THERMODYNAMIC

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THERMODYNAMIC

• Conducted by
• Efa Mai Inanignsih

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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

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The phenomenonWhat do you think of

• 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

• Balloon explosion??

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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
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External Work
The figure is a tube containing ideal gas closed
with a piston 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
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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.

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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
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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?

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Thermodynamics Processes

• Isothermal Process
• Isobaric Process
• Isochoric Process
• Adiabatic Process

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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 Boyles law
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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.

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Isobaric Process

• Isobaric process is the change process of gas
  state at constant pressure
• because p is constant and nR is always constant,
  then

the Gay-Lussacs law
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Isobaric Process(continued)

• Graph of isobaric process

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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)

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Isochoric Process

• Isochoric process is the change process of gas
  state at constant volume
• because V is constant and nR is always constant,
  then
• This process complies the Gay-Lussacs law.

graph of isochoric process
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External work in Isochoric Process

• The external work done by gas in isochoric
  process
• In other words, in isochoric process (constant
  volume), the gas does not do any external work.
  So the area under the p V graph only forms a
  point.

The area under the p V graph for isochoric
process is zero
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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 Poissons formula
• Where
• ? Laplace constant Cp/Cv
• Cp specific heat of gas at constant pressure
• CV specific heat of gas at constant volume

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Adiabatic Process(contd.)

• The equation above can also be expressed in
  another equation as follows

Curvature of p V adiabatic graph is steeper
than isothermal
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External work in Adiabatic Process

• The external work done by gas in adiabatic
  process is expressed as follows.

The external work done by gas is equal to the
area of shaded region under the p V graph
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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 W (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.
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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) Thus, the ratio of the final to the
initial pressure is
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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)

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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 external work (J) ?U the change
of internal energy (J)
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The First Law

• in non-adiabatic process, the gas will not only
  receive external work but also heat.
• 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.

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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

The rules of W and Q values
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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

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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

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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

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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

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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).

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Derivation

• The first law of thermodynamics can be derived as

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• For isobaric process (p constant)
• Monatomic gas Diatomic gas

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• For isochoric process (V constant).
• Monatomic gas Diatomic gas
• Relationship between Cp and CV

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Other parameters

• Molar heat capacity of gas

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• Specific heat of gas

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Sample Problem
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.

• 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)

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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!

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Thermodynamic Cycle

• Cycle means the process which runs from the
  initial state and returns to that initial state
  after gas does work

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.

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• The total external work done by the gas in one
  cycle a b c a is the 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

Wabca Wab Wbc Wca area of abV2V1
(-area of bcV1V2) 0 Wabca area of abca
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Thermodynamic Cycle Carnot engine

• Carnot engine is assumed as an ideal heat engine
  which works cyclically and reversible between two
  temperatures without any loss of energy.

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
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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

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Efficiency

• Efficiency of engine
• In the Carnot engine, holds W Q1 Q2

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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

• Determine
  
• the system internal energy in state C,
• the heat given to the system in A ? B process,
  and
• the heat given to the system or taken from in C ?
  A process.

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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.

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• 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.

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• c. Process from C ? A
• WCA -area of ACED
• -(area of ABED area ?ABC)
• -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.

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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.

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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 gt 0).
• entropy is a measurement of the amount of energy
  or heat which cannot changed into work
• the total change of entropy of the Carnot engine
  is

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• 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
• Where
• Q1 heat absorbed from low temperature
• Q2 heat given at high temperature

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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)
• What is time required to make 1 kg of ice if the
  power of the motor is 50 W?

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Solution

• The scheme of the cooler engine

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
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• Cp engine 20 x Cp ideal
• The work done by electric motor (W) is
• Thus, the work done by the electric motor (W) is
  9.3x104 J.
• b. The time required to make 1 kg of ice is
• Thus, the time required to make 1 kg of ice is
  31 minutes.

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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?

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Thats all!!!Thanks, bye bye