The study of relationships
involving
heat,
mechanical work, and
other aspects of energy
and energy transfer
for
the system.
Thermodynamic system
is
any collection of objects that is
convenient to regard as a unit,
and that
may have the
potential energy to exchange
energy with its surroundings
.
CHAPTER 15:
Thermodynamics
(4 Hours)
15.1 First Law of Thermodynamics
15.2 Thermodynamics Processes
15.3 Thermodynamics Work
SUBTOPIC
2
Learning Outcome:
At the end of this chapter, students should be able to:
Distinguish
between work done on the system and work
done by the system.
State and use
first law of thermodynamics,
3
15.1
First law of thermodynamics
(1 hour)
•
Thermodynamics
is the
study of energy relationships that
involve heat, mechanical work, and other aspects of energy
and energy transfer
.
15.1 First Law of Thermodynamics
•
3 quantities
involved in a thermodynamic system :
•
The first law of thermodynamics
is the extension of the
principle of conservation of energy
to include both heat
and mechanical energy.
1.
Heat ,
Q
2.
Internal energy ,
Δ
U
3.
Work ,
W
4
Gas
A
A
dx
Initial
Final
•
When a gas expands, its pushes out on its boundary
surfaces as they move outward; an
expanding gas
always
does positive work
.
Work done by the system (+)
5
•
Figure above shows a gas in a cylinder with a moveable
piston.
•
Suppose that the cylinder has a cross sectional area,
A
and the pressure exerted by the gas (system) at the
piston face is
P
.
•
The force
F
exerted on the piston by the system is
F=PA.
•
When the piston moves out a small distance
d
x
, the work
d
W
done by the force is
dW
=
F
dx
=
PA
dx
but
A
dx
=
dV
dV
is the small change of volume of the system (gas)
•
The
work done
by
the system
is
dW
=
P
dV
P
, pressure is constant
6
Work done on the system (

)
d
x
•
Suppose that the cylinder has
a cross sectional area,
A
and
the pressure of the gas is
P
.
•
The external force
F
exerted on the system is
F
F=PA
•
The
magnitude of external
force
F
exerted on the
system equal to
PA
because the piston
is
always in equilibrium
between the external force
and the force from the gas.
7
•
When the piston moves in a small distance
dx
, the work
dW
done by the force is
dW
=

F
dx
=

PA
dx
negative sign
because
initial value is greater than
final value
but

A
dx
=

dV
•
The
work done
on
the system
is
dW
=

P
dV
P
, pressure is constant
8
•
In both cases if the volume of the gas changes from
V
1
to
V
2
, the work done is given by
Work done
by
the system
•
Gas expands
•
Volume increases
•
+
W
Work done
on
the system
•
Gas is compressed
•
Volume decreases
•

W
9
The First Law of Thermodynamics (flot)
“ The change in internal energy of a closed
system,
Δ
U
, will be equal to the
heat added
to
the system
minus
the
work done by
the system “
system
+Q

Q

W
+W
10
Rearrange
•
When
Q
is added to a system (gas) , the temperature of the
gas increases, thus causing the internal energy to increase
by an amount of
Δ
U
joule.
•
At the same time, when its temperature increases, its
volume increases too.
•
When the volume of the gas increases, work is done by the
gas (
W
).
Translation :
11
12
Example 15.1
A 2500 J heat is added to a system and 1800 J work is
done on the system. Calculate the change in internal
energy of the system.
Solution
13
Example 15.2
The work done to compress one mole of a monoatomic
ideal gas is 6200 J. The temperature of the gas changes
from 350 to 550 K.
a)
How much heat flows between the gas and its
surroundings ?
b)
Determine whether the heat flows into or out of the gas.
Solution
14
Solution 15.2
a)
15
Example 15.3
A gas in a cylinder expands from a volume of 0.400 m
3
to
0.700 m
3
. Heat is added just rapidly enough to keep the
pressure constant at 2.00 x 10
5
Pa during the expansion.
The total heat added is 1.40 x 10
5
J.
Calculate the work done by the gas and the change in
internal energy of the gas.
Solution
16
Exercise
1.
A system absorbs 200 J of heat as the internal energy
increases by 150 J. What work is done by the gas ?
50 J
2. In a chemical laboratory, a technician applies 340 J of
energy to a gas while the system surrounding the gas
does 140 J of work on the gas. What is the change in
internal energy ?
480 J
17
3.
8000 J of heat is removed from a refrigerator by a
compressor which has done 5000 J of work. What is the
change in internal energy of the gas in the system?

