One mole of an ideal monatomic gas is taken through the cycle by dfhdhdhdhjr

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									***Work…
Work is the area under the curve
Work for an isobaric process
Work in an isochoric process
Work for an isometric process
Work in an isothermal expansion or compression
                of an ideal gas
Work for an isothermal process
What is the work in an isothermal process...
Work for an adiabatic process
  Work in an adiabatic expansion or
compression of an ideal monatomic gas
Work is greatest for idiabats
Work for a cyclic process




                       Wnet = Area
***Cycles…
 One mole of an ideal monatomic gas is taken through the cycle
 shown. Assume that p = 2p0, V = 2V0, p0 = 1.08x105 Pa, and V0 =
  0.0244 m3. What is the efficiency of this cycle? What is the
ideal efficiency of a heat engine operating between the highest
                  and the lowest temperature?
     2T0 , V0 ,2 P0
                            4T0 ,2V0 ,2 P0

                                             W = p0∙V0
                                             Qabc = nCv∙(Tb-Ta)+nCp∙(Tc-Tb) =
                                             Qabc = n∙(3R/2)∙(Tb-Ta)+n∙(5R/2)∙(Tc-Tb) =
                                             = nRTa∙[3/2∙(Tb/Ta-1)+5/2∙(Tc/Ta-Tb/Ta)] =
                                             = p0∙V0∙[3/2∙(2-1)+5/2∙(4-2)] = 13/2∙(p0∙V0)
     T0 , V0 , P0     2T0 ,2V0 , P0          ε = W/Qh = 2/13 = 15.4%
                                             εideal = 1-Ta/Tc = 1 – ¼ = 75%
***Suppose the working substance of this heat engine is 0.75
kg of an ideal monatomic gas. How would the efficiency of this
cycle compare with that of a Carnot engine operating between
      the highest and lowest temperatures of this cycle?




  W = Area = (250-150)·(2.25-1.75)·10 = 500 J
  Q412 = nCv∙(T1-T4)+nCp∙(T2-T1) =
  Q412 = n∙(3R/2)∙(T1-T4)+n∙(5R/2)∙(T2-T1) =
  = nRT4∙[3/2∙(T1/T4-1)+5/2∙(T2/T4-T1/T4)] =
  = p4∙V4∙[3/2∙(250/150-1)+5/2∙((250/150)(2.25/1.75)-(250/150)] = 2.19∙(p4∙V4) = 5749 J
  ε = W/Qh = 8.7%
  εideal = 1-T4/T2 = 1 – (150/250)·(1.75/2.25) = 53.3%
        ***Suppose 5 moles of a monatomic ideal gas follows the three-part
        cycle shown. What is the efficiency of the cycle? What is the ideal
         efficiency of an engine operating between the highest and lowest
                            temperature of this cycle?

                                                 TA  3  TC




                                                                                      TB  4  TC




Wnet  Wab  Wbc  Wca                                                              Qin  QCA  Q AB  nCv ·(T - TC )  15  nR  TC 
                                                                                                               A          2

 1 P  V  1  PA  PC   VB  VC   1  2 PC  3VC  3PC VC  3  nR  TC
  2           2                             2                                         n  3 ·R  2TC  15  nR  TC  9  nR  TC
                                                                                           2             2

U AB  3  nR  TB  TA   3  nR  TC                                                                   3  nR  TC
        2                     2
                                                                                     e real  Wnet /Qin                 33%
U AB  QAB  WAB  QAB  2 PC  3VC                                                                        9  nR  TC
QAB  15  nR  TC
       2
                                                                                     eideal  1 - TC /TB  1 - 1/4  75%
    Suppose 9 moles of a monatomic ideal gas follows the three-part cycle,
     the first part of which involves an isothermal expansion from point A,
     where V = 7.5x10-5 m3 and P=6x105 Pa, to point B, where the pressure
    is lower by a factor of 6. What is the efficiency of the cycle? What is
      the ideal efficiency of an engine operating between the highest and
                        lowest temperature of this cycle?




