1. Thermodynamics … the study of how thermal energy can do work
Thermal energy … can produce useful work
work can produce …
Thermal energy

2. Materials have internal energy U
Internal Energy is KE of random motions of atoms + PE due to forces between atoms
Can be modeled as vibrating springs joining atoms to each other in solids or within molecules
Energy is also stored as vibrational, rotational and translational motions

3. Internal energy
Materials have internal energy U,
(thermal energy + potential energy in bonds)
U is the sum of all KE and PE of atoms/molecules in the material
U is the change of internal energy
If U > 0 then internal energy has increased
If U < 0 then internal energy has decreased

4. Internal Energy of Ideal gases
For each “degree of freedom” (different direction in 3D space) an atom (or molecule) can store energy:  kT
k is Boltzmann’s constant, 1.38 (10-23) J/K, T is absolute temperature
 U =  NkT for a monatomic gas with N molecules, since there are 3 dimensions (directions)
In an Ideal Gas U  T (in K)
In polyatomic gases the molecules can store energy in rotations, and vibrations, as well as translations and this adds more degrees of freedom increasing the internal energy at a given temperature so that complex gases are slower to warm up

5. Heat and Internal Energy
Heat is not Internal Energy
Heat is the flow of Thermal Energy from one object to another and will increase the Internal Energy of the receiver and decrease the Internal Energy of the donor
HEAT more random motion (<KE>) higher temperature
HEAT stretched bonds  higher PE, without changing T
(Like Work is not Mechanical Energy
Work is the transfer of Mechanical Energy from one object to another)

6. Heat Engines and Refrigerators
The reasons for studying thermodynamics were mainly practical – engines and the Industrial Revolution
Efficient engines were needed which meant analyzing how fuel (thermal energy) may be harnessed to do useful work.
The earliest known engine is Hero’s – a Greek from Alexandria 2000 years ago.
Engines use a working fluid, often a gas, to create motion and drive equipment
Engines (and refrigerators) must repeat their cycles over and over to continue to do work

7. 0th Law of Thermodynamics
(this is the 0th law because it was added after 1,2, and 3)
Temperature exists and can be measured
When 2 objects are in thermal equilibrium separately with a 3rd object then they are in thermal equilibrium with each other
T1 = T2 and T2 = T3  T1 = T3
Thermal equilibrium means there is no net thermal energy flow between the objects

8. 1st Law of Thermodynamics
Energy is conserved – it is neither created not destroyed
Energy may be transferred from one object to another, or changed in form (KE to PE for example)
U = Q – W
For thermodynamic systems
The energy change of a system is the heat in less the work done by the system

9. EXAMPLE
1000 J of thermal energy flows into a system (Q = 1000 J).  At the same time, 400 J of work is done by the system (W = 400 J).What is the change in the system's internal energy U?
U = Q –W =
= 600 J 9

10. NB: work done on the system is +, work done by the system is -
EXAMPLE
800 J of work is done by a system (W = 800 J) as 500 J of thermal energy is  removed from the system (Q = -500 J).What is the change in the system's internal energy U?
U = Q –W
= -500 – 800
= J
NB: work done on the system is +, work done by the system is -
& heat into the system is +, heat out of the system is - 10

11. Thermodynamic Processes
A system can change its state
A state is a unique set of values for P, V, n, & T
(so PV = nRT is also called a “State Equation”)
When you know the state of a system you know U since U =  NkT =  nRT =  PV, for a monatomic gas
A “process” is a means of going from 1 state to another
There are 4 basic processes with n constant
Isobaric, a change at constant pressure
Isochoric or isovolumetric, a change at constant volume, W = 0
Isothermal, a change at constant temperature (U = 0, Q = W)
Adiabatic, a change at no heat (Q = 0)
“iso” means “same”

12. Thermodynamic Processes
Isobar P
(P1,V1) T1
(P2,V2) T2 1 2
Isochore
Adiabat 4
(P4,V4) T4
Q = 0
T3 = T4 3
(P3,V3) T3
Isotherm
The trip from 12341 is call a “thermodynamic cycle”
Each part of the cycle is a process V
All state changes can be broken down into the 4 basic processes

