Transcript Thermodynamics

Chapter 20 - Thermodynamics
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
THERMODYNAMICS
Thermodynamics is
the study of energy
relationships that
involve heat,
mechanical work,
and other aspects of
energy and heat
transfer.
Central Heating
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.
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
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.
TWO WAYS TO INCREASE THE
INTERNAL ENERGY, U.
+U
WORK DONE
ON A GAS
(Positive)
HEAT PUT INTO
A SYSTEM
(Positive)
TWO WAYS TO DECREASE THE
INTERNAL ENERGY, U.
Wout
Qout
-U
Decrease
hot
WORK DONE BY
EXPANDING GAS:
W is positive
hot
HEAT LEAVES A
SYSTEM
Q is negative
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
THERMODYNAMIC PROCESS
Increase in Internal Energy, U.
Wout
Qin
Initial State:
P1 V1 T1 n1
Heat input
Final State:
Work by gas
P2 V2 T2 n2
The Reverse Process
Decrease in Internal Energy, U.
Win
Qout
Initial State:
P1 V1 T1 n1
Work on gas
Loss of heat
Final State:
P2 V2 T2 n2
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.
SIGN CONVENTIONS
FOR FIRST LAW
• Heat Q input is positive
+Wout
+Qin
U
• Work BY a gas is positive
-Win
U
• Work ON a gas is negative
• Heat OUT is negative
Q = U + W
-Qout
final - initial)
APPLICATION OF FIRST
LAW OF THERMODYNAMICS
Example 1: In the figure, the
Wout =120 J
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?
Qin
Apply First Law:
Q = U + W
400 J
Example 1 (Cont.): Apply First Law
Q is positive: +400 J (Heat IN)
Wout =120 J
W is positive: +120 J (Work OUT)
Q = U + W
U = Q - W
Qin
400 J
U = Q - W
= (+400 J) - (+120 J)
= +280 J
U = +280 J
Example 1 (Cont.): Apply First Law
Energy is conserved:
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:
Wout =120 J
Qin
400 J
U = +280 J
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
ISOCHORIC PROCESS:
CONSTANT VOLUME, V = 0, W = 0
0
Q = U + W
so that
Q = U
QIN
+U
QOUT
No Work
Done
-U
HEAT IN = INCREASE IN INTERNAL ENERGY
HEAT OUT = DECREASE IN INTERNAL ENERGY
ISOCHORIC EXAMPLE:
No Change in
volume:
P2
B
P1
A
PA
TA
=
PB
TB
V1= V2
400 J
Heat input
increases P
with const. V
400 J heat input increases
internal energy by 400 J
and zero work is done.
ISOBARIC PROCESS:
CONSTANT PRESSURE, P = 0
Q = U + W
But
W = P V
QIN
QOUT
Work Out
+U
-U
Work
In
HEAT IN = Wout + INCREASE IN INTERNAL ENERGY
HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY
ISOBARIC EXAMPLE (Constant Pressure):
P
A
B
VA
TA
400 J
Heat input
increases V
with const. P
V1
=
VB
TB
V2
400 J heat does 120 J of
work, increasing the
internal energy by 280 J.
ISOBARIC WORK
P
A
B
VA
TA
400 J
V1
V2
=
TB
PA = PB
Work = Area under PV curve
W
orkPV
VB
ISOTHERMAL PROCESS:
CONST. TEMPERATURE, T = 0, U = 0
Q = U + W
AND
QIN
U = 0
Q = W
QOUT
Work Out
U = 0
Work
In
NET HEAT INPUT = WORK OUTPUT
WORK INPUT = NET HEAT OUT
ISOTHERMAL EXAMPLE (Constant T):
PA
A
B
PB
U = T = 0
PAVA = PBVB
V2
V1
Slow compression at
constant temperature:
----- No change in U.
ISOTHERMAL EXPANSION (Constant T):
PA
A
B
PB
U = T = 0
VA
VB
400 J of energy is absorbed
by gas as 400 J of work is
done on gas.
T = U = 0
PAVA = PBVB
TA = TB
Isothermal Work
VB
W  nRT ln
VA
ADIABATIC PROCESS:
NO HEAT EXCHANGE, Q = 0
Q = U + W ; W = -U or U = -W
U = -W
W = -U
U
Work Out
Q = 0
+U
Work
In
Work done at EXPENSE of internal energy
INPUT Work INCREASES internal energy
ADIABATIC EXAMPLE:
PA
A
B
PB
V1
Insulated
Walls: Q = 0
V2
Expanding gas does
work with zero heat
loss. Work = -U
ADIABATIC EXPANSION:
PA
A
B
PB
Q = 0
PAVA
TA
VA
400 J of WORK is done,
DECREASING the internal
energy by 400 J: Net heat
exchange is ZERO. Q = 0
=
PBVB
TB
VB


