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Physics SF 016 Chapter 15Chapter 16
1
CHAPTER 15:Thermodynamics
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Physics SF 016 Chapter 15
15.1 Learning Outcome Remarks :
Keypoint :
Distinguish between
thermodynamic work done on
the system and work done by
the system. State and use first law of
thermodynamics,
W U Q
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Physics SF 016 Chapter 15
0Q3
15.1.1 Signs for heat, Q and work, W
Sign convention for heat, Q :
Q = positive value
Q = negative value
Heat flow into the system
Heat flow out of the system
3
Surroundings
(environment)
System
0W
(a)
Surroundings
(environment)
System
0W
(b)
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Physics SF 016 Chapter 15
0Q
4
Sign convention for work, W :
W = positive value
W = negative value
Work done by the system
Work done on the system
Surroundings
(environment)
System
0W
Surroundings
(environment)
System
0W
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Physics SF 016 Chapter 15
0Q
5
Surroundings
(environment)
System
0W
Surroundings
(environment)
System
0W
Q = positive value
Q = negative value
W
= positive value
W = negative value
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Physics SF 016 Chapter 15
6
Air
Compression
Air
Expansion
Air
Initially
Motion of piston
Motion of
piston
Work done by gas (Expansion)
When the air is expanded , the
molecule loses kinetic energy and
does positive work on piston.
Work done on gas(Compression)
When the air is compressed , themolecule gains kinetic energy and
does negative work on piston.
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Physics SF 016 Chapter 15
15.1.3 First law of thermodynamics
It states that : “The heat (Q) supplied to a system is equal to the increase in theinternal energy (U ) of the system plus the work done (W ) by the system on its
surroundings.”
W U Q
suppliedheatof quantity:Q
energyinternalinitial:1U
where
energyinternalfinal:2U
donework :W
and12 U U U
energyinternalin thechange:U
(15.2)
For infinitesimal change in the energy,
dW dU dQ
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Physics SF 016 Chapter 15
•The first law of thermodynamics is
a generalization of the principle of conservation of energy to include
energy transfer through heat as wellas mechanical work.
•The change in the internal energy
(U ) of a system during anythermodynamic process is
independent of path. For example athermodynamics system goes fromstate 1 to state 2 as shown in Figure
16.5.
2P
1V
3
4
1P
2V
Figure 15.4
2
1
P
V 0
23124121 U U U
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Physics SF 016 Chapter 15
Calculate:
a. the work done in the process ABC,
b. the change in the internal energyof the gas in the process ABC,
c. the work done in the process ADC,
d. the total amount of heattransferred in the process ADC.
a. The work done in the process ABC is
given by :
BCABABC W W W
but 0BC W
ABAABC V V PW
223
ABC 100.2100.410150 W
J3000ABC W
W = P.dV
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Physics SF 016 Chapter 15
c. The work done in the process ADC isgiven by
DCADADC W W W
but 0AD W
DCDADC V V PW
223
ADC100.2100.410300
W
J6000ADC W
Calculate:
a. the work done in the process ABC,
b. the change in the internal energyof the gas in the process ABC,
c. the work done in the process ADC,
d. the total amount of heattransferred in the process ADC.
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Physics SF 016 Chapter 15
d. By applying the 1st law of thermodynamics for ADC, thus
ADCADCADC W U Q
ADCABCADC W U Q
J1067.1 4
ADC Q
60001007.1 4
ADCQ
andABCADC U U
Calculate:
a. the work done in the process ABC,
b. the change in the internal energyof the gas in the process ABC,
c. the work done in the process ADC,
d. the total amount of heattransferred in the process ADC.
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Physics SF 016 Chapter 15
15
Thermodynamics processes (1 hours)
Remarks :
Keypoint :
Define the following thermodynamicsprocesses:
i) Isothermal, ΔU = 0
ii) Isovolumetric, W = 0iii) Isobaric, Δ P = 0 iv) Adiabatic, Q = 0
Sketch P V graph to distinguishbetween isothermal process andadiabatic process.
15.2 Learning Outcome
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Physics SF 016 Chapter 15
There are four specific kinds of thermodynamicprocesses. It is :
Isothermal process
Isovolumetric @ Isochoric process
Isobaric process
Adiabatic process
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Physics SF 016 Chapter 15
15.2.1 Isothermal process
is defined as a process that occurs atconstant temperature.
0
U
W Q W U Q
Thus,
Isothermal changesWhen a gas expands or compresses
isothermally (constant temperature)
thus
constantPV (16.3)
Equation (16.3) can be expressed as
If the gas expand isothermally, thus
V2>V1
If the gas compress isothermally, thus
V2<V1
2211 V PV P
W = positive
W = negative
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Physics SF 016 Chapter 15
is defined as a process that occurswithout heat transfer into or out of a
system i.e.
For example, the compression stroke in
an internal combustion engine is an
approximately adiabatic process.
W U U U 12
0Q
W U Q thus
Notes :
For Adiabatic expansion
(V 2>V 1), W = positive value
but U =negative value hence
the internal energy of thesystem decreases.