                                                                                                                        P V P V 
                                                              Qin  Q ca  Q ab  nC v ·(T - Tc )  Wab  n(3/2)·R  a a c c  – Wab
                                                                                          a
Wab  nRTab ·ln(Vb /Va )  Pa Va ·ln(P /Pb )  Pa Va ·ln(6)
                                      a
                                                                                                                           nR    
Wbc  Pbc ·Vbc  Pbc   5Vc                                3/2  5/6·P Va  Pa Va ·ln(6)  Pa Va ·5/4  ln(6)
                                                                               a

Wnet  Wab  Wbc  Wca                                       e real  Wnet /Qin 
                                                                                     ln(6) - 5 / 6
                                                                                                    31.5%
 Pa Va ·ln(6)- Pbc  5Vc  0  Pa Va ·ln(6) - 5 / 6                              5/4  ln(6)
                                                              eideal  1 - Tc /Ta  1 - 1/6  83.3%
  9 moles of a monatomic ideal gas follows the 3-part cycle as shown.
The first part of which involves an isothermal expansion from point A,
 where V = 3x10-5 m3 and the pressure is 6x105 Pa, to point B, where
 the pressure is lower by a factor of 6. Fill in the missing entries and
              determine the actual and ideal efficiencies.




         Q(J)    W(J)   ΔU(J)

A B     32.3    32.3     0        e real  Wnet /Qin  17.3/54.8 31.6%
BC      -37.5   -15    -22.5      eideal  1 - Tc /Ta  1 - 1/6  83.3%

CA      22.5     0     22.5
Two moles of a monatomic ideal gas are taken through the reversible
cycle shown. Process bc is an adiabatic expansion, with Pb = 11.9 atm and
Vb = 10 L. Find the efficiency of the cycle and the maximum efficiency
for the temperature extremes of the cycle.




                              
                      V 
                                           5/3
                                 1
            Pa  Pc   b  Pb   
                      V                        11.9  0.372 atm
                       c       8
                                                 3        (P  P )  Vb 3
            Q in  Q ab  nC v  (Tb  Ta )  n  RC v  b a              (Pb  Pa )  Vb  17.5 kJ
                                                 2            nR         2
            Q out  Q ca  nCp T   Pa Va  Pc Vc    Pa  Va  Vb   6.6 kJ
                                      5                     5
                                      2                     2
                                                   W 17.5  6.6
            completecycle: W  Q  e act                            62%
                                                   Q in     17.5
                                         Pa
            eideal  1  Ta / Tb  1        96.9%
                                         Pb
 ***Five moles of an ideal monatomic gas expand adiabatically,
and its temperature decreases from 362 to 292 K. Determine
the work done by the gas and the change in its internal energy.




                              3
         W  Q  U  U   nRT  4363 J
                              2
  ***In the cycle shown, 1 mol of an ideal diatomic gas is initially at a
   pressure of 1 atm and a temperature of 0°C. The gas is heated at
 constant volume to T2 = 141°C and is then expanded adiabatically until
its pressure is again 1 atm. It is then compressed at constant pressure
   back to its original state. Find the efficiency of this cycle and the
    efficiency of a Carnot cycle operating between the temperature
                          extremes of this cycle.
   Two moles of a diatomic gas are taken through the cycle ABCA as
  shown on the PV diagram. At A, the pressure and temperature are 4
  atm and 592 K, respectively. The volume at B is twice that at A. The
      segment BC is an adiabatic expansion and the segment CA is an
    isothermal compression. Find the efficiency of this cycle and the
     efficiency of a Carnot cycle operating between the temperature
                          extremes of this cycle.

