1. Chapter 16: The Second Law of Thermodynamics
© 2016 Pearson Education, Inc.

2. The second law of thermodynamics

3. Goals To examine the directions of thermodynamic processes.
To study heat engines.
To overview internal combustion engines and refrigerators.
To learn and apply the second law of thermodynamics
To study the Carnot engine: the most efficient heat engine.
To overview entropy.
© 2016 Pearson Education, Inc.

4. The Direction of a Thermodynamic Process
Heat flows spontaneously from a "hot" object to a "cold" object.
A process can be
Spontaneous
Non-spontaneous
Reversible
Irreversible
In equilibrium
Overall, there can be an increase or decrease in order.
Devices can interconvert order/disorder/energy.
© 2016 Pearson Education, Inc.

5. Fundamental laws of thermodynamics
Zeroth Law
Two systems that are each in thermal equilibrium with a third system are in thermal equilibrium with each other.
First Law
If Q is added and W is done then the U will change by:
ΔU = Q-W
Second Law
Engine statement
It is impossible for any heat engine to undergo a cyclic process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work

6. Which of the following actions is not even approximately reversible?
Clicker - Questions
Which of the following actions is not even approximately reversible?
Jumping upward.
Bouncing on a pogo stick.
Skiing down an icy slope.
Swimming from one end of the pool to the other.

7. heat engines- cycle process
Why does heat always flow from hotter to colder places and not in reverse?
One-way processes are derived by the second law of thermodynamics.
It uses the concept of entropy = quantitative measure of the degree of disorder
Heat engines, convert heat into work
Refrigerator, transport heat from colder to hotter object
Heat engine:
Cyclic process with;
∆𝑈=𝑄−𝑊=0 →𝑄=𝑊
𝑄= 𝑄 𝐻 + 𝑄 𝐶 = 𝑄 𝐻 − 𝑄 𝐶
𝑊=𝑄=𝑖𝑑𝑒𝑎𝑙𝑦 𝑄 𝐻 =𝑊 𝑎𝑛𝑑 𝑄 𝐶 =0
Thermodynamical efficiency

8. The ottocycle = gasoline engine
𝑟=𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜𝑛= 𝑉 𝑚𝑎𝑥 𝑉 𝑚𝑖𝑛 = 𝑝𝑖𝑠𝑡𝑜𝑛 𝑑𝑜𝑤𝑛 𝑝𝑖𝑠𝑡𝑜𝑛 𝑢𝑝 ≈8
3 types of heat engines:
Gasoline or Otto engine e = 56%
Diesel engine e= 68%
Carnot engine most efficient heat engine e= 90% for large temp difference

9. Line bc: Adiabatic expansion e=0.56 r=8 otto e=0.70 r=15-20 Diesel
Line ab: At a air-gasoline mixture enters and is adiabatically compressed. At b ignition occurs.
Line bc: Adiabatic expansion
e= r=8 otto
e= r= Diesel
𝑒=1− 1 𝑟 𝛾−1
𝛾= 𝐶 𝑝 𝐶 𝑉
For Diesel there is no fuel in the compression stroke and r can be large
Pre-igintion cannot occur since only hot air is present before the fuel is injected immediately before the power stroke
Second law of thermodynamics
Experimental evidence suggests that it is not possible to build an engine that convert heat completely into work.
No cyclic process can convert heat completely into work

10. Refrigerator= heat engine in reverse
𝑄 𝐶 >0, 𝑊<0 𝑎𝑛𝑑 𝑄 𝐻 <0
For a cyclic process ∆𝑈=0
𝑄 𝐻 + 𝑄 𝐶 −𝑊=0 and 𝑄 𝐶 −𝑊=− 𝑄 𝐻
Or because W =−𝑊 and |𝑄 𝐻 |=− 𝑄 𝐻
|𝑄 𝐻 = |𝑄 𝐶 +|𝑊| ℎ𝑒𝑎𝑡 𝑙𝑒𝑎𝑣𝑖𝑛𝑔𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑎𝑛𝑑 𝑔𝑖𝑣𝑒𝑛 𝑡𝑜 𝑡ℎ𝑒 ℎ𝑜𝑡 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟
𝑖𝑠 𝑎𝑙𝑤𝑎𝑦𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡
𝑡𝑎𝑘𝑒𝑛 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑜𝑙𝑑 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟
Performance coefficient of a refrigerator;
𝐾= 𝑄 𝐶 |𝑊| = | 𝑄 𝐶 | 𝑄 𝐻 −| 𝑄 𝐶 | =[dimenionless]
Refrigerator
Air conditioner

11. No; the city would heat up. No; there would be no net change.
Clicker - Questions
It is a hot day in the city. Could you cool the place down by having everyone run their air conditioners with their window open?
Yes.
No; the city would heat up.
No; there would be no net change.

