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Kinetic Theory Of Gases




7. Kinetic interpretation of temperature


  • From equation (5) we have
         PV = (1/3)Nmvmq2
    where N is the number of molecules in the sample. Above equation can also be written as
         PV = (2/3)N(1/2)Nmvmq2                (7)
  • The quantity (1/2)Nmvmq2 in equation (7) is the kinetic energy of molecules in the gas. Since the internal energy of an ideal gas is purely kinetic we have,
         E=(1/2)Nmvmq2                     (8)
  • Combining equation 7 and 8 we get
         PV=(2/3)E
  • Comparing this result with the ideal gas equation (equation (4) ) we get
         E=(3/2)KBNT
    or,     E/N=(1/2)mvmq2 =(3/2)KBT          
    (9)
    Where, KB is known as Boltzmann constant and its value is KB=1.38 X 10-23 J/K
  • From equation (11) we conclude that the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas and is independent of the pressure , volume and nature of the gas.
  • Hence average KE per molecule is
         (1/2)mv2¯=(3/2)KBT
    from this since v2¯=(vrms)2, rms velocity of a molecule is
         vrms=√(3KBT/m)                    (10)
    This can also be written as
         vrms=√(3KBNT/Nm)
          =√(3RT/M)
                        (11)
    where, M=mN is the molecular mass of the gas.

    Assignment
    (1)Gas laws (Boyle's and charle's law) and perfect gas equation can be derived using kinetic theory of gases. try to derive them.




7. Law of Equipartition of energy


  • According to the principle of equipartition of energy, each velocity component has, on the average, an associated kinetic energy (1/2)KT.
  • The number of velocity components needs to describe the motion of a molecule completely is called the number of degrees of freedom.
  • For a mono atomic gas there are three degrees of freedom and the average total KE per molecule for any monotomic gas is 3/2 KBT.
         




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