                                          Wab  Pa  2Va  Va   Pa  Va  nRTa
A: 4 atm, 592 K
                                                              5 
                                          Wbc   U  n   R   Tc  Tb  
                                                              2 
                  B: 4 atm, 2VA, TB=2TA
                                                          5                       5 
                                                    n   R   Ta  2Ta   n   R   Ta
                                                          2                       2 
                                          Wca  nRTa  ln Va / Vc   nRTa  ln Vb / 2Vc 
                                          PbVb  PcVc or TbVb 1  TcVc 1
                                           Vc = Vb  (Tb /Tc )1/(  -1)  Vb  (2)1/(  -1)

                                          Qin  Qab  nC p  Tb  Ta    nR  2Ta  Ta   nR  Ta
                                                                         7                    7
                                                                         2                     2
                                                                       5                         1          
                                                                 nRTa  nRTa  nRTa  ln  2  (2)1/(  -1) 
                                                                                                             
                                          ereal    Wtot / Qin 
                                                                       2                                      30.7%
                                                                             7
                                                                                 nR  T
                                                                              2
 ***Two moles of a diatomic gas are carried through the cycle ABCDA shown
 in the PV diagram. The segment AB represents an isothermal expansion, the
segment BC an adiabatic expansion. The pressure and temperature at A are 5
   atm and 563 K. The volume at B is twice that at A. The pressure at D is 1
atm. What is the efficiency of this cycle? What is the efficiency of a Carnot
      cycle operating between the temperature extremes of this cycle?

                                                Wab  nRTa  ln Vb / Va   nRTa  ln 2 
      A: P=5 atm, TA=563 K
                                                                        5 
                                                ??? Wbc   U   n   R   Tc  Tb  
                       B: 4 atm, 5VA/4, TB=TA
                                                                        2 
                                                                5                       5 
                                                          n   R   Ta  2Ta   n   R   Ta
                                                                2                       2 

                                                                     
                                                Pb Vb  Pc Vc or Tb Vb 1  Tc Vc 1
                                                 Vc = Vb  (Tb /Tc )1/( -1)  Vb  (2)1/( -1)

                                                Q in  Q ab  nCp  Tb  Ta   nR  2Ta  Ta   nR  Ta
                                                                               7                    7
 D: P=1 atm, TD=TA/5                                                           2                    2
                                                                             5                         1         
                   C: P=1 atm, TC=???TA/5                              nRTa  nRTa  nRTa  ln  2  (2)1/( -1) 
                                                                                                                  
                                                e real  Wtot / Q in 
                                                                             2                                     30.7%
                                                                                   7
                                                                                       nR  T
                                                                                    2
 ***How much work must be done to 22 grams of CO
at standard temperature and pressure to compress it
     to 1/5th of its initial volume if the process is
   isothermal? How much work must be done if the
    process is adiabatic? What if the gas was CO2?
 *** An air pump has a cylinder 0.280 m long with a movable piston. The
     pump is used to compress air from the atmosphere (at absolute
    pressure 1.01 105 Pa) into a very large tank at 4.50x105 Pa gauge
      pressure. (For air , Cv = 20.8 J/mol · K.) The piston begins the
  compression stroke at the open end of the cylinder. How far down the
length of the cylinder has the piston moved when air first begins to flow
   from the cylinder into the tank? If the air is taken into the pump at
  27.0°C, what is the temperature of the compressed air? Assume that
                       the compression is adiabatic.
***A sample of an ideal gas goes through the process shown in
Figure P20.32. From A to B, the process is adiabatic; from B to
 C, it is isobaric with 98 kJ of energy entering the system by
heat. From C to D, the process is isothermal; from D to A, it is
  isobaric with 160 kJ of energy leaving the system by heat.
   Determine the difference in internal energy, Eint, B - Eint, A.
***Argon enters a turbine at a rate of 82.0 kg/min, a
temperature of 850°C, and a pressure of 1.40 MPa. It expands
adiabatically as it pushes on the turbine blades and exits at
pressure 350 kPa. Calculate the maximum efficiency of the
turbine.
    0.5 mole of a diatomic gas, γ = 1.4, is taken through the cycle
      shown. Given that V1 = 5 L, V2 = 15 L, Th = 570°C, and Tc =
   290°C, calculate the efficiency of the cycle from the definition
           e=W/Qin and compare to the maximum possible.
                                             V1  5 L, T1  570  C




                                                                                V2  15 L, T2  570  C


       
   T4 V4 1  T1V1 1

         V4  13.7 L, T4  290  C


                                                                                                 
                                                                                             T2 V2 1  T3 V3 1

                                                                      V3  41.2 L, T3  290  C

Q h  Q12  W12  nRTh  ln( V2 / V1 )  3848 J
Q c  Q34  W34  nRTc  ln( V4 / V3 )  2573 J
W  Q h  Q c  1275 J
                                                                             e  W / Qh  0.33
                                                                             e max  1  Tc / Th  0.33
    ***What is the efficiency of this Carnot engine?




