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Conducted by
Efa Mai Inanignsih
Objective
 to be able to formulate work, heat and internal
energy based on the principal law of
thermodynamics and apply it in problem solving
 to be able to apply general equation of ideal gases
in isothermal, isochoric and isobaric processes
 to be able to analyze ideal gases process based on
pressure against volume graph
 to be able to describe the principle of Carnot
engine
The phenomenon
What do you think of
Balloon explosion??
This is it!!
If a balloon is inflated and
kept in a hot place, it will
finally explode, because the
particles of gas in the
balloon continue to expand
and press the balloon wall so
that when the wall is unable
to stand the gas pressure,
the balloon will
explode
Introduction
 Thermodynamics discuss the relationship between
heat and mechanical work

also the basic knowledge about temperature and heat, the
influence of temperature and heat to the characteristics of
substances, and kinetic theory of gases
 System and Environment
is the
everything
becoming our
observation
object
is the
everything
which is
outside the
system
External Work
The figure is a tube
containing ideal gas
closed with a piston
p
∆s
gas
piston
F
tube
If the gas in is heated at a
constant pressure, then
the gas will expand and
push the piston with force
F so that the piston will
shift as far as s
External Work (continued)
W = F.s
Because F = pA
Then
W = p.A.s
Where
W = mechanical work of gas (J)
p = gas pressure
A = section area of tube (m2)
s = displacement of piston (m)
Because A s = V, then the work equation above becomes
W = p.V
Where V is change of volume (m3)
Work done by the gas is directly proportional to pressure (p)
and change of volume (V) of the gas.
Sample Problem
 A gas is compressed
so that the volume
decreases from 5.0 L
to 3.5 L at a constant
pressure of 1.0 x 105
Pa. Calculate the
external work
applied to the gas!
Solution
V1 = 5.0 L
V2 = 3.5 L
p = 1.0 x 105 Pa
because the gas is compressed, then
W
= p V
= p (V2 – V1)
= 1.0 x 105 Pa (3.5 L – 0.5 L)
= 1.0 x 105 Pa (-1.5 x 10-3 m3)
= -150 J
Thus, the external work applied to
the gas is -150 J
Exercises
 A number of ideal gases are heated at a constant
pressure of 2 x 105 N/m2 so that the volume changes
from 20 liters to 30 liters. Calculate the external work
done by the gas during expansion!
 A gas is compressed at constant pressure 2.00 x 105 Pa
from a volume of 2.00 m3 to a volume of 0.500 m3.
What is the work done to the gas? If the temperature
initially was 40oC, what is the final temperature of the
gas?
Thermodynamics Processes
 Isothermal Process
 Isobaric Process
 Isochoric Process
 Adiabatic Process
Isothermal Process
 Isothermal process is the change process
of gas state at constant temperature
 From the ideal gas state equation, pV =
nRT
obtained pV = constant, because nRT has
a constant value.
So
pV = constant
then
p1V1 = p2V2
Where
p1 = initial pressure
V1 = initial volume
p2 = final pressure
V2 = final volume
complies with
the Boyle’s law
p
V
External work in Isothermal Process
 The external work done by gas in the isothermal
process can be determined from the equation pV =
nRT and W = p V.
W  pΔV
nRT
W 
ΔV
V
ΔV
W  nRT
V
W  nRT ln V 
V2
V1
p
p1
1
W  nRT ln V2 - ln V1 
V
W  nRT ln 2
V1
2
p2
V1
V2
V
Isobaric Process
 Isobaric process is the change process of gas state at
constant pressure
pV  nRT
V
nR

T
p
because p is constant and nR is always constant, then
V1
V1 V2
 constant or

T1
T1 T2
the Gay-Lussac’s law
Isobaric Process(continued)
 Graph of isobaric
process
p
V
External work in Isobaric Process
 The external work done by gas in isobaric process can
be determined by equation
W = p V = p (V1 – V2)
it can also be determined by the area under the p – V
graph (shaded region)
p
p1 = p2
1
2
V1
V1
V
Isochoric Process
 Isochoric process is the change
process of gas state at constant
volume
pV  nRT
p
nR

T
V
 because V is constant and nR is
always constant, then
p
p
p1 p2
 constant or

T
T1 T2
 This process complies the Gay-
Lussac’s law.
V
graph of isochoric process
External work in Isochoric Process
 The external work done by gas in isochoric
process:
W  p ΔV
W 0
 In other words, in isochoric process
The area under the p – V
graph for isochoric
process is zero
p
(constant volume), the gas does not do
any external work. So the area under the p p
2
– V graph only forms a point.
p1
V1 = V2
V
Adiabatic Process
 Adiabatic process is the process change of a gas state which
does not experience any transfer of heat or there is no heat
entering or coming out of the system (gas)
 This process complies with the Poisson’s formula


p V  constant or p1V1  p 2 V2
Where
 = Laplace constant = Cp/Cv
Cp = specific heat of gas at constant pressure
CV = specific heat of gas at constant volume

