Transcript DU = Q + W The First Law of Thermodynamics During an expansion

The Laws of Thermodynamics
AP Physics B
Lecture Notes
The Laws of Thermodynamics
Topics
12-01 Work in Thermodynamic Processes
12-02 The First Law of Thermodynamics
12-03 Heat Engines and the Second Law of Thermodynamics
12-04 Entropy
Work in Thermodynamic Processes
P
F
A
F
W = -F Dx
F  PA
DV = A Dx
W= -PA Dx
W = -P DV
Area
DV
Dx
During a compression:
Work done on a gas is positive.
During an expansion:
Work done by a gas is negative.
The First Law of Thermodynamics
System
U
Q
W
Environment
DU = Q + W
First Law of Thermodynamics
The First Law of Thermodynamics
Pressure - Volume Graph
Isotherms
(lines of constant
temperature)
P
Pressure
T4
T3
T2
T1
V
Volume
Area under curve
represents work
Internal energy
is proportional
to temperature
The First Law of Thermodynamics
Common Processes
A.
B.
C.
D.
Isobaric
Isochoric (Isovolumetric)
Isothermal
Adiabatic
The First Law of Thermodynamics
Isobaric Expansion (Constant Pressure)
P
During an expansion:
Work done by a gas is negative.
Po
a) W = -Po DV
T4
T3
T2
DU is
positive
b) DU increases
c) Q = DU - W
T1
V
DV
First Law of Thermodynamics:
DU = Q + W
Heat must
be added
The First Law of Thermodynamics
Isobaric Compression (Constant Pressure)
During a compression:
P
Work done on a gas is positive.
Po
a) W = -Po DV
T4
b) DU decreases
T3
T2
c) Q = DU - W
DU is
Negative
T1
DV
V
First Law of Thermodynamics:
DU = Q + W
Heat must
be removed
The Laws of Thermodynamics 12-01
The process shown on the Pressure-Volume diagram is an
(A) adiabatic expansion.
(B) isothermal expansion.
P
(C) isometric expansion.
(D) isobaric expansion.
o
V
The First Law of Thermodynamics
Isochoric (Constant Volume)
Decrease in Pressure
P
a) W = 0
Po
DU is
Negative
b) DU decreases
c) Q = DU
Pf
T3
T2
T1
V
First Law of Thermodynamics:
DU = Q + W
Heat must
be removed
The First Law of Thermodynamics
Isochoric (Constant Volume)
Increase in Pressure
P
a) W = 0
Pf
DU is
positive
b) DU increases
c) Q = DU
Po
T3
T2
T1
V
First Law of Thermodynamics:
DU = Q + W
Heat must
be added
The Laws of Thermodynamics 12-02
The process shown on the PV diagram is
(A) adiabatic.
(B) isothermal.
P
(C) isochoric.
(D) isobaric.
o
V
The Laws of Thermodynamics 12-03
In an isochoric process, there is no change in
(A) pressure.
(B) temperature.
(C) volume.
(D) internal energy.
The First Law of Thermodynamics
Isothermal (Constant Temperature)
Expansion
During an expansion:
P
Work done by a gas is negative.
a) W is negative
Po
b) DU = 0
T3
T2
Pf
T1
Vo
Vf
V
First Law of Thermodynamics:
DU = Q + W
c) Q = -W
Heat must
be added
The First Law of Thermodynamics
Isothermal (Constant Temperature)
Compression
During a compression:
P
Work done on a gas is positive.
a) W is positive
Pf
b) DU = 0
Po
T3
T2
T1
V
Vo
Vf
First Law of Thermodynamics:
DU = Q + W
c) Q = -W
Heat must
be removed
The Laws of Thermodynamics 12-04
The process shown on the Temperature-Volume graph is an
(A) adiabatic compression.
(B) isothermal compression.
T
(C) isochoric compression.
(D) isobaric compression.
o
V
The Laws of Thermodynamics 12-05
When the first law of thermodynamics, Q = ΔU + W, is
applied to an ideal gas that is taken through an
isothermal process,
(A) ΔU = 0
(B) W = 0
(C) Q = 0
(D) none of the above
The Laws of Thermodynamics 12-06
An ideal gas is compressed to one-half its original volume
during an isothermal process. The final pressure of the gas
(A) increases to twice its original value.
(B) increases to less than twice its original value.
(C) increases to more than twice its original value.
(D) does not change.
The First Law of Thermodynamics
Adiabatic (No Heat Exchange)
Expansion
During an expansion:
P
Work done by a gas is negative.
a) W = Negative
Po
DU is
Negative
b) DU decreases
T2
Pf
T1
V
Vo
Vf
First Law of Thermodynamics:
DU = Q + W
c) Q = 0
W = DU
Negative work done by the
gas is equal to the decrease
in internal energy.
The First Law of Thermodynamics
Adiabatic (No Heat Exchange)
Compression
During a compression:
P
Work done on a gas is positive.
a) W = Positive
Pf
DU is
positive
b) DU increases
T2
Po
T1
V
Vf
Vo
First Law of Thermodynamics:
DU = Q + W
c) Q = 0
W = DU
Positive work done on the
gas is equal to the increase
in internal energy.
The Laws of Thermodynamics 12-07
When the first law of thermodynamics, Q = ΔU + W, is
applied to an ideal gas that is taken through an
adiabatic process,
(A) ΔU = 0.
(B) W = 0.
(C) Q = 0.
(D) none of the above
The First Law of Thermodynamics (Problem)
An ideal gas expands to 10 times its original volume,
maintaining a constant 440 K temperature. If the gas
does 3.3 kJ of work on its surroundings, how much heat
does it absorb?
First Law of Thermodynamics:
P
Isothermal process:
DU = 0
Po
W = -3000 J
DU  Q  W
Q  DU - W
Q  0 - - 3000 J 
Q  3000 J
Pf
T
Vo
Vf
V
The Laws of Thermodynamics 12-08
A gas is taken through the cycle illustrated here. During
one cycle, how much work is done by an engine operating
on this cycle?
(A) PV
2P
(B) 2PV
P
(C) 3PV
(D) 4PV
V
2V 3V 4V
The First Law of Thermodynamics (Problem)
II
An ideal gas initially has
pressure Po, at volume Vo
and absolute temperature
To. It then undergoes the
following series of
processes:
2P
o
III
I
IV
Po
V
2Vo
o
I. Heated, at constant volume to pressure 2Po
II. Heated, at constant pressure to pressure 3Vo
III. Cooled, at constant volume to pressure Po
IV. Cooled, at constant pressure to volume Vo
3Vo
The First Law of Thermodynamics (Problem) con’t
2P
o
2To
II