P
AV
A P
BV
B
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.
SPECIFIC HEAT CAPACITY
Remember the definition of specific heat
capacity as the heat per unit mass
required to change the temperature?
Q
c
m t
For example, copper: c = 390 J/kgK
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.
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 - 273 K)
Q = +4220 J
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
U = nCv T
Thus, U is determined by the
change of temperature and the
specific heat at constant volume.
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?
Q = U + W
Same
Q = 4220 J + J
Q = 5220 J
Cp > Cv
HEAT CAPACITY (Cont.)
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
REMEMBER, FOR ANY PROCESS
INVOLVING AN IDEAL GAS:
PV = nRT
Q = U +  W
PAVA
TA
=
PBVB
TB
U = nCv T
Example Problem:
A 2-L sample of Oxygen gas has an initial temperature 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.
PV-DIAGRAM FOR PROBLEM
How many moles
of O2 are present?
Consider point A:
PV = nRT
PB
1 atm
B
A
400 K
800 K
200 K
2L
PV (101, 300Pa)(0.002m3 )
n

 0.122 mol
RT (8.314J/mol  K)(200K)
PROCESS AB: ISOCHORIC
What is the pressure
at point B?
PA
TA
=
1 atm
200 K
B
PB
A
1 atm
PB
200 K
2L
TB
=
400 K
PB
400 K
P B = 2 atm
or
203 kPa
800 K
PROCESS AB: Q = U + W
Analyze first law
for ISOCHORIC
process AB.
W = 0
PB
1 atm
B
A
Q = U = nCv T
400 K
800 K
200 K
2L
U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K)
Q = +514 J
U = +514 J
W = 0
PROCESS BC: ISOBARIC
What is the volume
at point C (& D)?
VB
TB
=
2L
400 K
VC
1 atm
TC
=
PB
B
400 K
800 K
C
200 K
2L
VC
800 K
D
4L
VC = VD = 4 L
FINDING U FOR PROCESS BC.
Process BC is
ISOBARIC.
P = 0
2 atm
B
1 atm
U = nCv T
400 K
800 K
C
200 K
2L
4L
U = (0.122 mol)(21.1 J/mol K)(800 K - 400 K)
U = +1028 J
FINDING W FOR PROCESS BC.
Work depends
on change in V.
P = 0
Work = P V
2 atm
B
400 K
800 K
C
200 K
1 atm
2L
4L
W = (2 atm)(4 L - 2 L) = 4 atm L = 405 J
