Download What you absolutely have to know about Thermodynamics to pass

What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Thermodynamics is the most universally applicable branch of science.
It is used in Physics, Chemistry, Biology, Astronomy, Environmental Science….
Thermodynamics is the science of energy in its broadest sense.
It explains how energy transforms from one form to another and how it can be converted into useful work.
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Basic Thermo definitions for Physics:
A System – a collection of stuff.
The Environment – everything other then the stuff in the system.
Thermal Energy/Internal Energy (U) – the sum of all the Kinetic and Potential energies of each and every
atom in the system. Units = Joules (J)
Temperature (T) – the “average” kinetic energy per molecule in the system. Molecules move faster on
average in hot objects. At Absolute Zero the molecules would not be moving at all. Units = Kelvin (K)
Heat (Q) – is thermal energy transferred between objects. This transfer occurs when there is a temperature
difference. Hot (high temperature) objects exchange heat to cold (low temperature) objects. Units = (J)
Thermal Equilibrium – When objects reach the same temperature, heat ceases to pass between them.
Work (W) – A force that causes a displacement. W = F∆rcosθ. Work can change the energy of the system
Units = Joules (J)
Let’s review some Mechanics and end up with Thermodynamics!
Remember that little thing called conservation of energy?
0 as long as we did not have any external forces.
E1 = E2 or ∆E =
But… external forces on the system do work that change the energy of the system.
W = ∆K (work-energy theorem)
Remember that there were 3 types of external forces that can do work:
1. Conservative forces (gravity, spring, electromagnetic, etc…) do work that can be represented as a change in
the systems potential energy ( U g , U s , U E , etc…).
2. Dissipative forces do work that increases the thermal energy of the system ( ∆U ).
3. Other external forces (Tension, Normal Force, etc…) that produce work by moving things around.
Thus the work-energy theorem can be written like so: ∆K + ∆U Potential + ∆U ThermalEnergy =WExternalForces
Thermodynamics expands this idea one final step by adding the effect of Heat (Q) transfer into or out of the
system. ∆K + ∆U Potential + ∆U ThermalEnergy
= QHeatTransfer + WExternalForces
st
There you have it… The 1 Law of Thermodynamics which, of course, is nothing more than what we learned in
mechanics with the addition of Heat (Q).
To make life easier on us at this stage of the game, we are going to neglect Kinetic and Potential since we
already studied that in mechanics anyway. That gives us the classical form of the 1st Law of Thermodynamics:
∆U ThermalEnergy =QHeatTransfer + WExternalForces
∆U = Q + W
st
Remember that the 1 Law of Thermo tells us what is happening to the energy of the system.
Chris Bruhn
Page 1
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Here are two nebulous pictures of a system and how it might interact with the environment.
Case #1 The system energy goes up.
Work done on the system
by the environment. +W
“The Environment”
Heat added to the system
from the environment. +Q
Case #2 The system energy goes down.
“The System”
“The System”
In this case Work is done
on the system (+W) and
Heat is added to the
system. The 1st Law of
Thermo ∆U = Q + W
tells us that the internal
energy of the system will
go up (+∆U)
In this case Work is done
Work done by the system
by the system (-W) and
on the environment. -W
Heat is removed from the
system. The 1st Law of
“The Environment”
Thermo ∆U = Q + W
tells us that the internal
energy of the system will Heat removed from the system
go down (-∆U)
and sent to the environment. -Q
Note the sign conventions for work (W) and heat (Q). This is very important!
Here is a table of “code words” to look for so that you know when Heat and Work are positive or negative:
+W
-W
+Q
-Q
Work added to the system
Work done on the system
The system is compressed
Work removed from the system
Work done by the system
The system expands
Heat added to the system
Heat input
Heat absorbed
Heat removed from the system
Heat exhausted
Heat removed
If this is starting to seem rather vague… It’s really not that hard. Here are two examples:
Example #1: System – gecko
Environment – everything else but the gecko
The gecko wakes up and notices that its “system” is cold. It sees a hot rock in the environment and snuggles up
to it. The gecko absorbs heat from the environment/rock (+Q). The gecko/system has a positive internal energy
change (+∆U). The heat exchange ceases when the gecko/system and the rock/environment reach the same
temperature (thermal equilibrium).
