Transcript First Law of Thermodynamics

SPH3UW
An Introductory Course in
Thermodynamics
Waterloo Collegiate Institute
SPH3UW: Lecture 1, Pg 1
Welcome to Thermodynamics
Heat, Temperature, and Power
Thermal Expansion
Ideal Gas Law
Kinetic Theory of Gases
First Law of Thermodynamics
Work and PV Diagrams
Isothermal Processes
Adiabatic Processes
Isobaric Processes
Isochoric Process
Heat Engines and the Second Law of Thermodynamics
SPH3UW: Lecture 1, Pg 2
Thermal Physics
electrons and holes in semiconductors
converting energy into work
magnetism
thin films and surface chemistry
thermal radiation (global warming)
and much more…
SPH3UW: Lecture 1, Pg 3
Some Definitions






Absolute Zero: the lowest possible temperature, at which
all molecular motion would cease and a gas would have
zero volume.
Calorie: the amount of heat required to raise the
temperature of one gram of water by one Celsius degree.
Calorimeter: device which isolates objects to measure
temperature changes due to heat flow.
Celsius (C): temperature scale in which the freezing point
of water is 0 and the boiling point of water is 100
Convection: heat transfer by the movement of a heated
substance, due to the differences in density.
Conduction: heat transfer from molecule to molecule in
substances due to differences in temperature.
SPH3UW: Lecture 1, Pg 4
Heat
Heat, represented by the variable Q, is a type of energy that can
be transferred from one body to another
 Heat is measured in Joules
 Energy must be transferred in order to be called heat.
(So heat may be gained or lost, but not possessed. It
is incorrect to say, “a gas has 4000 J of heat”
Internal energy:
Temperature:
The sum of the energies of all of the molecules in a
substance. Represented by the variable U (for
example: the total of the Kinetic and Potential
energy at the molecular level is called the Internal
Energy of the system. U=KE+PE)
Related to the average kinetic energy per
molecule of a substance.
SPH3UW: Lecture 1, Pg 5
What is Heat?



Up to mid-1800’s heat was considered a substance -- a
“caloric fluid” that could be stored in an object and
transferred between objects. After 1850, kinetic theory.
A more recent and still common misconception is that heat
is the quantity of thermal energy in an object.
The term Heat (Q) is properly used to describe energy in
transit, thermal energy transferred into or out of a system
from a thermal reservoir …

Q
U
(like cash transfers into and out of your bank account)
Sign of Q :
Q > 0 system gains thermal energy
Q < 0 system loses thermal energy
W > 0 work done on system
W < 0 work done by the systems
So we give Q + W a name: The Internal Energy
SPH3UW: Lecture 1, Pg 6
Still More Heat
We will be discussing three states of matter (solid, liquid, and gas). The
molecules of a solid are fixed in a rigid structure. The molecules of a
liquid are loosely bound and may mix with one another freely. (While a
liquid has a definite volume, it still takes the shape of its container. The
molecules of a gas interact with each other slightly, but usually move at
higher speeds than that of solid of liquid.
In all three states of matter the molecule are moving and therefore have
Kinetic Energy. But, they also have Potential Energy due to the bonds
between them. The sum of the potential and kinetic energies of the
molecule of the substance is also known as its Internal Energy.
When a warmer substance comes in contact with a cooler substance,
some of the kinetic energy of the molecules in the warmer substance is
transferred to the cooler substance. The energy representing this kinetic
energy of the molecule that is transferred from the warmer to the cooler
substance is called heat energy.
SPH3UW: Lecture 1, Pg 7
What is temperature?
A mercury thermometer
The mercury rises up
the tube as it expands.
This is movement.
The mercury is
gaining (internal)
energy from
the hot chocolate.
An object (say, a cup of
hot chocolate)
This transfer of energy is what we call
heat.
When the transfer stops, the objects
are in thermal equilibrium.
SPH3UW: Lecture 1, Pg 8
Temperature Scales
•Celsius (0C)
• Zero defined by an ice-water bath at 1 atm.
• Unit defined by water-steam (100ºC) at 1 atm.
•Kelvin (absolute K)
• Zero defined by absolute zero, but we cannot
reach that temperature experimentally
• 273.16 K defined by the triple-point of water
(0.01ºC at 4.58 mm of mercury, water can exist
in all three states of matter)
• Unit is the same as the Celsius scale
• Fahrenheit (0F)
• Zero and unit based on salt-water, water freezes
at 32 0F and boils at 212 0F )
SPH3UW: Lecture 1, Pg 9
Converting Between the Scales
From Celsius to Kelvin:
TC  TK  273.15
From Fahrenheit to Celsius:
9
TF  TC  32ºF
5
SPH3UW: Lecture 1, Pg 10
Example
You place a small piece of melting ice in your mouth. Eventually, the
water all converts from the ice at T1=32.00 0F to body temperature,
T2 =98.60 0F. Express these temperatures as 0C and K.
Plan: We convert Fahrenheit to Celsius temperature, then from
Celsius to Kelvin
TC 
5
TF  32F 
9
TK  TC +273.15
5
 32.00F  32F 
9
=0.00C
5
  98.60 F  32 F 
9
=37.00C
T1C 
T2C
T1K  0.00C  273.15
=273.15K
T1K  37.00C  273.15
=310.15K
SPH3UW: Lecture 1, Pg 11
Mechanical Equivalent of Heat
James Joule showed that mechanical energy could be converted to
heat and arrived at the conclusion that heat was another form of
energy.
He showed that 1 calorie (c) of heat was equivalent to 4.184 J of
work. (that is 1 calorie is defined as the heat needed to raise the
temperature of 1 gram of water 1 0C)
1 cal = 4.184 J
Kilocalorie(C) – the amount of heat needed to raise the temperature
of 1,000 grams of water by 1 °C. (Used with food, Food calories (C)
are determined by burning the food and measuring the amount of
energy that is released.)
British Thermal Units (BTUs) are the amount of heat to raise one
pound of water by 1 °F.
SPH3UW: Lecture 1, Pg 12
Heat Transfer and Temperature Change
The change in temperature that a substance experiences depends
upon two things: its identity (specific heat) and the amount of
material (mass). The equation that connects the amount of heat, Q,
and the resulting temperature change , T in 0C, is:
Q  mcT
T  T f  Ti
Where Q is the quantity of heat (calories) m is the mass of the
sample in grams and c is the intrinsic property called specific heat
capacity in 1 cal/g°C.
Note: that positive Q is interpreted as heat coming in (T is
positive, so T increases), while negative Q corresponds to heat
going out (T is negative, so T decreases).
SPH3UW: Lecture 1, Pg 13
Heat Transfer Example
During a bout with the flu an 80. kg man ran a fever of 39.0 0C
instead of the normal body temperature of 37.0 0C. Assuming that
the human body is mostly water (c=1cal/g 0C), how much heat, in
calories and Joules, is required to raise his temperature by that
that amount?
J 

