c
Q TCOP
W T T
c cC
h c
Q TCOP
W T T
26. A heat pump, shown in Figure P22.26, is
essentially an air conditioner installed backward. It
extracts energy from colder air outside and deposits it in
a warmer room. Suppose that the ratio of the actual
energy entering the room to the work done by the
device’s motor is 10.0% of the theoretical maximum
ratio. Determine the energy entering the room per joule
of work done by the motor, given that the inside
temperature is 20.0°C and the outside temperature is –
5.00°C.
Carnot cycle
10.100 0.100
Carnot efficiency
h hQ Q
W W
293 K0.100 0.100 1.17
293 K 268 K
h h
h c
Q T
W T T
1.17 joules of energy enter the room by heat for each joule of work done.
Can this refrigerator be built?
c cC
h c
Q TCOP
W T T
COP
H C
c
W Q Q
Q
W