1. Chapter 20 - Thermodynamics
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. THERMODYNAMICS
Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer.
Central Heating

3. Objectives: After finishing this unit, you should be able to:
State and apply the first and second laws of thermodynamics.
Demonstrate your understanding of adiabatic, isochoric, isothermal, and isobaric processes.
Write and apply a relationship for determining the ideal efficiency of a heat engine.
Write and apply a relationship for determining coefficient of performance for a refrigeratior.

4. A THERMODYNAMIC SYSTEM
A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.)
Work done on gas or work done by gas

5. INTERNAL ENERGY OF SYSTEM
The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system.
Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.

6. TWO WAYS TO INCREASE THE INTERNAL ENERGY, U.
WORK DONE ON A GAS (Positive)
HEAT PUT INTO A SYSTEM (Positive)

7. TWO WAYS TO DECREASE THE INTERNAL ENERGY, U.
Wout
hot
HEAT LEAVES A SYSTEM
Q is negative
Qout
hot
-U
Decrease
WORK DONE BY EXPANDING GAS: W is positive

8. THERMODYNAMIC STATE
The STATE of a thermodynamic system is determined by four factors:
Absolute Pressure P in Pascals
Temperature T in Kelvins
Volume V in cubic meters
Number of moles, n, of working gas

9. THERMODYNAMIC PROCESS
Increase in Internal Energy, U.
Wout
Initial State:
P1 V1 T1 n1
Final State:
P2 V2 T2 n2
Heat input
Qin
Work by gas

10. The Reverse Process Decrease in Internal Energy, U. Win Qout
Work on gas
Loss of heat
Qout
Win
Initial State:
P1 V1 T1 n1
Final State:
P2 V2 T2 n2

11. THE FIRST LAW OF THERMODYAMICS:
The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system.
Q = U + W final - initial)
Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.

12. SIGN CONVENTIONS FOR FIRST LAW
+Wout
+Qin
Heat Q input is positive
U
Work BY a gas is positive
-Win
U
Work ON a gas is negative
-Qout
Heat OUT is negative
Q = U + W final - initial)

13. APPLICATION OF FIRST LAW OF THERMODYNAMICS
Example 1: In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of work on the piston. What is the change in internal energy of the system?
Wout =120 J
Qin
400 J
Q = U + W
Apply First Law:

14. Example 1 (Cont.): Apply First Law
Qin
400 J
Wout =120 J
DQ is positive: +400 J (Heat IN)
DW is positive: +120 J (Work OUT)
Q = U + W
U = Q - W
U = Q - W
= (+400 J) - (+120 J)
= +280 J

U = +280 J

15. Example 1 (Cont.): Apply First Law
Energy is conserved:
Qin
400 J
Wout =120 J
The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J
The increase in internal energy is:

U = +280 J

16. FOUR THERMODYNAMIC PROCESSES:
Isochoric Process: V = 0, W = 0
Isobaric Process: P = 0
Isothermal Process: T = 0, U = 0
Adiabatic Process: Q = 0
Q = U + W

17. ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0
Q = U + W so that Q = U
+U
-U
QIN
QOUT
No Work Done
HEAT IN = INCREASE IN INTERNAL ENERGY
HEAT OUT = DECREASE IN INTERNAL ENERGY

18. ISOCHORIC EXAMPLE: 400 J B A P2 V1= V2 P1 PA P B TA T B =
No Change in volume: B A P2
V1= V2 P1
PA P B
TA T B =
400 J heat input increases internal energy by 400 J and zero work is done.
Heat input increases P with const. V

19. ISOBARIC PROCESS: CONSTANT PRESSURE, P = 0
Q = U + W But W = P V
QIN
QOUT
Work Out
Work In
+U
-U
HEAT IN = Wout + INCREASE IN INTERNAL ENERGY
HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY

20. ISOBARIC EXAMPLE (Constant Pressure):
400 J B A P
V1 V2
VA VB
TA T B =
400 J heat does 120 J of work, increasing the internal energy by 280 J.
Heat input increases V with const. P

21. Work = Area under PV curve
ISOBARIC WORK B A P
V1 V2
VA VB
TA T B =
400 J
PA = PB
Work = Area under PV curve

22. ISOTHERMAL PROCESS: CONST. TEMPERATURE, T = 0, U = 0
Q = U + W AND Q = W
QIN
QOUT
Work Out
Work In
U = 0
U = 0
NET HEAT INPUT = WORK OUTPUT
WORK INPUT = NET HEAT OUT

