1. Thermodynamics -chap note for AP B physics: if it says ON gas, use W= -PDV if it says BY gas, use W= PDV always: U= Q + W

2. The Ideal Gas Law (review chap.14)
P V = N kB T
N = number of molecules
N = number of moles (n) x NA molecules/mole
kB = Boltzmann’s constant = 1.38 x J/K
P V = n R T
R = ideal gas constant = NAkB = 8.31 J/mol/K
NA = Avogrado’s number = 6.02 x 1023

3. = number of molecules x ave KE/molecule = N (3/2) kBT
Summary of Kinetic Theory: The relationship between energy and temperature (for monatomic ideal gas)
Root-mean-square speed :
Careful with units: R = 8.31 J/mol/K m = molar mass in kg/mol vrms = speed in m/s
Internal Energy U
= number of molecules x ave KE/molecule
= N (3/2) kBT
= (3/2) n RT = (3/2) P V (ideal gas)

4. Thermodynamics
a system (gas) interacts with surroundings via thermal processes (heat and work transferred into gas) :
system  surroundings
P,V,T
Example: gas piston (car engine, bike pump, syringe)
system = gas at pressure P and temperature T and volume V
surroundings = piston and walls of container
state of system: values for P,V, and T describe gas

5. The 4 Laws of Thermodynamics
0. Law: Two objects at thermal equilibrium will have no heat flow between them and the same temperature T1 = T2
1st law: Total internal energy = heat put in + work done on it U = Q + W
2nd law: Heat will spontaneously flow from hot to cold
Carnot engine: ideal efficiency= (Hot- cold temp) /hot temp (otherwise: efficiency… use heat’s Q, not temp’s T)
3rd law: You can’t get to absolute zero! (T= OK) (can’t remove all the energy w/o still adding some)

6. First Law of Thermodynamics
“The internal energy of a system tend to increase when HEAT is added and work is done ON the system.”
Suggests a CHANGE or subtraction
Q > 0 : heat is added to system
Q < 0 : heat is subtracted from system
W > 0 : work done on system by surroundings
W < 0 : work done on surroundings by system

7. Internal Energy (DU) and Heat Energy (Q)
All of the energy inside a system is called INTERNAL ENERGY, DU.
When you add HEAT(Q), you are adding energy and the internal energy INCREASES.
Both are measured in joules. But when you add heat, there is usually an increase in temperature associated with the change.

8. Thermodynamic Systems and P-V Diagrams
Ideal gas law: P V = n R T
For n fixed, P and V determine the “state” of the system
T = P V/ (n R)
U = (3/2) n RT = (3/2) P V
Examples:
which point has highest T ? 2
which point has lowest U ? 3
to change the system from 3 to 2,
energy must be added to system.
Work done in a thermal process is the area underneath the P,V graph. P P1 P3 1 2 3
V V2 V

9. Work done by a gas
Suppose you had a piston filled with a specific amount of gas. As you add heat, the temperature rises and thus the volume of the gas expands. The gas then applies a force on the piston wall pushing it a specific displacement. Thus the gas does NEGATIVE work
(uses its own internal energy to do work).

10. Work is the AREA of a P vs. V graph
W= -PDV = -P* (Vf- Vi) the work done on the gas
Compress gas: gas expands itself smaller volume (DV=-) larger volume (DV=+) positive work more (W=+) negative work (W=-) gas has more energy gas loses energy
move left on P-V diagram move right on P-V diagram
-Work: clockwise +Work: counterclockwise

11. Negative or positive Work Done ON gas? (P=constant)
System: Gas
Surroundings:
Piston, walls
Isobaric process y
Gas
P,V2,T2
Gas
P,V1,T1
W = -F s = -P A D y = -P V
V > 0
expanding gas system does work on surroundings
(isobaric process: pressure kept constant even though increase volume by adding heat – which offsets some but not all work done by gas)
answer= negative work on gas

