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Thermodynamics Homework 2

  • Student ID: 1W23CF13
  • Name: TAI Yungche
  • Date: 2025/06/29

Question 1

\(c_V = C_V \times \frac{\mu}{1000 \times N_A}\)

  • Water (H₂O): \(C_V = 4180 \frac{\text{J}}{\text{kg} \cdot \text{K}} \times \frac{18 \, \text{g/mol}}{1000 \, \text{g/kg} \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 1.25 \times 10^{-22} \frac{\text{J}}{\text{K}}\)

  • Copper (Cu): \(C_V = 385 \frac{\text{J}}{\text{kg} \cdot \text{K}} \times \frac{63.5 \, \text{g/mol}}{1000 \, \text{g/kg} \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 4.06 \times 10^{-23} \frac{\text{J}}{\text{K}}\)

  • Mercury (Hg): \(C_V = 140 \frac{\text{J}}{\text{kg} \cdot \text{K}} \times \frac{200.6 \, \text{g/mol}}{1000 \, \text{g/kg} \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 4.66 \times 10^{-23} \frac{\text{J}}{\text{K}}\)

  • Aluminium (Al): \(C_V = 897 \frac{\text{J}}{\text{kg} \cdot \text{K}} \times \frac{27 \, \text{g/mol}}{1000 \, \text{g/kg} \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 4.02 \times 10^{-23} \frac{\text{J}}{\text{K}}\)

  • Methanol (CH₃OH): \(C_V = 2140 \frac{\text{J}}{\text{kg} \cdot \text{K}} \times \frac{32 \, \text{g/mol}}{1000 \, \text{g/kg} \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 1.14 \times 10^{-22} \frac{\text{J}}{\text{K}}\)

  • Magnesium (Mg): \(C_V = 1020 \frac{\text{J}}{\text{kg} \cdot \text{K}} \times \frac{24.3 \, \text{g/mol}}{1000 \, \text{g/kg} \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 4.11 \times 10^{-23} \frac{\text{J}}{\text{K}}\)

Question 2

The equipartition theorem says that each quadratic degree of freedom of a molecule contributes (1/2)k_B T to its average internal energy, where k_B is the Boltzmann constant.

The internal energy per molecule is U_molecule = (f/2) k_B T.

\(c_V = \frac{dU_{\text{molecule}}}{dT} = \frac{f}{2} k_B\)

\(f = \frac{2 c_V}{k_B}\)

  • Water (H₂O): \(f_{\text{water}} = \frac{2 \times (1.25 \times 10^{-22} \, \text{J/K})}{1.381 \times 10^{-23} \, \text{J/K}} \approx 18.1\)

  • Copper (Cu): \(f_{\text{Cu}} = \frac{2 \times (4.06 \times 10^{-23} \, \text{J/K})}{1.381 \times 10^{-23} \, \text{J/K}} \approx 5.88\)

  • Mercury (Hg): \(f_{\text{Hg}} = \frac{2 \times (4.66 \times 10^{-23} \, \text{J/K})}{1.381 \times 10^{-23} \, \text{J/K}} \approx 6.75\)

  • Aluminium (Al): \(f_{\text{Al}} = \frac{2 \times (4.02 \times 10^{-23} \, \text{J/K})}{1.381 \times 10^{-23} \, \text{J/K}} \approx 5.82\)

  • Methanol (CH₃OH): \(f_{\text{CH}_3\text{OH}} = \frac{2 \times (1.14 \times 10^{-22} \, \text{J/K})}{1.381 \times 10^{-23} \, \text{J/K}} \approx 16.5\)

  • Magnesium (Mg): \(f_{\text{Mg}} = \frac{2 \times (4.11 \times 10^{-23} \, \text{J/K})}{1.381 \times 10^{-23} \, \text{J/K}} \approx 5.95\)

Question 3

Solid Materials:

  • Copper (Cu): f ≈ 5.88
  • Aluminium (Al): f ≈ 5.82
  • Magnesium (Mg): f ≈ 5.95

Liquid Materials:

  • Water (H₂O): f ≈ 18.1
  • Mercury (Hg): f ≈ 6.75
  • Methanol (CH₃OH): f ≈ 16.5

Solid Materials

For the three solid materials (Copper, Aluminium, Magnesium), the calculated number of degrees of freedom is very close to 6.