12. Second law of thermodynamics
Experimental evidence suggest that it is impossible to build a heat engine that converts heat completely into work (with efficiency of 100%)
Engine statement
It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work with the system ending in the same state in which it began.
Refrigerator statement
It is impossible for any process to have as its sole result of the transfer of heat from a colder to a hotter object.

13. Carnot Engine
Carnot engine=most efficient heat engine
Engine of maximum efficiency remain consistent with the second law of thermodynamics.
𝑒=1+ 𝑄 𝐶 𝑄 𝐻
For Carnot cycle engine;
𝑄 𝐶 𝑄 𝐻 =− 𝑇 𝐶 𝑇 𝐻
𝑒=1− 𝑇 𝐶 𝑇 𝐻 = 𝑇 𝐻 − 𝑇 𝐶 𝑇 𝐻
Efficiency is the larger when the temperature difference is also larger.
The conversion ofwork into heat is an irreversible process, therefore avoid all
irreversible processes
Bottom line: every process of the cycle must be isothermal or adiabatic

14. Any other type of heat engine.
Clicker - Questions
The thermal efficiency formula 𝑒= 𝑊 𝑄 𝐻 is valid for which of the following heat engines? (There may be more than one correct choice)
Carnot engine.
Otto engine.
Diesel engine.
Any other type of heat engine.

15. Set up: 𝑒= 𝑊 𝑄 𝐻 , W= 𝑄 𝐻 + 𝑄 𝐶 and 𝑊 𝑄 𝐻 >0,
Problem
If a heat engine is 33% efficient, how much work can you get by putting 150J of heat, and how much heat do you waste doing each cycle?
Set up: 𝑒= 𝑊 𝑄 𝐻 , W= 𝑄 𝐻 + 𝑄 𝐶 and 𝑊 𝑄 𝐻 >0,
𝑄 𝐶 =<0 and 𝑄 𝐻 =150𝐽
𝑊=𝑒 𝑄 𝐻 =0.33∗150𝐽=50 𝐽
𝑄 𝐶 =𝑊− 𝑄 𝐻 =50𝐽−150𝐽=−100 𝐽 (wasted)

16. Solve: the monoatomic gas gives a larger e. For a monoatomic gas;
Problem 16.12
For an Otto engine with a compression ratio of 7.50, you have your choice of using an
Ideal monatomic or ideal diatomic gas. Which one would give greater efficiency?
Set up: 𝑒=1− 1 𝑟 𝛾−1 . For a monoatomic ideal gas 𝛾=1.67 and for a diatomic gas 𝛾=1.40
Solve: the monoatomic gas gives a larger e. For a monoatomic gas;
𝑒=1− =0.741=74.1%
For a diatomic gas;
𝑒=1− =0.553=55.3%

17. Problem 16.10
In one cycle a freezer uses 785 J of electrical energy in order to remove 1750 J of heat
From its freezer compartment at 10 F. a) what is the coefficient of performance
Of this freezer? b) how much heat does it expel into the room during this cycle?
Set up: For a refrigerator, the coefficient of performance is 𝐾= | 𝑄 𝐶 | |𝑊| .
𝑄 𝐶 =1750 𝐽 and 𝑊 =785 𝐽 𝑄 𝐻 = 𝑄 𝐶 +|𝑊|
Solve: (a) 𝐾= 1750 𝐽 785 𝐽 =2.23
(b) 𝑄 𝐻 = 𝑄 𝐶 + 𝑊 =1750 J+785 J=2535 J

