Thermodynamics
Thermodynamics – What is it?

The branch of physics that is built upon
the fundamental laws that heat and work
obey.
Heat & Internal Energy

Heat is energy that flows from a higher-temperature object to
a lower temperature object due to the differences in
temperature.

Heat is positive when a system gains heat and negative
when it loses heat.

Internal energy is a measure of the temperature of a
substance due to the kinetic energy of the individual
molecules or atoms as well as potential energy of these
particles .
Zeroth Law of Thermodynamics

Thermal Equilibrium:
•
•
•
If two object have the same temperature, then there
will be no flow of heat between the two systems.
If two objects are in thermal equilibrium with a third
object, then they must be in thermal equilibrium with
each other.
When two objects are brought in contact with one
another, heat will flow from the warmer object to the
cooler one.
Temperature Scales

A common scale used for measuring and recording
temperature is Celsius.
•


Water freezes at 0°C and boils at 100°C.
To convert from °F to °C:
•
T (°C) = 5/9 (T(°F) – 32)
The Kelvin temperature scale is of more significances than
the Celsius scale.
•
•
It is measured in the same increments as the Celsius scale, but
is measured from absolute 0.
The temperature at which water freezes (ice point) is 273.15 K /
0°C, and the temperature at which water boils (steam point) is
373.15 K / 100°C.
Heat and Energy



What is Heat?
What is Energy?
Is there a difference?
•
Yes
• Heat is a measure of the transfer of energy from one
•
•
object that is at a higher temperature to another that is at
a lower energy level.
Energy is a measure of the average kinetic energy of all
the particles in a gas sample, the potential energy that
exists between the atoms and molecules, etc.
Objects do not contain heat, they contain energy.
How can heat be transferred?



Convection: The process of transferring heat
by the bulk movement of a gas or liquid.
Conduction: The process of transferring heat
through a body whereby the bulk motion of the
material plays no role in the transfer.
Radiation: Process by which energy is
transferred by means of electromagnetic
radiation.
Phases of Matter

Matter comes in three phases – gas, liquid and
solid.
•
•
•
•
When a substance undergoes a phase change, the
temperature does not change.
Solids melt to become liquid or sublimate to become a
vapor.
Liquids evaporate to become a vapor or freeze to
become a solid.
Gases condense to become a liquid, or go through
deposition to become a solid.
Heat Transfer – Change in T

Heat is measure of the amount of energy
that flows to or from a substance is
dependent on the temperature, mass of
the object and the specific heat (c).
Q = mcΔT
• c is called the specific heat and is unique for
•
every substance.
If ΔT is positive, then Q is positive and heat
flows into the substance.
Phase Changes
Heat Transfer – No Change in T

When a substance goes through a
phase change, the temperature does not
change, so the formula Q = mcΔT will
not hold true. Instead:
Q = mL
• Where
• L = Latent Heat of Transformation
• For a solid to liquid L is called the heat of fusion
• For a liquid to gas L is called the heat of
vaporization
Linear Thermal Expansion

Most materials will expand when heated and
shrink when cooled.
•
Railroad tracks, bridges and pipelines are a few
examples of things that expand when heated.
L = Lf – Li =  Li (Tf – Ti)
Where:
L is the length.
T is the temperature.
 is the coefficient of linear expansion.
Example 1: Linear Expansion

How much will the length of a 30 m beam change as the
temperature rises from 15°C to 35°C? The coefficient of
linear expansion of the steel beam is 1.2 x 10-5/°C.
L = Li (Tf – Ti)
L = (1.2 x 10-5/°C)(30m)(35°C - 15°C)
L = 7.2 x 10-3 m = 7.2 mm
Volume Thermal Expansion