12. What area is the total work done from 1 … to 4?V P 1 2 3 4
Wtot = ??
What area is the total work done from 1 … to 4?
Isobaric Process: P=constant (horizontal line)
Isochoric Process: V=constant (vertical line) V P
W =- PDV (<0) 1 2 3 4
DV > 0 V P
W = -PDV = 0 1 2 3 4
DV = 0 V P 1 2 3 4
Wtot < 0 V P
W =- PDV (>0) 1 2 3 4
DV < 0 V
W = -PDV = 0 1 2 3 4 P
DV = 0 V P 1 2 3 4
If we go the other way then Wtot > 0

13. Now try this: 1 to 2 to 3 to 1 (one cycle) a
Now try this: 1 to 2 to 3 to 1 (one cycle) a. What’s the total change in internal energy U? b. Is positive or negative work done? C. Is heat absorbed or released? D. Write an equation for work using P’s and V’s P P1 P3 1 2 3 V
V V2
DU = 0 since no change overall (DT=0)
overall net work is negative (clockwise)
Heat is absorbed (Q>0) since work <0 and U =0
Area = (V2-V1)x(P1-P3)/2

14. Concept Question
Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest?
1. Case 1
2. Case 2
3. Same A B 4 2 3 9
V(m3)
Case 1
P(atm)
Case 2
correct
Net Work = area under P-V curve
Area the same in both cases!

15. First Law of Thermodynamics ExampleV P 1 2
V V2
2 moles of monatomic ideal gas is taken
from state 1 to state 2 at constant pressure
P=1000 Pa, where V1 =2m3 and V2 =3m3. Find
T1, T2, DU, W, Q.
1. P V1 = n R T1  T1 = P V1/(nR) = 120K
2. P V2 = n RT2  T2 = P V2/(nR) = 180K
3. DU = (3/2) n R DT = 1500 J or DU = (3/2) P DV = 1500 J
4. W = -P DV = J (neg work done on gas)
5. Q = DU - W = 1500 J J = 2500 J > 0 heat gained by gas

16. First Law of Thermodynamics Example
2 moles of monatomic ideal gas is taken
from state 1 to state 2 at constant volume
V=2m3, where T1=120K and T2 =180K. Find Q. P P2 P1 2
1. P V1 = n R T1  P1 = n R T1/V = 1000 Pa
2. P V2 = n R T1  P2 = n R T2/V = 1500 Pa
3. DU = (3/2) n R DT = 1500 J
4. W = -P DV = 0 J (no change in volume)
5. Q = DU - W = 1500 – 0 = 1500 J
=> It requires less heat to raise T at const. volume than at const. pressure. 1 V V

17. Example
Sketch a PV diagram and find the work done on the gas during the following stages.
A gas is expanded from a volume of 1.0 L to 3.0 L at a constant pressure of 3.0 atm.
The gas is then cooled at a constant volume until the pressure falls to 2.0 atm
-600 J

18. Example continued
The gas is then compressed at a constant pressure of 2.0 atm from a volume of 3.0 L to 1.0 L.
The gas is then heated until its pressure increases from 2.0 atm to 3.0 atm at a constant volume.
+400 J

19. Example continued What is the net work ON gas? -600 J + 400 J = -200 J
NET work is
the area inside the shape.
-600 J J = -200 J
Rule of thumb: If the system rotates CW, the NET work is negative.
If the system rotates CCW, the NET work is positive.
Side note: work by gas = +200J

20. Example
A series of thermodynamic processes is shown in the pV-diagram. In process ab 150 J of heat is added to the system, and in process bd , 600J of heat is added. Fill in the chart. On gas.
150
150 J
600
-240
360 J
750
-240
510 J
-90
600
600
-90
510 J

21. Classification of Thermal Processes (using work on gas definition)
Isobaric : P = constant W = -P DV
Isochoric : V = constant: W = 0
Isothermal : T = constant( DU = 0): W = -Q
Adiabatic : no heat flow (Q = 0) W = - DU
* Remember: work is area under P-V curve (positive work if compress gas to less volume)

