Transcript Slide 1 - University of Virginia

To the Members of Mr. Jefferson’s University:
By the driven dispositions that brought us to this
school, we hold ourselves back from “sticky”
situations much too often, preferring to not ask about
potentially dangerous relationships, eating habits, or
abuse of any kind, lest we be cast as meddlesome.
While it is much easier to lower our blinds and
ignore what might not be “our business,” the wellbeing of every individual within this community is,
in fact, our business. We write today in the hopes of
instilling a better understanding of the available
resources, particularly CAPS and 295-TALK. We
hope that students will come to see these lines as
places they feel really comfortable calling just to ask
any questions about student well-being, not just a
number to report suspicions, and we encourage you
to take advantage of them as a means to nourish the
UVA community.
We wish you the best of luck as you commence your
examination preparations.
Lecture 25
The Laws of
Thermodynamics
Announcements
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Assignment #13 is not due for credit
Reversible Thermal Processes
We will assume that all processes we discuss are
“quasi-static” – they are slow enough that the
system is always “in equilibrium” (fluid volumes
have the same temperature throughout, etc.)
We also assume they are reversible (frictionless pistons, etc.):
For a process to be reversible, it must be possible to return both the
system and its surroundings to the same states they were in before the
process began.
We will discuss 4 idealized processes with Ideal Gases:
•Constant Pressure
•Constant Volume
•Constant Temperature
•Q= 0 (adiabatic)
Constant pressure
Isobaric process
Work done by an expanding gas, constant pressure:
Work is area
under the PV
graph
Examples: piston against atmosphere, or
vertical piston with constant weight on top
so changing volume implies
changing temperature
imagining any general process as approximated by
a number of constant pressure processes:
Work is area under
the PV graph
Constant Volume
Isovolumetric process
If the volume stays constant, nothing moves and
no work is done.
Change in internal energy is
related only to the net heat input
so changing pressure implies
changing temperature
Constant Temperature
Isothermal processes
If the temperature is constant, the
pressure varies inversely with the
volume.
Constant Temperature
A system connected to a large heat reservoir is
usually thought to be held at constant temperature.
Volume can change, pressure can change, but the
temperature remains that of the reservoir.
T = constant
if W < 0 (work done on the system)
than Q<0 (heat flows out of the system)
W=Q
if W > 0 (work done by the system)
than Q>0 (heat flows out into the system)
Work in an Constant-Temperature Process
The work done is the area under the curve:
For you calculus junkies:
Adiabatic Process
An adiabatic process is one in which no heat
flows into or out of the system.
One way to ensure that a process is adiabatic is
to insulate the system.
Q=0
The adiabatic P-V curve is
similar to the isothermal
one, but is steeper.
Rapid Adiabatic Process
Another way to ensure
that a process is
effectively adiabatic is
to have the volume
change occur very
quickly.
In this case, heat has no
time to flow in or out of
the system.
Thermal Processes
The different types of ideal thermal processes
Specific Heat for an Ideal Gas at Constant Volume
Specific heats for ideal gases must be quoted
either at constant pressure or at constant
volume. For a constant-volume process,
First Law of Thermodynamics
for an ideal gas (from the
kinetic theory)
Specific Heat for an Ideal Gas at Constant Pressure
At constant pressure, (some work is done)
Some of the heat energy goes into the mechanical work, so
more heat input is required to produce the same ΔT
First Law of Thermodynamics
for an ideal gas (from the
kinetic theory)
Specific Heats for an Ideal Gas
Both CV and CP can be
calculated for a
monatomic ideal gas
Although this calculation was
using the first law of
done for an ideal, monatomic gas,
thermodynamics.
the difference Cp - Cv works well
for real gases.
Specific Heats and Adiabats In Ideal Gas
The P-V curve for an adiabat is
given by
for
monotonic
gases
Work of a Thermal Cycle
In the closed
thermodynamic
cycle shown in the PV diagram, the work
done by the gas is:
a) positive
b) zero
c) negative
P
V
Work of a Thermal Cycle
In the closed
thermodynamic
cycle shown in the PV diagram, the work
done by the gas is:
a) positive
b) zero
c) negative
P
The gas expands at a higher pressure
and compresses at a lower pressure.
In general, clockwise = positive work;
counterclockwise = negative work.
V
a) 4 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much work is done by the gas in this
process, in terms of the initial pressure and
volume?
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 4 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much work is done by the gas in this
process, in terms of the initial pressure and
volume?
