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Thermal Physics
Topic 10.2 Processes
The First Law of Thermodynamics
Is a statement of the Law of Conservation of
Energy in which the equivalence of work and
thermal energy flow is taken into account.
It can be stated as
the heat added to a closed system equals the
change in the internal
energy of the system plus the work done by
the system.
That is,
 Q = U +  W = U + pV
or
U = Q - W
Explanation
Where `+ Q' is the thermal energy added to
the system and `+ W' is the work done by
the system.
If thermal energy leaves the system, then Q is
negative. If work is done on the system, then
W is negative.
For an isolated system, then W = Q = 0 and
U = 0.
Example
If 22 J of work is done on a system and
3.4 x 102 J of heat is added, what is the
change in internal energy of the
system?
Solution
Using the formula, Q = U + W
We have that 340 J = U + (-22) J
So that U = 340 J + 22 J
= 362 J
That is, the change in internal energy of
the system is 3.6 x 102 J.
Isobaric
A graph of pressure as a function of
volume change when the pressure is
kept constant is shown below
p -V diagram
p
p
Isobaric process
V1
V2 V
Area = work done
= p (V1 - V2)
V
Work Done
Such a process is said to be isobaric.
Note that the work done by the gas is
equal to the area under the curve
An isobaric transformation requires a
volume change at constant pressure
For this to occur, the temperature needs
to change to keep the pressure constant
Example

6.0 dm3 of an ideal gas is at a pressure
of 202.6 kPa. It is heated so that it
expands at constant pressure until its
volume is 12 dm3. Find the work done
by the gas.
Solution

Using the formula W = pV
 we have that
 W = 202.6 kPa x(12 - 6.0) dm3

= 202.6 x 103 Pa x (12 - 6.0) x 10-3m3
= 1.216 x 103 J
 That is, the work done by the gas in the
expansion is 1.2 x 103 J.

Isochoric / Isovolumetric
A graph of pressure as a function of
volume change when the volume is kept
constant is shown below
p
Isochoric / Isovolumetric
V
Work Done
Such a process is said to be isochoric.
When the volume is kept fixed, the curve
of the transformation is said to be an
isochore.
Note that the work done by the gas is
equal to zero as V = 0.
There is zero area under the curve on a
p-V diagram.
p - V diagram
p
Isochoric / Isovolumetric
V
However
The temperature and pressure can both
change and so such a transformation
will be accompanied by a thermal
energy change.
Example
A thermal system containing a gas is
taken around the cycle as in the next
slide
(Cyclic processes such as this one have
important applications in heat engines
that convert internal energy into useful
mechanical energy).
p
6
B
C
2
A
D
4
10 V
Starting at point A on the diagram,
describe the cycle, and, calculate the
work done by the system in the
completion of one cycle.
Solution
From A to B, the volume is kept constant
(isovolumetric) as the pressure
increases. This can be achieved by
heating the gas. Since V = 0, then no
work is done by the gas, W = 0.
From B to C, the gas expands (volume
increases) (isobaric expansion) while the
pressure is kept constant. The amount of
work done by the gas is given by the
area under the 6 kPa isobar.
Now, we have that W = p V
So that W = 6 kPa. (10 - 4) m3
= 36,000J
From C to D, the gas is cooled to
keep the volume constant as the
pressure is decreased (isovolumetric)
Again V = 0 and no work is done by the
gas, W = 0
From D to A, the gas is
compressed (volume decreases)
(isobaric compression) while the
pressure is kept constant. The amount of
work done on the gas is given by the
area under the 2 kPa isobar.
Again, using the fact that W = p V
we have that W = 2 kPa. (4 -10) m3
= - 12,000 J
That is, the net work done by the gas is
therefore 36,000 J – 12,000 J
=24,000 J.
Because the cycle is traced in a
clockwise direction, the net work done
on the surroundings is positive. If the
cycle was traced in a counterclockwise direction, the work done
would be negative
Isothermal Processes
A thermodynamic process in which the
pressure and the volume are varied
while the temperature is kept constant is
called an isothermal process

