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WEEK 6
THERMODYNAMICS
PHYSICS 3
FIS 0334
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Chapter 1
Thermodynamic system and its surrounding
Thermodynamic process
The first law of thermodynamics
Work done during volume changes P-V diagram
Kinds of thermodynamic process
Adiabatic, Isochoric, Isobaric and Isothermal
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Learning Outcomes
By the end of this chapter, students will be able to do the following:
Describe thermodynamics process by involving the change of state
of a thermodynamic system.
Use the first law of thermodynamics to relate heat transfer, work
done and internal energy change of a thermodynamics system withits surroundings.
Calculate the work done by a thermodynamics system when its
volume changes.
Distinguish among adiabatic, isochoric, isobaric and isothermal
processes. Sketch and analyze the P-V diagram for a thermodynamics process.
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IntroductionExample 1
Thermodynamics of Popping Popcorn
A quantity of popcorn kernels in a
pot with a lid is initially placed on a
stove.
Heat energy is added to the
popcorn by conduction.
As the popcorn pops and expands,
it does work on its surroundings
(by exerting an upward force on thelid and moves through a
displacement).
FsW
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IntroductionExample 1
Thermodynamics of Popping Popcorn
The stateof popcorn changes in
this process, since the volume,temperature and pressure of
popcorn all change as it pops.
Such process involves the change
in state of a thermodynamic
system (Popcorn)is calledthermodynamic process.
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IntroductionExample 2
Thermodynamics of Gasoline Car Engine
If you put a tiny amount of high energy
fuel (i.e. gasoline) in a small and
enclosed space and ignite it.
Turn to create an incredible amount of
energy released in the form of
expanding gas (heated gas).
The heated gas pushes on the piston
within cylinder, doing mechanical work
that used to propel the car in distance.
Here, the thermodynamic system is the
quantity of gas enclosed in a cylinder
with a movable piston.
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IntroductionExample 2
Principle behind Gasoline Car Engine
Four-Stroke Combustion Cycle (Intake, Compression, Combustion and Exhaust stroke)
* Get the online animation to enhance your understanding
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IntroductionExample 2
Step 1: Intake Stroke
The piston starts at top, when the intake valve open, the piston moves down to let the
engine take in a cylinder-full of air and gasoline.
Step 2: Compression Stroke
The piston moves back up to compress the mixture of air/gasoline mixture. As
compression makes the explosion more powerful.
Step 3: Combustion Stroke
When the piston reaches top, the spark plug emit a spark to ignite the gasoline. The
gasoline in the cylinder explodes, driving the piston down.
Step 4: Exhaust Stroke
Once the piston hits the bottom surface, exhaust valve opens and the exhaust leaves
the cylinder.
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Introduction
THERMODYNAMICS has its roots in many practical problems
essentially in the gasoline engine in an automobile which use the
heat of combustion of their fuel to perform mechanical work in
propelling the vehicles.
Thus, we able to describe the energy relationships in any
thermodynamic processin term of the quantity of heat Qadded to
the systemand the work Wdone by the system which
consequently resulting the change of internal energy Uof system.
Internal Energy, U is the total energy of all of molecules in the
system ( due to their individual random motions) excluding for the
kinetic and potential energy due to external force.
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The change in internal energy of a closed system will be equal
to the heat energy added to the system minus the work done by
the system on its surroundings.
** Be careful with the sign of heat energy, Q and work done, W
The First Law of Thermodynamics
WQU
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The First Law of Thermodynamics
WQU
Quantity Sign Situation
U+ Increasein internal energy
- Decreasein internal energy
Q+ Heat flows in tothe system
- Heat flows out of the system
W+ Work done bythe system (Expanding gas)
- Work done onthe system (Compression gas)
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The First Law of ThermodynamicsExample 1
Q = 150J W = 100J
U = Q W = + 50J
More heat is added to
system than systemdoes work.
Internal energy of
system increases.
system
surroundings
system
surroundings
Q = -150J W = -100J
U = Q W = - 50J
More heat flows out of
system than work isdone on system.
Internal energy of
system decreases.
U = Q W = 0J
Heat added to system
equals work done bysystem.
Internal energy of
system unchanged.
system
surroundings
Q = 150J W = 150J
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The First Law of ThermodynamicsExample 2
Your body is a thermodynamic system
When you exercise, your body does work. W > 0
(Since the work is done by your body)
Your body also warms up during exercise; and by perspiration and
other means the body releases the heat. Q < 0 Since Q is negative and W is positive,
The bodys internal energy decreases.
Thats why exercise helps you to lose weight since some of the
internal energy stored in your body in form of fat is used up.
0 WQU
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Question 1
Rank the following thermodynamic processes according to the
change in internal energy in each process, from most positive to
most negative.
(i) As you do 250J of work on a system, it transfers 250J of heat to
its surroundings;
(ii) as you do 250J of work on a system, it absorbs 250J of heat from
its surroundings;
(iii) as a system does 250J of work on you, it transfers 250J of heat
to its surroundings;
(iv) as a system does 250J of work on you, it absorbs 250J of heatfrom its surroundings.