18. Set up: For a refrigerator, 𝑄 𝐻 = 𝑄 𝐶 +|𝑊|.
Problem 16.15
A refrigerator has a coefficient of performance of Each cycle, it absorbs
3.4 E4 J of heat from the cold reservoir. A) how much mechanical energy is required
Each cycle to operate the refrigerator? B) During each cycle, how much heat is
Discarded to the high-temp reservoir?
Set up: For a refrigerator, 𝑄 𝐻 = 𝑄 𝐶 +|𝑊|.
𝑄 𝐶 >0, 𝑄 𝐻 <0, and 𝐾= | 𝑄 𝐶 | |𝑊| .
Solve: (a)𝑊= | 𝑄 𝐶 | 𝐾 = 3.40𝑥 𝐽 2.10 =1.62𝑥 𝐽
(b) 𝑄 𝐻 =3.40𝑥 𝐽+1.62𝑥 𝐽=5.02𝑥 J
Reflect: More heat is discarded to the high temperature reservoir than is absorbed from the cold reservoir.

19. Example 16.1 Fuel consumption in a truck
The gasoline engine in a truck takes 2500J of heat and delivers 500J of mechanical work per cycle. Assume the heat of combustion 𝐿 𝐶 =5𝑥 𝐽 𝑔
What is the thermal efficiency?
𝑒= 𝑊 𝑄 𝐻 = 500𝐽 2500𝐽 =0.2=20%
(b) How much heat is discarded in each cycle?
𝑊= 𝑄 𝐻 + 𝑄 𝐶
500𝐽=2500𝐽+ 𝑄 𝐶
(c) How much gasoline is burn in each cycle?
𝑄 𝐻 =𝑚 𝐿 𝐶 , m= 𝑄 𝐻 𝐿 𝐶 = 2500𝐽 5𝑥 𝐽 𝑔 =0.05𝑔
(d) Assume the engine goes through 100cycles/second, what is the power output in watts?
𝑃=500 𝐽 𝑐𝑦𝑐𝑙𝑒 ∗100 𝑐𝑦𝑐𝑙𝑒 𝑠 =50000 𝑊=50 𝑘𝑊
(e) How much gasoline is burnt per second and per hour?
Per second:
0.05 𝑔 𝑐𝑦𝑐𝑙𝑒 ∗100 𝑐𝑦𝑐𝑙𝑒𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 =5 𝑔 𝑠
Per hour:
5 𝑔 𝑠 ∗3600 𝑠𝑒𝑐𝑜𝑛𝑑 ℎ𝑜𝑢𝑟 =18000 𝑔 ℎ =18 𝑘𝑔 ℎ
Note: the density of gasoline 𝜌=0.7 𝑔 𝑐𝑚 3 Volume=18000 𝑔 ℎ ∗ 1 𝑐𝑚 𝑔 =25700 𝑐𝑚 3 =25.7 𝐿
𝑄 𝐶 =−2000 𝐽

20. Problem 16.21: Carnot engine
Consider the pV-diagram of the Carnot engine shown
If this engine is used as a heat engine, what is the direction of the cycle; clockwise or counter-clockwise?
Clockwise, because for a heat engine 𝑊 and 𝑄 𝐻 are positive and 𝑄 𝐶 is negative.
More positive work is done during ab and bc than the magnitude of the negative work done in cd and da. The net work done in the cycle is positive and equal to the area enclosed by the loop.

