Let's solve **each part step by step**: --- ##a. Number of Independent Configurations** **Given:** - Volume has \( V \) lattice sites. - Polymer of length \( N \) (i.e., \( N \) monomers). - Coordination number \( Z = 6 \) (each site has 6 neighbors in 3D cubic lattice). - Ideal random walk (not self-avoiding; monomers can overlap). - Ignore edge effects. - All monomers are identical. ### **Solution Steps** 1. **Random Walk Configurations:** - For each monomer after the first, it can go to any of \( Z \) neighbors. - The first monomer can be placed anywhere: \( V \) choices. - Each subsequent monomer (from 2 to \( N \)) can be placed in any neighbor site (including possibly overlapping previous monomers). 2. **Total Number of Configurations:** - First monomer: \( V \) choices. - Next \( N-1 \) monomers: each has \( Z \) choices. - Total: \( V \cdot Z^{N-1} \) 3. **Is this the density of states?** - **Density of states**: Number of microscopic configurations for a given macrostate. - Here, this is the number of possible configurations, assuming no restrictions except connectivity. - **Edge effects** would reduce the number due to boundaries, but are ignored here. - **Conclusion:** Yes, this is the same as the density of states under these assumptions. ### **Final Answer (a):** \[ \boxed{ \text{Number of independent configurations} = V \cdot Z^{N-1} } \] **Sources Used:** - Doi, M. and Edwards, S.F., *The Theory of Polymer Dynamics* - Standard statistical mechanics texts (e.g., Pathria & Beale) --- ## **b. Density of States and Entropy for Fully Adsorbed Chain** **Now:** - There is a wall at \( z = \). - Polymer is fully adsorbed: every monomer is on the wall (monomers confined to the 2D wall). ### **Solution Steps** 1. **Confinement to the Wall:** - The wall at \( z = \) is a 2D plane. - Each monomer is on the wall: only 2D lattice moves are possible. - In a 2D square lattice, coordination number is \( Z_{2D} = 4 \). 2. **Number of Configurations:** - First monomer: number of sites on wall = \( V_{wall} \). - If the wall is size \( L \times L \), \( V_{wall} = L^2 \). - Each subsequent monomer: 4 choices. - Total: \( V_{wall} \cdot 4^{N-1} \) 3. **Entropy:** - Entropy \( S = k_B \ln(\text{number of configurations}) \) - \( S_{adsorbed} = k_B \ln(V_{wall} \cdot 4^{N-1}) = k_B \ln V_{wall} + k_B (N-1) \ln 4 \) ### **Final Answer (b):** \[ \boxed{ \text{Density of states for fully adsorbed chain} = V_{wall} \cdot 4^{N-1} } \] \[ \boxed{ S_{adsorbed} = k_B \ln V_{wall} + k_B (N-1) \ln 4 } \] **Sources Used:** - See above references. --- ## **c. Canonical Partition Function with Polymer-Wall Attraction** **Now:** - Each adsorbed monomer gains energy \( -\epsilon \). - Fully adsorbed: all \( N \) monomers on wall. ### **Solution Steps** 1. **Energy of Fully Adsorbed Configuration:** - Total energy: \( E = -N\epsilon \) 2. **Partition Function:** - Partition function \( Z \) sums over all microstates: \[ Z = \sum_{\text{states}} e^{-\beta E} \] - All states have energy \( -N\epsilon \), and there are \( V_{wall} \cdot 4^{N-1} \) such states. \[ Z_{adsorbed} = (\text{number of states}) \times e^{-\beta E} \] \[ Z_{adsorbed} = V_{wall} \cdot 4^{N-1} \cdot e^{\beta N\epsilon} \] ### **Final Answer (c):** \[ \boxed{ Z_{adsorbed} = V_{wall} \cdot 4^{N-1} \cdot e^{\beta N\epsilon} } \] where \( \beta = 1/(k_B T) \). --- ## **d. Helmholtz Energy and Favorability Condition** **Now:** - Want the Helmholtz free energy difference between adsorbed and non-adsorbed (bulk) states. - Adsorption favorable if \( \Delta F < \). ### **Solution Steps** 1. **Helmholtz Free Energy:** - \( F = -k_B T \ln Z \) - For adsorbed: \( F_{ads} = -k_B T \ln Z_{adsorbed} \) - For free chain: \( F_{free} = -k_B T \ln Z_{free} \) - \( Z_{free} = V \cdot 6^{N-1} \) 2. **Difference:** \[ \Delta F = F_{ads} - F_{free} = -k_B T \left[ \ln Z_{adsorbed} - \ln Z_{free} \right] \] \[ = -k_B T \left[ \ln \left( \frac{V_{wall} \cdot 4^{N-1} \cdot e^{\beta N\epsilon}}{V \cdot 6^{N-1}} \right) \right] \] \[ = -k_B T \left[ \ln \left( \frac{V_{wall}}{V} \right) + (N-1) \ln \left( \frac{4}{6} \right) + N\beta\epsilon \right] \] Simplify for large \( N \): \[ \Delta F \approx -k_B T \left[ N \ln \left( \frac{2}{3} \right) + N\frac{\epsilon}{k_B T} \right] \] 3. **Minimum \(\epsilon\) for Favorable Adsorption (\(\Delta F < \)):** \[ N \ln \left( \frac{2}{3} \right) + N\frac{\epsilon}{k_B T} > \] \[ \ln \left( \frac{2}{3} \right) + \frac{\epsilon}{k_B T} > \] \[ \frac{\epsilon}{k_B T} > -\ln \left( \frac{2}{3} \right) \] \[ \epsilon > k_B T \ln \left( \frac{3}{2} \right) \] ### **Final Answer (d):** \[ \boxed{ F_{ads} = -k_B T \ln (V_{wall} \cdot 4^{N-1} \cdot e^{\beta N\epsilon}) } \] \[ \boxed{ \text{Minimum value of } \epsilon \text{ for favorable adsorption: } \epsilon > k_B T \ln \left( \frac{3}{2} \right) } \] --- ### **Summary Table** | Part | Answer | |------|-----------------------------------------------------------------------------------------| | a | \( V \cdot 6^{N-1} \) configurations (density of states, with Z=6) | | b | \( V_{wall} \cdot 4^{N-1} \) (density of states); \( S = k_B \ln(V_{wall}) + k_B (N-1) \ln 4 \) | | c | \( Z_{adsorbed} = V_{wall} \cdot 4^{N-1} \cdot e^{\beta N\epsilon} \) | | d | \( F_{ads} = -k_B T \ln Z_{adsorbed} \); adsorption favorable if \( \epsilon > k_BT \ln(3/2) \) | --- **References:** - Doi, M. and Edwards, S.F. *The Theory of Polymer Dynamics* - Rubinstein, M., and Colby, R.H. *Polymer Physics* - Pathria & Beale, *Statistical Mechanics* If you need further explanation for any step, let me know!