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Work Done in Volume Changes
Motion of piston Figure shows a system whose volume canchange in a cylinder with a movable piston.
The cylinder has cross-sectional areaA
and the pressure exerted by system at
piston is P. The total force exerted bysystem on piston is F = PA.
When the piston moves out an infinitesimal
distance , the work done by this force isd
PdVdW
PAdFddW
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P-V diagram (Pressure vs. volume)
If the gas expands / compressed from volume VAto VB, the work
done by the gas is equal the area under the curve;
Work doneby gas
Work done
on gas
VB
PB
PA
VA A
B
V
V
V
V
V
VnRTW
V
dVnRT
PdVW
B
A
B
A
ln
B
A
V
V
PdVAreaWork 0 B
A
V
V
PdVAreaWork 0
A
B
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P-V diagram (Pressure vs. volume)
If the volume is constant (does not change), VA= VB, hence no work
is done, W = 0.
For a gas that expands / compressed with constant pressure,
Work done
by gasWork done
on gas
)(AB
VVPVPW
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Last Updated:5 November 2014 LMS SEGi education group 18
A process in which the temperatureof the system remains constantis
called an isothermal process. On a p-V diagram, a path representing a
constant temperature process is called an isotherm.
Internal energy U depends only on temperature, so if the temperature is
constant, the internal energy is also constant;
According to Boyles law , PV = constant.
If the gas is expanded, heat flows intothe gas from the water.
If the gas is compressed, heat flows outof the gas into the water.
The First Law of Thermodynamics Applied:
Isothermal process (T=0)
QW
U
0
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The First Law of Thermodynamics Applied:
Isothermal process (T=0)
PV diagram for an ideal gasundergoing isothermal
processes at two different
temperatures.
The ideal gas in the cylinder
is expanding isothermally at
temperature T.
Isotherms
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The First Law of Thermodynamics Applied:
Adiabatic process (Q=0)
Defined as one with no heat transferred into or out of a system.
System expands adiabat ical ly, W > 0, U < 0.
System is com pressed adiabat ical ly, W < 0 , U > 0.
To do work , the gas uses up some of its internal energy as heat
is not allow to enter the gas. When its internal energy decrease,
its temperature decreases too.
WU
Q
0
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The First Law of Thermodynamics Applied:
Adiabatic process (Q=0)
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The First Law of Thermodynamics Applied:
Isochoric process (V=0)
QU
WQU
W
0
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The First Law of Thermodynamics Applied:
Isobaric process (P=0)
An isobaric process occurs at constant pressure.
The pressure P experience by the substance is always the same
and is determined by the external atmosphere and the weight of
the piston and the block resting on it.
Heating the substance makes it expands and do work W in lifting
the piston and block through the displacement s.
The expression of the work at constant pressure;
)( if VVPVPW
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The First Law of Thermodynamics Applied:
Isobaric process (P=0)
Positive W value for the work done by
the gas when it expands.Negative W value for the work done
on the gas to compress it.
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Describe the changes when an ideal gas changes from state WX Y Z
and back to W.
The First Law of Thermodynamics Applied
Net change in heat
Qnet= Q WX+ Q XY- Q YZ- Q ZW
Net change in work
Wnet= W WX- W YZ
Net change in internal energy
Unet
=0
(For a cyclic process, that eventually return a system
to its initial state, Uf= Ui; Unet=0)
Qnet= Wnet= shaded area
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The First Law of Thermodynamics Applied
Change of
state
Equation Explanation
WX U= Q - W Expansion at constant pressure, temperature increases.
Work done, heat absorbed.
XY U= Q Pressure increase at constant volume. Temperature increase.
Hence U>0 , Q>0. Heat absorbed.
YZ U= Q - W Compression at constant pressure, temperature decreases.
Work done on the gas, heat lost from the gas.
ZW U= Q Pressure decrease at constant volume, temperature decreases.
Hence U
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Describe the changes when an ideal gas changes from state AD B and
back to A.
The First Law of Thermodynamics Applied
Net change in heat
Qnet= Q DB- QAD- Q BA
Net change in internal energy
Unet= 0
(For a cyclic process, that eventually return a system
to its initial state, Uf= Ui; Unet=0)
Net work = shaded area
Isobaric
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The First Law of Thermodynamics Applied
Change of
state
Equation Explanation
AD W=0
U= Q - W
U= Q
As temperature decreases, the internal energy decreases too
(U0 and W>0.
Temperature increases. Work is done by the gas, heat is absorbed
by the gas.
BA U= Q - W
U=0, T=0
Isothermal compression at constant temperature.
Since W
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The following is a simple summary of the various thermodynamic
processes.
The First Law of Thermodynamics Applied
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The First Law of Thermodynamics Applied
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References
University Physics, H.D. Young and R.A. Freedman
College Physics,A. Giambattista, B.M. Richardson and R.C.
Richardson
SerwaysEssentias of College Physics, R.A. Serway and C. Vuille