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THERMODYNAMIC (PowerPoint)

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THERMODYNAMIC (PowerPoint) Powered By Docstoc
					     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                  is the
  everything              everything
  becoming our            which is
  observation             outside the
  object                  system
External Work
       p                   The figure is a tube
                           containing ideal gas
                           closed with a piston
           ∆s
                  piston
                           If the gas in is heated at a
       F                   constant pressure, then
 gas            tube       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     Solution
 so that the volume
                          V1 = 5.0 L
 decreases from 5.0 L
                          V2 = 3.5 L
 to 3.5 L at a constant   p = 1.0 x 105 Pa
 pressure of 1.0 x 105
 Pa. Calculate the        because the gas is compressed, then
 external work            W      = p V
 applied to the gas!             = 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                   complies with
  then            p1V1 = p2V2                    the Boyle’s law
                                             p
  Where
        p1 = initial pressure
        V1 = initial volume
        p2 = final pressure
        V2 = final volume
                                                               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
                                      p

W  nRT ln V 
                V2                   p1     1
                V1

W  nRT ln V2 - ln V1                             2
                                      p2
          V
W  nRT ln 2
          V1
                                           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


                                       1        2
                            p1 = p2




                                      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                 The area under the p – V
                           W 0                     graph for isochoric
                                                    process is zero
 In other words, in isochoric process         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  p2 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              γ    Curvature of p – V adiabatic
       p1V1  p2V2                graph is steeper than
                                  isothermal
  nRT1  γ  nRT2  γ
 
  V  V1   V  V2
                                  p
                 
  1        2                                  adiabatic
             γ-1            γ-1
      T1V1          T2V2                              isothermal




                                                              V
External work in Adiabatic Process
                                     p
 The external work done by gas in   p1

 adiabatic process is expressed as
 follows.
                                     p2
              1
          W      (p1V1 - p2V2 )
             γ -1                             V1          V2   V

                                          The external work done
                3                         by gas is equal to the
          or W  nR (T1 - T2 )            area of shaded region
                2                         under the p – V graph
Sample Problem
 Two moles of gas
 is compressed at
                        Solution
 a constant             Given
 temperature of -       n = 2 mol
 23oC so that its       R = 8.31 J/mol K
 volume becomes         T = (-23 + 273) K = 250 K
 half of the initial.
 Calculate the          V2 = ½ V1    so V2/V1 = 1/2
 external work
                        The external work done in isothermal process
 done by the gas!
                        W         =           V
 (R = 8.31 J/mol K,                   nRT ln 2
                                              V1
 ln 1 = 0, ln 2 =                 = (2 mol) (8.31 J/mol K) (250 K) (ln )
 0.69)                            = 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       Solution
 gas ( = 5/3) is
 compressed                Because  = 5/3 and V2/V1 = ½
 adiabatically and the     then
 volume decreases to its             p1V1 = p2V2
 half. Determine the
 ratio of the final        p2 /p1 = V1 /V2 = 2 (5/3)
 pressure to the initial              3
 pressure!                         =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               Q
                                              +

                                          W       -
                     The rules of W and                        -
                                                                   Q
                     Q values                     W
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           Where
     ΔT             C = heat capacity (J/K)
                    Q = absorbed heat (J)
                    T = the change of temperature (K)
      QV
 CV     or QV  CV ΔT         for isochoric process
      ΔT
        Qp
 Cp         or Qp  C p ΔT    for isobaric process
        ΔT
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:                         Diatomic gas:
       3
         nRΔR                                7
  Cp  2      
                nRΔR                    C p  nR
         ΔT      ΔT                          2
       3
  C p  nR  nR
       2
       5
  C p  nR
       2
 For isochoric process (V = constant).


 Monatomic gas:                           Diatomic gas:

  W     0                                      5
                                           C V  nR
          3                                     2
            nRΔR
          2            0
   CV             
             ΔT       ΔT
          3
   CV      nR
          2
 Relationship between Cp and CV :                   7     5
                                          C p - CV    nR - nR
                                                     2     2
                                          C p - CV  nR
Other parameters
                                              Q
 Molar heat capacity of gas            Cm 
                                             n ΔT
         QV
 CV,m       or QV  n CV ,m ΔT            isochoric process
        n ΔT
           Qp
 C p,m           or Q p  n C p,m ΔT      isobaric process
           n ΔT
                                  Q
 Specific heat of gas        C
                                 m ΔT
          QV
    CV       or QV  mCV T          for isochoric process
         m T


         3   R                                         5 R
    CV                  for monatomic gas      Cp 
         2   M                                         2M
         5   R                                       7 R
    CV                  for diatomic gas       Cp 
         2   M                                       2M


             Qp
    Cp           or Qp  mC p ΔT      for isobaric process
           m ΔT
Sample Problem
 One mole of gas is        Solution
 compressed at a
                            Because
 constant temperature                 n = 1 mol
 of -23oC so that its                 T = (-23 + 273) K = 250 K
 volume decreases to                  V2 = ½ V1
                                      R = 8.31 J/mol K
 half of its initial
 volume. Calculate the      Then
 work done by the gas!      W         = nRT ln V2/V1
                                      = 1 x 8.31 x 250 ln ½
 (R = 8.31 J/mole K; ln 1
                                      = -1.4 x 103 joule
 = 0, ln 2 = 0.69)
                            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                               p
                                                        a
Cycle
  Cycle means the process which                    c
   runs from the initial state and                                 b
   returns to that initial state after
   gas does work                                    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          =Wab + Wbc + Wca
                    area of abV2V1 + (-area of bcV1V2) + 0
    Wabca          =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
  High temperature reservoir T1
                                                   Carnot engine is assumed as an
                                                    ideal heat engine which works
                                                    cyclically and reversible
               Q1                                   between two temperatures
                                                    without any loss of energy.

                                                    Within one cycle, the gas returns to
          Heat engine             W = Q1 - Q2       its initial state, so there is no change
                                                    of internal energy (U = 0).

                                                    Q = U + W
               Q1                                   Q1 – Q2 = 0 + W
                                                    W = Q1 – Q2
  Low temperature reservoir T2
                                         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                         T2 
     η  1   100%
          Q                       1   100%
             1 
                                       T 
                                          1 
                                                 p

Sample Problem                                   90                   C


                                                 60
 The figure below           Pressure (N/m 3 )
  indicates the                                  30
                                                      A               B
  thermodynamics change                                                   V
                                                 O    D               E
  of system from iitial                               Volume (m 3 )
  state A to B and C and
  back to A. If VA = 0, VB   Determine
  = 30 joule and heat           a. the system internal energy
  given to the system in           in state C,
  process B  C = 50 J          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)
              AB  BC          2  60 
        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
                                Q2 Q1
                   ΔS 2  ΔS1    -   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.
  a. 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?
Solution
 The scheme of the cooler engine

    T1 = 15 + 273 = 288 K
                                                 m = 1 kg
                                                 Lice = 3.4 x 105 J/kg
             Q1
                                                 T1 = 15 + 273 = 298 K
                                                 T2 = 0 + 273 = 273 K
           Cooler           W from motor         Cp engine = 20% x Cp ideal
                                                 p         = 50 W
             Q2

                                                     T2
    T2 = 0 + 273 = 273 K              Cp ideal   
                                                   T1 - T2
                                                      273 K        273
                                                                
                                                   288 K - 273 K   15
                                  20 273 91
 Cp engine = 20% x Cp ideal          
                                 100 15 25
                                                                 Q2
  The work done by electric motor (W) is             Cp engine 
                                                                 W
           Q2
  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
        work    W
   t             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

				
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