Adiabatic Process(contd.)
 The equation above can also be expressed in another
equation as follows
γ
p1V1  p2V2
γ
 nRT1  γ  nRT2  γ

 V1  
 V2
 V1 
 V2 
T1V1
γ-1
 T2V2
Curvature of p – V adiabatic
graph is steeper than
isothermal
p
adiabatic
γ-1
isothermal
V
External work in Adiabatic Process
p
 The external work done by gas in
p1
adiabatic process is expressed as
follows.
1
W
(p1V1 - p2V2 )
γ -1
3
or W  nR (T1 - T2 )
2
p2
V1
V2
V
The external work done
by gas is equal to the
area of shaded region
under the p – V graph
Sample Problem
 Two moles of gas
is compressed at
a constant
temperature of 23oC so that its
volume becomes
half of the initial.
Calculate the
external work
done by the gas!
(R = 8.31 J/mol K,
ln 1 = 0, ln 2 =
0.69)
Solution
Given
n = 2 mol
R = 8.31 J/mol K
T = (-23 + 273) K = 250 K
V2 = ½ V1
so V2/V1 = 1/2
The external work done in isothermal process
V
W
=
nRT ln 2
V1
= (2 mol) (8.31 J/mol K) (250 K) (ln )
= 4155 J (ln 1 – ln 2)
= 4155 J (0 – 0.69)
= -2866.95 J
Thus, the external work done is -2866.95 J
(-) sign indicates that at the gas is applied work.
Sample Problem
 A monoatomic ideal
gas ( = 5/3) is
compressed
adiabatically and the
volume decreases to its
half. Determine the
ratio of the final
pressure to the initial
pressure!
Solution
Because  = 5/3 and V2/V1 = ½
then
p1V1 = p2V2
p2 /p1 = V1 /V2 = 2 (5/3)
3
=2 4
Thus, the ratio of the final to the
initial pressure is 23 4
Exercise
 A gas occupying a room of 40 cm3 is heated at a
constant pressure so that the volume becomes twice
the initial. The gas pressure is 105 Pa. Calculate the
external work done by the gas!
 Two moles of ideal gas initially has a temperature of
27oC, a volume V1 and pressure p1 = 6.0 atm. The gas
expands isothermally to V2 volume and pressure p2 =
3.0 atm. Calculate the external work done by the gas!
(R = 8.31 J/mole K)
The First Law of Thermodynamic
 If an external work is applied on a system, then the
temperature of the system will increase. This happens
because the system receives energy from the
environment. This increase of temperature relates to
the increase of the internal energy.
 In adiabatic process, the external work applied on
the ideal gas will be equal to the change of internal
energy of the ideal gas.
Where
W  ΔU or W - ΔU  0
W = external work (J)
U = the change of internal energy (J)
The First Law
 in non-adiabatic process, the gas will not only receive
external work but also heat.
Q  ΔU  W or ΔU  Q - W
Where Q = heat (J)
This is it!!.....The first law of thermodynamics
That states
“Though heat energy has turned into the change of internal
energy and external work, the amount of all energy is
always constant”.
The First Law
 If a system receives (absorbs) heat
from the environment
Q = U + (+W) = U + W
 If a system receives heat from the
environment
Q = U + (-W) = U – W
environment
+
system
+
W
The rules of W and
Q values
W
Q
-
Q
The First Law in isothermal process
 In isothermal process (constant temperature), the change
of internal energy U = 0, because the change of
temperature T = 0. So that the first law of
thermodynamics becomes
ΔU  Q - W
0  Q -W
V2
Q  W  nRT ln
V1
The First Law in isochoric process
 In isochoric process (constant volume), the work
applied by gas W = 0 because the change of volume
U = 0. So that the first law of thermodynamics
becomes
ΔU  Q - W
ΔU  Q - 0
ΔU  Q
The First Law in isobaric process
 In isobaric process (constant pressure), the work done
by gas W = p V = p (V2 – V1). So, the first law of
thermodynamics becomes
ΔU  Q - W
ΔU  Q - p (V2 - V1 )
The First Law in adiabatic process
 In adiabatic process, the system does not receive heat
or release heat, so Q = 0. Therefore, the first law of
thermodynamics becomes
ΔU  Q -W
ΔU  0 - W
ΔU  - W
Heat Capacity
 Heat capacity of gas is the amount of heat energy
needed to increase gas temperature by one Kelvin (1 K)
or one degree Celsius (1oC).
Q
C
ΔT
Where
C = heat capacity (J/K)
Q = absorbed heat (J)
T = the change of temperature (K)
QV
CV 
or QV  CV ΔT
ΔT
Cp 
Qp
ΔT
or Q p  C p ΔT
for isochoric process
for isobaric process
Derivation
 The first law of thermodynamics can be derived as
Q  ΔU  W
Q
ΔU
W


