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Final Exam May 3, 10:30 am-12:30 pm ELLT 116
Lecture 26
Purdue University, Physics 220

2. Purdue University, Physics 220
Lecture 26
Thermodynamics I
Lecture 26
Purdue University, Physics 220

3. Purdue University, Physics 220
Lecture 26
Purdue University, Physics 220

4. First Law of Thermodynamics
Energy Conservation
The change in internal energy of a system (U) is equal to the heat flow into the system (Q) minus the work done by the system (W)
DU = Q - W
Work done by system
Increase in internal energy of system
Heat flow into system
Equivalent ways of writing 1st Law:
Q = DU + W
Lecture 26
Purdue University, Physics 220

5. Purdue University, Physics 220
Lecture 26
Purdue University, Physics 220

6. Purdue University, Physics 220
Signs Example
You are heating some soup in a pan on the stove. To keep it from burning, you also stir the soup. Apply the 1st law of thermodynamics to the soup. What is the sign of
A) Q
B) W
C) DU
Positive, heat flows into soup
Negative, stirring does work on soup
Positive, soup gets warmer
Lecture 26
Purdue University, Physics 220

7. Purdue University, Physics 220
Work Done by a System M M y
W = F d cosq= P A d =
P A Dy = P DV
W = P V
W > 0 if V > 0
expanding system does positive work
W < 0 if V < 0
contracting system does negative work
W = 0 if V = 0
system with constant volume does no work
Lecture 26
Purdue University, Physics 220

8. Purdue University, Physics 220
PV Diagram
Ideal gas law: PV = nRT
For n fixed, P and V determine “state” of system
T = PV/nR
U = (3/2)nRT = (3/2)PV
Examples:
which point has highest T? B
which point has lowest U? C
to change the system from C to B,
energy must be added to system V P A B C
V V2 P1 P3
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Purdue University, Physics 220

9. Purdue University, Physics 220
Example (Isobaric)
2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure p=1000 Pa, where V1 = 2m3 and V2 = 3m3. Find T1, T2, DU, W, Q. V P 1 2
V V2
1. pV1 = nRT1  T1 = pV1/nR = 120K
2. pV2 = nRT2  T2 = pV2/nR = 180K
3. DU = (3/2) nR DT = 1500 J
DU = (3/2) D pV = (3/2) ( )J =1500 J
(has to be the same)
4. W = p DV = 1000 J
5. Q = DU + W = = 2500 J
Lecture 26
Purdue University, Physics 220

10. Purdue University, Physics 220
Example (Isochoric)
2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume V=2m3, where T1 = 120K and T2 = 180K. Find Q. P 2 P2 P1 1
1. pV1 = nRT1  p1 = nRT1/V = 1000 Pa
2. pV2 = nRT2  p2 = nRT2/V = 1500 Pa
3. DU = (3/2) nR DT = 1500 J
4. W = p DV = 0 J
5. Q = DU + W = = 1500 J
Requires less heat to raise T at const. volume than at const. pressure CV=CP-R V V
Lecture 26
Purdue University, Physics 220

11. Purdue University, Physics 220
Question
Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest?
A) Case 1
B) Case 2
C) Same A B 4 2 3 9
V(m3)
Case 1
P(atm)
Case 2
Net Work = area under P-V curve
Area the same in both cases!
Lecture 26
Purdue University, Physics 220

12. Purdue University, Physics 220
Question
Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the change in internal energy of the system the biggest?
A) Case 1
B) Case 2
C) Same A B 4 2 3 9
V(m3)
Case 1
P(atm)
Case 2
U = 3/2 (PV)
Case 1: (PV) = 4x9-2x3=30 atm*m3
Case 2: (PV) = 2x9-4x3= 6 atm*m3
Lecture 26
Purdue University, Physics 220

13. Purdue University, Physics 220
iClicker
Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the heat added to the system the biggest?
A) Case 1
B) Case 2
C) Same A B 4 2 3 9
V(m3)
Case 1
P(atm)
Case 2
Q = U + W
W is same for both
U is larger for Case 1
Therefore, Q is larger for Case 1
Lecture 26
Purdue University, Physics 220

14. Purdue University, Physics 220
Example (Isothermal)
W=nRTln(Vf/Vi)
for isothermal expansion
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Purdue University, Physics 220

15. Purdue University, Physics 220
Special PV Cases V P
W = PDV (>0) 1 2 3 4
DV > 0
Constant Pressure
isobaric
Constant Volume
isochoric
Constant Temp DU = 0
isothermal
Adiabatic Q=0 (no heat is transferred) V P
W = PDV = 0 1 2 3 4
DV = 0
Lecture 26
Purdue University, Physics 220

16. Purdue University, Physics 220
First Law Questions
Q = DU + W
Work done by system V P 1 2 3
V V2 P1 P3
Increase in internal energy of system
Heat flow into system
Some questions:
Which part of cycle has largest change in internal energy, U ?
2  3 (since U = 3/2 PV)
Which part of cycle involves the least work W ?
3  1 (since W = PV)
What is change in internal energy for full cycle?
U = 0 for closed cycle (since both P & V are back where they started)
What is net heat into system for full cycle (positive or negative)?
U = 0  Q = W = area of triangle (>0)
Lecture 26
Purdue University, Physics 220

17. Purdue University, Physics 220
1st Law Summary
1st Law of Thermodynamics: Energy Conservation
Q = DU + W
Heat flow into system
Increase in internal energy of system
Work done by system P
Point on P-V plot completely specifies state of system (PV = nRT)
Work done is area under curve
U depends only on T
(U = 3nRT/2 = 3PV/2)
For a complete cycle DU=0  Q=W V
Lecture 26
Purdue University, Physics 220

18. Purdue University, Physics 220
Reversible?
Most “physics” processes are reversible, you could play movie backwards and still looks fine. (drop ball vs throw ball up)
Exceptions:
Non-conservative forces (friction)
Heat Flow
Heat never flows spontaneously from a colder body to a hotter body.
Lecture 26
Purdue University, Physics 220

19. Example: 19 19

20. Example: 20 20

21. Example: 21 21

22. Example: 22 22

23. Purdue University, Physics 220
iClicker
Path A and path B both take system from initial (i) to final state (f).
Which of the following is true:
A) WA=WB and QA=QB
B) WA=WB and QAQB
C) WAWB and QA=QB
D) WAWB and QAQB
WA=Pi(Vf-Vi)
WB=Pf(Vf-Vi)
UA=QA-WA
UB=QB-WB
Lecture 26
Purdue University, Physics 220