Download Work done by the system

The study of relationships
involving heat,
mechanical work, and
other aspects of energy
and energy transfer for
the system.
CHAPTER 15:
Thermodynamics
(4 Hours)
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.
SUBTOPIC
15.1 First Law of Thermodynamics
15.2 Thermodynamics Processes
15.3 Thermodynamics Work
2
Learning Outcome:
15.1 First law of thermodynamics (1 hour)
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,
Q  U  W
3
15.1 First Law of Thermodynamics
• Thermodynamics is the study of energy relationships that
involve heat, mechanical work, and other aspects of energy
and energy transfer.
• The first law of thermodynamics is the extension of the
principle of conservation of energy to include both heat
and mechanical energy.
• 3 quantities involved in a thermodynamic system :
1. Heat , Q
2. Internal energy , ΔU
3. Work , W
4
Work done by the system (+)
Initial
A
Gas
dx
Final

F
A
• When a gas expands, its pushes out on its boundary
surfaces as they move outward; an expanding gas always
does positive work.
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 dx, the work
dW 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 (-)
• Suppose that the cylinder has
a cross sectional area, A and
the pressure of the gas is P.
F
• The external force F
exerted on the system is
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.
dx
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 V1 to
V2 , the work done is given by dW  P dV
V2
 dW  
V1
PdV
W  PV2  V1 
W  PV
Work done by the system
• Gas expands
• Volume increases
•+W
W : work done
P : gas pressure
V1 : initial volume
V2 : final volume
W = - PV
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, ΔU, will be equal to the heat added to
the system minus the work done by the system “
U  Q  W
+ U : increase in internal energy
- U : decrease in internal energy
+Q
+ Q : heat is added to the system
- Q : heat is removed from the system
-W
+ W : work is done by the system
- W : work is done on the system
-Q
system
+W
10
U  Q  W
Rearrange
Q = U + W
Translation :
• 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 ΔU 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).
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
n  1, f  3, W  6200 J
T1  350 K , T2  550 K
a)
15
Example 15.3
A gas in a cylinder expands from a volume of 0.400 m3 to
0.700 m3. Heat is added just rapidly enough to keep the
pressure constant at 2.00 x 105 Pa during the expansion.
The total heat added is 1.40 x 105 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
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
17
Learning Outcome:
15.2 Thermodynamics processes (1 hour)
At the end of this chapter, students should be able to:

Define the following thermodynamics processes:

Isothermal, ΔU= 0

Isovolumetric, W = 0

Isobaric, ΔP = 0
Adiabatic, Q = 0
Sketch PV graph for all the thermodynamic processes.


18
15.2 Thermodynamics processes
• There are 4 common processes of thermodynamics:
1)
2)
3)
4)
Isothermal process
Isochoric (isovolumetric) process
Isobaric process
Adiabatic process
(“iso” = same)
T V,P
19
1) Isothermal process
• Isothermal process is defined as a process that occurs
at constant temperature.
•
U  0
U = U 2 - U1
T1 = T2
f
f
U = nRT2 - nRT1 = 0
2
2
From flot, U  Q  W
If U = 0 then
Q =W
• From the Boyle’s law :
PV = constant
P1V1  P2V2
20
2) Isochoric (isovolumetric) process
• Isochoric (isovolumetric) process is defined as a
process that occurs at constant volume.
•
W =0
W = P(V2 - V1 ) = 0
V1 = V2
From flot, U  Q  W
If W = 0 then
U = Q
21
3) Isobaric process
• Isobaric process is defined as a process that occurs
at constant pressure.
From flot, U  Q  W
U  Q  PV2  V1 
W  PV
4) Adiabatic process
• Adiabatic process is defined as a process that occurs
without the transfer of heat (into or out of the system).
•
Q=0
From flot, U  Q  W
If Q = 0 then
U = - W
22
Pressure-Volume Diagram (graph) for
Thermodynamic Processes
P
T4 > T3 > T2 > T1
A
PA
D
0
Path AB
Path AC
Path AD
Path AE
VA
E
B
C
T4
T3
T2
T1
V
Isothermal process (TB=TA)
Adiabatic process (TC<TA)
Isochoric process (TD<TA)
Isobaric process (TE>TA)
23
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.
24
Learning Outcome:
15.3 Thermodynamics work (2 hours)
At the end of this chapter, students should be able to:




Derive expression for work , W 
 PdV
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
V2
p
 nRT ln 1
V1
p2
isobaric process, use W   PdV  p(V2  V1 )
W  nRT ln


isovolumetric process, use W 
 PdV  0
25
15.3 Thermodynamics work
Work done in the thermodynamics system

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.
Initial
A
Gas
dx
Final


F
A
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.
26

The work, dW done by the gas is given by

In a finite change of volume from V1 to V2,
dW  Fdx cos where   0 and F  PA
dW  PAdx and Adx  dV
dW  PdV
V2
 dW  
V1
V2
PdV
W   PdV
(16.1)
V1
where
W : work done
P : gas pressure
V1 : initial volume of the gas
V2 : final volume of the gas
27
PV diagram
Work done = area under the P-V graph
P
P
isothermal
expansion
P1 1
P2
0
W 0
V1
P2
2
V2
isobaric expansion
1
2
W  P1 V2  V1   0
0
0
1
W 0
V2
V1
V
P
P
P1
2
P1
V
isothermal
compression
V1
V2
V
P2
2
isochoric
W 0
P1
1
0
V1
V 28
Equation of work done in thermodynamic processes
1) Isothermal
U  0
U = Q - W = 0
Q =W
nRT
Q =W = ∫
PdV = ∫
dV
V1
V1
V
V2
From Boyle’s law : P1V1  P2V2
W = nRT ln
V1
 V2   P1 
    