       Nicholas Léonard Sadi Carnot
              (1796 - 1832)

                                                          Qh  nRTh ln(V2 V1 )
                                                          Qc  nRTc ln(V3 V4 )
                      
P2 V2  P3 V3 and P4 V4  P1V1
                            
T2 V2 1  T3 V3 1 and T4 V4 1  T1V1 1
                            
Th V2 1  Tc V3 1 and Tc V4 1  Th V1 1
(V2 / V1 ) 1  (V3 / V4 ) 1
 ln( V2 V1 )  ln( V3 V4 ) or Q c / Q h  Tc / Th     W / Qh  (Qh  Qc ) Qh  1  Tc Th
Carnot cycle in a P-V and T-S diagrams
P-V relationship for an adiabatic path



         dU  ncv dT   PdV
         PdV  VdP  nRdT
                       R
         PdV  VdP       PdV
                       cv
                   c p  cv  dV
                                   1   
         dV dP                               dV
               
                   c  V    
         V   P         v                   V


         dP      dV
                    0
          P       V
         ln P   ln V  constant
         PV   constant
 ***Legs 1 and 3 are adiabats; legs 2 and 4 are
isomets. Can this cycle achieve ideal efficiency?
What does this plunger have to do
     with a diesel engine?
The fuel-air mixture in the cylinder of a diesel engine at
20ºC is compressed from an initial pressure of 1 atm and
 volume of 800 cm3 to a volume of 60 cm3. Assuming the
mixture behaves as an ideal gas with γ=1.40 and that the
  compression is adiabatic, find the final pressure and
              temperature of the mixture.




           Pf = Pi·(Vi/Vf)γ = 1·(800/60)1.4 = 37.6 atm
                 Tf = (Pf·Vf)/(Pi·Vi)·Ti = 826 K
***Equipartition of energy...
Molecular model of ideal gas




                                             2
                    2mvxi    2mvxi        mvxi
               Fi                     
                     t     (2 L / vx )    L
                                           1     
                    N     2
                 F      mvxi 2 N
               P                      mv 2 
                 A i 1 AL 3 V             2     
                  2       1     
               PV   N   mv 2 
                  3       2     
                1 2 3
                mv  k B  T
                2        2
Equipartition and specific heat of a diatomic gas
                                    Specific heats of gases

TABLE 21.2              Molar Specific Heats of Various Gases
                                                  Molar Specific Heat [J/mol·K]
       Gas
                             CP                     CV                 CP - CV     γ = CP / CV
  Monatomic Gases
        He                   20.8                  12.5                  8.33         1.67
        Ar                   20.8                  12.5                  8.33         1.67
        Ne                   20.8                  12.7                  8.12         1.64
        Kr                   20.8                  12.3                  8.49         1.69
  Diatomic Gases
        H2                   28.8                  20.4                  8.33         1.41
        N2                   29.1                  20.8                  8.33         1.40
        O2                   29.4                  21.1                  8.33         1.40
        CO                   29.3                  21.0                  8.33         1.40
        Cl2                  34.7                  25.7                  8.96         1.35
  Polyatomic Gases
        CO2                  37.0                  28.5                  8.50         1.30
        SO2                  40.4                  31.4                  9.00         1.29
        H2 O                 35.4                  27.0                  8.37         1.30
        CH4                  35.5                  27.1                  8.41         1.31


               All specific heats except those for water were obtained at 300 K.
***Specific heat vs molecular structure
The difference between Cv and Cp




      dU  dQ  PdV
      dU p  nC p dT  nRdT
      dU v  nCv dT
      dU p  dU v (U  U (T ))
       nCv dT  nC p dT  nRdT
      C p  Cv  R
What is the specific heat of
          solids?
Actual heat capacities for some solids
Otto cycle

								
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