5
5
1.6

10
cal
4.184

6.7

10
J




Q  mcT
cal 

 1cal 
3
 80 10 g 1
  2.0C 
We could also use:
 g C 


 1.6 105 cal
 160kcal
 160 (food calories)
Q  mcT

1J 
  80kg   4190
  2.0 K 
kg  K 

 6.7 105 J
SPH3UW: Lecture 1, Pg 14
Heat Transfer and Phase Changes
Consider an ice cube. Since water freezes at 0 0C, the temperature
of the ice cube is 0 0C. If we add heat to the ice, its temperature
does not rise. Instead the thermal energy absorbed by the ice goes
into loosening the intermolecular bonds of the ice, thereby
transforming it into a liquid. The temperature remains at 0 0C.
In each phase change (solid to liquid, liquid to gas), absorbed heat
causes no temperature change so Q=mcT does not apply. The
equation we use is:
Q  mL
Where L is the latent heat of transformation (solid to liquid, or vice
versa L is latent heat of fusion. From liquid to gas, L is called latent
heat of vaporization). This equation tells us how much heat must be
transferred in order to cause a sample of mass m to undergo a
phase change.
SPH3UW: Lecture 1, Pg 15
Example on Temperature and Phase
You want to cool 0.25kg of water, initially at 25 0C, by adding ice, initially
at -20 0C. How much ice should you add so that the final temperature
will be 0 0C with all the ice melted [cwater = 4190 J/kg K, cice=2100 J/kg K,
L=334000J/kg]?
The ice and water are the objects that exchange heat. The water
undergoes a temperature change only, while the ice undergoes
both a temperature and phase change. We require the mass of the
ice.
Let’s first determine the
negative heat added to the
water.
For Ice, first we determine
the heat needed to warm
the ice.
Qwater  mwater cwater Twater

J 
  0.25kg   4190
  25K 
kg  K 

 26000 J
QiceT   micecice Tice

J 
  mice   2.1103
  20 K 
kg  K 


J 
  mice   4.2 104 
kg 

SPH3UW: Lecture 1, Pg 16
Example on Temperature and Phase
You want to cool 0.25kg of water, initially at 25 0C, by adding ice,
initially at -20 0C. How much ice should you add so that the final
temperature will be 0 0C with all the ice melted [cwater = 4190 J/kg K,
L=334000J/kg]?
Qwater  26000 J
For Ice, now we need the heat to
phase shift it from solid to liquid.

J 
QiceT    mice   4.2 104

kg 

Qice phase   mice L fusion
The sum of these quantities
must be zero

J 
 mice  3.34  105

kg 



J 
J 
Qwater  QiceT   Qice Phase   26000 J   mice   42000    mice   334000 
kg 
kg 




J 
J 
0  26000 J   mice   42000    mice   334000 
kg 
kg 


mice  0.069kg
 69 g
SPH3UW: Lecture 1, Pg 17
Heat Transfer and Thermal Expansion
When a substance undergoes a temperature change, it changes
in size. Changes in length due to temperature is governed by:
L f  Li   Li T f  Ti 
L   Li T
Where
is the coefficient of linear expansion of the material
Similarly, there are formulas for changes in Area and volumes
A   Ai T
V  Vi T
Water contracts as it is cooled until it reaches 4 °C. From then
on, cooling causes expansion. This is because freezing water has
a crystalline structure.
SPH3UW: Lecture 1, Pg 18
Understanding
You need to slide an aluminum ring onto a rod. At room temperature
(20oC), the internal diameter of the ring is 50.0 mm and the diameter of
the rod is 50.1 mm.
a) Should you heat or cool the ring to make it fit onto the rod?
b) If the coefficient of linear expansion for aluminum is 2.5 x 10-5 oC-1,
at what temperature will the ring just barely fit onto the rod?
a) Heat, even the hole will expand

5
1
0.0001
m

2.5

10

C

L


L

T
b)
0
  0.050m  T
T  80C
Therefore the final temperature is 20oC+80oC or 100oC
SPH3UW: Lecture 1, Pg 19
The Kinetic Theory of Gases
Unlike the condensed phases of matter (solid or liquid),
molecules of a gas move freely and rapidly. A confined gas
exerts a force on the walls of its container. The moving
molecules strike the walls and rebound. The magnitude of the
force per unit area is called pressure, and is denoted by P:
F
P
A
The SI unit for pressure is the N/m2, the pascal (Pa).
We also need a way to talk about the vast number of
molecules in a given sample of gas. One mole of atoms or
molecules (or teachers)contains N A  6.022 1023 of these
elementary quantities. NA is known as Avogadro’s Number.
SPH3UW: Lecture 1, Pg 20
Molecular Picture of Gas

Gas is made up of many individual molecules

Number density is Number of molecules/Volume:
N/V = r/m
r is the mass density
m is the mass for one molecule