W = +405 J
FINDING Q FOR PROCESS BC.
Analyze first
law for BC.
2 atm
Q = U + W
1 atm
Q = +1028 J + 405 J
B
400 K
800 K
C
200 K
2L
4L
Q = +1433 J
Q = 1433 J
U = 1028 J
W = +405 J
PROCESS CD: ISOCHORIC
What is temperature
at point D?
PC
TC
=
2 atm
800 K
1 atm
PD
A
400 K
200 K
2L
TD
=
PB
B
1 atm
TD
T D = 400 K
800 K
C
D
PROCESS CD: Q = U + W
Analyze first law
for ISOCHORIC
process CD.
PB
W = 0
400 K 800 K
200 K
1 atm
Q = U = nCv T
C
D
400 K
2L
U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K)
Q = -1028 J
U = -1028 J
W = 0
FINDING U FOR PROCESS DA.
Process DA is
ISOBARIC.
P = 0
U = nCv T
400 K
2 atm
1 atm
800 K
200 K
A
2L
400 K
D
4L
U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K)
U = -514 J
FINDING W FOR PROCESS DA.
Work depends
on change in V.
P = 0
400 K
2 atm
1 atm
A
Work = P V
200 K
2L
800 K
400 K
D
4L
W = (1 atm)(2 L - 4 L) = -2 atm L = -203 J
W = -203 J
FINDING Q FOR PROCESS DA.
Analyze first
law for DA.
400 K
2 atm
Q = U + W
1 atm
Q = -514 J - 203 J
A
800 K
200 K
2L
D
400 K
4L
Q = -717 J
Q = -717 J
U = -514 J
W = -203 J
PROBLEM SUMMARY
For all
Q
=
U
+
W
processes:
Process
Q
U
W
AB
514 J
514 J
0
BC
1433 J
1028 J
405 J
CD
-1028 J -1028 J
DA
-717 J
-514 J
-203 J
Totals
202 J
0
202 J
0
NET WORK FOR COMPLETE
CYCLE IS ENCLOSED AREA
2 atm
B
+404 J
C
1 atm
Neg
1 atm
2L
2 atm
2 atm
B -202 J C
B
4L
C
1 atm
2L
4L
Area = (1 atm)(2 L)
Net Work = 2 atm L = 202 J
2L
4L
ADIABATIC EXAMPLE:
Example 2: A diatomic gas at 300 K and
1 atm is compressed adiabatically, decreasing
its volume by 1/12. (VA = 12VB). What is the
new pressure and temperature? ( = 1.4)
PB
B
Q = 0
VB

PAVA
PBVB
PAVA = PBVB
A
PA

VA
TA
=
TB
ADIABATIC (Cont.): FIND PB

B
PB
PAVA = PBVB
300 K
1 atm
A
Q = 0
VB 12VB
Solve for PB:
 VA 
PB  PA 

V
 B
1.4
 12VB 
PB  PA 

 VB 
P
(
1a
tm
)(1
2
)
B
1
.4

PB = 32.4 atm
or 3284 kPa

ADIABATIC (Cont.): FIND TB
32.4 atm
1 atm
Q = 0
B TB=?
300 K
A
VB 12VB
(1 atm)(12VB)
(300 K)
=
PAVA PBVB