So yes, there is a clear universal behavior. In a crystalline solid, each atom is modeled as a three-dimensional harmonic oscillator.

For each of the three dimensions (x, y, z), the atom has one kinetic energy term and one potential energy term. This gives total degrees of freedom: f = 3 (kinetic) + 3 (potential) = 6. The experimental values (f ≈ 5.8-6.0) align with the theoretical prediction.

This means that at the temperatures where these heat capacities were measured, the equipartition theorem gives a very good model for simple monatomic solids.

Liquid Materials

For the three liquid materials (Water, Mercury, Methanol), the calculated degrees of freedom vary a lot. f for Mercury is ~6.8, while f for Water and Methanol are much larger at ~18.1 and ~16.5.

So no, there is no simple universal behavior for liquids. The wide range of f values (6.75 to 18.1) shows that unlike simple solids, there is no single number that characterizes the degrees of freedom for all liquids. The behavior is specific to the molecular structure of each substance.

This is because the number of degrees of freedom in a liquid is highly dependent on the complexity of its constituent molecules and the nature of intermolecular interactions.

Question 4

\(W = \int_{V_i}^{V_f} P \, dV\)

\(PV^{4/3} = \text{const}\)

Call this constant C:

\(P(V) = \frac{C}{V^{4/3}} = \frac{P_i V_i^{4/3}}{V^{4/3}}\)

\(W = \int_{V_i}^{V_f} \frac{P_i V_i^{4/3}}{V^{4/3}} \, dV\)

\(W = P_i V_i^{4/3} \int_{V_i}^{V_f} V^{-4/3} \, dV\)

\(\int V^{-4/3} \, dV = \frac{V^{-4/3 + 1}}{-4/3 + 1} = \frac{V^{-1/3}}{-1/3} = -3V^{-1/3}\)

\(W = P_i V_i^{4/3} \left[ -3V^{-1/3} \right]_{V_i}^{V_f}\) \(W = P_i V_i^{4/3} \left( -3V_f^{-1/3} - (-3V_i^{-1/3}) \right)\) \(W = 3 P_i V_i^{4/3} \left( V_i^{-1/3} - V_f^{-1/3} \right)\) \(W = 3 \left( P_i V_i^{4/3} V_i^{-1/3} - P_i V_i^{4/3} V_f^{-1/3} \right)\) \(W = 3 \left( P_i V_i - (P_i V_i^{4/3}) V_f^{-1/3} \right)\)

\(W = 3 \left( P_i V_i - (P_f V_f^{4/3}) V_f^{-1/3} \right)\) \(W = 3 \left( P_i V_i - P_f V_f \right)\)

This is the total mechanical work done by the gas.

Question 5

Work done by the gas is converted into the kinetic energy of the piston:

\(dW = d\left(\frac{Mu^2}{2}\right)\)

\(W = \Delta K = K_f - K_i\)

The piston is initially at rest, so its initial velocity u_i = 0 and its initial kinetic energy K_i = 0. The final kinetic energy is K_f = (1/2)Mu_f².

\(W = \frac{1}{2}Mu_f^2\)

\(W_{\text{total}} = \lim_{V_f \to \infty} 3(P_iV_i - P_fV_f)\)

From the adiabatic relation \(P_f V_f^{4/3} = P_i V_i^{4/3}\), we have \(P_f = P_i \left(\frac{V_i}{V_f}\right)^{4/3}\).

So, \(P_f V_f = P_i \left(\frac{V_i}{V_f}\right)^{4/3} V_f = P_i V_i \left(\frac{V_i}{V_f}\right)^{1/3}\).

As \(V_f \to \infty\), the ratio \(\left(\frac{V_i}{V_f}\right) \to 0\), so the term \(P_f V_f \to 0\).

