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Thermodynamics
AP Physics
Chapter 15
Thermodynamics
13.3 Zeroth Law of Thermodynamics
13.3 Zeroth Law of Thermodynamics
If two objects of different temperatures are
placed in thermal contact, they will eventually
reach the same temperature.
Thermal Eq Animation
They reach Thermal Equilibrium
Energy flowing in equals the energy flowing
out
13.3
13.3 Zeroth Law of Thermodynamics
Zeroth Law of Thermodynamics – if two
systems are in thermal equilibrium with a third
system, then they are in thermal equilibrium
with each other
Allows for a definition of temperature
(two objects have the same temperature when
they are in thermal equilibirum)
13.3
Thermodynamics
13.4 Thermal Expansion
13.4 Thermal Expansion
Objects usually expand when heated and
contract when cooled.
This can lead to some problems
So we include expansion joints
13.4
13.4 Thermal Expansion
Change in length is proportional to
temperature
L
0
T0
DL
T
So the equation is
L0
DL  aL0 DT
a is called the coefficient of linear expansion
13.4
13.4 Thermal Expansion
Common Misconception
When a object with a hole in it is heated, does
the hole get larger or smaller?
Imagine an infinitely thin ring
If it is heated length (circumference) of the
ring increase
13.4
13.4 Thermal Expansion
Volume follows the same relationship
DV  V0 DT
is the coefficient of volume expansion
 is usually equal to approximately 3a
13.4
Thermodynamics
13.7 The Ideal Gas Law
13.7 The Ideal Gas Law
PV  nRT
P – pressure in Pa (absolute pressure)
V – volume in m3
R = 8.314J/molK (in standard units)
T – temperature in K
n – quantity in mol
13.7
13.7 The Ideal Gas Law
If the equation is written in terms of molecules
The number of molecules (or atoms) in one
mole is
N A  6.022 x1023
So if
N
PV

