1. 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 well-being 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.

2. The Laws of Thermodynamics
Lecture 25
The Laws of Thermodynamics

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Assignment #13 is not due for credit

4. 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)

5. Constant pressure Work done by an expanding gas, 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

6. Work is area under the PV graph
imagining any general process as approximated by a number of constant pressure processes:
Work is area under the PV graph

7. 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

8. Constant Temperature
Isothermal processes
If the temperature is constant, the pressure varies inversely with the volume.

9. W = Q 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
W = Q
if W < 0 (work done on the system)
than Q<0 (heat flows out of the system)
if W > 0 (work done by the system)
than Q>0 (heat flows out into the system)

10. Work in an Constant-Temperature Process
The work done is the area under the curve:
For you calculus junkies:

11. 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.
The adiabatic P-V curve is similar to the isothermal one, but is steeper.
Q = 0

12. 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.

13. Thermal Processes
The different types of ideal thermal processes

14. 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,

15. for an ideal gas (from the kinetic theory)
First Law of Thermodynamics

16. 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

17. for an ideal gas (from the kinetic theory)
First Law of Thermodynamics

18. Specific Heats for an Ideal Gas
Both CV and CP can be calculated for a monatomic ideal gas using the first law of thermodynamics.
Although this calculation was done for an ideal, monatomic gas, the difference Cp - Cv works well for real gases.

19. Specific Heats and Adiabats In Ideal Gas
The P-V curve for an adiabat is given by
for monotonic gases

20. Work of a Thermal Cycle
In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is:
a) positive
b) zero
c) negative
Answer: a V P

21. Work of a Thermal Cycle
In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is:
a) positive
b) zero
c) negative V P
The gas expands at a higher pressure
and compresses at a lower pressure.
In general, clockwise = positive work;
counterclockwise = negative work.

22. a) 4 P1V1
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 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? P1 V1
V2 =5V1
P2 = 3P1

23. a) 4 P1V1
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 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? P1 V1
V2 =5V1
P2 = 3P1
Area under the curve:
(4 V1)(P1) + 1/2 (4V1)(2P1)
= 8 V1P1

24. a) 7 P1V1
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 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? P1 V1
V2 =5V1
P2 = 3P1

25. a) 7 P1V1
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 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?
Ideal monatomic gas: U = 3/2 nRT P1 V1
V2 =5V1
P2 = 3P1
Ideal gas law: PV = nRT
U = 3/2 PV
P2V2 = 15 P1V1
Δ(PV) = 14 P1V1
ΔU = 21 P1V1

26. a) 7 P1V1
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 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? P1 V1
V2 =5V1
P2 = 3P1

27. W = 8 P1V1 a) 7 P1V1 b) 8 P1V1 c) 15 P1V1 d) 21 P1V1 e) 29 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?
First Law of Thermodynamics P1 V1
V2 =5V1
P2 = 3P1
W = 8 P1V1

28. 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:
The change in internal energy for the process from C to D is:
a) zero
b) -153 J
c) -41 J
d) -26 J
e) 41 J

29. 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:
The change in internal energy for the process from C to D is:
a) zero
b) -153 J
c) -41 J
d) -26 J
e) 41 J
PV = nRT
so in a PV cycle, ΔT = 0
ΔT = 0 means that ΔU = 0
ΔUCD = -41 J

30. Internal Energy
a) at constant pressure
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
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?

31. Internal Energy
a) at constant pressure
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
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?
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!

32. 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.

33. 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.

34. 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 V P
area under the curve
W = W

35. 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.

36. 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

37. 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.

38. 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.

40. 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.

41. Show the efficiency of the Carnot Cycle:

43. 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 =?

44. 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
Answer: b

45. 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.

46. 4-stroke Automobile Engine

47. 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.

48. 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.

50. 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.

51. 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.

52. 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:

53. 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

54. 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.

55. Gasoline engines
(O)A AB BC CD DA
Idealized Diesel cycle

56. 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

57. 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.

58. The Laws of Thermodynamics
A continuous system (which is not consuming internal energy) cannot output more work than it takes in heat energy
I) ΔU = Q - W
YOU CAN’T WIN!
II)
III) TC = 0 is not achievable
YOU CAN’T BREAK EVEN!

59. 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.

60. In a reversible heat engine, the entropy does not change.
Like internal energy, entropy is a state function
Unlike energy, entropy is NOT conserved
In a reversible heat engine, the entropy does not change.

61. Any irreversible process results in an increase of 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.

62. 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.

63. 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.

64. Entropy
Figure (a) A royal flush is a highly ordered poker hand with low probability of occurring. (b) A disordered and worthless poker hand. The probability of this particular hand occurring is the same as that of the royal flush. There are so many worthless hands, however, that the probability of being dealt a worthless hand is much higher than that of a royal flush.

65. 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.

67. 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?