1. Worksheet
A thermodynamic system undergoes a three-step process. An adiabatic expansion takes it from state 1 to state 2; heat is added at constant pressure to move the system to state 3; and an isothermal compression returns the system to state 1. The system consists of a diatomic ideal gas with CV = 5R/2.
The number of moles is chosen so nR = 100 J/K.
The following information is known about states 2 and 3.
Pressure: P2 = P3 = 100 kPa
Volume: V3 = 0.5 m3
What is the temperature
of the system in state 3?

2. Apply the ideal gas law
The system does J of work in the constant pressure process that takes it from state 2 to state 3. What is the volume and temperature of the system in state 2?

3. The temperature in state 2
What is the temperature of the system in state 2? K K K K
5. None of the above

4. Finding work
The system does J of work in the constant pressure process that takes it from state 2 to state 3. What is the volume and temperature of the system in state 2?
For constant pressure, we can use:
Finding volume:
Finding temperature: (use the ideal gas law, or …)

5. Complete the table
For the same system, complete the table. The total work done by the system in the cycle is –19400 J.
First fill in all the terms that are zero.
Each row satisfies the First Law of Thermodynamics.
Also remember that
Process Q
ΔEint W
1 to 2
2 to 3 J
3 to 1
Entire cycle J

6. Complete the table
For the same system, complete the table. The total work done by the system in the cycle is –19400 J.
Q is zero for an adiabatic process.
The change in internal energy is zero for an isothermal process, and is always zero for a complete cycle.
Process Q
ΔEint W
1 to 2
2 to 3 J
3 to 1
Entire cycle J

7. Complete the table
For the same system, complete the table. The total work done by the system in the cycle is –19400 J.
Even the last row has to satisfy the first law:
Process Q
ΔEint W
1 to 2
2 to 3 J
3 to 1
Entire cycle J

8. Complete the table
For the same system, complete the table. The total work done by the system in the cycle is –19400 J.
Find the change in internal energy for the 2  3 process.
Process Q
ΔEint W
1 to 2
2 to 3 J J
3 to 1
Entire cycle J

9. Complete the table
For the same system, complete the table. The total work done by the system in the cycle is –19400 J.
Rows have to obey the first law.
Columns have to sum to the value for the entire cycle.
Process Q
ΔEint W
1 to 2 J
2 to 3 J J J
3 to 1
Entire cycle J

10. Complete the table
For the same system, complete the table. The total work done by the system in the cycle is –19400 J.
Rows have to obey the first law.
Columns have to sum to the value for the entire cycle.
Process Q
ΔEint W
1 to 2 J J
2 to 3 J J
3 to 1 J
Entire cycle J

11. A heat engine
A heat engine is a device that uses heat to do work. A gasoline-powered car engine is a good example.
To be useful, the engine must go through cycles, with work being done every cycle. Two temperatures are required. The higher temperature causes the system to expand, doing work, and the lower temperature re-sets the engine so another cycle can begin. In a full cycle, three things happen:
Heat QH is added at a relatively high temperature TH. Some of this energy is used to do work W. The rest is removed as heat QL at a lower temperature TL.
For the cycle: QH = W + QL (all positive quantities)

12. Efficiency
In general, efficiency is the ratio of the work done divided by the heat needed to do the work.
The net work done in one cycle
is the area enclosed by the cycle
on the P-V diagram.

13. Carnot’s principle
Sadi Carnot (1796 – 1832), a French engineer, discovered an interesting result that is a consequence of the Second Law of Thermodynamics.
Even in an ideal situation, the efficiency of a heat engine is limited by the temperatures between which the engine operates. 100% efficiency is not possible, and most engines, even in ideal cases, achieve much less than 100% efficiency.
Carnot’s principle:
Ideal (Carnot) efficiency:

14. The Carnot cycle Step Description Heat DS 1 Isothermal expansion at THQH
(+) 2
Adiabatic expansion to TL 3
Isothermal compression at TL QL
(-) 4
Adiabatic compression to TH
Whole cycle
QH + QL
???

15. Carnot’s principle The net change in entropy for a complete cycle is:
The first term is positive and the second is negative.

16. Carnot’s principle The net change in entropy for a complete cycle is:
The first term is positive and the second is negative.
After a complete cycle, the system is back to its original state, which means it is back to its original pressure, volume, temperature, internal energy, and entropy. Thus, DS for a complete cycle must be zero.
Thus,
Except for the minus sign, this is Carnot’s result.