13. Thermodynamic Processes
Isobar, expansion at constant pressure, work is done P 1 2
Isochoric pressure change, W = 0
Adiabatic expansion; no heat, Q = 0 4 3
Isothermal compression W = Q, U is constant
The area enclosed by the cycle is the total work done, W
The work done, W, in a cycle is + if you travel clockwise V

14. Heat Engines and Refrigerators
Engines use a working fluid, often a gas, to create motion and drive equipment; the gas moves from 1 state (P, V, n, & T define a state) to another in a cycle
The Stirling Cycle: 2 isotherms
2 isochores
Stirling designed this engine in the early 18th century – simple and effective
The Stirling Engine

15. Isobaric expansion of a piston in a cylinder
The work done W = Fd = PAd = PV
The work done is the area under the process W = PV
4 stroke engine

16. Isochoric expansion of a piston in a cylinder
The work done W = 0 since there is no change in volume
Thus U = Q – W = Q

17. Adiabatic expansion of an ideal gas
The work done W = 0 here because chamber B is empty and P = 0
Thus U = Q – W = 0, that is adiabatic expansion against no resistance does not change the internal energy of a system

18. Each rectangle on the graph represents 100 Pa-m³ = 100 J
EXAMPLE
How much work is done by the system when the system is taken from:
(a)  A to B  (900 J) (b)  B to C  (0 J) (c)  C to A  (-1500 J)
Each rectangle on the graph represents 100 Pa-m³ = 100 J
(a) From A B the area is 900 J, isobaric expansion
(b) From B  C, 0, isovolumetric change of pressure
(c) From C A the area is J 18

19. So heating the steam produces a higher internal energy and expansion
EXAMPLE
10 grams of steam at 100 C at constant pressure rises to 110 C:  P   = 4 x 105 Pa             T = 10 C    V = 30.0 x 10-6 m3        c = 2.01 J/g
What is the change in internal energy?
U = Q – W
U = mcT – PV
U = 189 J
So heating the steam produces a higher internal energy and expansion 19

20. Q = mcT m = V0 V0 = L3 W = PV V = V0T U = Q – W
EXAMPLE
Aluminum cube of side L is heated in a chamber at atmospheric  pressure. What is the change in the cube's internal energy if L = 10 cm and T = 5 °C?
Q = mcT  m = V0 V0 = L3 W = PV V = V0T
U = Q – W
U = mcT – PV
cAl = 0.90 J/g°C
Al = 72(10-6) °C-1
U = V0cT – PV0T
Patm = kPa
U = V0T (c – P)
Al = 2.7 g/cm³
U = L³T (c – P)
U = 0.10³(5)((2700)(900) – 101.5(10³)(72(10-6))
U = 12,150 J
NB: P is neglible 20

21. EXAMPLE P 1, (P1,V1) T1 2, (P2,V2) T2 3, (P3,V3) T3 4, (P4,V4) T4 V
Find the work done for a cycle if P1 = 1000 kPa, V1 = 0.01 m³,
V2 = m³, V3 = V4 = 0.04 m³, T1 = 400 K, T2 = 600K, n = 2 mol
W = Area enclosed
= P1V12
+  (P2+P3)V23
–  (P1+P4)V41 P
1, (P1,V1) T1
2, (P2,V2) T2
Isobar
Isotherm
3, (P3,V3) T3
Isotherm
Isochore
4, (P4,V4) T4 V
1. P2 = P1 = 1000 kPa
W = Area enclosed
= P1V12 +  (P2+P3)V23 +  (P1+P4)V41
= ( – 18.75)(10³) = kJ
2. T4 = T1 = 400 K
3. T3 = T2 = 600 K
4. P3 = P2V2/V3 = 625 kPa
5. P4 = P1V1/V4 = 250 kPa