∆U = W + Q
Since there were no external forced that did work on the gecko: W = 0
Thus: ∆U =
Q
The gecko’s internal energy increase is equal to the about of heat that was added to its system
Example #2
System – gas confined in a cylinder with a movable piston
Environment – everything else outside of the gas including the cylinder and the piston
A blowtorch is held below the cylinder. The gas/system absorbs heat (+Q) from the blowtorch/environment.
The gas expands as it heats up forcing the piston upward into the environment thus doing negative work (-W).
∆U =
−W + Q
The system gains heat energy from the blowtorch.
The system loses energy as it expands into the environment and does work.
The net change in energy of the system depends on which factor is largest (–W or +Q).
That wasn’t so hard! Don’t make thermo problems any harder then they need to be. It’s just energy and work.
Chris Bruhn
Page 2
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Gases and the 1st law of Thermo
A gas turns out to be a convenient tool to turn thermal energy in to useful work. Adding heat to a gas can make
it expand. An expanding gas can be harnessed to move things around. This has a several important implications:
Implication #1: Work (W)
The basic equation for Work is: W = Fs cos θ
F cos θ
Multiplying and dividing by Area (A) we get: W =
( sA)
A
F cos θ
But:
= Pressure and sA = ∆Volume
A
(The cosθ drops out of the equation because pressure is always perpendicular to the volume displacement.)
So: Wgas =
− paverage ∆V = -nR∆T
Note: We have a negative sign in there. That is because, by definition, an expanding gas does negative work.
So, don’t forget the negative sign or you will always get the wrong answer!
Implication #2: Internal Energy change (∆U) and Temperature change (∆T)
Remember that an Ideal gas has no internal potential energy. Gases only have internal kinetic energy because
the gas molecules are not connected to each other.
3
3
The average kinetic energy of a gas: K avg = k BT and the Internal Energy of an Ideal gas: U = nRT
2
2
This means that when the temperature of a gas goes up the kinetic energy increases and thus the internal energy
of the gas goes up as well.
3
So remember: ±∆T ∝ ±∆U and ∆U=
nR∆T
2
Implication #3: The Ideal Gas Law
Since we will be dealing with gases we will be using the Ideal Gas Law: pV = nRT where R = 8.314
J
molK
It wouldn’t be Physics unless we tried to graph it – the pV diagram
We need a way to represent what is going on in our thermodynamic gas. We do that with a pV diagram which is
nothing more than a graph of the changing pressure and volume of a gas.
3
Isothermals and how to find T and ∆U on a pV diagram using pV = nRT & ∆U=
nR∆T
2
How do we find the temperature on a pV diagram?
Chris Bruhn
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2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Which of these locations has the highest temperature? Where is internal energy highest for the gas?
Which way would you move on the pV diagram to get warmer or colder?
Every point that has the same pV value has the same temperature – Isothermal lines
For which path is ∆U positive, negative, and zero?
Chris Bruhn
Page 4
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
How to find Work on a pV diagram using Wgas =− p∆V
Movement to the right is negative work:
An expanding gas moves the environment thus transferring energy from itself to the environment.
Movement to the left is positive work:
A gas compressed by an outside force receives energy from the outside environment as a consequence.
What happens when you move up or down on the pV diagram?
Calculus & pV diagrams… (Area under the curve.)
Chris Bruhn
Page 5
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
How to find Heat (Q) from a pV diagram using ∆U = W + Q
How do we find Heat (Q) from a pV diagram? Well… we don’t. Not directly anyway. You may have been
taught these equations: =
Q n p ∆cT and =
Q ncV ∆T in Chemistry or Physics. However, you are not required to
know them for the AP Physics B exam. Besides, these two equations are only good for constant pressure and
constant volume processes anyway. So, how do we find Heat (Q)? We use the 1st law of Thermo ∆U = W + Q
Here is the procedure:
1. We either find W and ∆U from the pV diagram or they are given values in a problem.
2. We plug these values into ∆U = W + Q and calculate Q.
That’s it!
Here is a practice problem. For each path in the diagram below determine if the value is positive, negative, or
zero and fill in the table with a +, -, or 0. (The key is on the last page.)
Path
A
B
C
D
How to
determine
the sign
Chris Bruhn
∆T
∆U
∆T is found by seeing how the path
moves through the Isotherms.
∆T & ∆U always have the same sign.
