The Physics
Opus in profectus

Kinetic-Molecular Theory

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practice problem 1

How about a simple, straightforward problem?
  1. Compute the rms speed of an oxygen molecule at room temperature.
  2. Use the results of part a. to determine the rms speed of a hydrogen molecule at room temperature.
  3. Use the results of part b. to determine the rms speed of a mercury atom at 1200 K.


Here come the solutions…

  1. Use the formula. Recall that oxygen is a diatomic molecule in everyday situations. Let's not go nuts here with precision. Just use the approximate molecular mass of oxygen (2 × 16 u = 32 u) and the approximate value of room temperature (300 K). Let's try calculating it both ways: first with the mass of a molecule…

    vO2 = √ 3kT
    vO2 = √ 3(1.38 × 10−23 J/K)(300 K)
    (32 u)(1.66 × 10−27 kg/u)
    vO2 = 483.5 m/s  

    and then with the mass of a mole…

    vO2 = √ 3RT
    vO2 = √ 3(8.31 J/mol K)(300 K)
    (0.032 kg/mol)
    vO2 = 483.4 m/s  

    The two answers are slightly different in the fourth significant digit, but I said to be reasonable with the precision. Let's just say the answer is…

    vO2 = 480 m/s

  2. Exploit the simple ratio of the two molecular masses. Oxygen is 16 times heavier than hydrogen on a per atom or per molecule comparison (since both gases are diatomic in our everyday lives). RMS speed is inversely proportional to the square root of mass (molecular or molar). This means the rms speed of hydrogen should be √16 = 4 times faster. If you would like to see the mathematical reasoning presented formally, here it is…

    vH2  =  √3kT/mH2
    vO2 √3kT/mO2
    vH2  = √ mO2
    vO2 mH2
    vH2  = √ 32 u  = 4
    vO2 2 u
    vH2  = 4
    vH2 = 4 vO2
    vH2 = 4(480 m/s)
    vH2 = 1,920 m/s
  3. Another question with rigged numbers. The atomic mass of mercury (200 u) is 100 times that of molecular hydrogen (2 u). This difference reduces the speed by 1√100 = 110. In a similar vein, the temperature of these mercury atoms is 4 times that of the hydrogen molecules in part b. This change raises the rms speed by a factor of √4 = 2. Combining both changes gives a new rms speed that's 210 of the old one. Again, if you would like to see the mathematical reasoning presented formally, here it is…

    vHg  =  √3kTHg/mHg
    vH2 √3kTH2/mH2
    vHg  = √ THgmH2
    vH2 TH2mHg
    vHg  = √ (1200 K)(2 u)
    vH2 (300 K)(200 u)
    vHg  =  2  =  1
    vH2 10 5
    vHg = ⅕vH2
    vHg = ⅕(1,920 m/s)
    vHg = 384 m/s

practice problem 2

Write something else.


Answer it.

practice problem 3

Write something different.


Answer it.

practice problem 4

Derive the law of Dulong and Petit by applying the equipartition of energy to the atoms in a solid.


Atoms in a solid have six degrees of freedom. (Why?) Therefore, the energy per atom is

K⟩ =  6  kT = 3kT

Multiply by Avogadro's constant to get the internal energy in a mole of atoms.

U = ⟨KNA = 3kTNA

Specific heat is the rate of change in internal energy with respect to temperature. Molar specific heat is this derivative applied to the internal energy in one mole of atoms.

CV =   (3kTNA) = 3kNA

This shows that molar specific heat is a constant for all materials since Boltzmann's constant (k) and Avogadro's constant (NA) are both constant. Substitution using unusually precise values for the two constants yields the usually stated value of this constant.

CV = 3kNA  
CV = 3(1.380 649 × 10−23 J/K)(6.022 140 76 × 1023 1/mol)  
CV = 24.94 J/mol K