22. EXAMPLE P 1, (P1,V1) T1 2, (P2,V2) T2 3, (P3,V3) T3 4, (P4,V4) T4 V
Find the internal energy for each state if P1 = 1000 kPa, V1 = 0.01 m³,
V2 = m³, V3 = V4 = 0.04 m³, T1 = 400 K, T2 = 600K, n = 2 mol
1. P2 = P1 = 1000 kPa
2. T4 = T1 = 400 K P
3. T3 = T2 = 600 K
1, (P1,V1) T1
2, (P2,V2) T2
Isobar
4. P3 = P2V2/V3 = 625 kPa
5. P4 = P1V1/V4 = 250 kPa
Isotherm
3, (P3,V3) T3
Isotherm
Isochore
4, (P4,V4) T4 V
6. U1 =  nRT1 = 9972 J
7. U4 = U1 = 9972 J
8. U2 =  nRT2 = J
9. U3 = U2 = J

23. EXAMPLE Q12 P 1, (P1,V1) T1 2, (P2,V2) T2 Q34 Q41 Q34 3, (P3,V3) T3
Find the thermal energy change Q for each state if P1 = 1000 kPa,
V1 = 0.01 m³, V2 = m³, V3 = V4 = 0.04 m³, T1 = 400 K, T2 = 600K,
n = 2 mol
1. P2 = P1 = 1000 kPa
Q12
2. T4 = T1 = 400 K P
3. T3 = T2 = 600 K
1, (P1,V1) T1
2, (P2,V2) T2
Isobar
4. P3 = P2V2/V3 = 625 kPa
5. P4 = P1V1/V4 = 250 kPa
Q34
Q41
Isotherm
Q34
3, (P3,V3) T3
Isotherm
Isochore
4, (P4,V4) T4 V
10. Q12 = U12 + W12 = J
6. U1 =  nRT1 = 9972 J
11. Q23 = W23 (U23 = 0) W23 =  (P2+P3)V23 = kJ
7. U4 = U1 = 9972 J
12. Q34 = U34 = J
8. U2 =  nRT2 = J
13. Q41 = W41 (U41 = 0) W41 =  (P4+P1)V41 = kJ
9. U3 = U2 = J

24. Heat Engines and Refrigerators
The Wankel Rotary engine is a powerful and simple alternative to the piston engine used by Nissan and invented by the German, Wankel in the 1920s
The Wankel Cycle: 2 adiabats
2 isochores
The Wankel Engine

25. Recap 1st Law of Thermodynamics energy conservation Q = DU + W
Work done by system
Heat flow into system
Increase in internal energy of system P
U depends only on T (U = nRT = PV)
point on PV plot completely specifies
state of system (PV = nRT)
work done is area under curve
for complete cycle
DU = 0  Q = W V

26. What do the cycles apply to?TC QH QC W
HEAT ENGINE
REFRIGERATOR
system
system taken in closed cycle  Usystem = 0
therefore, net heat absorbed = work done
QH - QC = W (engine)
QC - QH = -W (refrigerator)
energy into blue blob = energy leaving bluegreen blob

27. Heat Engine: Efficiency
Goal: Get work from thermal energy in the hot reservoir
1st Law: QH - QC = W,
(U = 0 for cycle)
Define efficiency as work done per thermal energy used
e 
What is the best we can do?
Solved by Sadi Carnot in 1824 with the Carnot Cycle TH TC QH QC W
HEAT ENGINE W QH

28. / Carnot Cycle emax = 1 – TC TH P 1 QH 2 4 3 QC V
Designed by Sadi Carnot in 1824, maximally efficient
QH enters from 1-2 at constant TH and QC leaves from 3-4 at constant TL P 1 QH
Work done Wnet = WH – WC = QH – QC = W
Isotherm QH = WH
Efficiency is W / QH = ( QH – QC ) / QH 2
Adiabat Q = 0
Since U  T then Q – W is also proportional to T but from (1-2) and (3-4) Q = W so Q  T
Adiabat Q = 0 4
Efficiency is W / TH = ( TH – TC ) / TH
Isotherm QC = WC 3 QC TC TH / V
emax = 1 –

29. Entropy measures the disorder of a system
Heat Engine: Entropy
We can define a useful new quantity
Entropy, S TH TC QH QC W
HEAT ENGINE
Entropy measures the disorder of a system
Only changes in S matter to us T Q
S =
Change in entropy depends on thermal energy flow (heat) at temperature T