For Adiabatic compression
(V 2<V 1), W = negative value but U =positive value hencethe internal energy of the
system increases.
15.2.2 Adiabatic Process
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Physics SF 016 Chapter 15
15.2.3 Isovolumetric @
Isochoricis defined as a process that occurs atconstant volume i.e.
In an isochoric process, all the energyadded as heat remains in the systemas an increase in the internal energy thusthe temperature of the system increases.
For example, heating a gas in a closedconstant volume container is an isochoric
process.
W U Q
0W thus
12 U U U Q
15.2.4 Isobaric
is defined as a process that occurs atconstant pressure i.e.
For example, boiling water at constantpressure is an isobaric process.
W U Q
V PW
thus
V PU Q
0P and
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Physics SF 016 Chapter 15
15.2.4 Pressure-Volume diagram (graph) for thermodynamic processes
Figure 15.5 shows a P V diagram for eachthermodynamic process for a constant
amount of an ideal gas.
1V
2P
3P
3
V
P
V 0
1P
4T 3T
1T 2T
BD
2
V
E
1234 T T T T
A
Figure 16.8
C
Path AB
Isothermal process (T B=T A)
Path AC
Path AD
Path AE
Adiabatic process (T C<T A)
Isochoric process (T D<T A)
Isobaric process (T E>T A)
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Physics SF 016 Chapter 15
From the Figure 15.5,
For comparison between the
isothermal (AB) and adiabaticexpansions (AC):
The temperature fall (T C<T B) which
accompanies the adiabatic
expansion results in a lower final
pressure than that produced by the
isothermal expansion (P C<P B).
The area under the isothermal is
greater than that under the
adiabatic, i.e. more work is done by
the isothermal expansion than bythe adiabatic expansion.
The adiabatic through any point is
steeper than the isothermal through
that point.
2P
3P
P
V 0
1P
4T
3T
1T 2T
B
D
EA
C
Figure 15.5 shows a P V diagram for eachthermodynamic process for a constant
amount of an ideal gas.
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Physics SF 016 Chapter 15
Air is contained in a cylinder by a frictionless gas-tight piston.
a. Calculate the work done by the air as it expands from a
volume of 0.015 m3 to a volume of 0.027 m3 at a
constant pressure of 2.0 105 Pa.
b. Determine the final pressure of the air if it starts from thesame initial conditions as in (a) and expanding by the same
amount, the change occurs isothermally.
Example 3 :
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Physics SF 016 Chapter 15
Air is contained in a cylinder by africtionless gas-tight piston.
a. Calculate the work done by the air as
it expands from a volume of 0.015 m3 to
a volume of 0.027 m3 at a constant
pressure of 2.0 105 Pa.
b. Determine the final pressure of the air
if it starts from the same initialconditions as in (a) and expanding by the
same amount, the change occurs
isothermally
Example 3 : Solution :
a. Given
The work done by the air is:
Pa100.2
;m027.0;m015.0
5
1
3
2
3
1
P
V V
121 V V PW
015.0027.0100.2 5 W
J2400W
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Physics SF 016 Chapter 15
Example 3 :b. The final pressure for the isothermal
process is
2211 V PV P
027.0015.0100.2 2
5
PPa1011.1 5
2 P
Air is contained in a cylinder by africtionless gas-tight piston.
a. Calculate the work done by the air as
it expands from a volume of 0.015 m3 to
a volume of 0.027 m3 at a constant
pressure of 2.0 105 Pa.
b. Determine the final pressure of the air
if it starts from the same initialconditions as in (a) and expanding by the
same amount, the change occurs
isothermally
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Physics SF 016 Chapter 15
25
Remarks :
Keypoint :
• Derive expression for work, W =
• Determine work from the area under p-V graph.
• Derive the equation of work done in isothermal,
isovolumetric and isobaric processes.