Number of moles: n = N / NA
NA = Avogadro’s Number = 6.022x1023 mole-1

Atomic
mass unit
Mass of 1 mole of “stuff” in grams = molecular mass in u
e.g., 1 mole of N2 has mass of 2x14=28 grams
SPH3UW: Lecture 1, Pg 21
The Ideal Gas Law
Three physical properties – Pressure (P), volume (V), and
temperature (T) describe a gas. At low densities, all gasses
approach ideal behaviour; This means that these three variables
are related by the equation:
PV  nRT
Where n is the number of moles of gas and R is the constant
(8.31 J/mol K), called the universal gas constant. This equation
is known as the Ideal Gas Law.
SPH3UW: Lecture 1, Pg 22
The Ideal Gas Law
An important consequence of this equation is that, for a fixed volume of
gas, an increase in P gives a proportional increase in T. By applying
Newton’s Second Law we can find that the pressure – exerted by N
molecules of gas in a container of volume V is related to the average kinetic
energy of the molecules by the equation: PV  2 NK avg
3
2
NK avg  nRT
3
2
N A K avg  RT
3
3 R
K avg 
T
2 NA
Therefore
3
 k BT
2
; NA 
N
n
R
; kB 
NA
1 2
3
mvavg  k BT
2
2
3k BT
3RT
root-mean-square
vavg 

m
M
Boltzmann’s constant
J
k B  1.38 1023
K
m=molecular mass (kg)
M=molar mass (kg/mole)
SPH3UW: Lecture 1, Pg 23
The Ideal Gas Law
We recall that the sum of the Kinetic and Potential energy (at
the molecular level) is called the Internal energy of the system.
U  KE  PE
With the average kinetic energy of a molecule related to
temperature by:
3
K ave  kT
Atoms of a single element
2
Then for an ideal monatomic gas (No potential energy) containing
N molecules, the Internal energy will just be N time this equation
3
3
U  NkT  nRT
2
2
n is the number of Moles and R
the Ideal gas constant.
SPH3UW: Lecture 1, Pg 24
The Ideal Gas Law
Calculate the internal energy of the air in a typical room with a
volume of 50 m3. treat the air as if it were a monatomic ideal
gas at 1 atm = 1.01 x 105 PA
Hint:
PV  nRT
3
U  nRT
2
PV  nRT
3
U  nRT
2
3
 PV
2
3
 1.01105 PA  50m3 
2
 7.58 106 J
SPH3UW: Lecture 1, Pg 25
The Ideal Gas Law

P V = N kB T
P = pressure in N/m2 (or Pascals)
V = volume in m3
N = number of molecules
T = absolute temperature in K
k B = Boltzmann’s constant = 1.38 x 10-23 J/K
Note: P V has units of N-m or J (energy!)

PV=nRT
n = number of moles
R = ideal gas constant = NAkB = 8.31 J/(mol*K)
SPH3UW: Lecture 1, Pg 26
20
Ideal Gas Law ACT II
PV = nRT
You inflate the tires of your car so the pressure is
30 psi, when the air inside the tires is at 20
degrees C. After driving on the highway for a
while, the air inside the tires heats up to 38 C.
Which number is closest to the new air pressure?
1) 16 psi
2) 32 psi
3) 57 psi
Careful, you need to use the temperature in K
P = P0 (38+273)/(20+273)
SPH3UW: Lecture 1, Pg 27
23
Boyle’s Law
The quantities P, V, n, and T aren’t independent but are
related by an equation of state. You can perform various
experiments where two of these quantities are held fixed and
a relation between the other two is determined. These minilaws have names like Boyle’s Law and Charles’s Law.
PV=constant
n and T held fixed
SPH3UW: Lecture 1, Pg 28
Charles’s Law
The quantities P, V, n, and T aren’t independent but are
related by an equation of state. You can perform various
experiments where two of these quantities are held fixed and
a relation between the other two is determined. These minilaws have names like Boyle’s Law and Charles’s Law.
V/T=constant
n and P held fixed
SPH3UW: Lecture 1, Pg 29
Ideal Gas Law: ACT 1
PV = nRT

A piston has volume 20 ml, and pressure of 30 psi. If the volume is
decreased to 10 ml, what is the new pressure? (Assume T is
constant.)
1) 60
2) 30
3) 15
When n and T are constant, we
have PV = constant (Boyle’s Law)
PV
1 1  PV
2 2
 30 psi  20ml   P2 10ml 
V=20
P=30
V=10
P=??
P2  60 psi
SPH3UW: Lecture 1, Pg 30
Balloon ACT 1

What happens to the pressure of the air inside a hot-air
balloon when the air is heated? (Assume V is constant)
1) Increases
2) Same
3) Decreases
Balloon is still open to atmospheric pressure, so
it stays at 1 atm
SPH3UW: Lecture 1, Pg 31
Balloon ACT 2

What happens to the buoyant force on the balloon when
the air is heated? (Assume V remains constant)
1) Increases
2) Same
3) Decreases
FB = r V g, r is density of outside air!
SPH3UW: Lecture 1, Pg 32
32
Balloon ACT 3

What happens to the number of air molecules inside the
balloon when the air is heated? (Assume V remains constant)
1) Increases
2) Same
3) Decreases
PV = NkT
P and V are constant. If T increases N decreases.
SPH3UW: Lecture 1, Pg 33
34
Balloon Summary
In terms of the ideal gas law, explain briefly how a hot air balloon
works.
Hot air has less mole density than cool air. So less hot air is required
in order to achieve the same pressure as cool air. This makes the
density of hot air less allowing it to float.
When temperature increases the volume of the gas increases, thus
reducing the density of the gas making it lighter that then
surrounding air, which causes the balloon to rise.
Note! this is not a pressure effect, it is a density effect. As T
increases, the density decreases the balloon then floats due to
Archimedes principle. The pressure remains constant!
SPH3UW: Lecture 1, Pg 34
The Laws of Thermodynamics
We have learned about two ways in which energy may be
transferred between a system and its environment. One is work
(force act over a distance), the other is heat (energy
transferred due to a difference in temperature). The study of
energy transfers involving work and heat, and the resulting
changes in internal energy, temperature, volume, and pressure
is called thermodynamics.
The order of discovery: (A hint at the logic of Thermodynamics)
The 2nd Law was discovered First
The 1st Law was discovered Second
The 0th Law was the third to be discovered
The 4th Law is called the 3rd Law
SPH3UW: Lecture 1, Pg 35
The Zeroth Law of Thermodynamics
When two objects are brought into contact, heat will flow from
the warmer object to the cooler one until they reach thermal
equilibrium.
If Objects 1 and 2 are each in thermal equilibrium with
Object 3, then Objects 1 and 2 are in thermal equilibrium
with each other.
SPH3UW: Lecture 1, Pg 36
The First Law of Thermodynamics
Simply put, the first Law of Thermodynamics is a statement of
the Conservation of energy that includes heat.
So what does the First Law say? In words, the internal energy
of a body (such as a gas) can be increased by heating it or by
doing mechanical work on it
James Joule showed that mechanical energy could be
converted to heat and arrived at the conclusion that heat was
another form of energy. He showed that 1 calorie of heat was
equivalent to 4.184 J of work.
1 cal = 4.184 J
SPH3UW: Lecture 1, Pg 37
Changing the Internal Energy