21. Problem 16.7
Set up: ca is at constant volume, ab has 𝑄=0, and bc is at constant
pressure. For a constant pressure process 𝑊=𝑝∆𝑉 and 𝑄=𝑛 𝐶 𝑝 ∆𝑇=𝑛𝑅𝑇 gives 𝑛∆𝑇= 𝑉∆𝑝 𝑅
So, 𝑄= 𝐶 𝑉 𝑅 𝑉∆𝑝. For diatomic ideal 𝐶 𝑉 = 5 2 𝑅. 1atm=1.013x105 Pa
(a) Find the pressure in point a
Solve: (a) 𝑉 𝑏 =9.0𝑥 10 −3 𝑚 3 , 𝑝 𝑏 =1.5 𝑎𝑡𝑚 and 𝑉 𝑎 =2.0𝑥 10 −3 𝑚 3 . For adiabatic process 𝑝 𝑎 𝑉 𝑎 𝛾 = 𝑝 𝑏 𝑉 𝑏 𝛾 .
𝑝 𝑎 = 𝑝 𝑏 ( 𝑉 𝑏 𝑉 𝑎 ) 𝛾 =(1.5 𝑎𝑡𝑚)( 9.0𝑥 10 −3 𝑚 𝑥 10 −3 𝑚 3 ) 1.4 =12.3 𝑎𝑡𝑚
(b) How much heat enters this gas per cycle?
Solve: (b) Heat enters the gas in process ca., since T increases.
𝑄= 𝐶 𝑉 𝑅 𝑉∆𝑝= 𝑥 10 −3 𝑚 𝑎𝑡𝑚−1.5𝑎𝑡𝑚 𝑥 𝑃𝑎 𝑎𝑡𝑚 =5470 𝐽= 𝑄 𝑀
(c) How much heat leaves this gas per cycle?
Solve: (c) Heat leaves the gas in process bc., since T decreases.
𝑄= 𝐶 𝑝 𝑅 𝑝∆𝑉= 𝑎𝑡𝑚 𝑥 𝑃𝑎 𝑎𝑡𝑚 −7.0𝑥 10 −3 𝑚 3 =−3723 𝐽= 𝑄 𝐶
(d) How much work does this engine do per cycle?
Solve: (d) 𝑊= 𝑄 𝑀 + 𝑄 𝐶 =+5470𝐽+ −3723𝐽 =1747 𝐽
(e) What is the thermal efficiency?
Solve: (e) 𝑒= 𝑊 𝑄 𝐻 = 1747𝐽 5470𝐽 =0.319=31.9%
Reflect: We did not use the number of moles of the gas
Consider the pV-diagram of the heat engine that uses 0.25mol of an ideal gas with 𝛾=1.4. The curved path ab is adiabatic.

22. Entropy and Disorder
Entropy provides a quantitative measure of disorder
Consider an infinitesimal isothermal expansion of an ideal gas, in that process we add heat Q, such that the temperature remains constant. Then all Q is converted to W.
𝑄=𝑊=𝑝∆𝑉= 𝑛𝑅𝑇 𝑉 ∆𝑉 → 𝑄 𝑇 =𝑛𝑅 ∆𝑉 𝑉
The gas is in a more disorder state after the expansion since the molecules move in a larger volume and have more random mass in position.
Special most simple case
In a reversible isothermal process the entropy change ∆𝑆
∆𝑆= 𝑆 2 − 𝑆 1 = 𝑄 𝑇 unit [ 𝐽 𝐾 ]

23. Nature Favors Decrease in Order – Increase in Entropy
The cream spontaneously mixes with the coffee, never the opposite.
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24. The entropy of the milk increases
Clicker - Questions
If you mix cold milk with hot coffee in an insulated Styrofoam cup, which of the following things happen? (There may be more than one correct answer)
The entropy of the milk increases
The entropy of the coffee decreases by the same amount that the entropy of the milk increased.
The net entropy of the coffee-milk mixture does not change, because no heat was added to the system.
The entropy of the coffee-milk mixture increases.

25. Entropy and the Coffee
Creamer does disperse in coffee, never the opposite.
Rooms become dirty.
A house of cards naturally will fall.
A pressurized gas will spontaneously expand.
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26. Clicker - Questions
An insulated box has a barrier that confines a gas to only one side of the box. The barrier springs a leak, allowing the gas to flow and occupy both sides of the box. Which statement best describes the entropy of this system?
The entropy is greater in the first state, with all the gas on one side of the box.
The entropy is greater in the second state, with all the gas on both sides of the box.
The entropy is same in both states, since no heat was added to the gas and its temperature did not change.

27. Example 16.7 ∆𝑆= 𝑄 𝑇 Isothermal expansion=reversible process
Free expansion=Irreversible process
Figure 16.12
∆𝑆= 𝑄 𝑇
Q=0, W=0, ∆U=0 in spite of this as
we can see that entropy increases in a) and b)
Entropy change depends only on initial and final state. Since both processes have the same endpoints =same ∆𝑆
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28. Example 16.4: Entropy change in melting ice
Compute the change in entropy of 1kg ice at 0oC when it is melted and converted to water at 0oC.
𝑇=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡=273𝐾
𝑄=𝑚 𝐿 𝑓 =1𝑘𝑔∗3.34𝑥 𝐽 𝑘𝑔 =3.34𝑥 𝐽
∆𝑆= 𝑄 𝑇 = 3.34𝑥 =1220 𝐽 𝐾

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