In other words, when an ideal gas
expands or is compressed at constant
temperature, then the gas is said to
undergo an isothermal expansion or
compression.
The figure below shows three isotherms
for an ideal gas at
different temperatures where
Tl < T2 < T3
p
T1
T2
T3
Isothermal process
V
Isothermal Processes
The curve of an isothermal process
represents a Boyle's Law relation
pV =constant = nRT
• the moles of gas n,
• the molar gas constant R,
• the absolute temperature T are constant
How?
In order to keep the temperature constant
during an isothermal process
• the gas is assumed to be held in a thin container
with a high thermal conductivity
that is in contact with a heat reservoir - an ideal
body of large mass whose
temperature remains constant when heat is
exchanged with it.
e.g.. a constant-temperature water bath.
And
• the expansion or compression
should be done slowly so that no
eddies are produced to create hot
spots that would disrupt the
energy equilibrium of the gas.
An Ideal Gas
Consider an ideal gas enclosed in a thin
conducting vessel that is in contact with
a heat reservoir, and is fitted with a light,
frictionless, movable piston
If an amount of heat Q is added to the
system which is at point A of an
isotherm, then the system will move to
another point on the graph, B.
The heat taken in will cause the gas to
expand isothermally and will
be equivalent to the mechanical work
done by the gas
Because the temperature is constant,
there is no change in internal energy of
the gas
That is,
T = 0
and U = 0
Q=W
The work done by gas is equal to the Area
under the curve between A and B
Isothermally
If the gas expands isothermally from A to
B and then returns from B to A following
exactly the same path during
compression
Then the isothermal change is said to be
reversible.
Adiabatic Processes
An adiabatic expansion or contraction is
one in which no heat Q is allowed to
flow into or out of the system
For the entire adiabatic process, Q = 0
How
To ensure that no heat enters or leaves
the system during an adiabatic process
it is important to:
• make sure that the system is extremely
well insulated
• carry out the process rapidly so that the
heat does not have the time to leave the
system
p
T1
T2
Adiabatic compression
Adiabatic Expansion
V
Example
The compression stroke of an automobile
engine is essentially an adiabatic
compression of the air-fuel mixture
The compression occurs too rapidly for
appreciable heat transfer to take place
What happens
In an adiabatic compression the work
done on the gas will lead to an increase
in the internal energy resulting in an
increase in temperature.
U=Q - W
but Q=O
therefore U= - W
In an adiabatic expansion the work done
by the gas will lead to a decrease in the
internal energy
Resulting in a decrease in temperature.
Work done
p
p
V
Area = work done
by the gas
expanding
V
Area = work done
on the gas
as it is
compressed
p
V
Area between
curves
= Net work
done by
Example
For the compression stroke of an experimental diesel
engine, the air is rapidly decreased in volume by a
factor of 15, the compression ratio.
The work done on the air-fuel mixture for this
compression is measured to be 550 J
(a) What type of thermodynamic process is likely to
have occurred?
(b) What is the change in internal energy of the air-fuel
mixture?
(c) Is the temperature likely to increase or decrease?
Solution
(a) Because the compression occurs
rapidly appreciable heat transfer does
not take place, and the process can be
considered to be adiabatic, Q = 0
(b)
U=Q-W
= 0 - (-550) J
Therefore, the change in internal
energy is 550J
(c) The temperature rise will be
very large resulting in the
spontaneous ignition of the airfuel mix
Thermodynamic Cycle and Engines
A thermodynamic engine is a device that
transforms thermal energy to
mechanical energy
Cars, Steam trains, jets and rockets have
engines that transform fuel energy
(chemical energy) into kinetic energy of
their motion
Efficiency
In all engines the conversion is accompanied by
the emission of exhaust gases that waste
some of the thermal energy
These engines are not very efficient as only part
of the thermal energy is converted to
mechanical energy
Presenters
The Engine
By Vija and Bombo
Internal Combustion
By Adil and Suhayb
Schematic of a Heat Engine
By Ham and Heisei
Refrigerators
By Aleem and Bhavik
Heat Pumps
By Alex and Said
Schematic for a Refrigerator by
Karim and Chilla
Carnot´s Theorem
By Martin and Mitul
The Engine
By Vija and Bombo
Has two crucial features:
It must work in cycles to be useful
the cyclic engine must have more than one
heat reservoir
The Thermodynamic Cycle
Is the process in which the system is
returned to the same state from which it
started
That is, the initial and final states are the
same in the cyclic process
Combustion Engines
The steam engine and the steam turbine
are examples of external combustion
engines
The fuel is burnt outside the engine and
the thermal energy is transferred to the
piston or a turbine chamber by means
of steam
The Steam Engine
Water is heated in a boiler heat
reservoir (high temperature) to
produce steam
As the piston returns to its original
position, the intake valve is closed
and the exhaust valve opens
The Steam Engine cont...
The steam passes into an open intake
valve where it expands causing a piston
to move outwards
This allows the exhaust gas to be forced
out into a condenser heat reservoir (low
temperature)
The Steam Engine
Steam Turbines
Most electricity is produced using steam
engines
The piston is replaced with a rotating
turbine that contains blades
The rotating turbine converts mechanical
energy to electrical energy via a
generator
The steam is cooled in cooling towers
Internal Combustion
By Adil and Suhayb
The Intake Stroke
With the exhaust valve closed, a mixture
of petrol vapour and air from the
carburetor is sucked into the
combustion chamber through the inlet
valve as the piston moves down
The Compression Stroke
Both valves are closed and the piston