ΔT
ΔT
ΔT
ΔU
W
C 

ΔT
ΔT
 For isobaric process (p = constant)
W  p ΔV  nRΔR
3
3
U  nRT  ΔU  nRΔR
2
2
Monatomic gas:
3
nRΔR
nRΔR
2
Cp 

ΔT
ΔT
3
C p  nR  nR
2
5
C p  nR
2
Diatomic gas:
7
C p  nR
2
 For isochoric process (V = constant).
Monatomic gas:
 0
3
nRΔR
0
2
CV 

ΔT
ΔT
3
CV 
nR
2
 Relationship between Cp and CV :
W
Diatomic gas:
CV 
5
nR
2
7
5
C p - CV  nR - nR
2
2
C p - CV  nR
Other parameters
 Molar heat capacity of gas
CV,m
QV

or QV  n CV ,m ΔT
n ΔT
C p,m 
Qp
n ΔT
or Q p  n C p,m ΔT
Q
Cm 
n ΔT
isochoric process
isobaric process
 Specific heat of gas
Q
C
m ΔT
QV
CV 
or Q V  mC V T
m T
3 R
CV 
2M
5 R
CV 
2M
Cp 
Qp
m ΔT
for isochoric process
for monatomic gas
5 R
Cp 
2M
for diatomic gas
7 R
Cp 
2M
or Q p  mC p ΔT
for isobaric process
Sample Problem
 One mole of gas is
compressed at a
constant temperature
of -23oC so that its
volume decreases to
half of its initial
volume. Calculate the
work done by the gas!
(R = 8.31 J/mole K; ln 1
= 0, ln 2 = 0.69)
Solution
Because
n = 1 mol
T = (-23 + 273) K = 250 K
V2 = ½ V1
R = 8.31 J/mol K
Then
W
= nRT ln V2/V1
= 1 x 8.31 x 250 ln ½
= -1.4 x 103 joule
Thus, the work done by the gas is
-1.4 x 103 joule.
Exercise
 Two moles of ideal gas initially has temperature of 27oC,
volume V1 and pressure p1 = 6.0 atm. The gas expands in
isothermic process and reaches volume of V2 and
pressure p2 = 3.0 atm. Calculate the external work done
by the gas! (R = 8.3 J/mole K) (Answer : 11.5 J)
 2.5 m3 of neon gas with temperature 52oC is heated in
isobaric process to 91oC. If the pressure of the gas is
4.0 x 105 N/m2, determine the work done by the gas!
 56 x 10-3 kg of nitrogen is heated from -3oC to 27oC. If it is heated in
a free expanding vessel, then required heat of 2.33 kJ. If the
nitrogen is heated in a stiff vessel (cannot expand), then the heat
required is 1.66 kJ. If the relative mass of nitrogen molecules is 28
g/mole, calculated (a) the heat capacity of nitrogen, (b) the
general gas constant!
Thermodynamic
Cycle
 Cycle means the process which
runs from the initial state and
returns to that initial state after
gas does work
p
a
c
b
V1
V2
V
Random cycle in p – V diagram
 In the process a – b, the gas expands in adiabatic process and the
work done by the gas is the area of plane abV2V1, its value is negative.
In process b – c the compressed gas in isothermal process is the area
of plane bcV1V2, its value is positive. In process c – a the gas does not
do any work because its volume is constant. The process c – a is an
isochoric process which is done to make the gas returns to its initial
state.
 The total external work done by the gas in one cycle
a – b – c – a is the area of abca.
Wabca
Wabca
=Wab + Wbc + Wca
area of abV2V1 + (-area of bcV1V2) + 0
=area of abca
 A thermodynamics cycle can occurs in a heat engine,
such as otto engine (Otto cycle), diesel engine (diesel
cycle), steam engine (Rankine cycle), and carnot
engine
Thermodynamic Cycle
 Carnot engine is assumed as an
High temperature reservoir T1
ideal heat engine which works
cyclically and reversible
between two temperatures
without any loss of energy.
Q1
Heat engine
Q1
Low temperature reservoir T2
: Carnot engine
W = Q1 - Q2
Within one cycle, the gas returns to
its initial state, so there is no change
of internal energy (U = 0).
Q = U + W
Q1 – Q2 = 0 + W
W = Q1 – Q2
imaginary Carnot engine
Carnot cycle
Work process of Carnot
engine to produce Carnot
cycle
Entire of process in the
Carnot can be represented
in pressure (P) against
volume (V) graph:
Efficiency
 Efficiency of engine
W
η
 100%
Q1
 In the Carnot engine, holds W = Q1 – Q2
Q1 - Q2
η
 100%
Q1
 Q2 
η  1    100%
 Q1 
 T2 
  1   100%
 T1 
Sample Problem
 The figure below
indicates the
thermodynamics change
of system from iitial
state A to B and C and
back to A. If VA = 0, VB
= 30 joule and heat
given to the system in
process B  C = 50 J
Pressure (N/m 3 )
p
C
90
60
30
O
A
D
Volume (m 3 )
B
E
V
Determine
a. the system internal energy
in state C,
b. the heat given to the
system in A  B process,
and
c. the heat given to the
system or taken from in C
 A process.
Solution
a. The heat which is given away to the system in B  C
process, is QBC = +50 joule.
QBC = +50 joule
WBC = 0, because B  C process is isochoric
Use the first law of thermodynamics
QBC = UBC + WBC
UBC = QBC – WBC
UBC = 50 – 0 = 50 J
UBC = UC – UB
UC = UBC + UB
UC = 50 + 30 = 80 J
Thus, the system internal energy in state C is 80 J.
b. Process from A  B
WAB = area of ABED
= AB x BE
= 2 x 30 = 60 J
UAB = UB – UA
= 30 – 0 = 30 J
QAB is calculated by using the first law of thermodynamics
QAB = UAB + WAB
QAB = 30 + 60 = 90 J
Thus, the heat given in A  B process is 90 J.
c. Process from C  A
WCA = -area of ACED
= -(area of ABED + area ABC)
2  60 
AB  BC 


