Page 85 - SP015 Past Years PSPM Chapter 6 -14 Ver 2020
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PSPM SP015



               18.   (b)  Principle of equipartition of energy states that energy is shared equally among the active degrees
                                                                                     1
                        of freedom, each active degrees of freedom has an average energy equal to  kT .
                                                                                     2
                        Principle of equipartition of energy states that energy of a system is shared equally among its
                        available degree of freedom.
                        Principle of equipartition of energy states that an average energy of each degree of freedom of a
                        molecule is ½kT.
               19.   Heat is the energy that being transferred due to temperature difference.
                     Internal energy is the sum of all the energy of all the molecules in a gas.
               20.   Internal energy U of a gas equals the total amount of average kinetic energy, K and potential energy
                     which contains in any gasses.
                     The capacity of the internal energy depends on the state of the system. For ideal gases, potential energy
                     can be neglected because the force of exertion between molecules does not exist so the internal energy, U
                     equals the total of average kinetic energy of the molecule, K.
                           f               f
                       U    NkT      or     U   nRT

                           2               2
                         internal energy depends on temperature and degrees of freedom
                     Types of degree of freedom are translational, rotational, and vibration.
               21.   U = 227 J
               22.   U = 2837 J
                                    -1
               23.   (a)  v rms  = 483 m s      (b)  U = 18.7 kJ
                                     3
                                         -1
               24.   (a)  v rms  = 1.36  10  m s    (b)  K = 1857 J
                                4
               25.   U = 3.75  10  J,     T = 752 K
               26.   At the same temperature, diatomic gas a greater energy per mole because U  f.
                     f diatomic  = 5   and   f monoatomic  = 3   so   U diatomic  > U monoatomic


               FINAL ANSWER FOR CHAPTER 14:  THERMODYNAMICS

               1.    First law of thermodynamics:   Q = U + W
                     where  Q = thermal energy, U = internal energy, W = work done by the gas
               2.    U = –400 J
               3.    Isothermal process is a thermodynamic process where the temperature is constant.
               5.    Adiabatic process is a process where heat is not absorbed or released.
               6.    Adiabatic compression process is a process where heat is not absorbed or released during compression.
               7.    (a)  P to Q: Isobaric      (b)  Q to R: Isovolumetric
                     (c)  Q to T: Isothermal    (d)  Q to S: Adiabatic
               9.    W   V f  p  dV
                           V i
               11.   (a)  Q KLM  = 500 J        (b)  W MJ  = –20 J   (c)  Q MJK  = –420 J
               12.   (a)  U ABC  = 0 J         (b)  Q ABC  = 2250 J
               13.   (a)  W ABC  = –1621 J      (b)  p f  = 10 atm
               14.   (a)  T 2  = 357.6 K = 84.6 C   (b)  W = 1.01 J
               15.   (b)  Q = –1657 J           (c)  Heat is released.
               16.   (b)  T 2  = 600 K          (c)  W 1  = 4986 J
                                 3
               17.   W = –1.32  10  J
               18.   (a)  Piston X moves a lot faster than piston Y during expansion.   (c)  W = 52.4 J
               19.   (b)  W = 196.3 J
               20.   (a)  Internal energy of gas decreases.
                     (b)  Surrounding does the work. Gas volume decreases hence negative work done.
               21.   (a)  W = 747.9 J       (b)  Q = 2610.9 J
               22.   (b)  Q = 1200 J
                                    5
                                                                                 4
               23.   (b)  p 2  = 2.03  10  Pa   (c)  T 3  = 600 K   (d)  W = 3.11  10  J
               24.   Q = 6.62 kJ
               25.   (a)   U = 250 J       (b)   W = 0.297 J
               26.   U ABC  = 440 J
               27.   (a)  m = 2.44 g        (b)  U = 500 J       (c)  m = 10.32 g
               28.   Q is absorbed.   Q = 6912 J
               29.   U = 410 J



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