RT
PV NnRT
A
And the number of molecules is
N  nN A
The ideal gas law can be written
13.7
13.7 The Ideal Gas Law
R
The quantity
k
NA
N
PV
RT
PV NkT
NA
This is known as Boltzmann’s constant
Our equation becomes
The constant has a value k  1.38 x10 23 J / K
13.7
Thermodynamics
14.1 Heat as Energy Transfer
14.1 Heat as Energy Transfer
Heat – energy transferred from one object to
another because of a difference in
temperature
Unit – joule
14.1
Thermodynamics
14.2 Internal Energy
14.2 Internal Energy
Temperature (K)– measurement of average
kinetic energy of the particles
Temperature
Internal Energy – total energy of all the
particles in the object
Heat - transfer
14.2
14.2 Internal Energy
Internal Energy equation
First internal Energy (U) is equal to the
number of particles (N) times average kinetic
energy
2
31
NkT
UU N 2( 2nRT
mv )
Since
K  mv  kT
1
2
2
3
2
The equation changes to
And since
Nk  nR
14.2
Thermodynamics
14.3 Specific Heat and Calorimetry
14.3 Specific Heat and Calorimetry
Remember from chemistry
Q  mCDT
Q – heat transfer in Joules
m – mass in kg
C – specific heat (Cp – constant pressure, Cv
– constant volume)
DT – change in temperature (oC or K)
14.3
14.3 Specific Heat and Calorimetry
Also when energy is transfered
Qlost  Qgained
Expands to
mACADTA  mBCB DTB
14.3
Thermodynamics
14.6 Heat Transfer: Conduction
14.6 Heat Transfer: Conduction
Conduction – by molecular collisions
If heat is transferred through a substance
l
TH
A
Tc
The rate of heat transfer (Q/t) depends on
TH  TC
Q
 kA
t
l
14.6
14.6 Heat Transfer: Conduction
TH  TC
Q
 kA
t
l
Q = heat (J)
t = time (s)
k = thermal conductivity (J/smCo)
T = temperature (K or Co)
l = length
14.6
Thermodynamics
15.1 The First Law of Thermodynamics
15.1 The First Law of Thermodynamics
The change in internal energy of a closed
system will be equal to the energy added to
the system by heat minus the work done by
the system on the surroundings.
DU  Q  W
A broad statement of the law of conservation
of energy
15.1
15.1 The First Law of Thermodynamics
Internal energy, U, is a property of the system
Work and heat are not
First Law proven by Joule in an experiment
The work done by
The weight as if fell
(done by gravity)
Equaled the energy increase of the liquid in
the sealed chamber
15.1
15.1 The First Law of Thermodynamics
The first law can be expanded
If the system is moving and has potential
energy, then
DK  DU g  DU  Q  W
Remember
Q is positive when work flows in
W is positive when the system does work
15.1
Thermodynamics
15.2 Thermodynamic Processes & the First Law
15.2 Thermodynamic Processes & the First Law
You need to remember the names of these
processes
Isothermal – constant temperature
PV
PV const
nRT .
If temperature is held constant, then
A graph would look like
The curves are called
isotherms
15.2
15.2 Thermodynamic Processes & the First Law
If DT is zero, then DU is zero because
D0U nR
nRT
nR(0D)T
33
22
We can then show
D0UQ
Q
QW
WW
The work done by the gas in an isothermal
process equals the heat added to the gas
15.2
15.2 Thermodynamic Processes & the First Law
Adiabatic – no heat is allowed to flow into or
out of the system
Q  0W
DU
That leaves First Law as
If the gas expands, the internal energy
decreases, and so does the temperature
15.2
15.2 Thermodynamic Processes & the First Law
Isobaric – pressure is constant
Work = PV
Isovolumetric – volume is
Constant
Work = 0
15.2
15.2 Thermodynamic Processes & the First Law
Example 1: Isobaric Process
A gas is placed in a piston with an area of
.1m2. Pressure is maintained at a constant
8000 Pa while heat energy is added. The
piston moves upward 4 cm. If 42 J of energy
is added to the system what is the change in
internal energy?
15.2
15.2 Thermodynamic Processes & the First Law
Example 1: Isobaric Process
A gas is placed in a piston with an area of .1m2. Pressure is maintained at a
constant 8000 Pa while heat energy is added. The piston moves upward 4 cm. If
42 J of energy is added to the system what is the change in internal energy?
DU  Q  W
W W
(W
8000
)(.
1)(.
W
PA
P
32
DD
JV
x 04)
DU  42  32  10J
15.2
15.2 Thermodynamic Processes & the First Law
Example 2: Adiabatic Expansion
How much work is done the adiabatic
expansion of a car piston if it contains 0.10
mole an ideal monatomic gas that goes from
1200 K to 400 K?
15.2
15.2 Thermodynamic Processes & the First Law
Example 2: Adiabatic Expansion
How much work is done the adiabatic expansion of a car piston if it contains 0.10
mole an ideal monatomic gas that goes from 1200 K to 400 K?
Adiabatic Q=0
DD
UUQWW
3
3
3
DU  U
1W
)(U
8.
314
)(nRT
nR
D
998
22998
J400
J T 1200)
2 (.D
15.2
15.2 Thermodynamic Processes & the First Law
Example 3: Isovolumetric Process
Water with a mass of 2 kg is held at a
constant volume in a container, while 10 kJ of
energy is slowly added. 2 kJ of energy leaks
out to the surroundings. What is the
temperature change of the water?
15.2
15.2 Thermodynamic Processes & the First Law
Example 3: Isovolumetric Process
Water with a mass of 2 kg is held at a constant volume in a container, while 10 kJ
of energy is slowly added. 2 kJ of energy leaks out to the surroundings. What is
the temperature change of the water?
DDU
UDU
Q
8000
Q
W
3
3
8000 D
D
)(.88K
.314
U2T(111
25nR
D
T )DT
Constant volume and we don’t have Cv
n
m
mmole

