THERMODYNAMIC SYSTEMS

TO KNOW WORK , HEAT TRANSFER

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 thefirstand
second lawsofthermodynamics.
•Demonstrate your understanding
ofadiabatic, isochoric, isothermal,
and isobaric processes.
•Write and apply a relationship for determining
theideal efficiencyof a heat engine.
•Write and apply a relationship for determining
coefficient of performancefor 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 Uof 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 INCREASETHE
INTERNAL ENERGY, U.
HEAT PUT INTO
A SYSTEM
(Positive)
+U
WORK DONE
ONA GAS
(Positive)

7. WORK DONE BY
EXPANDING GAS:
W is positive
-U
Decrease
TWO WAYS TO DECREASETHE
INTERNAL ENERGY, U.
HEAT LEAVESA
SYSTEM
Q is negative
Q
out
hot
W
out
hot

8. THERMODYNAMIC STATE
The STATE of a thermodynamic
system is determined by four
factors:
•Absolute Pressure Pin
Pascals
•Temperature Tin Kelvins
•Volume Vin cubic meters
•Number of moles,n, of working gas

9. THERMODYNAMIC PROCESS
Increase in Internal Energy, U.
Initial State:
P
1V
1 T
1n
1
Final State:
P
2V
2 T
2n
2
Heat input
Q
in
W
out
Work by gas

10. The Reverse Process
Decrease in Internal Energy, U.
Initial State:
P
1V
1 T
1n
1
Final State:
P
2V
2 T
2n
2
Work on gas
Loss of heat
Q
out
W
in

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 BYthe system.
Q = U + W final -initial)
•Conversely, the work done ONa system is
equal to the change in internal energy plus
the heat lost in the process.

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

13. APPLICATION OF FIRST
LAW OF THERMODYNAMICS
Example 1:In the figure, the
gas absorbs400 Jof heat and
at the same time does120 J
of work on the piston. What
is the change in internal
energy of the system?
Q = U + W
Apply First Law:
Q
in
400 J
W
out=120 J

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

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

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

18. ISOCHORIC EXAMPLE:
Heat input
increases P
with const. V
400 Jheat input increases
internal energy by 400 J
and zero work is done.
B
A
P
2
V
1= V
2
P
1
P
A P
B
T
A T
B
=
400 J
No Change in
volume:

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

20. ISOBARIC EXAMPLE (Constant Pressure):
Heat input
increases V
with const. P
400 Jheat does 120 Jof
work, increasing the
internal energy by 280 J.
400 J
BA
P
V
1V
2
V
A V
B
T
A T
B
=

21. ISOBARIC WORK
400 J
Work = Area under PV curveWorkPV
BA
P
V
1V
2
V
A V
B
T
A T
B
=
P
A = P
B

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

23. ISOTHERMAL EXAMPLE (Constant T):
P
AV
A =P
BV
B
Slow compression at
constant temperature:
-----No change in U.
U = T= 0
B
A
P
A
V
2 V
1
P
B

24. ISOTHERMAL EXPANSION (Constant T):
400 Jof energy is absorbed
by gas as 400 Jof work is
done on gas.
T = U= 0
U= T= 0
B
A
P
A
V
AV
B
P
B
P
AV
A = P
BV
B
T
A = T
Bln
B
A
V
W nRT
V

Isothermal Work

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

26. ADIABATIC EXAMPLE:
Insulated
Walls: Q = 0
B
A
P
A
V
1 V
2
P
B
Expanding gas does
work with zero heat
loss. Work = -U

27. ADIABATIC EXPANSION:
400 Jof WORK is done,
DECREASING the internal
energy by 400 J:Net heat
exchange is ZERO. Q = 0
Q = 0
B
A
P
A
V
AV
B
P
B
P
AV
A P
BV
B
T
A T
B
=AABB
PVPV



28. MOLAR HEAT CAPACITY
OPTIONAL TREATMENT
The molar heat capacity Cis 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KQ
c
mt



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:
For example, a constant volume of oxygen
requires 21.1 Jto raise the temperature of
one moleby one kelvin degree.
C =
Q
n T

31. SPECIFIC HEAT CAPACITY
CONSTANT VOLUME
How much heat is required to
raise the temperature of 2 moles
of O
2from 0
o
C to 100
o
C?
Q= (2 mol)(21.1 J/mol K)(373 K -273 K)
Q= nCv T
Q= +4220 J

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

33. SPECIFIC HEAT CAPACITY
CONSTANT PRESSURE
We have just seen that 4220 Jof
heat were needed at constant
volume. Suppose we want to also
do1000 Jof work at constant
pressure?
Q = U + W
Q = 4220 J + J
Q =5220 J C
p> C
v
Same

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

35. REMEMBER, FOR ANYPROCESS
INVOLVING AN IDEAL GAS:
PV = nRT
U = nCv TQ = U + W
P
AV
A P
BV
B
T
A T
B
=

36. Example Problem:
•AB: Heated at constant V to 400 K.
A 2-L sample of Oxygen gas has an initial temp-
erature and pressure of 200 Kand 1 atm. The
gas undergoes four processes:
•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 PROBLEM
B
A
P
B
2 L
1 atm
200 K
400 K800 K
How many moles
of O2are present?
Consider point A:
PV = nRT3
(101,300Pa)(0.002m )
0.122 mol
(8.314J/mol K)(200K)
PV
n
RT
  