∆U=
3
nR∆T
2
W
Q
Move to the right = Move to the left = +
Up or down = 0
Find ∆U and W first.
Then use the 1st Law of
Thermo to calculate Q.
Wgas =− p∆V
Page 6
∆U = W + Q
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Round and round we go - Cycles
A cycle is a path on a pV diagram that starts and ends at the same spot. Here are some cycles. Note that each
cycle starts at point A and ends at point A. Also, cycle #1 is sometimes referred to as cycle ABCDA.
Lets apply what we have learned:
• Temp depends on the pV location. Since we start and end at the same spot ∆Tcycle =
0
•
Since ∆U & ∆T are related… When ∆Tcycle =
0 , ∆U cycle =
0
•
That means ∆U drops out of the 1st Law of Thermo and the equation becomes Q = −W !
Now look at cycle #1: WAB is negative, WBC &
WDA are both zero, and WCD is positive. Note that
the positive work of WCD is larger than the negative
work of WAB . Thus the net work of cycle #1 is
positive! Remember that work = the area under the
graph. Look at cycle #1 again. The area under path
CD is more than the area under the path AB. Note
that the positive and negative areas under the cycle
cancel out leaving only the area inside the cycle.
The net work of a
cycle is simply the
area inside the cycle!
This area under the
cycle cancels out!
The net work of a cycle = the area inside the cycle. Counterclockwise cycles have a net +W and –Q.
Clockwise cycles have a net –W and +Q. Note that we can not find the value of the net work for cycle #3
because of its odd shape. (We would need calculus to find the area of cycle #3.)
Chris Bruhn
Page 7
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
4 Special Processes (paths) on a pV diagram
(When you learn to spot these on a pV diagram your life & AP grade get much better!)
Constant Pressure – Isobaric
Wgas =− p∆V is easy to calculate since p = constant. These processes more right & left on a pV diagram
Constant Volume – Isochoric / Isovolumetric
W = 0 because ∆V =
0 thus ∆U =
Q . These processes move up & down on a pV diagram.
Constant Temperature – Isothermal
∆T =
0 therefore ∆U =
0 and Q = −W . These processes move along the hyperbolic constant pV lines.
No Heat transfer between the system and the environment – Adiabatic
W . This is a curved path similar to an Isothermal but steeper.
Q = 0 thus ∆U =
What do they look like on the pV diagram? (Note that each of the processes shown below could move in the
opposite direction! They just happen to be drawn moving to the right and downward for the problem below.)
Fill in the table below with a +, -, or 0 for each of the 4 special processes as shown above. (Key on last page.)
Path
W
Q
∆T
∆U
#1
#2
#3
#4
How to
determine
the sign
Chris Bruhn
∆T is found by seeing how the path moves
through the Isotherms.
∆T & ∆U always have the same sign.
∆U=
3
nR∆T
2
Page 8
Move to the right = Move to the left = +
Up or down = 0
Wgas =− p∆V
Find ∆U and W first.
Then use the 1st Law of
Thermo to calculate Q.
∆U = W + Q
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Why Heat flows from Hot to Cold, Entropy, and the 2nd Law of Thermodynamics
Molecules are in constant random motion. On average “hot” objects have faster
moving molecules than “cold” objects. As you can see in the graph at the right, it is
possible for some of the “cold” molecules to be moving faster than the “hot”
molecules. However, on average, the “hot” molecules are moving faster.
• Why does heat always move from hot to cold?
Molecules in a hot object tend to collide and transfer more energy to the molecules in a
cold object because they are moving faster. (Conservation of Momentum.)
• Is it possible for a “cold” molecule to collide and transfer energy to the “hot” molecule?
Sure! But on average it is much more likely for energy to transfer from “hot” to “cold”.
Just like it is much more likely for a speeding car to transfer energy to a slow moving car
in a collision than the other way around. For net heat to transfer from a cold object to a hot
object most of the “cold” molecules would have to transfer energy to the “hot” molecules.
While this might be theoretically possible in the magic would of physics, it is statistically
and practically impossible.
• When does the heat transfer between objects stop?
In reality the heat transfer between objects never really stops. Hot objects transfer lots of
heat to cold objects. But remember that “cold” objects have a few fast moving molecules
that can transfer heat to the “hot” object. Overall, the net heat transfer is from hot to cold.