P1
W = nRT ln
 V1   P2 
P2
V2
V2
29
2) Isochoric (isovolumetric)

Since the volume of the system in isovolumetric process
remains unchanged, thus
dV  0

Therefore the work done in the isovolumetric process is
W   PdV  0
Work done at constant
volume
3) Isobaric
The work done during the isobaric process which change of
volume from V1 to V2 is given by
V2
W   PdV
V1
V2
and
P  constant
W  P  dV W = PV = P(V2 - V1 )
V1
Work done at
constant
pressure
30
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
V1 = 3.00 liters, V2 = 24.0 liters ,
P = 20.0 atm
31
Example 15.5
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 ?
Solution T1 = 300 K, T2 = 500 K , P = 20.0 atm, V1 =2 liters
32
Example 15.6
(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]
(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:
or
where
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.
Px104 / Pa
8.0 B
3.0
0
A
2.0
D
C
5.0
3
Vx10 / m
3
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
Solution 15.7
QAB= 150 J, QBD= 600 J, VA=VB= 2.0x10-3 m3,
VC=VD= 5.0x10-3 m3, PA=PC= 3x104 Pa, PB=PD= 8x104 Pa
37
4
Solution 15.7
Px10 / Pa
QAB= 150 J, QBD= 600 J, 8.0 B
VA=VB= 2.0x10-3 m3,
VC=VD= 5.0x10-3 m3,
PA=PC= 3x104 Pa,
PB=PD= 8x104 Pa
3.0
0
b. The work done in process
ABD is
A
2.0
D
C
5.0
Vx103 / m3
The total heat transferred
in process ABD is given
by
Therefore, the change in internal energy :
38
4
Solution 15.7
Px10 / Pa
QAB= 150 J, QBD= 600 J, 8.0 B
VA=VB= 2.0x10-3 m3,
VC=VD= 5.0x10-3 m3,
PA=PC= 3x104 Pa,
PB=PD= 8x104 Pa
3.0
0
c. The change in internal
energy in process ACD is
The work done in
process ACD is given by
A
2.0
D
C
5.0
Vx103 / m
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.
P / x10 5 Pa
B
16.0
C
D
7.8
A
1.0
0
1.40
6.00
10.00
V / x10 4 m 3
40
Solution 15.8
PB=PC= 16.0x105 Pa, PA= 1.0x105 Pa, PD= 7.8x105 Pa,
VA=VD= 10.0x10-4 m3, VB= 1.40x10-4 m3, VC= 6.00x10-4 m3
a. Given TA= 300 K and TB= 660 K
P / x10 5 Pa
B
16.0
C
D
7.8
A
1.0
0
1.40
6.00
10.00
V / x10414 m 3
Solution 15.8
PB=PC= 16.0x105 Pa, PA= 1.0x105 Pa,
PD= 7.8x105 Pa, VA=VD= 10.0x10-4 m3, VB= 1.40x10-4 m3,
VC= 6.00x10-4 m3
5
P / x10 Pa
B
16.0
C
D
7.8
A
1.0
0
1.40
6.00
10.00
V / x10 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 P0 and V0,
respectively. At initial stage, the gas undergoes isothermal
expansion and its volume increase to 2V0. 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,
Q  nRT0 ln 2  P0V0 
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)
43
(Exam.Ques.intake 2001/2002)
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 P0 and V0,
respectively. At initial stage, the gas undergoes isothermal
expansion and its volume increase to 2V0. 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
Solution 15.9
(i) Sketch the pressure against volume graph for the entire
process.
Pressure, P
P0
A
C
T1
B
P1
0
T0
V0
2V0
Volume, V
46
Solution 15.9
ii. From the PV diagram,
In process AB
(Isothermal process):
Pressure, P
P0
A
C
T1
B
P1
0
T0
V0
2V0
Volume, V
In process CA (isobaric
process):
Using Charles’s law, hence
47
Solution15.9
The work done in
isobaric process:
Pressure, P
P0
A
C
T1
B
P1
0
T0
V0
2V0
Volume, V
In process BC (isochoric process):
48
Solution 15.9
Pressure, P
A
P0
The total heat, Q for
entire process is given
by
C
T1
B
P1
0
T0
V0
2V0
Volume, V
….. proved
49
Example 15.10
(a) State
i. the isobaric process.
ii. the isothermal process.
iii. the adiabatic process.
(b) State the 1st law of thermodynamics.
[3 marks]
[2 marks]
50
Solution 15.10
b. 1st law of thermodynamics :
51
Exercise
Two moles of ideal gas are at a temperature of 300K and
pressure 2.5 x 105 Pa. The gas expands isothermally to
twice its initial volume, and then undergoes isobaric
compression to its initial volume.
i)
ii)
Calculate the initial volume of the gas.
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 m3,1.3 x 105 Pa, 150 K, 3.5 x 103 J
52
Good luck
For
1st semester examination
53