moves up to squeeze the mixture of
petrol vapour and air to about 1/8th its
original volume
The Power Stroke
Both valves are closed the mixture is
ignited by a spark from the spark plug
The mixture burns rapidly and the hot
gases then expand against the piston
The Exhaust Stroke
1. The
exhaust valve opens as the piston
moves upwards
2. Exhaust gases are expelled
3. Then the cycle begins again
Schematic of a Heat Engine B
By Ham and Heisei
Fluids enter
carrying energy
QH
Hot reservoir at TH
QH
Engine
Fluids leave
carrying energy
QL = QH - W
QL
Cold reservoir at TL
Engine does
useful work, W
W
Efficiency
The thermal efficiency of a heat engine, ,
is defined as the ratio of work it does, W
to the heat input QH
 = W / QH
 = (QH - QL) / QH
It is clear that the efficiency will be greater
if Q can be made small
Question
An engine absorbs 230J of thermal
energy from a high temperature
reservoir, does work and exhausts 140J
to a cold temperature reservoir.
What is the efficiency?
Answer 39% and 90J
How much work is done?
Refrigerators
By Aleem and Bhavik
A refrigerator is a device operating in a
cycle that is designed to extract heat
from its interior so as to achieve or
maintain a lower temperature inside
The heat is exhausted to the surroundings
normally at a higher temperature
How the Refrigerator Works
A motor driving a compressor pump
provides the means by which a net
amount of work can be done by the
system for a cycle
Even though the cabinets are well
insulated, heat from the surroundings
leaks back inside
Heat Pumps
By Alex and Said
1. Any
device that can pump heat from a low
temperature reservoir to a high
temperature reservoir is called a heat
pump
2. Examples of heat pumps include
refrigerator and reverse cycle airconditioning devices used for space
heating and cooling
How..
1. A volatile
liquid called Freon is
circulated in a closed system of pipes
by the compressor pump
2. By the process of evaporative cooling,
the vaporised Freon is used to remove
the heat
3. The compressor maintains a high
pressure difference across a throttling
How..
Evaporation of the Freon occurs in
several loops called the evaporator
pipes that are usually the coldest part of
the fridge
As the liquid evaporates on the low
pressure, low temperature side, heat is
added to the system
How..
•
In order to turn from a liquid to a gas the
Freon requires thermal energy equal to
the latent heat of vaporization
•
This energy is obtained from the fridge
How..
•
•
•
On the high pressure, high temperature
side of the throttling valve, thermal
energy is removed from the system
The vaporized Freon in the compressor
pipes is compressed by the compressor
pump
It gives up its latent heat of vaporization
to the air surrounding the compressor
pipes
How..
•
The heat fins act as a heat sink to
radiate the thermal energy to the
surroundings at a faster rate
•
•
The fins are painted black and have a
relatively large surface area for their
size
A Typical Small Refrigerator
Schematic for a Refrigerator
Karim and Chilla
Hot reservoir at TH
QH
W
QL
Cold reservoir at TL
Refrigerator
•
•
•
The previous figure shows the energy
transfers that occurs in a refrigerator
cycle
By doing work on the system, heat QL is
added from the low temperature TL
reservoir
A greater amount of energy QH is
Reverse Cycle Heat Pump
WINTER
SUMMER
room
room
condenser
TH
evaporator
evaporator
TL
TL
condenser
TH
Winter
•
•
The evaporator heat exchanger is
outside the room
It exhausts its heat to the inside air
Summer
•
•
The evaporator heat exchanger is inside
the room
It exhausts its heat to the outside air
In Both Cases
•
•
Thermal energy is pumped from a lowtemperature reservoir to a hightemperature reservoir
The efficiency of refrigerators and heat
pumps is defined in a different manner
to heat engines
Coefficient of Performance, K
•
For a refrigerator and a heat pump used
for cooling
K = QL / W
• Ratio of heat absorbed from inside the
refrigerator and the work supplied to the
refrigerator in one cycle
Coefficient of Performance, K
•
For a heat pump used for heating
•
K = QH / W
•
Ratio of the total positive heat
exhausted to the inside and the work
supplied to the heat pump in one cycle
Carnot´s Theorem
By Martin and Mitul
•
•
•
No engine working between two heat
reservoirs can be more efficient than a
reversible engine between those
reservoirs
He argued that if thermal energy does
flow from a cold body to a hot body then
work must be done
Therefore no engine can be more
And...
•
•
•
•
All such engines have the same
efficiency
Therefore only a simple equation is
needed to calculate the efficiency
Consider an ideal perfectly insulated,
frictionless engine that can work
backwards as well as forwards
The p-V diagram would have the form...
The Carnot Engine
Explain
•
The net work is the area enclosed by
ABCDA
•
Thermal energy is absorbed by the
system at the high temperature TH and
is expelled at the low temperature
reservoir TL
and
•
Work is done by the system as it expands
along the top isotherm from A to B
• And also along the adiabat from B to C
• Work is then done on the system to compress
it along the bottom isotherm from C to D
• And also along the left adiabat from D to A
Carnot’s Efficiency
•
•
Carnot was able to obtain an expression
for the efficiency in terms of temperature
 = 1 - TL / TH
• Therefore the efficiency of the Carnot cycle
depends only on the absolute temperatures
of the high and low temperature reservoirs
• The greater the temperature difference the
greater the efficiency will be
•
•
•
•
•
But engine efficiency is
 = 1 - QL / QH
Therefore TC/TH = QC/QH
Or QC/TC = QH/TH
What would happen if the
temperature difference was not
maintained?
Real Life
•
In practice, friction and other
mechanical losses and thermal losses
reduce the overall efficiency
•
Nearly 2/3 of the heat generated by
modern power stations is released into
the environment
Thermal
Physics
Topic 10.3 Second Law of Thermodynamics and
Entropy