(See the diagrams to the right.) Once the two objects reach the same temperature, the
average molecular motion is the same for both. So, they transfer heat back and forth
between each other at equal rates. When two objects have the same temperature they are
in Thermal Equilibrium and the net heat transfer between them is zero.
• What is Entropy?
Entropy is a measure of disorder. Objects that are colder have less entropy because they have less random
motion in their molecules. Hot objects have higher entropy. When a hot “high entropy” object is placed next to
a cold “low entropy” object heat is exchanged. The hot object looses entropy and the cold object gains entropy.
Q
The equation for change in entropy is: ∆Entropy =
. Calculating the entropy lost by the hot object and gained
T
by the cold object, you will find that the overall entropy of the system has actually gone up! The disorder of the
system as a whole has increased because the “hot area” is no longer separated from the “cold area”. The thermal
energy has been “mixed up” into an overall disorderly “warm”. Remember for the AP Exam: When heat flows
into a system entropy increases and when heat flows out of a system entropy decreases. Also, the entropy
change of a cycle is zero. Heat flows into the gas during part of the cycle and then out of the gas during the rest
of the cycle, returning the gas back to its original state conditions of P, V, and T. Thus ∆Entropy = 0 for cycles.
The 2nd Law of Thermodynamics is a statement that tells us the direction energy will move:
The entropy of an isolated system either remains the same or increases until equilibrium is achieved.
Here are two of the important consequences of the 2nd Law:
1) Net heat always moves from hot to cold until thermal equilibrium is achieved.
2) Kinetic and potential energies are “ordered” forms of energy. They can spontaneously and completely be
converted into “random” thermal energy. Thermal energy won’t spontaneously convert into other forms of
energy. In fact, it is impossible to completely convert thermal energy into other forms of energy.
Chris Bruhn
Page 9
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
So what the heck to do we use all this stuff for? The Heat Engine
Heat naturally moves from hot places to cold places. As the hot place cools off and the cold place warms up the
heat transfer slows. When the two locations finally reach the same temperature the heat transfer stops.
This process is represented below in an Energy-Transfer Diagram.
Notice something very important: The heat flow out of the hot place = the heat flow into the cold place!
What we want to do is steal or “siphon off” some of this
energy as it moves from the hot place to the cold place. This
stolen energy can be used to do useful work like generating
electricity or moving our car down the street. The problem is
that we can only siphon off energy while heat is being
transferred. Once the hot place cools off the heat flow stops
and we can’t steal any more energy.
We solve this problem by either finding or making an energy reservoir. An energy reservoir is an object or part
of the environment that is so big that its temperature and thermal energy don’t change very much when heat
flows into or out of it. For instance:
1. When you jump into a pool to cool off you don’t change the temp of the pool very much because it is so big.
The pool is an example of a Cold Reservoir. Cold reservoirs absorb heat without increasing in temperature.
2. If you place your hand in a fire, your hand will heat up. However, the fire does not cool down very much
because it has a fuel source to burn that keeps it hot. The flame is a Hot Reservoir. A hot reservoir gives off
heat without loosing its hot temperature.
Here is an example: Build a fire (hot reservoir). As the heat (Q) from the
fire naturally flows to the atmosphere (cold reservoir) you steal some of
the energy as use it to do work. As long as you have fuel to keep the fire
hot, you can siphon off energy.
The device that is used to siphon off the energy is called a Heat Engine.
The energy-transfer diagram of a heat engine is shown at the right. The
Heat that flow out of the hot place is labeled QH . The Heat that flows
into the cold place is labeled QC . The energy that is siphoned off to do
useful work is labeled W .
Chris Bruhn
Page 10
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Notice several things about a Heat Engine. The heat that exits the hot place QH is no
longer equal to the heat entering the cold place QC .
Conservation of Energy tell us that the work stolen from the system must be
W
= QH − QC or QH= W + QC .
Look at the energy-transfer diagram at the right. If 100J of heat energy leaves the hot
place and we steal 40J of it to run a machine only 60J of heat energy is let to flow into
the cold place.