17. Carnot’s principle Except for the minus sign, this is Carnot’s result.
When we analyze heat engines, we generally just worry about the magnitude of the heat terms. If we treat the Q’s as just the magnitudes of the heats, then we agree with Carnot.

18. The temperature-entropy diagram
When thinking about the second law, a temperature-entropy diagram is often useful (as opposed to a P-V diagram). The T-S diagram for the Carnot cycle is particularly simple.
T (K)
isothermal
What does the area under the curve represent on the T-S diagram? TH
adiabatic TL
S (J/K) S1 S2

19. The temperature-entropy diagram
When thinking about the second law, a temperature-entropy diagram is often useful (as opposed to a P-V diagram). The T-S diagram for the Carnot cycle is particularly simple.
T (K)
isothermal
What does the area under the curve represent on the T-S diagram?
The heat. TH
adiabatic TL
S (J/K) S1 S2

20. A refrigerator
If you had a refrigerator in a closed, well-insulated room and you left the fridge door open for a long time, what would happen to the temperature in the room?
1. It would increase
2. It would decrease
3. It would stay the same

21. Heat engines running backwards
Refrigerators and air conditioners are heat engines that run backward. Work is done on the system to pump some heat QL from a low temperature region TL. An amount of heat QH = QL + W is then removed from the system at a higher temperature TH.
(a) Represents the cylinder in a car engine; while (b) represents a refrigerator.

22. A heat pump
If you heat your home using electric heat, 1000 J of electrical energy can be transformed into 1000 J of heat. An alternate way of heating is to use a heat pump, which extracts heat from a lower-temperature region (outside the house) and transfers it to the higher-temperature region (inside the house). Let's say the work done in the process is 1000 J, and the temperatures are TH = 27°C = 300 K and TL = -13 °C = 260 K. What is the maximum amount of heat that can be transferred into the house?
Something less than 1000 J
1000 J
Something more than 1000 J

23. A heat pump
The best we can do is determined by the Carnot relationship.
Using this in the energy equation gives:
For our numerical example this gives:
This is why heat pumps are much better than electric heaters. Instead of 1000 J of work going to 1000 J of heat we have 1000 J of work producing 7500 J of heat.

24. Entropy Entropy is in some sense a measure of disorder.
The symbol for entropy is S, and the units are J/K.
A container of ideal gas has an entropy value, just as it has a pressure, a volume, and a temperature. Unlike P, V, and T, which are quite easy to measure, the entropy of a system is difficult to calculate.
On the other, a change in entropy is easy to determine.

25. Change in entropy
Entropy changes whenever there is a transfer of heat. The change in entropy is the heat added divided by the temperature at which the transfer took place.
If the heat transfer takes place at a single temperature, the change in entropy is simply:
isothermal process:  
If the heat transfer takes place over a range of temperatures then, as long as ΔT is small compared to the absolute temperature T, the change in entropy is approximately:

26. The Second Law of Thermodynamics
The entropy of a closed system is constant for reversible processes and increases for irreversible processes. Entropy never decreases (for a closed system).
For a closed system,

27. Dropping a glass
You drop a glass of milk and the glass smashes into 27 pieces and milk spills all over the floor. If you videotaped this and ran the film backwards, it would be obvious to you that the film was running backwards. Why?
The process violates:
1. The Law of Conservation of Energy
2. The Law of Conservation of Momentum
3. The Second Law of Thermodynamics
4. All of the above

28. Reversible and irreversible processes
Entropy: time’s arrow
In the process of smashing the glass of milk, both energy and momentum are conserved. However, the entropy is increased The direction of time is the direction of increasing entropy.
Reversible and irreversible processes
In an irreversible process, the entropy of a closed system increases.
In a reversible process, the entropy stays the same.

29. Reversible or not?
You have two styrofoam containers of water. Each holds 1 kg of water. In one the water temperature is 17°C, while in the other it is 37°C. The colder water is then poured into the warmer water, and the system is allowed to come to equilibrium.
Is this process reversible or irreversible?
1. Reversible
2. Irreversible

30. Irreversible!
The container of water will not spontaneously separate into two parts that differ in temperature by 20°, so this process is irreversible. Let’s calculate the change in entropy.
Find the heat:
Use the change in entropy equation, using an average temperature of 22°C = 295 K for the cooler water and 32°C = 305 K for the warmer water.

31. Whiteboard