what you absolutely have to know about thermodynamics to pass the

What you absolutely have to know about Thermodynamics to pass the AP Physics B test!

Chris Bruhn 2/26/2010

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.

• A System – a collection of stuff. Basic Thermo definitions for Physics:

• 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)

Remember that little thing called conservation of energy? Let’s review some Mechanics and end up with Thermodynamics!

1 2E E= or 0E∆ = as long as we did not have any external forces. 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 ( , ,g s EU U U , 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: Potential ThermalEnergy ExternalForcesK U U W∆ + ∆ + ∆ = Thermodynamics expands this idea one final step by adding the effect of Heat (Q) transfer into or out of the system. Potential ThermalEnergy HeatTransfer ExternalForcesK U U Q W∆ + ∆ + ∆ = + There you have it… The 1st 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

ThermalEnergy HeatTransfer ExternalForcesU Q W∆ = +:

U Q W∆ = + Remember that the 1st

Law of Thermo tells us what is happening to the energy of the system.



Here are two nebulous pictures of a system and how it might interact with the environment.

Case #1 The system energy goes up. Case #2 The system energy goes down.

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: 0W =

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.

“The System” In this case Work is done by the system (-W) and

Heat is removed from the system. The 1st Law of Thermo U Q W∆ = +

tells us that the internal energy of the system will

go down (-∆U)

“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)

Work done on the system by the environment. +W

Heat removed from the system and sent to the environment. -Q

Work done by the system on the environment. -W

Heat added to the system from the environment. +Q

“The Environment” “The Environment”



Gases and the 1st

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:

law of Thermo

Implication #1: Work (W)

The basic equation for Work is: cosW Fs θ=

Multiplying and dividing by Area (A) we get: cos ( )FW sAA

θ=

But: cosF PressureA

θ= and sA Volume= ∆

(The cosθ drops out of the equation because pressure is always perpendicular to the volume displacement.) So: gas averageW p 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.

The average kinetic energy of a gas: 32avg BK k T= and the Internal Energy of an Ideal gas: 3

2U nRT=

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.

So remember: T U±∆ ∝ ±∆ and 32

U nR T∆ = ∆

Implication #3: The Ideal Gas Law

Since we will be dealing with gases we will be using the Ideal Gas Law: pV nRT= where 8.314 JRmolK

=

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.

Isothermals and how to find T and ∆U on a pV diagram using pV nRT= & 32

U nR T∆ = ∆

How do we find the temperature on a pV diagram?



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?



How to find Work on a pV diagram using gasW 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.)



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: pQ n cT= ∆ and VQ nc 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 U W Q∆ = + law of Thermo 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 ∆T ∆U W Q A B C D

How to determine the sign

∆T is found by seeing how the path moves through the Isotherms.

∆T & ∆U always have the same sign. 32

U nR T∆ = ∆

Move to the right = - Move to the left = +

Up or down = 0

gasW p V= − ∆

Find ∆U and W first. Then use the 1st

U W Q∆ = +

Law of Thermo to calculate Q.



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 0cycleT∆ = • Since ∆U & ∆T are related… When 0cycleT∆ = , 0cycleU∆ = • That means ∆U drops out of the 1st Q W= − Law of Thermo and the equation becomes ! Now look at cycle #1: ABW is negative, BCW &

DAW are both zero, and CDW is positive. Note that the positive work of CDW is larger than the negative work of ABW . 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 = 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.)

This area under the cycle cancels out!

The net work of a cycle is simply the

area inside the cycle!



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

gasW p V= − ∆ is easy to calculate since p = constant. These processes more right & left on a pV diagram Constant Volume – Isochoric / Isovolumetric

0W = because 0V∆ = thus U Q∆ = . These processes move up & down on a pV diagram. Constant Temperature – Isothermal

0T∆ = therefore 0U∆ = and Q W= − . These processes move along the hyperbolic constant pV lines. No Heat transfer between the system and the environment – Adiabatic

0Q = thus U W∆ = . This is a curved path similar to an Isothermal but steeper. 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 ∆T ∆U W Q #1 #2 #3 #4

How to

determine the sign



U nR T∆ = ∆


Up or down = 0

gasW p V= − ∆


U W Q∆ = +




Why Heat flows from Hot to Cold, Entropy, and the 2nd

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.