38. PROCESS AB: ISOCHORIC
What is the pressure
at point B?
P
A P
B
T
A T
B
=
1 atmP
B
200 K400 K
=
P
B= 2 atm
or 203 kPa
B
A
P
B
2 L
1 atm
200 K
400 K800 K

39. PROCESS AB: Q = U + W
Analyze first law
for ISOCHORIC
process AB.
W = 0
Q = U = nCv T
U =(0.122 mol)(21.1 J/mol K)(400 K -200 K)
B
A
P
B
2 L
1 atm
200 K
400 K800 K
Q= +514 J W= 0U= +514 J

40. PROCESS BC: ISOBARIC
What is the volume
at point C (& D)?
V
B V
C
T
B T
C
=
2 LV
C
400 K800 K
=
B
CP
B
2 L
1 atm
200 K
400 K800 K
D
4 L
V
C= V
D= 4 L

41. FINDING U FOR PROCESS BC.
Process BC is
ISOBARIC.
P = 0
U = nCv T
U= (0.122 mol)(21.1 J/mol K)(800 K -400 K)
U= +1028 J
B
C
2 L
1 atm
200 K
400 K800 K
4 L
2 atm

42. FINDING W FOR PROCESS BC.
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
B
C
2 L
1 atm
200 K
400 K800 K
4 L
2 atm

43. FINDING Q FOR PROCESS BC.
Analyze first
law for BC.
Q = U + W
Q = +1028 J + 405 J
Q = +1433 J
Q= 1433 J W= +405 J
B
C
2 L
1 atm
200 K
400 K800 K
4 L
2 atm
U= 1028 J

44. PROCESS CD: ISOCHORIC
What is temperature
at point D?
P
C P
D
T
C T
D
=
2 atm1 atm
800 KT
D
=
T
D= 400 K
B
A
P
B
2 L
1 atm
200 K
400 K800 K
C
D

45. PROCESS CD: Q = U + W
Analyze first law
for ISOCHORIC
process CD.
W = 0
Q = U = nCv T
U= (0.122 mol)(21.1 J/mol K)(400 K -800 K)
Q = -1028 J W = 0U = -1028 J
C
D
P
B
2 L
1 atm
200 K
400 K
800 K
400 K

46. FINDING U FOR PROCESS DA.
Process DA is
ISOBARIC.
P = 0
U = nCv T
U = (0.122 mol)(21.1 J/mol K)(400 K -200 K)
U = -514 J
A
D
2 L
1 atm
200 K
400 K800 K
4 L
2 atm
400 K

47. FINDING W FOR PROCESS DA.
Work depends
on change inV.
P = 0
Work = PV
W= (1 atm)(2 L -4 L) = -2 atm L = -203 J
W = -203 J
A
D
2 L
1 atm
200 K
400 K800 K
4 L
2 atm
400 K

48. FINDING Q FOR PROCESS DA.
Analyze first
law for DA.
Q = U + W
Q = -514 J -203 J
Q = -717 J
Q = -717 J W = -203 JU = -514 J
A
D
2 L
1 atm
200 K
400 K800 K
4 L
2 atm
400 K

49. PROBLEM SUMMARY
Q = U + W
For all
processes:Process Q U W
AB 514 J 514 J 0
BC 1433 J 1028 J 405 J
CD -1028 J -1028 J 0
DA -717 J -514 J -203 J
Totals 202 J 0 202 J

50. NET WORK FOR COMPLETE
CYCLE IS ENCLOSED AREA
B C
2 L
1 atm
4 L
2 atm
+404 J
B C
2 L
1 atm
4 L
2 atm
Neg
-202 J
Area = (1 atm)(2 L)
Net Work = 2 atm L = 202 J
2 L4 L
BC
1 atm
2 atm

51. ADIABATIC EXAMPLE:
Q = 0
A
B
P
B
V
BV
A
P
A P
AV
A P
BV
B
T
A T
B
=
P
AV
A= P
BV
B
 
Example 2:A diatomic gas at 300 K and
1 atm is compressed adiabatically, decreasing
its volume by 1/12. (V
A= 12V
B). What is the
new pressure and temperature? (= 1.4)

52. ADIABATIC (Cont.): FIND P
B
Q = 0
P
B= 32.4 atm
or 3284 kPa1.4
12
B
BA
B
V
PP
V


 1.4
(1 atm)(12)
B
P
P
AV
A= P
BV
B
 
A
B
P
B
V
B12V
B
1 atm
300 KSolve for P
B:A
BA
B
V
PP
V





53. ADIABATIC (Cont.): FIND T
B
Q = 0
T
B= 810 K
(1 atm)(12V
B)(32.4 atm)(1 V
B)
(300 K) T
B
=
A
B
32.4 atm
V
B12V
B
1 atm
300 K
Solve for T
B
T
B=?A A B B
AB
P V P V
TT