Since expansion should not be limited to one dimension,
a similar expression to linear expansion exists for
volumetric expansion.
V = Vi (Tf – Ti)
•
For many solids,  ~ 3
14.2 The Ideal Gas Law
An ideal gas is an idealized model for real
gases that have sufficiently low densities.
The condition of low density means that the
molecules are so far apart that they do not
interact except during collisions, which are
effectively elastic.
At constant volume the pressure
is proportional to the temperature.
P T
14.2 The Ideal Gas Law
At constant temperature, the
pressure is inversely
proportional to the volume.
P 1 V
The pressure is also
proportional to the amount
of gas.
Pn
14.2 The Ideal Gas Law
The absolute pressure of an ideal gas is directly proportional to
the Kelvin temperature and the number of moles of the gas and
is inversely proportional to the volume of the gas.
nRT
P
V
R  Universal Gas Constant  8.314 J /( mol  K )
PV  nRT
For an alternate representation:
Since : N  nN A and k 
PV  NkT
R
NA
k = Boltsmann’s constant = 1.38 x 10-23 J/K:
14.2 The Ideal Gas Law
Consider a sample of an ideal gas that is taken from an
initial to a final state, with the amount of the gas remaining
constant.
PV  nRT
PV
 nR  constant
T
Pf V f
Tf
PiVi

Ti
14.2 The Ideal Gas Law
Pf V f
Tf
Constant T, constant n:
Constant P, constant n:
PiVi

Ti
Pf V f  PiVi
Boyle’s law
Vf
Vi

T f Ti
Charles’ law
14.3 Kinetic Theory of Gases
•The particles are in constant, random
motion, colliding with each other and
with the walls of the container.
•Each collision changes the particle’s
speed.
•As a result, the atoms and molecules
have different speeds.
14.3 Kinetic Theory of Gases
Using the Impulse-Momentum Theorem:
v mv 
 F  ma  m t  t
Average force 
Final momentum - Initial momentum
Time between successive collisions

 mv   mv  mv2


2L v
L
14.3 Kinetic Theory of Gases
For a single molecule, the average force is:
mv2
F
L
For N molecules moving randomly in
three dimensions, the average force
is:
2
 N  mv 
F  
 3  L  root-mean-square
speed
F F  N  mv 2 
P  2   3
A L  3  L 
volume
14.3 Kinetic Theory of Gases
2
 N  mv 
P 
 3  V 
NkT
KE


2
PV  13 N mvrms
 23 N

1
2
2
mvrms
2
KE  12 mvrms
 32 kT
• This formula shows that the kinetic energy associated
with the translational motion of a particle can be
determined if we know the temperature of the system.

Average Speed of a Particle
3
KEavg = kBT
2

Since KEavg = ½ mv2, we can find the average speed of a
particle.
3kBT
vrms 
m
Internal Energy of a Monatomic Ideal
Gas

Internal Energy: The sum of the molecular kinetic energy, the
molecular potential energy, and other kinds of energy.
U = 3/2 nRT
Where:
n = number of moles.
R = gas constant
T = Temperature (K)
Note: A substance has internal energy, not “Heat”.
Note: Internal energy depends only on the state of a system, and not
on the method by which the system got their.
Note: Formula applies to monatomic ideal gases.
The 1st Law of Thermodynamics

Conservation of energy says that energy cannot be
created nor destroyed, but may be converted from one
form to another.
The 1st Law of Thermodynamics
(cont.)

If a system only gains heat:
U = Uf – Ui = Q

Heat is positive when a system gains heat and negative
when it loses heat.
The 1st Law of Thermodynamics
(cont.)

If a system only does work on its surroundings and no
heat is transferred:
U = Uf – Ui = -W
•
Work is positive when it is done by the system on the
surroundings and negative when it is done on the system by
the surroundings.
W = PV
where:
P = pressure
V = volume
The 1st Law of Thermodynamics
(cont.)

A system that consists of changes to both heat (internal
energy) and work simultaneously is represented by:
ΔU = Uf – Ui = Q - W
First Law of Thermodynamics = Law of Conservation of
energy.
Example 2: 1st Law of Thermo.

The temperature of 3 moles of a monatomic ideal gas is
reduced from Ti = 540K to Tf = 350K when 5500 J of heat flows
into the gas. What is the change in internal energy (ΔU) and
work (W) done by the gas?
U 
3
nR (Tf  Ti )
2
3
U  (3.0moles )(8.314 J / mole  K )(350 K  540 K )
2
U  7100 J

Using 1st Law of Thermodynamics: ΔU = Q – W
• W = Q – ΔU = 5500J – (-7100J) = 12,600J
Thermal Processes – Constant
Pressure

An isobaric process is one that occurs at constant
pressure.
ΔU = Q - W
Since the pressure is constant, the amount of work done is:
W = PV = P(Vf – Vi)
ΔU = Q – PΔV
Example 3: Isobaric Process

1 Gram of water is placed inside a cylinder where the pressure
is maintained at 2.0 x 105 Pa. The temperature of the water is
raised by 31 C°. The water is in the gaseous phase and
expands by 7.1 x 10-5 m3, and has a specific heat capacity of
2020J/(kg·C°). What is the change in internal energy of the
water?