III
I


Po
 6To
To
V
IV
2Vo
3To
3Vo
o
Find the temperature at each end point in terms of To
PV  nRT
To 
Po Vo
nR
The First Law of Thermodynamics (Problem) con’t
II

2P
Find the net work done by
the gas in terms of Po and Vo
o
III
I
W  2Po Vo


IV
Po
V
Net work equals net
area under curve

o
2Vo
3Vo
The First Law of Thermodynamics (Problem) con’t
II

2P
o
Find the net change in internal
energy in terms of Po and Vo
Po
DU  0

III
I


IV
V
o
2Vo
3Vo
Heat Engines and the Second Law of Thermodynamics
The second law of thermodynamics is a statement about
which processes occur and which do not. There are many
ways to state the second law; here is one:
Heat will flow spontaneously from
a hot object to a cold object.
It will not flow spontaneously from
a cold object to a hot object.
Heat Engines and the Second Law of Thermodynamics
Direction of Time
Heat Engines and the Second Law of Thermodynamics
In a heat engine; mechanical energy can be obtained from
thermal energy when heat flows from a higher temperature
to a lower temperature.
Work done by engine:
High Temp. Th
W  Qh - Qc
Thermal efficiency:
Qh
Engine
Qc
Low Temp. Tc
W
e
W
Qh
e
Qh - Qc
Qh
e  1-
Qc
Qh
Heat Engines and the Second Law of Thermodynamics
Th= 550 K
For the engine
= 920 J
Qh
Work done by engine each cycle
W  Qh - Qc
Engine
Qc
W  920 J - 630 J  290 J
= 630 J
Tc
The efficiency of the engine
e
W
Qh

290 J
920 J
 0.315
 31.5 %
W
Heat Engines and the Second Law of Thermodynamics
The Carnot Cycle
P
Isothermal Expansion
Th
Adiabatic Compression
Adiabatic Expansion
Tc
V
Isothermal Compression
Heat Engines and the Second Law of Thermodynamics
Qh
P
High Temp. Th
Qh
Engine
Th
W
Work
Qc
Low Temp. Tc
Carnot efficiency:
e  1-
Tc
Th
Tc

Qc
Qh

Tc
Th
Qc
V
Heat Engines and the Second Law of Thermodynamics
Th= 550 K
For the engine
= 890 J
Qh
Work done by engine each cycle
Engine
W  Qh - Qc
Qc
W  890 J - 470 J  420 J
= 470 J
Tc
The efficiency of the engine
e
W
Qh