U is a “state” function --- depends uniquely on the state of
the system in terms of p, V, T etc.
(e.g. For a classical ideal gas, U = NkT )

There are two ways to change the internal energy of a system:
WORK done by the system on
the environment
Wby = -Won
HEAT is the transfer of thermal
energy into the system from the
surroundings Q
Thermal reservoir
Work and Heat are process energies, not state functions.
SPH3UW: Lecture 1, Pg 38
The First Law of Thermodynamics
The first law deals with changes in the internal energy of the system.
It is important to note that changes in the internal energy of a system
can only occur if the system is not isolated. This tells us that the
system is imbedded in its surroundings in a way that there can be
energy transfer. There are two types of energy transfer (Work and
Heat), and the difference between these two is determined only by
what occurs in the surroundings of the system.
Work: is an energy transfer between a system and its surroundings
that is the result of an organized motion in the surroundings.
For example:
•You can increase the internal energy of a piece of plastic by
vigorously rubbing it.
•You can decrease the internal energy of a gas by letting it expand
against some external pressure (such as a piston).
Piston examples are commonly used to illustrate many concepts of
Thermodynamics, so Let’s look into this more deeply.
SPH3UW: Lecture 1, Pg 39
The First Law of Thermodynamics

Consider an example
system of a piston and
cylinder with an enclosed
dilute gas characterized by
P,V,T & n.
SPH3UW: Lecture 1, Pg 40
The First Law of Thermodynamics
• What happens to
the gas if the
piston is moved
inwards?
SPH3UW: Lecture 1, Pg 41
The First Law of Thermodynamics
• If the container is
insulated the
temperature will
rise, the atoms
move faster and the
pressure rises.
• Is there more
internal energy in
the gas?
Yes!
SPH3UW: Lecture 1, Pg 42
The First Law of Thermodynamics
• External agent did
work in pushing the
piston inward.
• W = Fd
•
=(PA)x
x
• W =PV
SPH3UW: Lecture 1, Pg 43
The First Law of Thermodynamics
• Work done on
the gas equals
the change in the
gases internal
energy,
x
W = U
SPH3UW: Lecture 1, Pg 44
The First Law of Thermodynamics
We can represent the force the gas
exerts on the piston or the external
force that compresses the piston
by a PV diagram.
Wby gas  PV
This tells us that the work done by the
gas during the expansion (from 1 to 2) is
just the area under the curve. By
convention, this area is positive for
positive work done by the gas. This also
corresponds to negative work done on
the gas. We must be careful to
distinguish the work done on the gas and
the work done by the gas.
Direction of arrow
(left to right is very
important
P
P
1
2
(+) area
(+) work done
by gas
(-) work done
on gas
V1
V2
V
P
P
1
2
(-) area
(-) work done
by gas
(+) work done
on gas
V1
V2
V
SPH3UW: Lecture 1, Pg 45
The First Law of Thermodynamics
Example:
One mole of monatomic ideal gas is enclosed under a frictionless
piston. A series of processes occur, and finally the state of this ideal
gas returns to its initial state (see PV diagram). Answer the following
questions in terms of P0, V0, and R.
(a) Determine the temperature at each vertex.
(b) Determine the change in internal energy ( U) for each process.
(c) Determine the work done by the gas for each process.
(a) Use PV
 nRT
PV
0 0  nRT1
P
1
P0
1/ 2 P0 4V0
P 4V
PV
0 0
T2  0 0 T3 
1/2P0
nR
nR
nR
2 P0V0
4 P0V0
P0V0



V0
R
R
R
T1 
2
3
2V0 4V0
SPH3UW: Lecture 1, Pg 46
V
The First Law of Thermodynamics
Example:
One mole of monatomic ideal gas is enclosed under a frictionless
piston. A series of processes occur, and finally the state of this ideal
gas returns to its initial state (see PV diagram). Answer the following
questions in terms of P0, V0, and R.
(a) Determine the temperature at each vertex.
(b) Determine the change in internal energy ( U) for each process.
(c) Determine the work done by the gas for each process.
3
P
(a) Use U  nRT
2
1
P0
PV
3  4 PV
9
0 0
0 0 
U12  nR 

0 0
  PV
2  R
R  2
1/2P0
3  2 PV 4 PV 
U 23  nR  0 0  0 0   3P0V0
2  R
R 
V0
2 PV
3  PV
3
0 0
0 0 
U 31  nR 

0 0
   PV
2  R
R 
2
2
3
2V0 4V0
SPH3UW: Lecture 1, Pg 47
V
The First Law of Thermodynamics
Example:
One mole of monatomic ideal gas is enclosed under a frictionless
piston. A series of processes occur, and finally the state of this ideal
gas returns to its initial state (see PV diagram). Answer the following
questions in terms of P0, V0, and R.
(a) Determine the temperature at each vertex.
(b) Determine the change in internal energy ( U) for each process.
(c) Determine the work done by the gas for each process.
(a) Use area under graph for each segment
P
1
W12 by  3PV
0 0
W23by  0
W31by
9
  PV
0 0
4
Note:
•No work for constant volume.
•Total work 3/4P0V0 is same
as area enclosed by triangle.
P0
1/2P0
2
3
V0
2V0 4V0
SPH3UW: Lecture 1, Pg 48
V
The First Law of Thermodynamics
Summary:
T
U
4.5PoVo
Wby
Won
3PoVo
-3PoVo
1 2
3(PoVo)/R
2 3
-2(PoVo)/R -3PoVo
0
0
3 1
-1(PoVo)/R -1.5PoVo
-2.25PoVo
+2.25PoVo
P
1
P0
1/2P0
2
3
V0
2V0 4V0
V
SPH3UW: Lecture 1, Pg 49
First Law of Thermodynamics




