Let's solve the problem step by step: --- ### (a) **Classical Oscillators & Equipartition Theorem** #### **Step 1: Model and Equipartition Theorem** - Each carbon atom in graphene can oscillate in two dimensions (x and y). - Each direction of oscillation is a harmonic oscillator. The equipartition theorem states **each quadratic degree of freedom contributes \(\frac{1}{2}k_B T\)** to the energy. For a 2D oscillator: - Each direction has kinetic and potential energy (\(x\) and \(y\)). - So, each atom has **4 quadratic degrees of freedom** (2 kinetic + 2 potential, one per direction). #### **Step 2: Total Energy Per Atom** \[ E_{\text{atom}} = 4 \times \frac{1}{2} k_B T = 2 k_B T \] #### **Step 3: Molar Energy** For one mole, the total energy: \[ E_{\text{mole}} = N_a \times 2k_B T = 2N_a k_B T = 2RT \] where \( R = N_a k_B \) is the gas constant. #### **Step 4: Molar Specific Heat at Constant Volume** \[ C_{V, \text{mole}} = \left(\frac{\partial E_{\text{mole}}}{\partial T}\right)_V = 2R \] #### **Step 5: Discussion and Third Law** - **Why is this not correct for all \(T\)?** At very low temperatures, quantum effects become important, and not all degrees of freedom are accessible. - **Third Law of Thermodynamics:** The third law states that as \(T \to 0\), the entropy (and specific heat) should also approach zero. Here, \(C_V = 2R\) even as \(T \to 0\), which violates the third law. --- #### **Summary for (a):** \[ \boxed{C_{V, \text{mole}}^{\text{classical}} = 2R} \] - This result is inconsistent with the third law of thermodynamics because it does not go to zero as \(T \to 0\). --- ### (b) **Quantum Harmonic Oscillator** #### **Step 1: Quantum Energy Levels** The energy levels for a 2D quantum harmonic oscillator: \[ E_n = \hbar \omega \left(n + \frac{1}{2}\right) \] For one oscillator, the average energy: \[ \langle E \rangle = \frac{\hbar \omega}{e^{\beta \hbar \omega} - 1} + \frac{1}{2} \hbar \omega \] But the zero-point energy (\(\frac{1}{2}\hbar\omega\)) does not contribute to specific heat. For 2D: Each atom has two oscillators: \[ \langle E_{\text{atom}} \rangle = 2 \cdot \frac{\hbar \omega}{e^{\beta \hbar \omega} - 1} \] where \(\beta = 1/(k_B T)\). #### **Step 2: Molar Energy** \[ E_{\text{mole}} = N_a \langle E_{\text{atom}} \rangle = 2N_a \frac{\hbar \omega}{e^{\beta \hbar \omega} - 1} \] #### **Step 3: Molar Specific Heat** \[ C_{V, \text{mole}} = \left(\frac{\partial E_{\text{mole}}}{\partial T}\right)_V \] Let \(x = \frac{\hbar \omega}{k_B T}\): \[ C_{V, \text{mole}} = 2N_a k_B \left( \frac{x^2 e^x}{(e^x - 1)^2} \right) \] Substituting \(N_a k_B = R\): \[ \boxed{C_{V, \text{mole}}^{\text{quantum}} = 2R \left( \frac{x^2 e^x}{(e^x - 1)^2} \right), \quad x = \frac{\hbar \omega}{k_B T}} \] #### **Step 4: Discussion** - **Consistency with Third Law:** As \(T \to 0\), \(x \to \infty\), \(C_V \to 0\). Thus, this obeys the third law. - **Low-Temperature Behavior:** For very low \(T\), this result drops off exponentially. However, it does not capture all the low-temperature physics of real graphene (such as phonon dispersion, collective excitations, etc.), but it is a significant improvement over the classical model. --- ## **Final Answers** ### **(a)** \[ \boxed{C_{V, \text{mole}}^{\text{classical}} = 2R} \] - Not valid at low \(T\), violates the third law. ### **(b)** \[ \boxed{C_{V, \text{mole}}^{\text{quantum}} = 2R \left( \frac{x^2 e^x}{(e^x - 1)^2} \right), \quad x = \frac{\hbar \omega}{k_B T}} \] - Consistent with the third law; at low \(T\), \(C_V \to 0\), but does not account for all real low-temperature effects.