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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.
They reach Thermal Equilibrium
Energy flowing in equals the energy flowing out
13.3
Thermal Eq Animation
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 joints13.4
13.4 Thermal Expansion
Change in length is proportional to temperature
So the equation is
is called the coefficient of linear expansion13.4
T0
T
L0
L0
L
TLL 0
13.4 Thermal Expansion
Common MisconceptionWhen 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
is the coefficient of volume expansion is usually equal to approximately 3
13.4
TVV 0
Thermodynamics
13.7 The Ideal Gas Law
13.7 The Ideal Gas Law
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
nRTPV
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
So if
And the number of molecules is
The ideal gas law can be written13.7
nRTPV
2310022.6 xN A
AnNN
RTN
NPV
A
13.7 The Ideal Gas Law
The quantity
This is known as Boltzmann’s constant
Our equation becomes
The constant has a value 13.7
RTN
NPV
A
kN
R
A
NkTPV
KJxk /1038.1 23
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
Internal Energy – total energy of all the particles in the object
Heat - transfer
14.2
Temperature
14.2 Internal Energy
Internal Energy equation
First internal Energy (U) is equal to the number of particles (N) times average kinetic energy
Since
The equation changes to
And since14.2
)( 221 vmNU
kTvmK 232
21
NkTU 23
nRNk
nRTU 23
Thermodynamics
14.3 Specific Heat and Calorimetry
14.3 Specific Heat and Calorimetry
Remember from chemistry
Q – heat transfer in Joules
m – mass in kg
C – specific heat (Cp – constant pressure, Cv – constant volume)
T – change in temperature (oC or K)
14.3
TmCQ
14.3 Specific Heat and Calorimetry
Also when energy is transfered
Expands to
14.3
gainedlost QQ
BBBAAA TCmTCm
Thermodynamics
14.6 Heat Transfer: Conduction
14.6 Heat Transfer: Conduction
Conduction – by molecular collisions
If heat is transferred through a substance
The rate of heat transfer (Q/t) depends on
14.6
A
l
Tc
TH
l
TTkA
t
Q CH
14.6 Heat Transfer: Conduction
Q = heat (J)
t = time (s)
k = thermal conductivity (J/smCo)
T = temperature (K or Co)
l = length
14.6
l
TTkA
t
Q CH
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.
A broad statement of the law of conservation of energy
15.1
WQU
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
Remember
Q is positive when work flows in
W is positive when the system does work
15.1
WQUUK g
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
If temperature is held constant, then
A graph would look like
The curves are called
isotherms15.2
nRTPV .constPV
15.2 Thermodynamic Processes & the First Law
If T is zero, then U is zero because
We can then show
The work done by the gas in an isothermal process equals the heat added to the gas
15.2
nRTU 23 TnRU 23 )0(0 23 nR
WQU WQ 0 WQ
15.2 Thermodynamic Processes & the First Law
Adiabatic – no heat is allowed to flow into or out of the system
That leaves First Law as
If the gas expands, the internal energy decreases, and so does the temperature
15.2
0Q WU
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?
15.2
WQU
VPW xPAW )04)(.1)(.8000(W JW 32
JU 103242
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
15.2
WQU WU nRTU 2
3 TnRU 23 )1200400)(314.8)(1(.2
3 U JU 998 JW 998
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?
Constant volume and we don’t have Cv
15.2
WQU TnRU 2
3
QU 8000U
molnmolemm 11118
2000
T )314.8)(111(8000 23 KT 8.5
15.2 Thermodynamic Processes & the First Law
When a process is cyclical
And the work done is the area bound by the curves
15.2
0U
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic ProcessAn 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.
BC Isothermal Expansion (Pave = 172.2 kPa)CA IsobaricCalculate U, Q, and W for each step and for the entire cycle
15.2
15.2 Thermodynamic Processes & the First Law
15.2
A
B
C
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic ProcessAn 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.
15.2
0)0( PVPW BA
WQU 0 QU QU QTnR 23
nRTPV )300)(314.8()005)(.101300( n nmol 203.0
QT )300)(314.8)(203.0(23
nRTPV T)314.8)(203.0()005)(.303900( T900
Q )300900)(314.8)(203.0(23 JQ 1519
JU 1519
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic ProcessAn 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
BC Isothermal Expansion (Pave = 172.2 kPa)
15.2
0U
WQU WQ 0 WQ
nRTPV
VPQ
)900)(314.8)(203.0()101300( V 3015.0 mV
)005.0015.0)(172200( Q JQ 1722
JW 1722
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic ProcessAn 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
CA Isobaric
15.2
WQU
VPW )015.0005.0)(101300( W JW 1013
TnRU 23 )900300)(314.8)(203.0(2
3 U JU 1519
)1013(1591 Q JQ 2532
15.2 Thermodynamic Processes & the First Law
Example 4: First law in a Cyclic ProcessAn 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
15.2
JW 709101317220
JU 0151901519
JQ 709253217221519
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
Never Q to K to Ug
2. Heat flows from TH to TC
Never TC to TH
15.4
Falling and Energy
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 Engine1. Heat flows into
the engine2. Energy is convert-
ed to work3. Remaining heat is
exhausted to cold
15.5
15.5 Heat Engines
Steps in an internal combustion engine1. 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 engine2. 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 engine3. 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 engine4. 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 engineIf the steam were the same temperature throughout-exhaust pressurewould be the sameas the intake pressure-then exhaust work would be the same as intake
workFor net work there must be a T
15.5
15.5 Heat Engines
Actual Efficiency of an engine is defined as
The ratio of work to heat inputSinceWe can write the efficiency as
H
WeQ
15.5
H CQ W Q
H C
H
Q Qe
Q
1 C
H
Qe
Q
15.5 Heat Engines
Carnot Engine (ideal) – no actual Carnot engineA four cycle engine1. Isothermal expansion (=0, Q=W)2. Adiabatic expansion (Q=0, U=-W)3. Isothermal compression 4. Adiabatic compressionEach process was considered reversibleThat is that each step is done very slowlyReal reactions occur quickly – there would be
turbulence, friction - irreversible
15.5
15.5 Heat Engines
The Carnot efficiency is defined as
15.5
1 Cideal
H
Te
T
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
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 must be added as a reversible process at a constant temperature
15.7
QS
T
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 S=0Or – 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
Hot Water
15.7
PQ mC T (50)(4186)(22 20)Q 418,600Q J418600294S 1424 /CS J K
(50)(4186)(22 24)Q 418600Q J418600296S 1414 /HS J K 1414 1424S 10 /S J K
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