Heat Engine Efficiency:
The more heat we siphon off the more efficient our heat engine is. The efficiency of our heat engine is equal to
W QH − QC
the work we can get it to do divided by how much available heat we had to steal from:=
e =
QH
QH
At this point some of you are thinking… “Why don’t we just turn all of the heat ( QH ) into work ( W ) and
produce a heat engine with 100% efficiency?” This is indeed a great idea but unfortunately, it can not be done
because it violates the 2nd Law of Thermo by turning disordered thermal energy completely into ordered work.
There is always wasted thermal energy in a heat engine. Efficiency ( e ) is always less than 1 or 100%.
What does a Heat Engine look like on a pV diagram?
On a pV diagram, a heat engine will be a cycle that moves in a clockwise direction. Here is an example:
Continued on the next page!
Chris Bruhn
Page 11
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
How does all this relate to our heat engine? First of all, remember our energy-transfer
diagram for a heat engine shown to the right. Remember that is shows heat QH
flowing in from a hot place and them work W being siphoned off and the rest of the
heat QC moving away to a cold place.
Now lets look at our pV diagram again and the data table we just calculated:
Where is the heat flowing into the cycle on the pV diagram? In other words, where is QH ? It is QAB = 12483 J .
Where is the heat flowing out of the cycle? Where is QC ? It is the combination of QBC + QCA =
11233 J .
How much work is siphoned off in the cycle? It is the net work of the cycle: Wcycle =
WAB + WBC + WCA =
−1250 J
Remember that for a heat engine: W
= QH − QC and this is true: 1250
=
J 12483 J − 11233 J
1250 J
W QH − QC
What is the efficiency of this heat engine?=
e
=
=
= 0.1
= 10%
12, 483 J
QH
QH
An energy-transfer diagram shows only the net energy flow for a heat engine in a very general picture.
A pV diagram shows all the details of what is going on inside the gas as the heat engine is operating.
Both diagrams show the same heat engine in a different format.
A heat engine takes advantage of the natural flow of heat from hot to cold and uses it to produce useful work.
Chris Bruhn
Page 12
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Refrigerator – Just a Heat Engine in reverse!
Is it possible to move heat in the wrong direction from a cold place to a hot place? Sure, but the 2nd Law of
Thermo tells us that it won’t happen without outside intervention. Work is required! The device that
accomplishes this task is called a refrigerator.
The energy-transfer diagram at the right shows how heat can be artificially moved from
cold to hot. Notice something strange and very important. Due to Conservation of
Energy: QH= W + QC the heat exhausted to the hot place is actually greater than the heat
removed from the cold place! This is why you can’t cool down your kitchen by leaving
the refrigerator door open.
What does a Refrigerator look like on a pV diagram?
On a pV diagram, a refrigerator will be a cycle that moves in a counterclockwise direction. Here is an example:
Path
AB
BC
CA
Net
∆T
∆U
W
Q
Continued on the next page!
Chris Bruhn
Page 13
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Now lets compare the refrigerator on the pV diagram with the energy-transfer diagram of a refrigerator.
Notice:
• Work W input is required to move heat QC out of a cold place and into a hot place.
•
The exhausted heat QH is larger than the removed heat QC .
•
Conservation of Energy tell us that QH= W + QC
Things to remember about heat engines and refrigerators:
1. Heat Engines and Refrigerators are both cycles on a pV diagram. ∆Tcycle =
0 for cycles
∆U cycle =
2. Heat Engines move in a clockwise cycle on a pV diagram.
3. Refrigerators move counterclockwise on a pV diagram.
4. pV diagrams show you the particulars of what is going on in a gas while the energy transfer diagram only
shows you the net effect of the energy movement.
5. The efficiency of a Heat Engine will be 0 < e < 1 .
6. The energy transfer diagram and efficiency equations will work equally well with units of Joules or Watts.
For example a problem on the AP exam might say: A heat engine takes in heat QH at a rate of 1000 W and
exhausts waist energy QC to the environment at a rate of 600 W. Using the equation: QH= W + QC , you can
W QH − QC
calculate the rate at which work W is done to be 400W! Using the equation:=
, you can
e =
QH
QH
calculate the efficiency of the heat engine to be 0.4 or 40%. As long as all of your units are the either Joules
or Watts, the equations function exactly the same.
We only have one more thing to talk about…
Chris Bruhn
Page 14
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
The most perfect cycle there never was. - The Carnot cycle
As you might have guessed, making heat engines and refrigerators more efficient is a good thing. However, the
2nd Law of Thermo tells us that neither one can be perfect. Heat engines can not convert 100% of thermal
energy into work. Refrigerators can’t move heat from a cold place to a hot place without inputting some work.