23. ISOTHERMAL EXAMPLE (Constant T):B A PA
V V1 PB
U = T = 0
Slow compression at constant temperature: No change in U.
PAVA = PBVB

24. ISOTHERMAL EXPANSION (Constant T):B A PA
VA VB PB
PAVA = PBVB
TA = TB
U = T = 0
400 J of energy is absorbed by gas as 400 J of work is done on gas.
T = U = 0
Isothermal Work

25. ADIABATIC PROCESS: NO HEAT EXCHANGE, Q = 0
Q = U + W ; W = -U or U = -W
Work Out
Work In
U
+U
Q = 0
W = -U
U = -W
Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy

26. ADIABATIC EXAMPLE: B A PA
V V2 PB
Expanding gas does work with zero heat loss. Work = -DU
Insulated Walls: Q = 0

27. ADIABATIC EXPANSION: Q = 0 A PA PAVA PBVB B = PB TA T B VA VB
400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0

28. MOLAR HEAT CAPACITY
OPTIONAL TREATMENT
The molar heat capacity C is defined as the heat per unit mole per Celsius degree.
Check with your instructor to see if this more thorough treatment of thermodynamic processes is required.

29. SPECIFIC HEAT CAPACITY
Remember the definition of specific heat capacity as the heat per unit mass required to change the temperature?
For example, copper: c = 390 J/kgK

30. MOLAR SPECIFIC HEAT CAPACITY
The “mole” is a better reference for gases than is the “kilogram.” Thus the molar specific heat capacity is defined by:
C = Q
n T
For example, a constant volume of oxygen requires 21.1 J to raise the temperature of one mole by one kelvin degree.

31. SPECIFIC HEAT CAPACITY CONSTANT VOLUME
How much heat is required to raise the temperature of 2 moles of O2 from 0oC to 100oC?
Q = nCv T
Q = (2 mol)(21.1 J/mol K)(373 K K)
Q = J

32. SPECIFIC HEAT CAPACITY CONSTANT VOLUME (Cont.)
Since the volume has not changed, no work is done. The entire 4220 J goes to increase the internal energy, U.
Q = U = nCv T = 4220 J
Thus, U is determined by the change of temperature and the specific heat at constant volume.
U = nCv T

33. SPECIFIC HEAT CAPACITY CONSTANT PRESSURE
We have just seen that 4220 J of heat were needed at constant volume. Suppose we want to also do 1000 J of work at constant pressure?
Same
Q = U + W
Q = 4220 J + J
Cp > Cv
Q = 5220 J

34. HEAT CAPACITY (Cont.)  U = nCvT For constant pressure
Heat to raise temperature of an ideal gas, U, is the same for any process.
U = nCvT
For constant pressure
Q = U + W
nCpT = nCvT + P V
Cp > Cv Cp Cv


35. REMEMBER, FOR ANY PROCESS INVOLVING AN IDEAL GAS:
PAVA PBVB
TA T B =
PV = nRT
Q = U + W
U = nCv T

36. Example Problem:
A 2-L sample of Oxygen gas has an initial temp- erature and pressure of 200 K and 1 atm. The gas undergoes four processes:
AB: Heated at constant V to 400 K.
BC: Heated at constant P to 800 K.
CD: Cooled at constant V back to 1 atm.
DA: Cooled at constant P back to 200 K.

37. PV-DIAGRAM FOR PROBLEMPB
2 L
1 atm
200 K
400 K
800 K
How many moles of O2 are present?
Consider point A: PV = nRT

38. PROCESS AB: ISOCHORIC B PB What is the pressure at point B? A PA P B =
2 L
1 atm
200 K
400 K
800 K
What is the pressure at point B?
PA P B
TA T B =
1 atm P B
200 K K =
P B = 2 atm
or kPa

39. PROCESS AB: Q = U + W W = 0 Q = U = nCv T
Analyze first law for ISOCHORIC process AB. B A PB
2 L
1 atm
200 K
400 K
800 K
W = 0
Q = U = nCv T
U = (0.122 mol)(21.1 J/mol K)(400 K K)
Q = +514 J
U = +514 J
W = 0

40. PROCESS BC: ISOBARIC B C PB D What is the volume at point C (& D)?
1 atm
200 K
400 K
800 K D
4 L
What is the volume at point C (& D)?
VB V C
TB T C =
2 L V C
400 K K =
V C = V D = 4 L