Let’s change the situation:
Keep the piston fixed at its
original location.
Place the cylinder on a hot plate.
What happens to gas?
SPH3UW: Lecture 1, Pg 50
The First Law of Thermodynamics
Heat flows into the gas.
Atoms move faster,
internal energy
increases.
Q = heat in Joules
U = change in internal
energy in Joules.
Q = U
SPH3UW: Lecture 1, Pg 51
The First Law of Thermodynamics
Heat:
Heat is an energy transfer between a system and its surroundings
that is the result of random motion in the surroundings. (Note: there is
a difference between a work process [organized motion in
surroundings] and a heat process [random motion in surroundings]).
Heat will always flow spontaneously from the system at higher
temperature to the system of lower temperature, but heat can be
made to flow in the opposite direction as well if work is done in the
process (a refrigerator).
SPH3UW: Lecture 1, Pg 52
The First Law of Thermodynamics

What if we added heat and
pushed the piston in at the
same time?
F
SPH3UW: Lecture 1, Pg 53
The First Law of Thermodynamics

Work is done on the gas, heat is
added to the gas and the internal
energy of the gas increases!
F
U=Q+W
SPH3UW: Lecture 1, Pg 54
First Law of Thermodynamics
Energy Conservation :The change in internal energy of
a system ( U) is equal to the heat flow into the system
(Q) minus the work done by the system (W)
U = Q - W
Increase in internal
energy of system
Heat flow
into system
Work done by system
Equivalent way of writing 1st Law:
U = Q + W
Increase in internal
energy of system
Heat flow
into system
P
P1
Work done
on system
ideal
gas
P3
1
2
3
V1
V2
SPH3UW: Lecture 1, Pg 55
V
07
First Law of Thermodynamics
If we re-examine our PV diagram and now calculate the heat transfer
(Q) for each process.
Since the work done by the gas and the change in internal
energy has already been calculated, it is very easy to calculate
the heat transfer from the First Law: U = Q
+W
into
on
2
U12  Won12  Qi1nto
 Qi1nto2  U12  Wby12
12
into
Q
P
1
P0
1/2P0
2
9
 PV
0 0  3PV
0 0
2
15
 PV
0 0
2
3
V0
2V0 4V0
V
SPH3UW: Lecture 1, Pg 56
First Law of Thermodynamics
If we re-examine our PV diagram and now calculate the heat transfer
(Q) for each process.
Since the work done by the gas and the change in internal
energy has already been calculated, it is very easy to calculate
the heat transfer from the First Law: U = Q
+W
into
on
3
3
23
U 23  Won23  Qi2nto
 Qi2n


U

W
to
23
by
3
Qin2
to  3PV
0 0 0
P
1
P0
1/2P0
 3PV
0 0
2
3
V0
2V0 4V0
V
SPH3UW: Lecture 1, Pg 57
First Law of Thermodynamics
If we re-examine our PV diagram and now calculate the heat transfer
(Q) for each process.
Since the work done by the gas and the change in internal
energy has already been calculated, it is very easy to calculate
the heat transfer from the First Law: U = Q
+W
into
on
1
1
31
U 31  Won31  Qi3nto
 Qi3n


U

W
to
31
by
3
9
2
Qi1nto
  PV

PV
0 0
0 0
2
4
15
  PV
0 0
4
P
1
P0
1/2P0
2
3
V0
2V0 4V0
V
SPH3UW: Lecture 1, Pg 58
First Law of Thermodynamics
T
U
4.5PoVo
Wby
Qin
3PoVo
7.5PoVo
1 2
3(PoVo)/R
2 3
-2(PoVo)/R -3PoVo
0
-3PoVo
3 1
-1(PoVo)/R -1.5PoVo
-2.25PoVo
-3.75PoVo
P
1
P0
1/2P0
2
3
V0
2V0 4V0
V
SPH3UW: Lecture 1, Pg 59
Signs Example

You are heating some soup in a pan on the stove. To keep it from
burning, you also stir the soup. Apply the 1st law of thermodynamics
to the soup. What is the sign of :
1) Q
2) W
3) U
(1) Positive, heat flows into soup
(2) Zero (is close enough)
(3) Positive, Soup gets warmer
SPH3UW: Lecture 1, Pg 60
Signs
Energy is also used to mean the potential to do work; the greater the ability
of something to change the things around it, the more energy it has.
Mathematically, the first law is expressed: U= Q+W.
Where Q is the change in the quantity of heat added to the system, U is
the change in the internal energy and W is the work done on or by the
system.
The sign convention here is that if U is positive the amount of internal
energy increases. This means that Q stands for the heat energy put into
the system and W for the work done on the system. This is known as the
‘physicists’ convention’.
Work
W>0: Work is done on the system by the surroundings
W<0: Work is done by the system on the surroundings
Heat
Q>0: Heat is added to the system from the surroundings
Q<0: Heat is released by the system to the surroundings
SPH3UW: Lecture 1, Pg 61
First Law of Thermodynamics
For each process in this cycle, indicate in the table below whether
the quantities W, Q, and U are positive (+), negative (-), or zero
(0). W is the work done BY the GAS.
Process
W
Q
1 -> 2
+
2 -> 3
0
-
+
-
3 -> 1
U
+
-
P
1
P0
1/2P0
2
3
V0
2V0 4V0
V
SPH3UW: Lecture 1, Pg 62
Work Done by a Gas ACT
M
y
M
The work done by a gas as it contracts is
A) Positive
B) Zero
C) Negative
W = F d cosq < 0
= P A d = P A y = P V
W = P V :For constant Pressure
W > 0 if V > 0 expanding system does positive work
W < 0 if V < 0 contracting system does negative work
W = 0 if V = 0 system with constant volume does no work
SPH3UW: Lecture 1, Pg 63
Work Done by a Gas ACT
A gas is kept in a cylinder that can be compressed by pushing
down on a piston. You add 2500 J of heat into the system, and
then you push the piston 1.0 m down with a constant force of
1800 N. What is the change in the gas’s internal energy.
We define the variables in the First Law, being careful of the signs:
Heat added : Q=+2500 J
We did work on gas, so W is positive:
W  F d
 1800 N 1.0m
 1800 J
Therefore: U  Q  W
  2500 J    1800 J 
 4300 J
SPH3UW: Lecture 1, Pg 64
Pressure as a Function of Volume
Work is the area under the curve of a PV-diagram.
Work depends on the path
taken in “PV space.”
When a process is depicted on a PV
diagram, directional arrows are
used on the graph. In a process that
moves to the right, the gas will do
positive work on the surroundings;
in a process that moves to the left,
the gas will do negative work on the
surroundings (ie the surroundings
do work on the gas)
The precise path serves to describe the kind of process that took place.
SPH3UW: Lecture 1, Pg 65
Work Done by a Gas
Work done by gas equals area inside graph
SPH3UW: Lecture 1, Pg 66
P-V Diagrams
When you are describing a process that involves a gas expanding
or compressing, you will need to use specific nomenclature. A
process is called isobaric if the pressure remains constant
throughout. It is called isothermal if the temperature remains
constant. If no heat energy enters or leaves the gas, it is called
adiabatic.
SPH3UW: Lecture 1, Pg 67
Different Thermodynamic Paths
The work done depends on the initial and final
states and the path taken between these states.
SPH3UW: Lecture 1, Pg 68
Thermodynamic Systems and P-V Diagrams
 ideal gas law: PV = nRT (nR = NkB)
 for n fixed, P and V determine “state” of monatomic ideal gas system
T = PV/nR
U = (3/2)nRT = (3/2)PV
 Examples:
 which point has highest T?
P
B
A
B
P1
 which point has lowest U?
C
C
P3
 to change the system from C to B,
energy must be added to system
V1
V2 V
SPH3UW: Lecture 1, Pg 69
First Law of Thermodynamics
Isobaric Example
2 moles of monatomic ideal gas is taken
P
from state 1 to state 2 at constant pressure
1
2
P
3
3
P=1000 Pa, where V1 =2m and V2 =3m
(R=8.31 J/k mole) Find:
1) T1
2) T2
V
V1
V2
3)U
1. PV1 = nRT1  T1 = PV1/nR = 120K
4) W
2. PV2 = nRT2  T2 = PV2/nR = 180K
5) Q
3. U = (3/2) nR T = 1500 J
U = (3/2) P V = 1500 J (has to be the same)
3
J 
60
K
 2 moles   8.31