Law of Thermodynamics

• 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.

The equation for change in entropy is: QEntropyT

∆ = . Calculating the entropy lost by the hot object and gained

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

The entropy of an isolated system either remains the same or increases until equilibrium is achieved. Law of Thermodynamics is a statement that tells us the direction energy will move:

Here are two of the important consequences of the 2nd

1) Net heat always moves from hot to cold until thermal equilibrium is achieved. Law:

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.



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 HQ . The Heat that flows into the cold place is labeled CQ . The energy that is siphoned off to do useful work is labeled W .



Notice several things about a Heat Engine. The heat that exits the hot place HQ is no longer equal to the heat entering the cold place CQ . Conservation of Energy tell us that the work stolen from the system must be

H CW Q Q= − or H CQ W Q= + . 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

the work we can get it to do divided by how much available heat we had to steal from: H C

H H

Q QWeQ Q

−= =

At this point some of you are thinking… “Why don’t we just turn all of the heat ( HQ ) 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

e Law of Thermo by turning disordered thermal energy completely into ordered work.

There is always wasted thermal energy in a heat engine. Efficiency ( ) 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!



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 HQ flowing in from a hot place and them work W being siphoned off and the rest of the heat CQ 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 HQ ? It is 12483ABQ J= . Where is the heat flowing out of the cycle? Where is CQ ? It is the combination of 11233BC CAQ Q J+ = . How much work is siphoned off in the cycle? It is the net work of the cycle: 1250cycle AB BC CAW W W W J= + + = − Remember that for a heat engine: H CW Q Q= − and this is true: 1250 12483 11233J J J= −

What is the efficiency of this heat engine? 1250 0.1 10%12,483

H C

H H

Q QW JeQ Q J

−= = = = =

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.



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: H CQ W Q= + 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 ∆T ∆U W Q AB BC CA Net

Continued on the next page!



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 CQ out of a cold place and into a hot place. • The exhausted heat HQ is larger than the removed heat CQ . • Conservation of Energy tell us that H CQ W Q= + Things to remember about heat engines and refrigerators: 1. Heat Engines and Refrigerators are both cycles on a pV diagram. 0cycle cycleT U∆ = ∆ = for cycles 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 1e< < . 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 HQ at a rate of 1000 W and exhausts waist energy CQ to the environment at a rate of 600 W. Using the equation: H CQ W Q= + , you can

calculate the rate at which work W is done to be 400W! Using the equation: H C

H H

Q QWeQ Q

−= = , you can

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…



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 0Q = during its mechanical operation. • Heat would have to be transfer to and from the machine in an isothermal process 0U∆ = . • 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 Type Significance Entropy 1-2 Isothermal 0T U∆ = ∆ = QH /TH = +∆S 2-3 Adiabatic 0Q∆ = ∆S = 0 3-4 Isothermal 0T U∆ = ∆ = QC /TC = -∆S 4-1 Adiabatic 0Q∆ = ∆S = 0

Net Entropy change of gas for the entire cycle → ∆Scycle = 0 Notice that a Carnot Cycle operates between two isothermals or a hot temperature HT and a cold temperature CT . 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

heat engine is: H CC

H

T TeT−

=

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:

600 300 0.5 50%600

H CC

H

T T K KeT K− −

= = = = . As you can see, our heat engine was not very good.

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.



Key for table on page 6:

Path ∆T ∆U W Q A + + - + (Big +) B 0 0 + - (same sign as W) C - - 0 - (same sign as ∆U) D - - + - (Big -)

How to determine the sign



U nR T∆ = ∆


Up or down = 0

gasW p V= − ∆


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 ∆T ∆U W Q #1 - - 0 - #2 - - - 0 #3 0 0 - + #4 + + - + (Big +)

How to

determine the sign



U nR T∆ = ∆


Up or down = 0

gasW p V= − ∆


U W Q∆ = +


Key for table on page 13: (Remember that for a cycle the 0cycle cycleT U∆ = ∆ = and that W Q= − . Q = 0 for Adiabatic processes and W = 0 when the process moved up and down on the pV diagram.)

Path ∆T ∆U W Q 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.

473 373 0.21 21%473

H CC

H

T T K KeT K− −

= = = =

what you absolutely have to know about thermodynamics to pass the

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