TA
TB
Solve for TB
(32.4 atm)(1 VB)
TB = 810 K
TB
ADIABATIC (Cont.): If VA= 96 cm3
and VA= 8 cm3, FIND W
32.4 atm
B
810 K
300 K
1 atm
Q = 0
A
8 cm3
W = - U = - nCV T
Find n from
point A
Since Q = 0,
W = - U
96 cm3
&
PV = nRT
CV= 21.1 j/mol K
n=
PV
RT
ADIABATIC (Cont.): If VA= 96 cm3
and VA= 8 cm3, FIND W
n=
PV
RT
=
(101,300 Pa)(8 x10-6 m3)
(8.314 J/mol K)(300 K)
n = 0.000325 mol &
T = 810 - 300 = 510 K
W = - U = - nCV T
W = - 3.50 J
CV= 21.1 j/mol K
32.4 atm
B
810 K
300 K
1 atm
A
8 cm3
96 cm3
HEAT ENGINES
Hot Res. TH
Qhot
Engine
Qcold
Cold Res. TC
Wout
A heat engine is any
device which through
a cyclic process:
• Absorbs heat Qhot
• Performs work Wout
• Rejects heat Qcold
THE SECOND LAW OF
THERMODYNAMICS
Hot Res. TH
Qhot
Engine
Wout
Qcold
Cold Res. TC
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)!
THE SECOND LAW OF
THERMODYNAMICS
Hot Res. TH
400 J
100 J
Engine
Hot Res. TH
400 J
Engine
400 J
300 J
Cold Res. TC
• A possible engine.
Cold Res. TC
• An IMPOSSIBLE
engine.
EFFICIENCY OF AN ENGINE
Hot Res. TH
QH
W
Engine
QC
The efficiency of a heat engine
is the ratio of the net work
done W to the heat input QH.
e=
W
QH
=
Cold Res. TC
e=1-
QH- QC
QH
QC
QH
EFFICIENCY EXAMPLE
Hot Res. TH
800 J
Engine
600 J
Cold Res. TC
W
An engine absorbs 800 J and
wastes 600 J every cycle. What
is the efficiency?
QC
e=1QH
e=1-
600 J
800 J
e = 25%
Question: How many joules of work is done?
EFFICIENCY OF AN IDEAL
ENGINE (Carnot Engine)
Hot Res. TH
QH
Engine
QC
W
For a perfect engine, the
quantities Q of heat gained
and lost are proportional to
the absolute temperatures T.
e=
TH- TC
Cold Res. TC
e=1-
TH
TC
TH
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?
e=1e=1-
TC
Actual e = 0.5ei = 20%
TH
W
300 K
500 K
e = 40%
e=
QH
W = eQH = 0.20 (600 J)
Work = 120 J
REFRIGERATORS
Hot Res. TH
Qhot
Win
Engine
Qcold
Cold Res. TC
A refrigerator is an engine
operating in reverse:
Work is done on gas
extracting heat from cold
reservoir and depositing
heat into hot reservoir.
Win + Qcold = Qhot
WIN = Qhot - Qcold
THE SECOND LAW FOR
REFRIGERATORS
Hot Res. TH
Qhot
Engine
Qcold
Cold Res. TC
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!
COEFFICIENT OF PERFORMANCE
Hot Res. TH
QH
W
Engine
QC
The COP (K) of a heat
engine is the ratio of the
HEAT Qc extracted to the
net WORK done W.
K=
Cold Res. TC
For an IDEAL
refrigerator:
QC
W
K=
=
QH
QH- QC
TH
TH- TC
COP EXAMPLE
500 K
Hot Res. TH
QH
W
Eng
ine
800 J
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 ?
K=
TC
TH- TC
=
400 K
500 K - 400 K
Cold Res. TC
400 K
C.O.P. (K) = 4.0
COP EXAMPLE (Cont.)
500 K
Hot Res. TH
QH
W
Eng
ine
Next we will find QH by
assuming same K for actual
refrigerator (Carnot).
K=
800 J
Cold Res. TC
400 K
4.0 =
QC
QH- QC
800 J
QH - 800 J
QH = 1000 J
COP EXAMPLE (Cont.)
500 K
Hot Res. TH
1000 J
W
Engine
800 J
Cold Res. TC
400 K
Now, can you say how much
work is done in each cycle?
Work = 1000 J - 800 J
Work = 200 J
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
Summary (Cont.)
The Molar
Specific Heat
capacity, C:
Units are:Joules
per mole per
Kelvin degree
Q
c = n T
The following are true for ANY process:
Q = U + W
PAVA PBVB

TA
TB
U = nCv T
PV = nRT
Summary (Cont.)
Hot Res. TH
Qhot
Wout
Engine
Qcold
Cold Res. TC
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)!
Summary (Cont.)
The efficiency of a heat engine:
QC
e=1- Q
H
e=1-
TC
TH
The coefficient of performance of a refrigerator:
QC
QC
K

Win QH  QC
TC
K
TH  TC
CONCLUSION: Chapter 20
Thermodynamics