3000 J
Learning Outcome:
At the end of this chapter, students should be able to:
Define the following
thermodynamics processes:
Isothermal,
Δ
U
= 0
Isovolumetric,
W
= 0
Isobaric,
Δ
P
= 0
Adiabatic,
Q
= 0
Sketch
P
V
graph for all the thermodynamic processes.
18
15.2 Thermodynamics processes (1 hour)
15.2 Thermodynamics processes
1)
Isothermal
process
2)
Isochoric (isovolumetric)
process
3)
Isobaric
process
4)
Adiabatic
process
•
There are 4 common processes of thermodynamics:
(“iso” = same)
T V,P
19
1)
Isothermal
process
•
Isothermal
process is defined as
a process that occurs
at
constant temperature.
•
T
1
=
T
2
From flot,
PV
= constant
•
From the Boyle’s law :
20
•
Isochoric (isovolumetric)
process is defined as
a
process that occurs at
constant volume
.
•
V
1
=
V
2
From flot,
2)
Isochoric
(
isovolumetric
)
process
21
•
Adiabatic
process is defined as
a process that occurs
without the transfer of heat (into or out of the system)
.
•
From flot,
•
Isobaric
process is defined as
a process that occurs
at
constant pressure.
From flot,
3) Isobaric
process
4) Adiabatic
process
22
Pressure

Volume Diagram (graph) for
Thermodynamic Processes
Path A
B
Isothermal process (
T
B
=T
A
)
Path A
C
Path A
D
Path A
E
Adiabatic process (
T
C
<T
A
)
Isochoric process (
T
D
<T
A
)
Isobaric process (
T
E
>T
A
)
23
24
Exercise
1.
A gas system which undergoes an adiabatic process
does 5.0kJ of work against an external force. What is
the change in its internal energy?
5000 J
2.
A gas is compressed under constant pressure,
i)
Sketch the pressure
–
volume graph.
ii)
How is the work done in compressing the gas
calculated?
iii)
Explain what will happen to the final temperature of
the gas.
3.
A gas undergoes the following thermodynamics
processes: isobaric expansion, heated at constant
volume, compressed isothermally, and finally expands
adiabatically back its initial pressure and volume. Sketch
all the processes given on the same P

V graph.
Learning Outcome:
At the end of this chapter, students should be able to:
Derive
expression for work ,
Determine
work from the area under the 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
25
15.3
Thermodynamics work
(
2
hours)
26
Gas
A
A
dx
Initial
Final
Consider the infinitesimal work done by the gas (system) during
the small expansion,
dx
in a cylinder with a movable piston as
shown in Figure 15.3.
Suppose that the cylinder has a cross sectional area,
A
and the
pressure exerted by the gas (system) at the piston face is
P
.
Work done in the thermodynamics system
Figure 15.3
15.3 Thermodynamics work
27
The work,
dW
done by the gas is given by
In a finite change of volume from
V
1
to
V
2
,
where
and
and
where
(16.1)
PV
diagram
Work done = area under the
P

V
graph
isothermal
expansion
isothermal
compression
isobaric expansion
isochoric
28
1) Isothermal
From Boyle’s law :
Equation of work done in thermodynamic processes
29
2) Isochoric (isovolumetric)
3) Isobaric
Since the volume of the system in
isovolumetric
process
remains unchanged, thus
Therefore the work done in the
isovolumetric
process is
The work done during the isobaric process which change of
volume from
V
1
to
V
2
is given by
and
30
Work done at constant
volume
Work done at
constant
pressure
Example 15.4
How much work is done by an ideal gas in expanding
isothermally from an initial volume of 3.00 liters at 20.0
atm to a final volume of 24.0 liters?
Solution
V
1
= 3.00 liters,
V
2
= 24.0 liters ,
P
= 20.0 atm
31
Two liters of an ideal gas have a temperature of 300 K
and a pressure of 20.0 atm. The gas undergoes an
isobaric expansion while its temperature is increased to
500 K. What work is done by the gas ?
Example 15.5
Solution
T
1
= 300 K,
T
2
= 500 K ,
P
= 20.0 atm,
V
1
=2 liters
32
Example 15.6
(a) Write an expression representing
i. the 1
st
law of thermodynamics and state the meaning of
all the symbols.
ii. the work done by an ideal gas at variable pressure.
[3 marks]
(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]
(Exam.Ques.intake 2003/2004)
33
Solution 15.6
a)
i.
1st law of thermodynamics:
ii.
Work done at variable pressure:
where
or
34
b)
PV
diagram below represents four thermodynamic
processes:
35
Example 15.7
In a thermodynamic system, the changing of state for that
system shows by the
PV

diagram below.
In process AB, 150 J of heat is added to the system
and in process BD, 600 J of heat is added. Determine
a. the change in internal energy in process AB.
b. the change in internal energy in process ABD.
c. the total heat added in process ACD.
36
Q
AB
=
150 J,
Q
BD
=
600 J
, V
A
=V
B
=
2.0x10

3
m
3
,
V
C
=V
D
=
5.0x10

3
m
3
, P
A
=P
C
=
3x10
4
Pa
, P
B
=P
D
=
8x10
4
Pa
Solution 15.7
37
b. The work done in process
ABD is
Therefore, the change in internal energy :
The total heat transferred
in process ABD is given
by
Solution 15.7
Q
AB
=
150 J,
Q
BD
=
600 J
,
V
A
=V
B
=
2.0x10