41. FINDING U FOR PROCESS BC.
2 L
1 atm
200 K
400 K
800 K
4 L
2 atm
Process BC is ISOBARIC.
P = 0
U = nCv T
U = (0.122 mol)(21.1 J/mol K)(800 K K)
U = J

42. FINDING W FOR PROCESS BC.
2 L
1 atm
200 K
400 K
800 K
4 L
2 atm
Work depends on change in V.
P = 0
Work = P V
W = (2 atm)(4 L - 2 L) = 4 atm L = 405 J
W = +405 J

43. FINDING Q FOR PROCESS BC.
2 L
1 atm
200 K
400 K
800 K
4 L
2 atm
Analyze first law for BC.
Q = U + W
Q = J J
Q = J
Q = 1433 J
U = 1028 J
W = +405 J

44. PROCESS CD: ISOCHORIC B A PB What is temperature at point D? PC P D
2 L
1 atm
200 K
400 K
800 K C D
What is temperature at point D?
PC P D
TC T D =
2 atm atm
800 K TD =
T D = 400 K

45. PROCESS CD: Q = U + W W = 0 Q = U = nCv TPB
2 L
1 atm
200 K
400 K
800 K
Analyze first law for ISOCHORIC process CD.
W = 0
Q = U = nCv T
U = (0.122 mol)(21.1 J/mol K)(400 K K)
Q = J
U = J
W = 0

46. FINDING U FOR PROCESS DA.
Process DA is ISOBARIC. A D
2 L
1 atm
200 K
400 K
800 K
4 L
2 atm
P = 0
U = nCv T
U = (0.122 mol)(21.1 J/mol K)(400 K K)
U = -514 J

47. FINDING W FOR PROCESS DA.
2 L
1 atm
200 K
400 K
800 K
4 L
2 atm
Work depends on change in V.
P = 0
Work = P V
W = (1 atm)(2 L - 4 L) = -2 atm L = -203 J
W = -203 J

48. FINDING Q FOR PROCESS DA.
2 L
1 atm
200 K
400 K
800 K
4 L
2 atm
Analyze first law for DA.
Q = U + W
Q = -514 J J
Q = -717 J
Q = -717 J
U = -514 J
W = -203 J

49. PROBLEM SUMMARY
DQ = DU + DW
For all processes:

50. NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREAB C
2 L
1 atm
4 L
2 atm
+404 J B C
2 L
1 atm
4 L
2 atm
Neg
-202 J
2 L
4 L B C
1 atm
2 atm
Area = (1 atm)(2 L)
Net Work = 2 atm L = 202 J

51. ADIABATIC EXAMPLE:
Example 2: A diatomic gas at 300 K and atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4) A B PB
VB VA PA
PAVA PBVB
TA T B =
PAVA = PBVB 
Q = 0

52. ADIABATIC (Cont.): FIND PB
Q = 0
PAVA = PBVB  A B PB
VB 12VB
1 atm
300 K
Solve for PB:
PB = 32.4 atm or kPa

53. ADIABATIC (Cont.): FIND TB
Q = 0 A B
32.4 atm
VB 12VB
1 atm
300 K
Solve for TB
TB=?
(1 atm)(12VB) (32.4 atm)(1 VB)
(300 K) T B =
TB = 810 K

54. ADIABATIC (Cont.): If VA= 96 cm3 and VA= 8 cm3, FIND W
Q = 0 A B
32.4 atm
1 atm
300 K
810 K
Since Q = 0,
W = - U
8 cm cm3
W = - U = - nCV T & CV= 21.1 j/mol K PV RT
n =
Find n from point A
PV = nRT

55. ADIABATIC (Cont.): If VA= 96 cm3 and VA= 8 cm3, FIND WPV RT
n = =
(101,300 Pa)(8 x10-6 m3)
(8.314 J/mol K)(300 K)
n = mol & CV= 21.1 j/mol K A B
32.4 atm
1 atm
300 K
810 K
8 cm cm3
T = = 510 K
W = - U = - nCV T
W = J

56. HEAT ENGINES Qhot Wout Qcold
A heat engine is any device which through a cyclic process:
Cold Res. TC
Engine
Hot Res. TH
Qhot
Wout
Qcold
Absorbs heat Qhot
Performs work Wout
Rejects heat Qcold

57. THE SECOND LAW OF THERMODYNAMICS
Wout
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.
Not only can you not win (1st law); you can’t even break even (2nd law)!