\(W_{\text{total}} = 3(P_iV_i - 0) = 3 P_i V_i\)

Since \(P_i V_i = NkT\)

\(W_{\text{total}} = 3NkT\)

\(\frac{1}{2}Mu_f^2 = 3NkT\)

Solve for u_f:

\(u_f^2 = \frac{6NkT}{M}\)

\(u_f = \sqrt{\frac{6NkT}{M}}\)

Question 6

A molecule composed of N atoms has a total of 3N degrees of freedom. For this molecule, N=4, so the total number of degrees of freedom is 12.

These total degrees of freedom are distributed among translational, rotational, and vibrational motions.

Rotational Degrees of Freedom is the number of rotational degrees of freedom depends on the geometry of the molecule.

The problem says the four identical atoms are arranged in a square. This is a non-linear, planar molecule.

A non-linear molecule can rotate about all three principal axes of inertia.

\(f_{\text{rot}} = 3\)

Oscillational (Vibrational) Degrees of Freedom is the remaining degrees of freedom must be vibrational.

\(f_{\text{osc}} = (\text{Total DoF}) - f_{\text{trans}} - f_{\text{rot}}\)

\(f_{\text{osc}} = 3N - 3 - 3\)

\(f_{\text{osc}} = 12 - 3 - 3 = 6\)

\(f_{\text{osc}} = 6\)

Question 7

In thermal equilibrium, the total energy is shared equally among all its various forms. For each quadratic degree of freedom, the average energy is (1/2)k_B T, where k_B is the Boltzmann constant. This means each such degree of freedom contributes (1/2)k_B to the heat capacity per molecule (c_V).

  • Number of translational degrees of freedom = 3
  • Number of rotational degrees of freedom = 3

Total number of degrees of freedom:

\(f_{\text{rot+linear}} = f_{\text{trans}} + f_{\text{rot}} = 3 + 3 = 6\)

Heat capacity contribution per molecule:

\(c_V^{\text{rot+linear}} = f_{\text{rot+linear}} \times \frac{1}{2} k_B\)

\(c_V^{\text{rot+linear}} = 6 \times \frac{1}{2} k_B = 3k_B\)

\(c_V^{\text{rot+linear}} = 3k_B\)

Question 8

Each vibrational degree of freedom contributes 2 × (1/2)k_B = k_B to the heat capacity.

  • Contribution from translation: \(c_V^{\text{trans}} = f_{\text{trans}} \times \frac{1}{2} k_B = 3 \times \frac{1}{2} k_B = \frac{3}{2} k_B\)
  • Contribution from rotation: \(c_V^{\text{rot}} = f_{\text{rot}} \times \frac{1}{2} k_B = 3 \times \frac{1}{2} k_B = \frac{3}{2} k_B\)
  • Contribution from oscillation: \(c_V^{\text{osc}} = f_{\text{osc}} \times k_B = 6 \times k_B = 6k_B\)

Total heat capacity per molecule is the sum:

\(c_V^{\text{total}} = \frac{3}{2} k_B + \frac{3}{2} k_B + 6k_B = 3k_B + 6k_B = 9k_B\)

Question 9

The heat capacity ratio γ is defined as the ratio of the heat capacity at constant pressure (c_P) to the heat capacity at constant volume (c_V):

\(\gamma = \frac{c_P}{c_V}\)

For an ideal gas, Mayer's relation for a single molecule is c_P - c_V = k_B, which means c_P = c_V + k_B:

\(\gamma = \frac{c_V + k_B}{c_V} = 1 + \frac{k_B}{c_V}\)

Calculate γ for both cases:

Case 1: Without oscillations (γ^(rot+linear))

Use the heat capacity calculated in question 7:

\(c_V^{\text{rot+linear}} = 3k_B\)

\(\gamma^{\text{rot+linear}} = 1 + \frac{k_B}{c_V^{\text{rot+linear}}} = 1 + \frac{k_B}{3k_B} = 1 + \frac{1}{3} = \frac{4}{3}\)

Case 2: With oscillations included (γ^(total))

Use the heat capacity calculated in question 8:

\(c_V^{\text{total}} = 9k_B\)

\(\gamma^{\text{total}} = 1 + \frac{k_B}{c_V^{\text{total}}} = 1 + \frac{k_B}{9k_B} = 1 + \frac{1}{9} = \frac{10}{9}\)