2000
18
 111mol
15.2
15.2 Thermodynamic Processes & the First Law
When a process is cyclical
DU  0
And the work done is the area bound by
the curves
15.2
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic Process
An ideal monatomic gas is confined in a cylinder
by a movable piston. The gas starts at A with P
= 101.3 kPa, V = .005 m3 and T = 300 K.
The cycle is
A  B is isovolumetric and raises P to 3 atm.
BC Isothermal Expansion (Pave = 172.2 kPa)
CA Isobaric
Calculate DU, Q, and W for each step and for the
entire cycle
15.2
15.2 Thermodynamic Processes & the First Law
B
A
C
15.2
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic Process
An ideal monatomic gas is confined in a cylinder by a movable piston. The gas starts
at A with P = 101.3 kPa, V = .005 m3 and T = 300 K.
The cycle is
A  B is isovolumetric and raises P to 3 atm.
WAB  PDV  P(0)  0
Q
W
U
D
1519
)(
Q900
Q
UD
JQ
nR
(0(0.203
.203)()(8DDQ
8.U
314
.314
)(T
T
0300
300) )QQ
33
22
3
2
(303900
(101300)(.)(.0D
005
))
nRT
(0nT.(
203
8J.n314
)(8)(.314
300))T
PV
.005
203
900
mol
U
1519
15.2
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic Process
An ideal monatomic gas is confined in a cylinder by a movable piston. The gas starts
at A with P = 101.3 kPa, V = .005 m3 and T = 300 K.
The cycle is
BC Isothermal Expansion (Pave = 172.2 kPa)
DU  0
Q  (172200
DQ
0Q
UQ
)(1722
Q
PQ
0W
D
.015
W
VW
J  0.005)
(101300)VVW
PV
)(m
nRT
1722
J8.314)(900)
(00.
.203
015
3
15.2
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic Process
An ideal monatomic gas is confined in a cylinder by a movable piston. The gas starts
at A with P = 101.3 kPa, V = .005 m3 and T = 300 K.
The cycle is
CA Isobaric
 1591
Q
DU Q2532
Q (W1013
J )
W  (101300
)(1013
.005
WW
P0D
V J  0.015)
DU  (0.D203
)(8.1519
314D)(TJ300  900)
U
nR
3
2
3
2
15.2
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic Process
An ideal monatomic gas is confined in a cylinder by a movable piston. The gas starts
at A with P = 101.3 kPa, V = .005 m3 and T = 300 K.
The cycle is
Totals
Q  1519  1722  2532  709 J
W  0 1722 1013  709J
DU  1519  0 1519  0J
15.2
S-69
Adiabatic Expansion
How much work is done in the adiabatic
expansion of a car piston if it contains 0.20
mole an ideal monatomic gas that goes from
500 K to 372 K?
Thermodynamics
15.4 The Second Law of Thermodynamics-Intro
15.3 The Second Law of Thermodynamics-Intro
The first law deals with conservation of energy.
However there are situations that would
conserve energy, but do not occur.
1. Falling objects convert from Ug to K to Q
Falling and Energy
Never Q to K to Ug
2. Heat flows from TH to TC
Never TC to TH
15.4
15.3 The Second Law of Thermodynamics-Intro
The second law explains why some processes
occur and some don’t
In terms of heat, the second law could be stated
-Heat can flow spontaneously from a hot object
to a cold object; heat will not flow
spontaneously from a cold object to a hot
object
15.4
Thermodynamics
15.5 Heat Engines
15.5 Heat Engines
The Heat Engine
1. Heat flows into
the engine
2. Energy is converted to work
3. Remaining heat is
exhausted to cold
15.5
15.5 Heat Engines
Steps in an internal combustion engine
1. The intake valve is open, and fuel and air are
drawn past the valve and into the combustion
chamber and cylinder from the intake
manifold located on top of the combustion
chamber (Intake Stroke)
15.5
15.5 Heat Engines
Steps in an internal combustion engine
2. With both valves closed, the combination of the
cylinder and combustion chamber form a completely
closed vessel containing the fuel/air mixture. As the
piston is pushed to the right, volume is reduced & the
fuel/air mixture is compressed (Compression Stroke)
15.5
15.5 Heat Engines
Steps in an internal combustion engine
3. the electrical contact is opened. The sudden opening
of the contact produces a spark in the combustion
chamber which ignites the fuel/air mixture. Rapid
combustion of fuel releases heat, produces exhaust
gases in the combustion chamber. (Power Stroke)
15.5
15.5 Heat Engines
Steps in an internal combustion engine
4. The purpose of the exhaust stroke is to clear
the cylinder of the spent exhaust in
preparation for another ignition cycle.
(Exhaust Stroke)
15.5
15.5 Heat Engines
Complete cycle
15.5
15.5 Heat Engines
Looking at a steam engine
If the steam were the
same temperature
throughout
-exhaust pressure
would be the same
as the intake pressure
-then exhaust work would be the same as intake
work
For net work there must be a DT
15.5
15.5 Heat Engines
Actual Efficiency of an engine is defined as
W
e
QH
The ratio of work to heat input
Since
We can write the efficiency as
QH  W  QC
QH QQ
CC
ee 1 
QQ
HH
15.5
15.5 Heat Engines
Carnot Engine (ideal) – no actual Carnot engine
A four cycle engine
1. Isothermal expansion (DT=0, Q=W)
2. Adiabatic expansion (Q=0, DU=-W)
3. Isothermal compression
4. Adiabatic compression
Each process was considered reversible
That is that each step is done very slowly
Real reactions occur quickly – there would be
turbulence, friction - irreversible
15.5
15.5 Heat Engines
The Carnot efficiency is defined as
eideal
TC
 1
TH
15.5
Thermodynamics
15.6 Refrigerators, Air Conditioners
15.6 Refrigerators, Air Conditioners
The reverse of Heat Engines
Work must be done – because heat flows from
hot to cold
15.6
S-70
List the four steps of the Carnot Engine.
What would be the Carnot efficiency of an
Engine running at 512 K when it was 40oC
outside?
What would be it’s efficiency if the air
temperature was 0oc?
Thermodynamics
15.7 Entropy and the 2nd Law of Thermodynamics
15.7 Entropy and the 2nd Law of Thermodynamics
Entropy – a measure of the order or disorder of a
system
Change in entropy is defined as
Q
DS 
T
Q must be added as a reversible process at a
constant temperature
15.7
15.7 Entropy and the 2nd Law of Thermodynamics
The Second Law of Thermodynamics – the
entropy of an isolated system never
decreases. It can only stay the same or
increase.
Only idealized processes have a DS=0
Or – the total entropy of any system plus that of
its environment increases as a result of any
natural process
15.7
15.7 Entropy and the 2nd Law of Thermodynamics
Example – A sample of 50 kg of water at 20oC is
mixed with 50 kg at 24oC. The final
temperature is 22oC. Estimate the change in
Entropy.
The reaction does not occur at constant
temperature, so use the average temperature
to estimate the entropy change.
15.7
15.7 Entropy and the 2nd Law of Thermodynamics
Example – A sample of 50 kg of water at 20oC is mixed with 50 kg at 24oC. The final
temperature is 22oC. Estimate the change in Entropy.
Cold Water
418600
SQ

