1. General Physics L02_paths.ppt energy transfers
Thermodynamic Paths
energy transfers
§ 15.1–15.4

2. Definitions System: bodies and surroundings exchanging energy
State: unique set of p, V, T, (n or N) (state variables)
Process: change in state of a system

3. Internal Energy U U  SKi + SSVij
j<i
Ki = kinetic energy of molecule i wrt com
Vij = intermolecular potential energy of i and j
Does not include potential or kinetic energy of bulk object
Each thermodynamic state has a unique U (U is a state function)

4. Monatomic Ideal Gas
U = S ½ mivi2 = S 3/2 kT = 3/2 NkT = 3/2 nRT

5. Question
All other things being equal, adding heat to a system increases its internal energy U.
True.
False.

6. Question
All other things being equal, lifting a system to a greater height increases its internal energy U.
True.
False.

7. Question
All other things being equal, accelerating a system to a greater speed increases its internal energy U.
True.
False.

8. Question
All other things being equal, doing work to compress a system increases its internal energy U.
True.
False.

9. Clarification Quizzes will generally be on Wednesdays
First quiz Feb 8/9 (Thursday/Friday: Monday late start)

10. Energy Transfers
Q: heat added to the system surroundings  system From a temperature difference
W: work done by the system system  surroundings Achieved by a volume change

11. Work W The surroundings exert pressure on the system.
If the system expands, it does work on the surroundings.
So, W > 0,
and the surroundings do negative work on the system.

12. First law of Thermodynamics
DU = Q – W
DU is path-independent

13. Conservation of Energy
DU of a system =
work done on the system +
heat added to the system

14. Work and Heat
Depend on the path taken between initial and final states.

15. pV Diagrams
W = area under pV curve
Source: Y&F, Figure 19.6a

16. Question What is this system doing? Expanding Contracting
Absorbing heat at constant volume
Absorbing heat at constant pressure
Source: Y&F, Figure 19.6b

17. Question What is the sign of the work W for this process? + –
Cannot be determined
Source: Y&F, Figure 19.6b

18. Question What is this system doing? Expanding at constant volume
Expanding at constant temperature
Expanding at constant pressure
Source: Y&F, Figure 19.6c

19. Question How is the temperature of this ideal gas changing? Increasing
Decreasing
Remaining constant
Cannot be determined
Source: Y&F, Figure 19.6c

20. Simple Case Expansion at constant pressure W = pDV
Source: Y&F, Figure 19.6c

21. Question
The work done by a thermodynamic system in a cyclic process (final state is also the initial state) is zero.
True.
False.
Source: Y&F, Figure 19.12

22. Cyclic Process W  0 Is the system a limitless source of work?
(Of course not.) W
Source: Y&F, Figure 19.12

23. Cyclic Processes DU = U1 – U1 = 0 so Q – W = 0 Q = W
Total work output = total heat input

24. Work out = Heat in
Does this mean cyclic processes convert heat to work with 100% efficiency?
(Of course not.)
Waste heat is expelled, not recovered.

25. Types of Processes
cool names, easy rules

26. Reversible
An infinitesimal change in conditions reverses the direction
Requires no non-conservative processes
no friction
no contact between different temperatures
An ideal concept
not actually possible
some processes can get close

27. Constant pressure
“Isobaric”
W = PDV

28. Constant Volume
“Isochoric”
W = 0

29. Constant Temperature
“Isothermal”
Ideal gas: W = nRT ln(Vf/Vi)

30. General Physics L02_paths.ppt
No Heat Flow
“Adiabatic”
Q = 0
W: more complicated
PfVfg = PiVig
g = heat capacity ratio CP/CV
CP for constant pressure
CV for constant volume

31. Specific Heats of Ideal Gases
§ 15.6

32. Constant Volume or Pressure
Constant volume: heating simply makes the molecules go faster
Constant pressure: As the molecules speed up, the system expands against the surroundings, doing work
It takes more heat to get the same DT at constant pressure than at constant volume

33. Constant Volume U = 3/2 nRT for a monatomic gas DU = 3/2 nRDT
DU = Q – W = Q
Q = 3/2 nRDT
Molar heat capacity CV = Q/(nDT)
CV = 3/2 R

34. Constant Pressure U = 3/2 nRT DU = 3/2 nRDT DU = Q – W W = PDV = nRDT
Q = DU + W = 3/2 nRDT + nRDT
= 5/2 nRDT
Molar heat capacity CP = Q/(nDT)
CP = 5/2 R

35. Heat Capacity Ratio
g = CP/CV = (5/2)/(3/2) = 5/3 For a monatomic ideal gas

36. Homework 1 Due Feb 1/Feb 2 pp. 442–443 Conceptual Questions 6 and 7
p. 444 Problems 1, 8, and 10
p. 444 Problem 15
Hint: Find DT in terms of DV using PV = nRT Find DU using U = 3/2 nRT
p. 445 Problem 25
Hint: What is DU for A→C?
p. 448 Problem 80