  60 

60




2
2 



= -120 J
UCA = UA – UC = 0 – 80 = 80 J
QCA is calculated by using the first law of
thermodynamics.
QCA = UCA + WCA = -80 + (-120)
= -200 J
Thus, the energy taken in C  A process is -200 J.
The Second Law of Thermodynamics
 is a restriction of the first law of
thermodynamics which expresses the energy
conservation
 It is states: “energy cannot be created or
destroyed but can only change from one form
to another”
 Rudolf Clausius: Heat flows spontaneously from
an object of high temperature to an object of lower
temperature and it does not flow spontaneously in
the opposite direction without external work.
Entropy
 The total entropy of the universe does not change
when a reversible process occurs (Suniverse = 0) and
increase when the irreversible process occurs (Suniverse
> 0).
 entropy is a measurement of the amount of energy
or heat which cannot changed into work
Q
ΔS 
with S = the change of entropy (J/K)
T
 the total change of entropy of the Carnot engine is
ΔS 2  ΔS 1 
Q2 Q1
0
T2 T1
 Kelvin and Planck formulate the second law of
thermodynamics about heat engine that it is
impossible to make an engine with 100% efficiency
 Principle of cooler engines: is flowing heat from the
cool reservoir T2 to the hot reservoir T1 by exerting
external effort on the system.
 The magnitude of external work needed in a cooler
engine is formulated as
W  Q1 - Q2
Where :
Q1 = heat absorbed from low temperature
Q2 = heat given at high temperature
Sample Problem
 A motor operates a cooler engine for producing ice.
Q2 heat is taken from a cooling room which
contains an amount of water at 0oC and Q1 heat is
given away to the air around it at 15oC. Suppose the
cooler engine has a coefficient of performance of
20% of the coefficient of performance of an ideal
cooler engine.
Calculate the work done by the motor to make 1 kg of
ice? (ice latent heat is 3.4 x 105 J/kg)
b. What is time required to make 1 kg of ice if the power
of the motor is 50 W?
a.
Solution
 The scheme of the cooler engine
T1 = 15 + 273 = 288 K
Q1
Cooler
W from motor
Q2
T2 = 0 + 273 = 273 K
Cp ideal
m = 1 kg
Lice = 3.4 x 105 J/kg
T1 = 15 + 273 = 298 K
T2 = 0 + 273 = 273 K
Cp engine = 20% x Cp ideal
p
= 50 W
T2

T1 - T2
273 K
273


288 K - 273 K
15
 Cp engine = 20% x Cp ideal 
20 273 91


100 15 25
The work done by electric motor (W) is
W 
Q2
Q2
Cp engine 
W
 m Lice / Cp engine  9.3 x 104 J
C p engine
Thus, the work done by the electric motor (W) is 9.3x104 J.
b. The time required to make 1 kg of ice is
t
work
W

 1860 seconds  31 minutes
power
P
Thus, the time required to make 1 kg of ice is 31 minutes.
Exercises
 An ideal refrigerator has coefficient of performance of
5.0. If the room temperature outside the refrigerator is
27oC, what is the lowest temperature in the refrigerator
which can be obtained?
 The coefficient of performance of a refrigerator is 4.0.
What is the electric energy used to transfer 4000 joule
of heat from the food in the refrigerator?
That’s all!!!
Thanks, ……bye bye