For an isobaric process we will use:

•
•
Q = cmΔT for heat added.
W = PΔV for work done by the gas.
Using 1st Law of Thermodynamics:
•
•
ΔU = Q - W
ΔU = cmΔT - PΔV
Table of Known & Unknowns
Mass of water
m
0.0010 kg
Pressure on water
P
2.0 x 105 Pa
Increase in T
ΔT
31C°
Increase in V
ΔV
7.1 x 10-5 m3
Specific Heat of H2O gas
cgas
2020 J/(kg·C°)
ΔU of gas
ΔUgas
?
ΔU = cmΔT – PΔV
ΔU = (2020 J/(kg·C°)(0.0010 kg)(31 C°) – (2.0 x 105 Pa)(7.1 x 10-5 m3)
ΔU = 63 J – 14 J
ΔU = 49 J
Thermal Processes – Constant
Volume


An isochoric process is one that occurs at
constant volume.
Since the volume (ΔV) is 0, the work is zero.
U = Q - W = Q
U = 3/2nR(Tf – Ti)
Thermal Processes – Constant
Temperature





An isothermal process is one that occurs at constant
temperature.
U = Q - W
Where W = nRT*ln (Vf/Vi)
U = Q - nRT*ln (Vf/Vi)
For an ideal gas:
U = 3/2nR(Tf – Ti)
Since the temperature does
not change, U = 0.
Therefore:
W = Q = nRT*ln (Vf/Vi)
Example 4: Isothermal
Expansion

2 moles of the monatomic gas argon expand isothermally at
298 K, from the initial volume of Vi = 0.025 m3 to a final volume
of Vf = 0.050 m3. Assuming that argon is an ideal gas, find the
work done by the gas, the change in internal energy of the gas
and the heat supplied by the gas.

For an isothermal process we will use:
 Vf 
W  nRT ln  
 Vi 

Since the temperature does not change,
the change in internal energy will be 0.
3
U  nR(Tf  Ti )  0
2
Table of Known & Unknowns

Initial Volume
Vi
0.025 m3
Final Volume
Vf
0.050 m3
Temperature
T
298 K
Gas Constant
R
8.314 J/(kg·K)
Heat supplied to gas
Q
?
Using the 1st Law of Thermodynamics: ΔU = Q - W
• Since ΔU = 0:
Q=W
 Vf 
Q  W  nRT ln  
 Vi 
 0.050m3 
Q  (2moles)(8.314 J /(mole  K )(298K ) ln 
 3400 J
3 
 0.025m 
Thermal Processes – Adiabatic

An adiabatic process is one that
occurs without the transfer of heat
(Q = 0).
U = Q - W = -W
and since
U = 3/2nRT
W = -U = 3/2nR(Ti – Tf)

If a gas expands adiabatically, then the final
temperature will be lower than the initial
temperature.
1st Law Relationships
Specific Heat Capacities

Specific Heat: A measure of the amount of heat
(Q) that must be supplied to change the
temperature of a substance by an amount T.
• First Law:
• Solving for Heat:
• Using Specific Heat:
• Where mc = nC
U = Q – W
Q = U + W
Q = nCT
(1)
(2)
Specific Heat Capacities (const. P)
•
From 1st Law:
U = Q – W
Q = U + PV
3
Q  nR (Tf  Ti )  nR (Tf  Ti )
2
5
Q  nR (Tf  Ti )
2
•
(3)
Substituting (2) for Q in (3) and solving for C gives:
5
nCp (Tf  Ti )  nR(Tf  Ti )
2
5
Cp  R
2
Note: This relationship holds true for monatomic ideal gases.
Specific Heat Capacities (const. V)
•
From 1st Law:
U = Q – W
Q = U + PV (Since V = 0, W = 0)
3
Q  nR(Tf  Ti )  0
2
3
Q  nR (Tf  Ti )
2
•
(3)
Substituting (2) for Q in (3) and solving for C gives:
3
nCv (Tf  Ti )  nR(Tf  Ti )
2
3
Cv  R
2
Note: This relationship holds true for monatomic ideal gases.
Specific Heat