Q D(50)(4186)(22
Q

418,
600
J

20)

mC
D
T
1424
J
/
K
P 294
C
Hot Water

418600
Q DS
(50)(4186)(22
QD


418600
J

24)



1414
J
/
K
S


1
414
10
J

/
1424
K
H
296
15.7
Thermodynamics
15.8 Order to Disorder
15.8 Order to Disorder
2nd Law can be stated – natural processes tend
to move toward a state of greater disorder
15.7
S-71
A cycle starts with 3 moles of an ideal gas at a
pressure of 300,000 Pa, and a volume of
0.06 m3. Calculate the change in internal
energy, heat flow, and work for each step.
A. Isobaric expansion to 0.10m3.
B. Adiabatic expansion to 0.12m3. Final
pressure is 100,000 Pa.
C. Isobaric compression to 0.09 m3.
D. Adiabatic compression.
E. Estimate the change in Entropy for each
step.
S-72
A cycle starts with 1.5 moles of an ideal gas at a
temperature of 400K, and a volume of 0.12
m3. Calculate the change in internal energy,
heat flow, and work for each step.
A. Isothermal expansion to 0.19m3. (use
average pressure)
B. Isobaric compression back to the original
volume.
C. Isovolumetric change back to original state.
D. Estimate the change in Entropy for each
step.
S-73
A cycle starts with 8 moles of an ideal gas at a
temperature of 87K, and a volume of 0.27
m3. Calculate the change in internal energy,
heat flow, and work for each step.
A. Isothermal expansion to 0.44m3. (use
average pressure)
B. Isothermal compression to 0.27m3.
C. Estimate the change in Entropy for each
step.
S-74
A cycle starts with 20 moles of an ideal gas at a
temperature of 718K, and a volume of 0.10
m3. Calculate the change in internal energy,
heat flow, and work for each step.
A. Adiabatic expansion to 0.20m3.
(Pave=1150000Pa)
B. Isothermal compression to 0.15m3.
C. Adiabatic compression to 0.05m3.
(1050000Pa)
D. Isothermal expansion to 0.10m3.
E. What is the Carnot efficiency of this engine?
F. What is the entropy change for the cycle?
S-75
A cycle starts with 21 moles of an ideal gas at a
temperature of 452K, and a volume of 0.24
m3. Calculate the change in internal energy,
heat flow, and work for each step.
A. Isothermal expansion to 0.42m3. (use
average pressure)
B. Isobaric compression back to the original
volume.
C. Isovolumetric change back to original state.
D. Estimate the change in Entropy for each
step.