30. measures the disorder of a system
Heat Engine: Entropy
Entropy, S
measures the disorder of a system T Q TH TC QH QC W
HEAT ENGINE
changes in S matter
S = TH QH TC QC
If = as in the Carnot Cycle
… then there is no net change in entropy for the cycle and efficiency is a maximum,
… because we do as much work as is possible

31. 2nd Law of Thermodynamics
Heat flows from hot to cold naturally
“One cannot convert a quantity of thermal energy entirely to useful work” (Kelvin)
“One cannot transfer thermal energy from a cold reservoir to hot reservoir without doing work” (Clausius)
In closed systems, S > 0 for all real processes
The entropy, disorder, always increases in closed systems
Only in the ideal case of maximum efficiency would S = 0

32. EXAMPLE
Does the apparent order of life on Earth imply the 2nd law is wrong or that some supernatural being is directing things?
No. The second law applies to closed systems, those with no energy coming in or going out. As long as the Sun shines more energy falls on the Earth, and more work can be done by the plants to build new mass, release oxygen, grow, metabolize.

33. What is happening to the Universe?
EXAMPLE
What is happening to the Universe?
The universe is slowly coming to an end. When the entire universe is at the same temperature, then no work will be possible, and no life and no change … billions and billions and billions of years from now … Heat Death

34. EXAMPLE
Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Is this possible?
The maximum efficiency is emax = 1 – TL/TH = 67%, but the proposed efficiency is eprop = W/QH = 80%. This violates the 2nd law – do not buy shares in the company designing this engine!

35. EXAMPLE
Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. Is this possible?
The entropy of the cold reservoir decreases by SC = 1000 J / 100 K = 10 J/K
The entropy of the heat reservoir increases by SH = 1200 J / 300 K = 4 J/K
There would be a net decrease in entropy which would violate the 2nd Law, so this refrigerator is not possible
What is the minimum work needed?
 2000 J, so that SH becomes at least 10 J/K

36. Air Conditioners
Uses a “working fluid” (freon or other nicer gas) to carry heat from cool room to hot surroundings – same as a refrigerator, moving Q from inside fridge to your kitchen, which you must then air condition!

37. Air Conditioners

38. Air Conditioners
Evaporator located in room air transfers heat from room air to fluid
Compressor located in outside air does work on fluid and heats it further
Condenser located in outside air transfers heat from fluid to outside air
Then the fluid reenters room for next cycle

39. Evaporator Heat exchanger made from a long metal pipe
Fluid nears evaporator as a high pressure liquid near room temperature
A constriction reduces the fluid pressure
Fluid enters evaporator as a low pressure liquid near room temperature
Working fluid evaporates in the evaporator – requires energy LV to separate molecules, so fluid cools & Q flows from room to fluid
Fluid leaves evaporator as a low pressure gas near room temperature, taking thermal energy with it, leaving the room cooler!

40. Compressor
Working fluid enters compressor as a low pressure gas near room temperature
Gas is compressed (PV work) so gas T rises (1st Law, T  U & U ↑ when PV work is done)
Compressing gas forces Q out of it into surroundings (open air)
Demonstration: Pump Air into a Gallon Jug with Thermometer
Fluid leaves compressor as hot, high pressure gas

41. Condenser
Fluid enters condenser (heat exchanger made from long metal pipe) as a hot, high pressure gas Q flows from fluid to outside air
Gas releases energy across heat exchanger to air and condenses forming bonds releases energy LV – thermal energy & fluid becomes hotter liquid so even more heat flows from fluid into outside air
Fluid leaves condenser as high pressure
liquid near room temperature to repeat the cycle

42. Summary
Evaporator – in room transfers heat from room air to working fluid
Compressor – outside does work on fluid, so fluid gets hotter
Condenser – in outside air transfers heat from fluid to outside air, including thermal energy extracted from inside air and thermal energy added by compressor
Demonstration: Pump Air into a Gallon Jug with Thermometer and Pop the Top
Demonstration: Air Conditioner/Refrigerator/Dehumidifier Opened and Operating
Entropy of room has decreased but entropy of outside has increased by more than enough to compensate – order to disorder