22. Thermodynamic Processes - Isothermal
To keep the temperature constant both the pressure and volume change to compensate. (Volume goes up, pressure goes down)
“BOYLES’ LAW”
Same temp so DU=0
DU= Q + W= 0 so
Q= -W
DV = + since expands
W = -PV = neg work
Q= -W = positive heat added to maintain temp while gas uses own energy to expand to larger volume

23. Thermodynamic Processes - Isobaric
More heat Q is added to the gas than work energy used by the gas internal energy (U) decreases since PV is smaller. W= + (compress) Q= - release heat DU = - (colder at end)
U= Q + W Q < W

24. Thermodynamic Processes - Isovolumetric
No work since constant volume
W= 0
DU = Q
Q = + since higher temp
DU = + (higher int. energy)

25. Thermodynamic Processes - Adiabatic
ADIABATIC- (GREEK- adiabatos- "impassable")
In other words, NO HEAT can leave or enter the system.
Q= 0 by definition U= Q + W U = W only
W = - since gas expands hence U < 0 (tem drops) gas loses temp as expands since don’t provide heat to offset work done by gas

26. In Summary
Q=-W
Q+W W

27. Second Law of Thermodynamics
“Heat will not flow spontaneously from a colder body to a warmer body AND heat energy cannot be transformed completely into mechanical work.”
The bottom line:
Heat always flows from a hot body to a cold body
Nothing is 100% efficient

28. Engines Heat flows from a HOT reservoir to a COLD reservoir
QH = remove from, absorbs = hot
QC= exhausts to, expels = cold

29. Engine Efficiency
In order to determine the thermal efficiency of an engine you have to look at how much ENERGY you get OUT based on how much you energy you take IN. In other words:

30. Rates of Energy Usage
Sometimes it is useful to express the energy usage of an engine as a RATE.
For example:
The RATE at which heat is absorbed!
The RATE at which heat is expelled.
The RATE at which WORK is DONE

31. Efficiency in terms of rates

32. Is there an IDEAL engine model?
Our goal is to figure out just how efficient such a heat engine can be: what’s the most work we can possibly get for a given amount of fuel?
The efficiency question was first posed—and solved—by Sadi Carnot in 1820, not long after steam engines had become efficient enough to begin replacing water wheels, at that time the main power sources for industry.  Not surprisingly, perhaps, Carnot visualized the heat engine as a kind of water wheel in which heat (the “fluid”) dropped from a high temperature to a low temperature, losing “potential energy” which the engine turned into work done, just like a water wheel. 

33. Carnot Efficiency
Carnot a believed that there was an absolute zero of temperature, from which he figured out that on being cooled to absolute zero, the fluid would give up all its heat energy.  Therefore, if it falls only half way to absolute zero from its beginning temperature, it will give up half its heat, and an engine taking in heat at T and shedding it at ½T will be utilizing half the possible heat, and be 50% efficient.  Picture a water wheel that takes in water at the top of a waterfall, but lets it out halfway down.  So, the efficiency of an ideal engine operating between two temperatures will be equal to the fraction of the temperature drop towards absolute zero that the heat undergoes.

34. Carnot Efficiency
Carnot temperatures must be expressed in KELVIN!!!!!!
The Carnot model has 4 parts
An Isothermal Expansion
An Adiabatic Expansion
An Isothermal Compression
An Adiabatic Compression
The PV diagram in a way shows us that the ratio of the heats are symbolic to the ratio of the 2 temperatures

35. Example
A particular engine has a power output of 5000 W and an efficiency of 25%. If the engine expels 8000 J of heat in each cycle, find (a) the heat absorbed in each cycle and (b) the time for each cycle
10,667 J
2667 J
0.53 s

36. Example
The efficiency of a Carnot engine is 30%. The engine absorbs 800 J of heat per cycle from a hot temperature reservoir at 500 K. Determine (a) the heat expelled per cycle and (b) the temperature of the cold reservoir
240 J
560 J
350 K