58. THE SECOND LAW OF THERMODYNAMICS
Cold Res. TC
Engine
Hot Res. TH
400 J
300 J
100 J
A possible engine.
An IMPOSSIBLE engine.
Cold Res. TC
Engine
Hot Res. TH
400 J

59. EFFICIENCY OF AN ENGINE
The efficiency of a heat engine is the ratio of the net work done W to the heat input QH.
Cold Res. TC
Engine
Hot Res. TH QH W QC
e = = W QH
QH- QC
e = 1 - QC QH

60. EFFICIENCY EXAMPLE 800 J W 600 J
An engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency?
Cold Res. TC
Engine
Hot Res. TH
800 J W
600 J
e = 1 - QC QH
e = 1 -
600 J
800 J
e = 25%
Question: How many joules of work is done?

61. EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine)
Cold Res. TC
Engine
Hot Res. TH QH W QC
For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T.
e =
TH- TC TH
e = 1 - TC TH

62. Example 3: A steam engine absorbs 600 J of heat at 500 K and the exhaust temperature is 300 K. If the actual efficiency is only half of the ideal efficiency, how much work is done during each cycle?
Actual e = 0.5ei = 20%
e = 1 - TC TH
e = W QH
e = 1 -
300 K
500 K
W = eQH = 0.20 (600 J)
e = 40%
Work = 120 J

63. REFRIGERATORS Qhot Win Win + Qcold = Qhot Qcold WIN = Qhot - Qcold
A refrigerator is an engine operating in reverse: Work is done on gas extracting heat from cold reservoir and depositing heat into hot reservoir.
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
Win
Win + Qcold = Qhot
WIN = Qhot - Qcold

64. THE SECOND LAW FOR REFRIGERATORS
It is impossible to construct a refrigerator that absorbs heat from a cold reservoir and deposits equal heat to a hot reservoir with W = 0.
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
If this were possible, we could establish perpetual motion!

65. COEFFICIENT OF PERFORMANCE
Cold Res. TC
Engine
Hot Res. TH QH W QC
The COP (K) of a heat engine is the ratio of the HEAT Qc extracted to the net WORK done W. QC W
K = = QH
QH- QC
K = TH
TH- TC
For an IDEAL refrigerator:

66. COP EXAMPLE
A Carnot refrigerator operates between 500 K and 400 K. It extracts 800 J from a cold reservoir during each cycle. What is C.O.P., W and QH ?
Cold Res. TC
Eng ine
Hot Res. TH
800 J W QH
500 K
400 K
K =
400 K
500 K K TC
TH- TC =
C.O.P. (K) = 4.0

67. COP EXAMPLE (Cont.) QH W 800 J 500 K 400 K
Cold Res. TC
Eng ine
Hot Res. TH
800 J W QH
500 K
400 K
Next we will find QH by assuming same K for actual refrigerator (Carnot).
K = QC
QH- QC
800 J
QH J =
4.0
QH = 1000 J

68. COP EXAMPLE (Cont.) 1000 J W 800 J 500 K
Cold Res. TC
Engine
Hot Res. TH
800 J W
1000 J
500 K
400 K
Now, can you say how much work is done in each cycle?
Work = 1000 J J
Work = 200 J

69. Q = U + W final - initial)
Summary
The First Law of Thermodynamics: The net heat taken in by a system is equal to the sum of the change in internal energy and the work done by the system.
Q = U + W final - initial)
Isochoric Process: V = 0, W = 0
Isobaric Process: P = 0
Isothermal Process: T = 0, U = 0
Adiabatic Process: Q = 0

70. The following are true for ANY process:
Summary (Cont.)
The Molar Specific Heat capacity, C:
Units are:Joules per mole per Kelvin degree
c = Q
n T
The following are true for ANY process:
Q = U + W
U = nCv T
PV = nRT

71. Summary (Cont.) Qhot Wout Qcold
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
Wout
The Second Law of Thermo: It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.
Not only can you not win (1st law); you can’t even break even (2nd law)!

72. Summary (Cont.) The efficiency of a heat engine: e = 1 - QC QH e = 1 -TC TH
The coefficient of performance of a refrigerator:

73. CONCLUSION: Chapter 20 Thermodynamics