Area under the curve:
(4 V1)(P1) + 1/2 (4V1)(2P1)
= 8 V 1P 1
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much internal energy is gained by the
gas in this process, in terms of the initial
pressure and volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much internal energy is gained by the
gas in this process, in terms of the initial
pressure and volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
Ideal monatomic gas: U = 3/2 nRT
Ideal gas law: PV = nRT
U = 3/2 PV
P2V2 = 15 P1V1
P2 = 3P1
Δ(PV) = 14 P1V1
P1
ΔU = 21 P1V1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much heat is gained by the gas in this
process, in terms of the initial pressure and
volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much heat is gained by the gas in this
process, in terms of the initial pressure and
volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
First Law of Thermodynamics
W = 8 P 1 V1
P2 = 3P1
P1
V1
V2 =5V1
Reading Quiz Internal Energy
An ideal gas is taken through
the four processes shown. The
changes in internal energy for
three of these processes is as
follows:
a) zero
b) -153 J
c) -41 J
d) -26 J
e) 41 J
The change in internal energy
for the process from C to D is:
Reading Quiz Internal Energy
An ideal gas is taken through
the four processes shown. The
changes in internal energy for
three of these processes is as
follows:
a) zero
b) -153 J
c) -41 J
d) -26 J
e) 41 J
The change in internal energy
for the process from C to D is:
PV = nRT
so in a PV cycle, ΔT = 0
ΔT = 0 means that ΔU = 0
ΔUCD = -41 J
Internal Energy
a) at constant pressure
An ideal gas undergoes a
reversible expansion to 2 times
its original volume. In which of
these processes does the gas
have the largest loss of internal
energy?
b) if the pressure increases in
proportion to the volume
c) if the pressure decreases in
proportion to the volume
d) at constant temperature
e) adiabatically
Internal Energy
a) at constant pressure
An ideal gas undergoes a
reversible expansion to 2 times
its original volume. In which of
these processes does the gas
have the largest loss of internal
energy?
b) if the pressure increases in
proportion to the volume
c) if the pressure decreases in
proportion to the volume
d) at constant temperature
e) adiabatically
Since U = 3/2 nRT, and PV=nRT, the largest loss in internal energy
corresponds to the largest drop in temperature, and so the largest
drop in the product PV.
a) PV doubles. Ufinal = 2Uinitial
b) (PV)final = 4 (PV)initial Ufinal = 4Uinitial
c) PV is constant, so U is constant
d) U is constant
e) Adiabatic, so ΔU = -W. This is the only process which reduces U!
The Zeroth Law of Thermodynamics
If object A is in thermal equilibrium with object B,
and object C is also in thermal equilibrium with
object B, then objects A and C will be in thermal
equilibrium if brought into thermal contact.
Object B can then be a thermometer,
providing a scale to compare objects:
Temperature
That is, temperature is the only factor that
determines whether two objects in thermal
contact are in thermal equilibrium or not.
The First Law of Thermodynamics
Combining these gives the first law of thermodynamics.
The change in a system’s internal energy is related to the
heat Q and the work W by conservation of energy:
It is vital to keep track of the signs of Q and W.
Reversible (frictionless pistons, etc.) and quasi-static processes
For a process to be reversible, it must be possible to return both the system
and its surroundings to the same states they were in before the process
began.
Quasi-static = slow enough that system
is always effectively in equilibrium
P
area under
W=
the curve
V
W
The Second Law of Thermodynamics
We observe that heat always
flows spontaneously from a
warmer object to a cooler one,
although the opposite would
not violate the conservation of
energy.
The Second Law of Thermodynamics:
When objects of different temperatures are brought
into thermal contact, the spontaneous flow of heat
that results is always from the high temperature
object to the low temperature object. Spontaneous
heat flow never proceeds in the reverse direction.
Heat Engines
A heat engine is a device that converts heat into work. A classic
example is the steam engine. Fuel heats the water; the vapor
expands and does work against the piston; the vapor condenses
back into water again and the cycle repeats.
All heat engines have:
a working substance
a high-temperature reservoir
a low-temperature reservoir
a cyclical engine
Efficiency of a Heat Engine
Assumption:
ΔU = 0 for each cycle, else the engine would
get hotter (or colder) with every cycle
An amount of heat Qh is supplied from the hot
reservoir to the engine during each cycle. Of that
heat, some appears as work, and the rest, Qc, is
given off as waste heat to the cold reservoir.
The efficiency is the fraction of the heat supplied
to the engine that appears as work.
Efficiency of a Heat Engine
The efficiency can also be written:
In order for the engine to run, there must be
a temperature difference; otherwise heat
will not be transferred.
The maximum-efficiency heat engine is
described in Carnot’s theorem:
If an engine operating between two
constant-temperature reservoirs is to
have maximum efficiency, it must be an
engine in which all processes are
reversible. In addition, all reversible
engines operating between the same two
temperatures, Tc and Th, have the same
efficiency.
This is an idealization; no
real engine can be perfectly
reversible.
Show the efficiency of the Carnot Cycle:
Maximum Work from a Heat Engine Cycle
The maximum work a heat engine can do is then:
If the two reservoirs are at the same
temperature, the efficiency is zero.
The smaller the ratio of the cold temperature to
the hot temperature, the closer the efficiency
will be to 1.
Perfect efficiency for Tc =?
Heat Engine
The heat engine below is:
a) a reversible (Carnot) heat engine
b) an irreversible heat engine
c) a hoax
d) none of the above
Heat Engine
The heat engine below is:
a) a reversible (Carnot) heat engine
b) an irreversible heat engine
c) a hoax
d) none of the above
Carnot e = 1 − TC/TH = 1 − 270/600 = 0.55.