An interesting outcome results when you
evaluate the difference between Cp and
Cv:
• The reason is due to the fact that work is
done under constant pressure where no work
is done when the volume is constant.
Cp  Cv  R
5
3
R R  R
2
2
2nd Law of Thermodynamics


Heat Flows Spontaneously from a substance at
higher temperature to a substance at a lower
temperature, while the reverse is not true.
In other words, natural processes go in a
direction that maintains or increases the
entropy of the universe.
 Entropy: A measure of disorder or
randomness of a system.
Note: Reversible processes do not change the
entropy of a system,….or universe.
Entropy
15.8 Heat Engines
A heat engine is any device that uses heat to
perform work. It has three essential features.
1. Heat is supplied to the engine at a relatively high
temperature from a place called the hot reservoir.
2. Part of the input heat is used to perform work by
the working substance of the engine (gasoline,
steam, etc).
3. The remainder of the input heat is rejected to a
place called the cold reservoir.
QH  magnitude of input heat
QC  magnitude of rejected heat
W  magnitude of the work done
15.8 Heat Engines
The efficiency of a heat engine is defined as
the ratio of the work done to the input heat:
e
W
QH
If there are no other losses, then
QH  W  QC
e  1
QC
QH
15.8 Heat Engines
Example 5 An Automobile Engine
An automobile engine has an efficiency of 22.0% and produces
2510 J of work. How much heat is rejected by the engine?
e
W
QH
QH 
QH  W  QC
W
e
15.8 Heat Engines
QH  W  QC
QH 
W
QC  QH  W
1 
QC 
 W  W   1
e
e 
W
 1

 2510 J 
 1  8900 J
 0.220 
e
15.9 Carnot’s Principle and the Carnot Engine
A reversible process is one in which both the system and the
environment can be returned to exactly the states they were in
before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND
LAW OF THERMODYNAMICS
• No irreversible engine operating between two reservoirs at constant
temperatures can have a greater efficiency than a reversible engine operating
between the same temperatures.
• Furthermore, all reversible engines operating between the same temperatures
have the same efficiency.
15.9 Carnot’s Principle and the Carnot Engine
The Carnot engine is useful as an idealized
model.
All of the heat input originates from a single
temperature, and all the rejected heat goes
into a cold reservoir at a single temperature.
Since the efficiency can only depend on
the reservoir temperatures, the ratio of
heats can only depend on those temperatures.
QC
QH
e  1
QC
QH
TC
 1
TH

TC
TH
15.9 Carnot’s Principle and the Carnot Engine
Example 6 A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a temperature of 298.2 K, whereas
the water 700 meters beneath the surface has a temperature of 280.2 K. It has
been proposed that the warm water be used as the hot reservoir and the cool water
as the cold reservoir of a heat engine. Find the maximum possible efficiency for
such an engine.
ecarnot
ecarnot  1 
TC
 1
TH
TC
280.2 K
 1
 0.060
TH
298.2 K
15.9 Carnot’s Principle and the Carnot Engine
Conceptual Example 7 Natural Limits on the Efficiency of a Heat Engine
Consider a hypothetical engine that receives 1000 J of heat as input from a
hot reservoir and delivers 1000J of work, rejecting no heat to a cold reservoir
whose temperature is above 0 K. Decide whether this engine violates the first
or second law of thermodynamics.
ecarnot
TC
 1
TH
• As far as the first law is concerned, it is not violated
since in theory, all the input heat can contribute to
work being done by the system on the environment.
• However, the second law is violated since you can
never lower the temperature to true absolute 0.
Hence the efficiency cannot be truly equal to 1.
Key Ideas





Thermal equilibrium exists between two bodies when they
both have the same temperature.
Heat can be transferred through convection, conduction and
radiation.
The 1st Law of Thermodynamics is based upon the principle
of Conservation of Energy.
The 2nd Law of Thermodynamics says that heat will flow
spontaneously from a substance at a higher temperature to a
substance at a lower temperature.
Irreversible processes lead to an increase in entropy of the
universe.