2
mol  K 

4. W = P V = +1000 J
 1495.8 J
5. Q = U + W = 1500 + 1000 = 2500 J
 1500 J
SPH3UW: Lecture 1, Pg 70
21
First Law of Thermodynamics
Isochoric Example
2 moles of monatomic ideal gas is taken
from state 1 to state 2 at constant volume
V=2m3, where T1=120K and T2 =180K. Find Q. P
P2
Q = U + W
 U = (3/2) nR T = 1500 J
W = P V = 0 J
 Q = U + W = 1500 + 0 = 1500 J

P1
2
1
V
Requires less heat to raise T at const. volume than at const.
pressure (no energy used for work)
SPH3UW: Lecture 1, Pg 71
V
Heat Transfer and Temperature Change for Gases
The value of the heat transfer in a system, Q, is also pathdependent. We recall for a solid or liquid Q=mcT. For gases,
however, the system is a little more complicated, because the
value of the proportionality constant between Q and T depends
on whether the volume or pressure is kept constant.
If the volume remains constant (isochoric) during heat transfer,
then
CV is the molar
Q  nCV T
heat capacity at
constant volume
If the pressure remains constant (isobaric) during the heat transfer
Q  nCP T
CP is the molar
heat capacity at
constant pressure
SPH3UW: Lecture 1, Pg 72
Total Work Done
1
P
2
Wtot = ??
P
P
1
4
W = PV (>0)
1
2
3
W = PV = 0
2
V
4
V > 0
P
1
4
3
2
4
3
V < 0
V = 0 V
V
W = PV (<0)
V
3
P
1
4
W = PV = 0
2
3
V = 0 V
P
1
2
Wtot > 0
4
3
V
General rule: work
done is area under P-V
curve (even if not
horizontal).
SPH3UW: Lecture 1, Pg 73
Question
25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of
pressure and 300 K. If 100 kg of mass is placed on the piston
causing the gas to compress to 20 L at constant pressure. This is
done by allowing heat to flow out of the gas. What is the work
done on the gas? What is the change in internal energy of the
gas? How much heat flowed out of the gas?
Po = 202,600 Pa, Vo = 0.025 m3, To = 300 K, Pf = 202,600 Pa, Vf=0.020 m3,
W =-PV = -202,600 Pa (0.020 – 0.025)m3
=1013 J energy added to the gas.
SPH3UW: Lecture 1, Pg 74
Question
25 L of monatomic gas is enclosed in a cylinder/piston apparatus
at 2 atm of pressure and 300 K. If 100 kg of mass is placed on
the piston causing the gas to compress to 20 L at constant
pressure. This is done by allowing heat to flow out of the gas.
What is the work done on the gas? What is the change in internal
energy of the gas? How much heat flowed out of the gas?
U 
n

PV
RT
 202600Pa   0.025m3 
J 

 8.31
  300 K 
mol

K


 2.03
U 
Ti  300K
3
nRT
2
n
2.03 
PV
RT
 202600 Pa   0.020m3 
J 

8.31

 T f 
mol  K 

T f  240 K
3
J 
 2.03  8.31
  60 K   1518 J
2
mol  K 

SPH3UW: Lecture 1, Pg 75
Question
25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of
pressure and 300 K. If 100 kg of mass is placed on the piston
causing the gas to compress to 20 L at constant pressure. This is
done by allowing heat to flow out of the gas. What is the work
done on the gas? What is the change in internal energy of the
gas? How much heat flowed out of the gas?
U = Won - Qout
Qout = Won - U = 1013J – (-1518J) = 2531 J heat out
SPH3UW: Lecture 1, Pg 76
Second Law of Thermodynamics
Entropy, S, is a measure of the disorder, or randomness of a
system. The greater the disorder of a system, the greater the
entropy. If a system is highly ordered (like particles is a solid) we
say that the entropy is low.
The second law of thermodynamics states that all spontaneous
processes proceeding in an isolated system lead to an increase in
entropy.
The increase or decrease in entropy can be found by
Q
S 
T
Where Q is heat flow into or out of a system and T is the average
Kelvin temperature
SPH3UW: Lecture 1, Pg 77
Thermal Efficiency of a Heat Engine
The thermal efficiency, e, of the heat engine is equal to
the ratio of the heat we get out to the heat we put in.
e
W Qin  Qout