• Calculate work done in :-
isothermal process and use
isobaric process, use
isovolumetric process, use
Thermodynamics work (4 hour)
2
1
1
2lnln
P
PnRT
V
V nRT W
12V V PPdV W
0 PdV W
pdV
15.3 Learning Outcome
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Physics SF 016 Chapter 15
15.3.1 Work done in the thermodynamics system
26
Gas
A
A
dx
Initial
Final
Figure 15.6
F
The work, dW done by the gas is
given by
In a finite change of volume fromV 1 to V 2,
PAF 0
cosFdxdW where and
PAdxdW and dV Adx PdV dW
2
1
V
V PdV W
2
1
V
V PdV dW
(15.3)
donework :W
where
pressuregas:P
gastheof volumeinitial:1
V gastheof volumefinal:2V
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Physics SF 016 Chapter 15
15.3.2 Work done in the thermodynamics system
27
1P
P
V 0
1 2
0121
V V PW
P
V 0
2P
1P
1V
1
2
0W
For a change in volume at constantpressure, P
12
V V PW
V PW Work done atconstant
pressure
For any process in the system which the
volume is constant (no change in volume),
the work done is
0W Work done at constantvolume
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Physics SF 016 Chapter 15
1V 2V
1P
2P
P
V 0
1
2
0W
Area under graph
= work done by gas
2V 1V
2P
1P
P
V 0
2
1
0W
Compression
Area under graph= work done on gas
Expansion
When a gas is expanded from V1 to V2
Work done by gas, 2
1
V
V W pdV
2
1ln
V
nRT V
2
1
1V
V W nRT dV
V
When a gas is compressed from V1=> V2
Work done on gas, '2
'1
V
V W pdV
'2
'1
1V
V W nRT dV
V
'
2
'
1
lnV
nRT V
Since V2< V1 the value of work done is (-)
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Physics SF 016 Chapter 15
From the equation of state for an ideal
gas,
Therefore the work done in the
isothermal process which change of volume from V 1 to V 2, is given
nRT PV V
nRT P then
2
1
V
V PdV W
2
1
V
V dV
V
nRT W
2
1
1V
V dV
V nRT W
1
2lnV
V nRT W
12 lnln V V nRT W
(15.9)
2
1ln
V
V V nRT W
15.3.3 Work done in Isothermal Process
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Physics SF 016 Chapter 15
For isothermal process, the temperature
of the system remains unchanged, thus
2211 V PV P 2
1
1
2
P
P
V
V
2
1ln P
PnRT W (15.10)
The equation (16.9) can be expressed as
By applying the 1st law of
Thermodynamics,thus
W U Q 0U and
W Q
2
1
1
2 lnlnP
PnRT
V
V nRT Q
15.3.3 Work done in Isothermal Process
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Physics SF 016 Chapter 15
15.3.3 Work done in isobaric process
The work done during the isobaricprocess which change of volume from V 1
to V 2 is given by
2
1
V
V PdV W
and constantP
2
1
V
V dV PW
12 V V PW
OR
V PW (15.10)
15.3.3 Work done in isovolumetric
process
Since the volume of the system in
isovolumetric process remains
unchanged, thus
Therefore the work done in the
isovolumetric process is
0dV
0 PdV W (15.11)
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Physics SF 016 Chapter 15
A quantity of ideal gas whose ratio of
molar heat capacities is 5/3 has a
temperature of 300 K, volume of 64
103 m3 and pressure of 243 kPa. It is
made to undergo the following three
changes in order:
1 : adiabatic compression to a volume
27 103 m3,
2 : isothermal expansion to 64 103 m3 ,
3 : a return to its original state.
Example 4 :
a. Describe the process 3.
b. Sketch and label a graph of pressure
against volume for the changes
described.
a. Process 3 is a process at constant
volume known as isovolumetric(isochoric).
b. The graph of gas pressure (P ) against
gas volume (V ) for the changes
described is shown in Figure 15.7.
3P
27
Pa)10( 4P
)m10( 33V 0
102
K533
K3003.24
64
1
2
3
Process 2
Process 3Process 1
Figure 15.7
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Physics SF 016 Chapter 15
Solution :
When the gas expands adiabatically, it does positive work.
Thus
The internal energy of the gas is reduced to provide the
necessary energy to do work. Since the internal energy is
proportional to the absolute temperature hence the
temperature decreases and resulting a temperature change.
W U Q W U
0Qand
A vessel of volume 8.00 103 m3 contains an ideal gas at a pressure of 1.14 105
Pa. A stopcock in the vessel is opened and the gas expands adiabatically, expellingsome of its original mass until its pressure is equal to that outside the vessel (1.01
105 Pa). The stopcock is then closed and the vessel is allowed to stand until the
temperature returns to its original value. In this equilibrium state, the pressure is
1.06 105 Pa. Explain why there was a temperature change as a result of the
adiabatic expansion?
Example 5 :
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Physics SF 016 Chapter 15
a. Write an expression representing
i. the 1st law of thermodynamics and state the meaning of all the
symbols.
ii. the work done by an ideal gas at variable pressure. [3 marks]
Example 6 :
Solution :
a. i. 1st law of thermodynamics:
ii. Work done at variable pressure:
W U Q
ferredheat transof quantity:Qenergyinternalinchange:U where
donework :W
1
2lnV
V nRT W
2
1
V
V
PdV W OR
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Physics SF 016 Chapter 15
Example 6 :
Solution :
b. Sketch a graph of pressure P versus volume V of 1 mole of ideal gas. Label and
show clearly the four thermodynamics process.
[5 marks]
b. PV diagram below represents four thermodynamic processes:
3T
1T
P
V
AP
0 AV
4T
2T B
E
DC
A
Isobaric process
Isochoric process
Isothermal process
adiabatic process
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Physics SF 016 Chapter 15
Example 6 :
Solution :
c. A monatomic ideal gas at pressure P and volume V is compressed isothermally
until its new pressure is 3P . The gas is then allowed to expand adiabatically until its
new volume is 9V . If P , V and for the gas is 1.2 105 Pa,1.0 102 m3 and 5/3
respectively, calculate
i. the work done on the gas during isothermal compression. [7 marks]
V
V nRT W 1ln
J1032.1 3W
V
V
PV W 3ln
PV nRT and
3
1ln100.1102.1
25W
i. The work done during the isothermal compression is
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Physics SF 016 Chapter 15
THE END…
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