Let's solve each part step by step: --- ### **Given:** - 1D infinite potential well: \( V(x) = 0 \) for \( 0 < x < L \), \( \infty \) otherwise. - Energy levels: \( E_n = \epsilon n^2 \), where \( \epsilon = \frac{\hbar^2 \pi^2}{2mL^2} \), \( n = 1, 2, \ldots \) - Only lowest two energy levels important (\( \epsilon \gg kT \)), so \( E_1 = \epsilon \), \( E_2 = 4\epsilon \). --- ## **Part (a): Partition function** ### **Single Particle Partition Function** \[ z = \sum_{n=1}^{2} e^{-E_n / kT} = e^{-\epsilon / kT} + e^{-4\epsilon / kT} \] ### **N Distinguishable Particles** \[ Z = z^N = \left( e^{-\epsilon / kT} + e^{-4\epsilon / kT} \right)^N \] ### **Logarithm of Z** \[ \ln Z = N \ln \left( e^{-\epsilon / kT} + e^{-4\epsilon / kT} \right) \] Factor \( e^{-\epsilon / kT} \) out: \[ = N \ln \left[ e^{-\epsilon / kT} \left( 1 + e^{-3\epsilon / kT} \right) \right] \] \[ = N \left[ -\frac{\epsilon}{kT} + \ln\left( 1 + e^{-3\epsilon / kT} \right) \right] \] When \( e^{-3\epsilon / kT} \ll 1 \) (since \( \epsilon \gg kT \)), use \( \ln(1+x) \approx x \): \[ \ln Z \approx N \left( -\frac{\epsilon}{kT} + e^{-3\epsilon / kT} \right) \] --- ### **Final answer for (a):** \[ \boxed{ \ln Z = N \left( -\frac{\epsilon}{kT} + e^{-3\epsilon / kT} \right) } \] --- ## **Part (b): Internal energy near \( T = 0 \)** Recall, \[ U = -\frac{\partial}{\partial \beta} \ln Z, \quad \beta = \frac{1}{kT} \] From (a): \[ \ln Z = N \left( -\epsilon \beta + e^{-3\epsilon \beta} \right) \] \[ U = -\frac{\partial}{\partial \beta} \ln Z = -N \left( -\epsilon + \frac{\partial}{\partial \beta} e^{-3\epsilon \beta} \right) \] \[ \frac{\partial}{\partial \beta} e^{-3\epsilon \beta} = -3\epsilon e^{-3\epsilon \beta} \] So, \[ U = N \left[ \epsilon + 3\epsilon e^{-3\epsilon \beta} \right] \] \[ = N\epsilon \left( 1 + 3 e^{-3\epsilon / kT} \right) \] --- ### **Final answer for (b):** \[ \boxed{ U = N\epsilon \left( 1 + 3 e^{-3\epsilon / kT} \right) } \] --- ## **Part (c): Show \( pL = 2U \) ** Recall: \[ p = kT \frac{\partial}{\partial L} \ln Z \] But \( \epsilon = \frac{\hbar^2 \pi^2}{2mL^2} \implies \frac{\partial \epsilon}{\partial L} = -\frac{2\epsilon}{L} \) From (a): \[ \ln Z = N \left( -\frac{\epsilon}{kT} + e^{-3\epsilon / kT} \right) \] So, \[ \frac{\partial \ln Z}{\partial L} = N \left( -\frac{1}{kT} \frac{\partial \epsilon}{\partial L} + \frac{\partial}{\partial L} e^{-3\epsilon / kT} \right) \] Calculate: \[ \frac{\partial}{\partial L} e^{-3\epsilon/kT} = e^{-3\epsilon/kT} \left( -\frac{3}{kT} \frac{\partial \epsilon}{\partial L} \right) \] \[ = -\frac{3}{kT} \frac{\partial \epsilon}{\partial L} e^{-3\epsilon/kT} \] So, \[ \frac{\partial \ln Z}{\partial L} = N \left( -\frac{1}{kT} \frac{\partial \epsilon}{\partial L} - \frac{3}{kT} \frac{\partial \epsilon}{\partial L} e^{-3\epsilon / kT} \right) \] \[ = -N \frac{1}{kT} \frac{\partial \epsilon}{\partial L} \left( 1 + 3 e^{-3\epsilon / kT} \right) \] Therefore \[ p = kT \frac{\partial \ln Z}{\partial L} = -N \frac{\partial \epsilon}{\partial L} \left( 1 + 3 e^{-3\epsilon / kT} \right) \] Recall: \( \frac{\partial \epsilon}{\partial L} = -\frac{2\epsilon}{L} \) So: \[ p = N \frac{2\epsilon}{L} \left( 1 + 3 e^{-3\epsilon / kT} \right) \] Multiply both sides by \( L \): \[ pL = 2N\epsilon \left( 1 + 3 e^{-3\epsilon / kT} \right) \] But from (b), \( U = N\epsilon (1 + 3 e^{-3\epsilon / kT}) \). Thus, \[ \boxed{ pL = 2U } \] --- ## **Part (d): T → 0 limit** As \( T \to 0 \), \( e^{-3\epsilon / kT} \to 0 \): - \( U \to N\epsilon \) (ground state energy, nonzero) - \( pL \to 2N\epsilon \) (nonzero) ### **Why do energy and pressure not vanish?** In quantum systems, even at \( T = 0 \), particles occupy the lowest allowed quantum state (ground state), which has nonzero energy due to quantum confinement (zero-point energy). In a classical ideal gas, particles can have zero energy at \( T = 0 \), but in quantum systems, the uncertainty principle forbids this. The pressure is nonzero because the particles still exert force on the walls due to their zero-point motion. --- ### **Summary of Answers:** **(a):** \(\ln Z = N \left( -\frac{\epsilon}{kT} + e^{-3\epsilon / kT} \right)\) **(b):** \(U = N\epsilon (1 + 3 e^{-3\epsilon / kT})\) **(c):** \(pL = 2U\) **(d):** Neither internal energy nor pressure vanishes as \(T \to 0\) because of quantum zero-point energy: distinguishable quantum particles cannot have zero energy due to confinement in the box (uncertainty principle), so both remain nonzero. In contrast, a classical ideal gas could have zero energy and pressure at \(T = 0\).