3
m
3
,
V
C
=V
D
=
5.0x10

3
m
3
,
P
A
=P
C
=
3x10
4
Pa
,
P
B
=P
D
=
8x10
4
Pa
38
c.
The change in internal
energy in process ACD is
Solution 15.7
Q
AB
=
150 J,
Q
BD
=
600 J
,
V
A
=V
B
=
2.0x10

3
m
3
,
V
C
=V
D
=
5.0x10

3
m
3
,
P
A
=P
C
=
3x10
4
Pa
,
P
B
=P
D
=
8x10
4
Pa
The work done in
process ACD is given by
Therefore, the total
heat transferred :
39
Example 15.8
A gas in the cylinder of a diesel engine can undergo cyclic
processes. Figure below shows one cycle ABCDA that is
executed by an ideal gas in the engine mentioned.
a.
If the temperature of the gas in states A and B are 300 K
and 660 K, respectively. Calculate the temperature in
states C and D.
b. Determine the work done by the gas in process BC.
40
Solution 15.8
P
B
=P
C
=
16.0x10
5
Pa,
P
A
=
1.0x10
5
Pa
, P
D
=
7.8x10
5
Pa
,
V
A
=V
D
=
10.0x10

4
m
3
, V
B
=
1.40x10

4
m
3
, V
C
=
6.00x10

4
m
3
a. Given
T
A
=
300 K
and
T
B
=
660 K
41
Solution 15.8
P
B
=P
C
=
16.0x10
5
Pa,
P
A
=
1.0x10
5
Pa
,
P
D
=
7.8x10
5
Pa
, V
A
=V
D
=
10.0x10

4
m
3
, V
B
=
1.40x10

4
m
3
,
V
C
=
6.00x10

4
m
3
b. Process BC occurs at constant pressure, thus the work
done by the gas is given by
42
Example 15.9
(a) Define (i) the adiabatic compression process
(ii) the reversible process
[2 marks]
(b) One mole of an ideal monatomic gas is at the initial temperature of
650 K. The initial pressure and volume of the gas is
P
0
and
V
0
,
respectively. At initial stage, the gas undergoes isothermal
expansion and its volume increase to
2V
0
. Then, this gas through
the isochoric process and return to its initial pressure. Finally, the
gas undergoes isobaric compression so that it return to its initial
temperature, pressure and volume.
(i) Sketch the pressure against volume graph for the entire
process.
[4 marks]
(ii) By using the 1st law of thermodynamics, proved that the
heat,
where
n
is the number of moles,
R
is molar gas
constant and
T
is the absolute temperature. Then, calculate
the total heat for the entire process.
[8 marks]
(iii) State whether the heat is absorbed or released. [1 mark]
(Use
R
= 8.31 J K

1 mol

1)
(Exam.Ques.intake 2001/2002)
43
Solution 15.9
44
Solution 15.9
(b) One mole of an ideal monatomic gas is at the initial temperature of
650 K. The initial pressure and volume of the gas is
P
0
and
V
0
,
respectively. At initial stage, the gas undergoes isothermal
expansion and its volume increase to
2
V
0
. Then, this gas through
the isochoric process and return to its initial pressure. Finally, the
gas undergoes isobaric compression so that it return to its initial
temperature, pressure and volume.
45
(i)
Sketch the pressure against volume graph for the entire
process.
Solution 15.9
46
ii. From the
PV
diagram,
Solution 15.9
In process AB
(Isothermal process):
In process CA (isobaric
process):
Using Charles’s law, hence
47
The work done in
isobaric process:
Solution15.9
In process BC (isochoric process):
48
Solution 15.9
The total heat,
Q
for
entire process is given
by
….. proved
49
Example 15.10
(a)
State i. the isobaric process.
ii. the isothermal process.
iii. the adiabatic process.
[3 marks]
(b) State the 1
st
law of thermodynamics.
[2 marks]
50
Solution 15.10
b.
1
st
law of thermodynamics :
51
52
Exercise
Two moles of ideal gas are at a temperature of 300K and
pressure 2.5 x 10
5
Pa. The gas expands isothermally to
twice its initial volume, and then undergoes isobaric
compression to its initial volume.
i)
Calculate the initial volume of the gas.
ii)
What is the pressure of the gas after the gas
expands isothermally to twice its initial volume?
iii)
What is the final temperature of the gas after being
compressed isobarically?
iv)
Calculate the work done in the isothermal
expansion.
v)
Draw the P

V graph for the processes above.
0.02 m
3
,1.3 x 10
5
Pa, 150 K, 3.5 x 10
3
J
53
Good luck
For
1
st
semester examination
Enter the password to open this PDF file:
File name:

File size:

Title:

Author:

Subject:

Keywords:

Creation Date:

Modification Date:

Creator:

PDF Producer:

PDF Version:

Page Count:

Preparing document for printing…
0%
Comments 0
Log in to post a comment