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    Kinetic Theory
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    Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).

    Mean free path = 1.11 × 10–7 m

    Collision frequency = 4.58 × 109 s–1 

    Successive collision time ≈ 500 × (Collision time) 

    Pressure inside the cylinder containing nitrogen,


     P = 2.0 atm

        = 2.026 × 105 Pa 

    Temperature inside the cylinder, T = 17°C =290 K 

    Radius of a nitrogen molecule, r = 1.0 Å

                                                    = 1 × 1010 m 

    Diameter, d = 2 × 1 × 1010 = 2 × 1010 m 

    Molecular mass of nitrogen, M = 28.0 g

                                                 = 28 × 10–3 kg 

    The root mean square speed of nitrogen is given by the relation,



    Collision frequency =  

                                 = 508.26 / (1.11 × 10-7) 

                                 =  4.58 × 10    9 s-1 

    Collision time is given as, 

    T = 

      =  

      =  3.93 × 10    -13 s

    Time taken between successive collisions, 

    T ' = l / vrms 

        = 1.11 × 10-7 / 508.26 

        =  2.18 × 10    -10 s

    Therefore, 

     

    Hence, the time taken between successive collisions is 500 times the time taken for a collision.

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    Answer
    At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u). 
    Temperature of the helium atom, THe = –20°C= 253 K

    Atomic mass of argon, MAr = 39.9 u

    Atomic mass of helium, MHe = 4.0 u

    Let, (vrms)Ar be the rms speed of argon.

    Let (vrms)He be the rms speed of helium.

    The rms speed of argon is given by, 
     
          = 2523.675

          = 2.52 × 10    3 K 

    Therefore, the temperature of the argon atom is 2.52 × 103 K.
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    Answer
    Estimate the average thermal energy of a helium atom at:
    (i) room temperature (27 °C),
    (ii) the temperature on the surface of the Sun (6000 K),
    (iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).
    (i) At room temperature, T = 27°C = 300 K

    Average thermal energy = (3/2)kT

    Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1

    ∴ (3/2)kT = (3/2) × 1.38 × 10-38 × 300

                   = 6.21 × 10–21J

    Hence, the average thermal energy of a helium atom at room temperature (27°C) = 6.21 × 10–21 J.

    (ii) On the surface of the sun, T = 6000 K

    Average thermal energy = (3/2)kT

                                         = (3/2) × 1.38 × 10-38 × 6000

                                         = 1.241 × 10-19 J

    Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.

    (iii) At temperature, T = 107 K

    Average thermal energy = (3/2)kT

                                         = (3/2) × 1.38 × 10-23 × 107

                                         = 2.07 × 10-16 J

    Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.
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    Answer
    Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
    All the three vessels have the same capacity, they have the same volume. Hence, each gas has the same pressure, volume, and temperature.

    According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.

    Number is equal to Avogadro’s number,     N = 6.023 × 1023.

    The root mean square speed (    vrms) of a gas of mass m, and temperature T, is given by the relation,
      
                         vrms = (3kT/m)1/2  

    where, 

    k is Boltzmann constant 

    For the given gases, k and T are constants. 

    Hence vrms depends only on the mass of the atoms.

    i.e.,                          vrms ∝ (1/m)1/2 

    Therefore, the root mean square speed of the molecules in the three cases is not the same.

    Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest.

    Hence, neon has the largest root mean square speed among the given gases.
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    Answer
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    An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

    Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3 

    Bubble rises to height, d = 40 m 

    Temperature at a depth of 40 m, T1 = 12°C = 285 K 

    Temperature at the surface of the lake, T2 = 35°C = 308 K 

    The pressure on the surface of the lake, 

    P2 = 1 atm = 1 ×1.013 × 105 Pa

    The pressure at the depth of 40 m, 

    P1 = 1 atm + dρg 

    where, 

    ρ is the density of water = 103 kg/m3 

    g is the acceleration due to gravity = 9.8 m/s2 

    Therefore, 

    P1 = 1.013 × 105 + 40 × 103 × 9.8

         = 493300 Pa 

    We have,

         P1V1 / T1 = P2V2 / T2 

    where,

         V2 is the volume of the air bubble when it reaches the surface. 

    V2 = P1V1T2 / T1P2 

          = 493300 × 1 × 10-6 × 308 / (285 × 1.013 × 105) 

          = 5.263 × 10–6 m3 
     
           = 5.263 cm    3 

    Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.
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