54. ADIABATIC (Cont.): If V
A= 96 cm
3
and V
A= 8 cm
3
, FIND W
Q = 0
W = -U = -nC
VT& C
V=21.1 j/mol K
A
B
32.4 atm
1 atm
300 K
810 K
Since Q = 0,
W = -U
8 cm
3
96 cm
3
Find n from
point A
PV = nRT
PV
RT
n =

55. ADIABATIC (Cont.): If V
A= 96 cm
3
and V
A= 8 cm
3
, FIND W
A
B
32.4 atm
1 atm
300 K
810 K
8 cm
3
96 cm
3
PV
RT
n = =
(101,300 Pa)(8 x10
-6
m
3
)
(8.314 J/mol K)(300 K)
n= 0.000325 mol & C
V= 21.1 j/mol K
T= 810 -300 = 510 K
W = -U = -nC
VT
W= -3.50 J

56. •Absorbs heat Q
hot
•Performs work W
out
•Rejects heat Q
cold
A heat engine is any
device which through
a cyclic process:
Cold Res. T
C
Engine
Hot Res. T
H
Q
hot
W
out
Q
cold
HEAT ENGINES

57. THE SECOND LAW OF
THERMODYNAMICS
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)!
W
out
Cold Res. T
C
Engine
Hot Res. T
H
Q
hot
Q
cold

58. THE SECOND LAW OF
THERMODYNAMICS
Cold Res. T
C
Engine
Hot Res. T
H
400 J
300 J
100 J
•A possible engine.•An IMPOSSIBLE
engine.
Cold Res. T
C
Engine
Hot Res. T
H
400 J
400 J

59. EFFICIENCY OF AN ENGINE
Cold Res. T
C
Engine
Hot Res. T
H
Q
H W
Q
C
The efficiency of a heat engine
is the ratio of the net work
done W to the heat input Q
H.
e = 1 -
Q
C
Q
H
e = =
W
Q
H
Q
H-Q
C
Q
H

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

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

62. Example 3:A steam engine absorbs 600 J
of heat at 500 Kand the exhaust
temperature is 300 K. If the actual
efficiency is only half of the ideal efficiency,
how much workis done during each cycle?
e = 1 -
T
C
T
H
e = 1 -
300 K
500 K
e = 40%
Actual e = 0.5e
i= 20%
e =
W
Q
H
W = eQ
H= 0.20 (600 J)
Work = 120 J

63. REFRIGERATORS
A refrigerator is an engine
operating in reverse:
Work is done ongas
extracting heat fromcold
reservoir and depositing
heat intohot reservoir.
W
in + Q
cold = Q
hot
W
IN = Q
hot -Q
cold
Cold Res. T
C
Engine
Hot Res. T
H
Q
hot
Q
cold
W
in

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.
If this were possible, we could
establish perpetual motion!
Cold Res. T
C
Engine
Hot Res. T
H
Q
hot
Q
cold

65. COEFFICIENT OF PERFORMANCE
Cold Res. T
C
Engine
Hot Res. T
H
Q
H W
Q
C
The COP (K)of a heat
engine is the ratio of the
HEATQ
cextracted to the
net WORKdone W.
K =
T
H
T
H-T
C
For an IDEAL
refrigerator:
Q
C
W
K = =
Q
H
Q
H-Q
C

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 Q
H?
Cold Res. T
C
Eng
ine
Hot Res. T
H
800 J
WQ
H
500 K
400 K
K =
400 K
500 K -400 K
T
C
T
H-T
C
=
C.O.P. (K) = 4.0

67. COP EXAMPLE (Cont.)
Next we will find Q
Hby
assuming same K for actual
refrigerator (Carnot).
Cold Res. T
C
Eng
ine
Hot Res. T
H
800 J
WQ
H
500 K
400 K
K =
Q
C
Q
H-Q
C
Q
H= 1000 J
800 J
Q
H-800 J
=4.0

68. COP EXAMPLE (Cont.)
Now, can you say how much
work is done in each cycle?
Cold Res. T
C
Engine
Hot Res. T
H
800 J
W1000 J
500 K
400 K
Work = 1000 J -800 J
Work = 200 J

69. Summary
Q = U + W final -initial)
TheFirst 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.
•Isochoric Process: V = 0, W = 0
•Isobaric Process: P = 0
•Isothermal Process: T = 0, U = 0
•Adiabatic Process: Q = 0

70. Summary (Cont.)
c=
Q
n T
U = nC
v T
The Molar
Specific Heat
capacity, C:
Units are:Joules
per mole per
Kelvin degree
The following are true for ANY process:
Q = U + W
PV = nRTA A B B
AB
P V P V
TT


71. Summary (Cont.)
TheSecond 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.
Cold Res. T
C
Engine
Hot Res. T
H
Q
hot
Q
cold
W
out
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 -
Q
C
Q
H
e = 1 -
T
C
T
H
The coefficient of performance of a refrigerator:CC
in H C
QQ
K
W Q Q

 C
HC
T
K
TT



73. CONCLUSION: Chapter 20
Thermodynamics