But by definition e = 1 − QL/QH
= 1 − 4000/8000 = 0.5, smaller
than Carnot e, thus irreversible.
4-stroke Automobile Engine
Refrigerators, Air Conditioners,
and Heat Pumps
While heat will flow spontaneously only from a
higher temperature to a lower one, it can be made
to flow the other way if work is done on the
system. Refrigerators, air conditioners, and heat
pumps all use work to transfer heat from a cold
object to a hot object.
Refrigerators
If we compare the
heat engine and the
refrigerator, we see
that the refrigerator is
basically a heat
engine running
backwards – it uses
work to extract heat
from the cold
reservoir (the inside of the refrigerator) and exhausts
to the kitchen. Note that
- more heat is exhausted to the kitchen than is
removed from the refrigerator.
Refrigerators
An ideal refrigerator would remove the most heat
from the interior while requiring the smallest
amount of work. This ratio is called the coefficient
of performance, COP:
Typical refrigerators have COP values between 2
and 6. Bigger is better!
An air conditioner is essentially identical to a refrigerator;
the cold reservoir is the interior of the house, and the hot
reservoir is outdoors.
Heat Pumps
Finally, a heat pump is the
same as an air conditioner,
except with the reservoirs
reversed.
Heat is removed from the
cold reservoir outside, and
exhausted into the house,
keeping it warm.
Note that the work the pump
does actually contributes to
the desired result (a warmer
house) in this case.
Heat Pump Efficiency
In an ideal heat pump with two operating temperatures
(cold and hot), the Carnot relationship holds; the work
needed to add heat Qh to a room is:
The COP for a heat pump:
Room Temperature
You haven’t had time to install
your new air condition in the
window yet, so as a short-term
measure you decide to place it
on the dining-room table and
turn it on to cool off a bit. As a
result, does the air in the dining
room:
a) get warmer
b) get cooler
c) stay the same
Room Temperature
You haven’t had time to install
your new air condition in the
window yet, so as a short-term
measure you decide to place it
on the dining-room table and
turn it on to cool off a bit. As a
result, does the air in the dining
room:
a) get warmer
b) get cooler
c) stay the same
The AC motor must do work to pull heat from one
side to the other. The heat that is exhausted is the
heat drawn from the room plus the work done by
the motor. The net effect is that the motor of the
AC is adding heat to the room.
Gasoline engines
(O)A
AB
BC
CD
DA
Idealized
Diesel
cycle
Approaching absolute zero
The efficiency of a reversible engine:
So a reversible engine has the following relation
between the heat transferred and the reservoir
temperatures:
(irreversible engines
so... how cold can we make something?
As a system approaches absolute zero,
heat becomes harder to extract
)
The Third Law of Thermodynamics
Absolute zero is a temperature that an object can
get arbitrarily close to, but never attain.
Temperatures as low as 2.0 x 10-8 K have been
achieved in the laboratory, but absolute zero will
remain ever elusive – there is simply nowhere to
“put” that last little bit of energy.
This is the third law of thermodynamics:
It is impossible to lower the temperature of an object to
absolute zero in a finite number of steps.
The Laws of Thermodynamics
I) ΔU = Q - W
A continuous system (which is not
consuming internal energy) cannot output
more work than it takes in heat energy
YOU CAN’T WIN!
II)
III) TC = 0 is not achievable
YOU CAN’T BREAK EVEN!
Entropy
A reversible engine has the following relation
between the heat transferred and the reservoir
temperatures:
Rewriting,
This quantity, Q/T, is the same for both
reservoirs. This conserved quantity is defined
as the change in entropy.
Entropy
Like internal energy, entropy is a state function
Unlike energy, entropy is NOT conserved
In a reversible heat engine, the
entropy does not change.
Entropy
A real engine will operate at a lower efficiency than
a reversible engine; this means that less heat is
converted to work.
for irreversible
processes
Any irreversible process results
in an increase of entropy.
Entropy
To generalize:
• The total entropy of the universe increases whenever an
irreversible process occurs.
• The total entropy of the universe is unchanged
whenever a reversible process occurs.
Since all real processes are irreversible, the
entropy of the universe continually increases. If
entropy decreases in a system due to work being
done on it, a greater increase in entropy occurs
outside the system.
Order, Disorder, and Entropy
Entropy can be thought of as the increase in
disorder in the universe.
In this diagram, the end state is less ordered than
the initial state – the separation between low and
high temperature areas has been lost.
Entropy
Entropy
As the total entropy of the universe
increases, its ability to do work decreases.
The excess heat exhausted during an
irreversible process cannot be recovered into
a more organized form of energy, or
temperature difference.
Doing that would require a net decrease in
entropy, which is not possible.
So far, we have focused on the rather
gloomy prospect of the universe
constantly evolving toward greater
disorder. Is it possible, however, that
life is an exception to this rule?