Qin
Qin
Where Qin is heat absorbed, and Qout is heat discharged
Unless Qout=0, the engines efficiency is always less than 1.
SPH3UW: Lecture 1, Pg 78
Efficiency
If we re-examine our PV diagram and now calculate the efficiency of
the system.
The efficiency of a cycle is defined as the ratio of the work
done by the gas to the heat Qin that flows into the system. Any
heat that is expelled into the surroundings is not included in
the calculation of Qin. From the point of view of efficiency, this
expelled heat is lost and its energy is not used by the system:
W
e
Qin
P
1
P0
1/2P0
2
3
V0
2V0 4V0
V
1 2
2 3
3
PV
0 0
3 1
4

 0.1
15
Total
PV
0 0
2
Wby
Qin
3PoVo
7.5PoVo
0
-3PoVo
-2.25PoVo
-3.75PoVo
0.75 PoVo
SPH3UW: Lecture 1, Pg 79
Heat Engine: Efficiency
The objective: turn heat from hot reservoir
HEAT ENGINE
into work
TH
The cost: “waste heat”
1st Law: Qin -Qout = W
efficiency e  W/Qin
=W/Qin
= (Qin-Qout)/Qin
= 1-Qout/Qin
Qin
W
Qout
TC
SPH3UW: Lecture 1, Pg 80
13
ACT
A hot (98 C) slab of metal is placed in a cool (5C)
bucket of water.
S = Q/T
What happens to the entropy of the metal?
A) Increase
B) Same C) Decreases
Heat leaves metal: Q<0
What happens to the entropy of the water?
A) Increase
B) Same C) Decreases
Heat enters water: Q>0
SPH3UW: Lecture 1, Pg 81
Description of Disorder

Isolated systems tend toward greater disorder, and
entropy is a measure of that disorder
S = kB ln (W)
» kB is Boltzmann’s constant
» W is a number proportional to the probability that the
system has a particular configuration


This version is a statement of what is most probable
rather than what must be
The Second Law also defines the direction of time of
all events as the direction in which the entropy of the
universe increases
SPH3UW: Lecture 1, Pg 82
Second Law of Thermodynamics

The entropy change (Q/T) of the
system+environment  0
never < 0
order to disorder

Consequences
A “disordered” state cannot spontaneously
transform into an “ordered” state
No engine operating between two reservoirs can
be more efficient than one that produces 0 change
in entropy. This is called a “Carnot engine”
SPH3UW: Lecture 1, Pg 83
31
Carnot Cycle
SPH3UW: Lecture 1, Pg 84
Carnot


Now suppose you want to make an engine that runs on
energy from a source of high temperature heat (at
temperature TH) such as a burner. And, furthermore,
suppose you want your heat engine to work in the most
efficient way possible.
The first thing you have to do is to get energy into your
working fluid. Obviously, the best process to do this will be
a constant temperature (isothermal) heat transfer, which
will result in an expansion of the working gas at TH (shown
on the graph as A-->B). Any other expansion process that
results in a reduction of the temperature of the working gas
below TH, will also result in a reduction of efficiency. [An
analogy can be made here between temperature and
pressure. Imagine a water turbine running from a dam -ideally you want to keep the water level at the maximum so
that you have the maximum pressure to turn the turbine. If
you draw off water too fast then the water level (and
therefore pressure) will drop and the power output of the
turbine will be reduced.]
SPH3UW: Lecture 1, Pg 85
Carnot

The isothermal expansion (A-->B) produces a work output - yay! -- but at the end of the process (point B) you are
faced with a dilemma. You want to get your piston back to
it's original position, so that you can repeat the
expansion/work output process again. But if you simply
compress the working gas, then you have to do exactly the
same amount of work to get back to the original state as
you produced during the expansion process (and, of
course, you end up rejecting your thermal energy back to
the high temperature heat source). This would mean that
you would get no net work output from your heat engine,
which would be very unsatisfactory.
SPH3UW: Lecture 1, Pg 86
Carnot


The answer, of course, is to perform another expansion process
with the high temperature heat source disconnected from the
system (shown on the graph as B-->C). The temperature (and
pressure) of the working gas will fall, but because the system is
not exchanging heat then you are not losing any of the
'potential' of the high temperature heat from the heat source.
This then allows you to compress the working gas at low
temperature and pressure (shown on the graph as C-->D),
which requires much less work to return the piston almost to its
original position. Even though you have to put work back into
the engine to do this, it is much less than the work output during
the isothermal expansion process (A-->B) -- and so, overall, the
engine produces a NET work output. This compression process
must, of course, reject heat -- which it does to a low
temperature heat sink at temperature TC (usually the ambient
environment). Again, an isothermal process is the best way to
do this, in order to keep the temperature difference between the
heat source and sink always at the maximum value.
SPH3UW: Lecture 1, Pg 87
Carnot


