CHAPTER 15: Thermodynamics (4 Hours)

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Oct 27, 2013 (3 years and 5 months ago)

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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