So… just how efficient can me make them? What would the most efficient cycle look like?
The most efficient cycle is a reversible cycle. Reversibility means quite a few things but here is the bottom line:
• The machine would have to be frictionless with no heat transfer Q = 0 during its mechanical operation.
• Heat would have to be transfer to and from the machine in an isothermal process ∆U =
0.
• The machine would have to work equally well as a heat engine running clockwise on a pV diagram or in
reverse, counterclockwise on a pV diagram, as a refrigerator!
This presents two problems:
1. You can’t really build a frictionless machine.
2. Isothermal processes move very slowly. So, if you could build it, it would run too slow to be of much use!
Why do we worry about this imaginary machine? It shows what maximum possibly efficiency can be achieved.
The perfect cycle is called the Carnot Cycle.
Here is what it looks like on a pV diagram when it is running
clockwise as a heat engine:
Process
1-2
2-3
3-4
4-1
Type
Isothermal
Adiabatic
Isothermal
Adiabatic
Significance
∆T =∆U =0
∆Q =
0
∆T =∆U =0
∆Q =
0
Net Entropy change of gas for the entire cycle →
Entropy
Q H /T H = +∆S
∆S = 0
Q C /T C = -∆S
∆S = 0
∆S cycle = 0
Notice that a Carnot Cycle operates between two isothermals or
a hot temperature TH and a cold temperature TC . The Carnot
cycle is the most efficient engine that can possibly operate
between any two temperatures.
The efficiency of a Carnot cycle when running clockwise as a
T −T
heat engine is: eC = H C
TH
If we go back to our heat engine on pages 11-12, we will see that its actual efficiency was 10%. That heat
engine operated between a highest temperature of 600K and a lowest temp or 300K. The max possible
efficiency that the heat engine could ever possibly achieve operating between those two temperatures would be:
TH − TC 600 K − 300 K
e=
=
= 0.5
= 50% . As you can see, our heat engine was not very good.
C
600 K
TH
Trick question: What is the max possible efficiency of a heat engine operating between 100°C and 200°C ?
The answer is on the last page.
Chris Bruhn
Page 15
2/26/2010
What you absolutely have to know about
Thermodynamics to pass the AP Physics B test!
Key for table on page 6:
Path
∆T
A
+
B
0
C
D
How to
determine
the sign
∆U
+
0
-
∆T is found by seeing how the path
moves through the Isotherms.
∆T & ∆U always have the same sign.
∆U=
W
+
0
+
Q
+ (Big +)
- (same sign as W)
- (same sign as ∆U)
- (Big -)
Move to the right = Move to the left = +
Up or down = 0
Find ∆U and W first.
Then use the 1st Law of
Thermo to calculate Q.
Wgas =− p∆V
3
nR∆T
2
∆U = W + Q
Key for table on page 8: (If the processes were moving to the left and upward instead of right and downward,
all the signs in the table below would be revered!)
Path
W
Q
∆T
∆U
#1
0
#2
0
#3
0
0
+
#4
+
+
+ (Big +)
How to
determine
the sign
∆T is found by seeing how the path moves
through the Isotherms.
∆T & ∆U always have the same sign.
∆U=
3
nR∆T
2
Move to the right = Move to the left = +
Up or down = 0
Wgas =− p∆V
Find ∆U and W first.
Then use the 1st Law of
Thermo to calculate Q.
∆U = W + Q
Key for table on page 13: (Remember that for a cycle the ∆Tcycle =
0 and that W = −Q . Q = 0 for
∆U cycle =
Adiabatic processes and W = 0 when the process moved up and down on the pV diagram.)
Path
W
Q
∆T
∆U
AB
+415 K
+1035 J
0
+1035 J
BC
-312 K
-778 J
+520 J
-1298 J
CA
-103 K
-257 J
-257 J
0
Net
0
0
+263 J
-263 J
Key to question on page 15:
Was your answer 0.5 or 50%? I hope not! Remember that temperature has to be in units of Kelvin not Celsius.
TH − TC 473K − 373K
=
eC
=
= 0.21
= 21%
473K
TH
Chris Bruhn
Page 16
2/26/2010