Of course, the engine is not quite back to its original state, as
the temperature of the working gas is still at TC. To return the
temperature to TH, we disconnect the system from the low
temperature heat sink, and perform another compression
process (shown on the graph as D-->A). This requires a work
input, which will be of exactly the same value as the work
output in B-->C. But, of course -- to reiterate -- the engine will
still have an overall net work output, since the isothermal work
output A-->B is much greater than the isothermal work input
C-->D.
Thus the engine has produced a work output, and returned to
its original state, ready to do the whole thing all over again.
This is how the Carnot cycle works.
SPH3UW: Lecture 1, Pg 88
Carnot Cycle
Idealized Heat Engine
 No Friction
 S = Q/T = 0
 Reversible Process
 Process A is expansion
 Process B is compression
1->2
Isothermal
expansion
QH
absorbed
T=TH
2->3
Adiabatic
expansion
No heat
exchange,
Temp drops
to TC
3->4
Isothermal
compressi
on
Heat QC
discarded
T=TC
4->1
Adiabatic
compressi
on
No heat
exchange,
Temp rises
to TH
A
B
T
U
Q
W
0
0
QH>0
QH
TC - TH
-W3
0
0
0
QC<0
TH - TC
W4
0
 Vf 
W  nRT ln  
 Vi 
WA 
3
nR TC  TH 
2
-QC
 Vf 
W  nRT ln  
 Vi 
3
WB  nR TH  Tc 
2
Note: work from (2->3)
and (4->1) balance, yet
more work by during
(1->2) than on during
(3->4).
SPH3UW: Lecture 1, Pg 89
The Carnot Cycle in Slow Motion
Stage 1
In the first stage the piston moves upward while
the chamber absorbs heat from a source and
the gas begins to expand. The portion of the
graphic from point 1 to point 2 represents this
behaviour. Because the temperature of the gas
does not change, all the heat drawn in from the
source goes into work performed by the
expansion of the gas.
There is only heat QH flowing into the
system, there is no change in
Potential Energy, therefore since
U = Q + W, the work by the system
is –Q (the gas does positive work).
SPH3UW: Lecture 1, Pg 90
The Carnot Cycle in Slow Motion
Stage 2
In the second stage the heat source is removed;
and the piston continues to move upward and
the gas is still expanding while cooling (lowering
in temperature from TH to TC). The portion of the
graphic from point 2 to point 3 represents this
behaviour. This stage is adiabatic expansion the
gas (no heat transfer).
Because the system expands it does
negative work (the gas does positive
work), its internal energy and
temperature decrease because it
receives no influx of heat from the
surroundings
SPH3UW: Lecture 1, Pg 91
The Carnot Cycle in Slow Motion
Stage 3
In the third stage the piston begins to move
downward and the cool gas (TC) is placed in
thermal contact with heat reservoir at
temperature TC, the gas is isothermally
compressed at this temperature TC , thus the
gas expels heat QC to the reservoir (the engine
gives energy to the environment).
Because the system contracts
(decrease in volume and increase in
pressure) the system does positive
work on the gas (gas does negative
work) and, rather than increasing its
internal energy, the gas discards heat
to the low temperature reservoir.
SPH3UW: Lecture 1, Pg 92
The Carnot Cycle in Slow Motion
Stage 4
In the final stage the piston moves downward
and the cool gas is compressed to its original
state. Its temperature also rises to its orignal
state (point 4 to point 1). No exchange of heat
with the surroundings take place.
Because the system is compressed,
its internal energy and temperature
increase (since no heat is discarded),
and the work done on the gas by the
environment is W42 (The gas does
negative work)
SPH3UW: Lecture 1, Pg 93
Carnot Cycle Review

During step one the gas does a positive amount of
work. Step two is adiabatic, with the gas doing a positive
amount of work. During step three there is a negative
amount of work done by the gas. Finally in step four, which
is adiabatic, the work done by the gas is negative. Notice
that the total work done (the remaining area) is positive
because positive work is done at high temperatures and
negative work is done at lower temperatures. During step
one heat is absorbed (Q > 0) and during step three heat is
released (Q < 0). More heat is absorbed than is released
for the entire cycle. This is the basis of how engines work:
Heat (from the hot reservoir) is transferred into mechanical
work (piston moving).
SPH3UW: Lecture 1, Pg 94
Engines and the 2nd Law
The objective: turn heat from hot reservoir HEAT ENGINE
into work
TH
The cost: “waste heat”
QH
1st Law: QH -QC = W
efficiency e  W/QH =W/QH = 1-QC/QH
S = QC/TC - QH/TH  0
S = 0 for Carnot
Therefore, QC/QH  TC/ TH
QC/QH = TC/ TH for Carnot
Therefore e = 1 - QC/QH  1 - TC/ TH
e = 1 - TC/ TH for Carnot
e = 1 is forbidden!
e largest if TC << TH
W
QC
TC
SPH3UW: Lecture 1, Pg 95
36
Example
Consider a hypothetical refrigerator that takes 1000 J of heat from a
cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at
300K.
TH
1. How much work does the refrigerator do?
2. What happens to the entropy of the universe?
3. Does this violate the 2nd law of thermodynamics?
QC = 1000 J
Since QC + W = QH, W = 200 J
QH = 1200 J
QH
W
QC
TC
SH = QH/TH = (1200 J) / (300 K) = 4 J/K
SC = -QC/TC = (-1000 J) / (100 K) = -10 J/K
STOTAL = SH + SC = -6 J/K  decreases (violates 2nd law)
SPH3UW: Lecture 1, Pg 96
Review
Which of the following is forbidden by the second
law of thermodynamics?
1.
2.
3.
4.
Heat flows into a gas and the temperature falls
The temperature of a gas rises without any heat flowing into it
Heat flows spontaneously from a cold to a hot reservoir
All of the above
Not 1, because volume could increase and lower temperature
Not 2, because increase of pressure can increase temperature
SPH3UW: Lecture 1, Pg 97
43
Summary
First Law of Thermodynamics
U  Q  W
Ideal Gas Law
PV
 R,
nT
Average KE of Gas Molecule
Internal energy of ideal Gas
PV
PV
1 1
 2 2
n1T1 n2T2
3
kT
2
3
5
U monotomic  nRT , U diatomic  nRT
2
2
EK avg 
Expansion Work
Wby  PV
Heat Engine Efficiency
QC
e  1
QH
Carnot Efficiency
TC
e  1
TH
SPH3UW: Lecture 1, Pg 98
Summary II
Isobaric Work:
W  P  V
Isochoric Work:
W 0
Isothermal Work:
Adiabatic Work:
Isothermal
Internal Energy
 Vf 
W  nRT ln  
 Vi 
3
W  nRT
2
U  0
The internal energy of an ideal gas depends only on temperature
SPH3UW: Lecture 1, Pg 99
Summary III
Cyclic process (originates and ends at same state)
U=0
Wby=Qinto
Isovolumetric (constant volume)
Qinto= U
Isobaric (constant pressure)
Isothermal (constant temperature)
U=0
Wby=Qinto
Adiabatic (no heat exchange)
Q=0
U=Won
U=-Wby
SPH3UW: Lecture 1, Pg 100