A brief note on Equipartition, Specific Heats and Mean Free Path . Read it before a test. Try each question at the end before you open its answer.
Degrees of freedom Moving: 3 for every molecule. Turning: 0 for one atom, 2 for a straight molecule, 3 for a bent one. Vibrating: 2 for each active vibration, one kinetic and one potential term. Examples: neon f = 3 f = 3 f = 3 , rigid carbon dioxide f = 5 f = 5 f = 5 , rigid methane f = 6 f = 6 f = 6 . Equipartition Each squared term has 1 2 k B T \tfrac12 k_BT 2 1 k B T on average. One molecule: E = f 2 k B T E = \tfrac{f}{2}k_BT E = 2 f k B T . A gas: U = f 2 n R T U = \tfrac{f}{2}nRT U = 2 f n R T . Translation alone: 3 2 k B T \tfrac32 k_BT 2 3 k B T for every gas. Heat capacities Gas: C v = f 2 R C_v = \tfrac{f}{2}R C v = 2 f R , C p = C v + R C_p = C_v + R C p = C v + R , γ = 1 + 2 f \gamma = 1 + \tfrac{2}{f} γ = 1 + f 2 . Polyatomic with v v v vibrations: C v = ( 3 + v ) R C_v = (3 + v)R C v = ( 3 + v ) R . Mixture: C v = n 1 C v 1 + n 2 C v 2 n 1 + n 2 C_v = \dfrac{n_1C_{v1} + n_2C_{v2}}{n_1 + n_2} C v = n 1 + n 2 n 1 C v 1 + n 2 C v 2 . Never average γ \gamma γ . Solid: C = 3 R C = 3R C = 3 R (Dulong and Petit), so c = 3 R / M c = 3R/M c = 3 R / M . Water: about 9 R 9R 9 R . Low temperatures: turning and vibrating freeze, and C C C falls. Quantum theory explains it. Mean free path λ = 1 2 π d 2 n \lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n} λ = 2 π d 2 n 1 , where d d d is the diameter. With n = p / k B T n = p/k_BT n = p / k B T : λ = k B T 2 π d 2 p \lambda = \dfrac{k_BT}{\sqrt{2}\,\pi d^2 p} λ = 2 π d 2 p k B T . Collisions each second: ν = v ˉ / λ \nu = \bar v/\lambda ν = v ˉ / λ . At a fixed n n n , λ \lambda λ does not change with T T T . A worked example 2.0 mol 2.0\ \text{mol} 2.0 mol of neon is mixed with 1.0 mol 1.0\ \text{mol} 1.0 mol of rigid carbon dioxide. Find γ \gamma γ .
Each gas. Neon: C v = 3 2 R C_v = \tfrac32 R C v = 2 3 R . Carbon dioxide: C v = 5 2 R C_v = \tfrac52 R C v = 2 5 R . The mixture. The top line is 2.0 mol × 1.5 R + 1.0 mol × 2.5 R = 5.5 mol × R 2.0\ \text{mol} \times 1.5R + 1.0\ \text{mol} \times 2.5R = 5.5\ \text{mol} \times R 2.0 mol × 1.5 R + 1.0 mol × 2.5 R = 5.5 mol × R . So C v = 5.5 mol × R 3.0 mol = 1.83 R C_v = \dfrac{5.5\ \text{mol} \times R}{3.0\ \text{mol}} = 1.83R C v = 3.0 mol 5.5 mol × R = 1.83 R . Gamma. C p = 2.83 R C_p = 2.83R C p = 2.83 R , so γ = 2.83 1.83 = 1.55 \gamma = \dfrac{2.83}{1.83} = 1.55 γ = 1.83 2.83 = 1.55 . Practice with new numbers
A program to try The program works through the example above.
Run it. Compare each line with the worked example.
1. Add (1.0, 6) for one mole of rigid methane. Does gamma rise or fall?
Mistakes to avoid Giving each term k B T k_BT k B T . It is 1 2 k B T \tfrac12 k_BT 2 1 k B T . Counting a vibration once. It counts twice. Averaging γ \gamma γ for a mixture. Average C v C_v C v . Using the radius in λ \lambda λ . Use the diameter. Check yourself How many degrees of freedom does a rigid straight molecule have? Stuck? Show a hint Show the answerWhat is C v C_v C v of a solid, for each mole? Stuck? Show a hint Show the answerThe pressure of a gas halves at a fixed temperature. What happens to λ \lambda λ ? Stuck? Show a hint Show the answerWhy does C v C_v C v of hydrogen fall at low temperatures? Stuck? Show a hint Show the answer