420 J
890 J
 0.472
 47.2 %
W
Heat Engines and the Second Law of Thermodynamics
Th= 550 K
Temperature of the cool reservoir
= 890 J
Qh
Engine
Qc

Qh
Tc
Th
 Tc  Th
 470 J 
Tc  550 K 

 890 J 
Qc
Qh
Qc
 290 K
= 470 J
Tc
The engine undergoes 22 cycles per second,
its mechanical power output
P
W
t
J 
cycles 

 22

 Wf   420
cycle 
s 

W = 420 J
 9.24 kW
Heat Engines and the Second Law of Thermodynamics (Problem)
Th
A carnot engine absorbs 900 J of
heat each cycle and provides 350 J
of work
Qh
The efficiency of the engine
Qc
e
W
Qh

350 J
900 J
 0.389
Engine
 38.9 %
The heat ejected each cycle
W  Qh - Qc
Qc  Qh - W  900 J - 350 J
= 900 J
 550 J
Tc
W = 350 J
Heat Engines and the Second Law of Thermodynamics (Problem)
Th
A carnot engine absorbs 900 J of
heat each cycle and provides 350 J
of work
Qh
The engine ejects heat at 10 oC
The temperature of the hot
reservoir
Qc
Qc
Qh

Tc
Th
 Th  Tc
 900 J 
T  283 K

h
 550 J 
= 900 J
Engine
W = 350 J
=550 J
Tc = 283 K
Qh
Qc
o
 463 K  190 C
Heat Engines and the Second Law of Thermodynamics (Problem)
Th= 650 K
A carnot engine operates
between a hot reservoir at 650 K
and a cold reservoir at 300 K. If
it absorbs 400 J of heat at the
hot reservoir, how much work
does it deliver?
Qh
Engine
W=?
Qc
Th - Tc
e

Qh
Th
Tc= 300 K
W
 Th - Tc 
W  Qh 

 Th 
 650 K - 300 K 
W  400 J 

650 K


= 400 J
 215 J
The Laws of Thermodynamics 12-09
If the theoretical efficiency of a Carnot engine is to be 100%,
the heat sink must be
(A) at absolute zero.
(B) at 0°C.
(C) at 100°C.
(D) infinitely hot.
The Laws of Thermodynamics 12-10
A Carnot cycle consists of
(A) two adiabats and two isobars.
(B) two isobars and two isotherms.
(C) two isotherms and two isomets.
(D) two adiabats and two isotherms.
Entropy
Definition of the change in entropy S when an amount of
heat Q is added:
ΔE 
Q
T
Another statement of the second law of thermodynamics:
The total entropy of an isolated system never decreases.
When an irreversible process occurs in a closed system,
the entropy S of the system always increases: it never
decreases.
Entropy
Entropy is a measure of the disorder of a system. This gives us
yet another statement of the second law:
Natural processes tend to move toward
a state of greater disorder.
Example: If you put milk and sugar in your coffee and stir it,
you wind up with coffee that is uniformly milky and sweet.
No amount of stirring will get the milk and sugar to come
back out of solution.
Entropy
Another example: when a tornado hits a building, there is
major damage.
You never see a tornado approach a pile of rubble and leave
a building behind when it passes.
Thermal equilibrium is a similar process –
the uniform final state has more disorder than
the separate temperatures in the initial state.
Entropy
Another consequence of the second law:
In any natural process, some energy
becomes unavailable to do useful work.
If we look at the universe as a whole, it seems inevitable that,
as more and more energy is converted to unavailable forms,
the ability to do work anywhere will gradually vanish. This is
called the heat death of the universe.
The Laws of Thermodynamics 12-11
The second law of thermodynamics leads us to conclude
that
(A) the total energy of the universe is constant.
(B) disorder in the universe is increasing with the passage
of time.
(C) it is theoretically possible to convert heat into work
with 100% efficiency.
(D) the average temperature of the universe is increasing
with the passage of time.
The Law of Thermodynamics Summary
First law of thermodynamics: DU  Q  W
Isothermal process: temperature is constant.
Adiabatic process: no heat is exchanged.
Work done by gas at constant pressure:
W  - PDV
Heat engine changes heat into useful work
(requires temperature difference).
Efficiency of a heat engine:
Carnot efficiency:
e
W
QH
ec  1 -
TL
TH
 1-
QL
QH
The Law of Thermodynamics Summary
Second law of thermodynamics:
heat flows spontaneously from a hot object
to a cold one, but not the reverse
Thermal energy cannot be changed entirely to work
natural processes tend to increase entropy.
Change in entropy:
ΔS 
Q
T
Entropy is a measure of disorder.
As time goes on, less and less energy is
available to do useful work.