practice problem 1
- Compute the rms speed of an oxygen molecule at room temperature.
- Use the results of part a. to determine the rms speed of a hydrogen molecule at room temperature.
- Use the results of part b. to determine the rms speed of a mercury atom at 1200 K.
Here come the solutions…
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 m 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 M vO2 = √ 3(8.31 J/molK)(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
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 vO2 vH2 = 4 vO2
vH2 = 4(480 m/s)
vH2 = 1,920 m/s
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
practice problem 3
practice problem 4
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 = ⟨K⟩NA = 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(1.3806503 × 10−23 J/K)
|CV = 24.94 J/molK|