Let's solve each part step by step **with explanations after each step** and a final answer at the end of each part. --- ## **Part A: Vibrational Partition Function (\(q_{vib}\))** The vibrational partition function for a diatomic molecule is: \[ q_{vib} = \frac{1}{1 - \exp\left(-\frac{\theta_{vib}}{T}\right)} \] Where: - \(\theta_{vib}\) = vibrational temperature = **2719 K** - \(T\) = temperature = **300 K** ### Step-by-step Solution 1. **Calculate exponent:** \[ \frac{\theta_{vib}}{T} = \frac{2719}{300} \approx 9.063 \] 2. **Calculate exponential term:** \[ \exp(-9.063) \approx 1.16 \times 10^{-4} \] 3. **Calculate denominator:** \[ 1 - 1.16 \times 10^{-4} \approx 0.999884 \] 4. **Calculate partition function:** \[ q_{vib} = \frac{1}{0.999884} \approx 1.00012 \] **Explanation**: The vibrational partition function is just a bit above 1 at room temperature because \(\theta_{vib} \gg T\), so most molecules are in the ground vibrational state. --- ### **Final Answer, Part A:** \[ \boxed{q_{vib} = 1.00012} \] --- ## **Part B: Molar Vibrational Internal Energy (\(U_{vib}\))** The average vibrational energy per molecule is: \[ \langle \varepsilon_{vib} \rangle = \frac{\theta_{vib} k_B}{\exp(\theta_{vib}/T) - 1} \] For 1 mole, multiply by Avogadro's number (\(N_A\)): \[ U_{vib} = N_A \langle \varepsilon_{vib} \rangle = R \frac{\theta_{vib}}{\exp(\theta_{vib}/T) - 1} \] Where: - \(R = 8.314\, \text{J/mol·K}\) ### Step-by-step Solution 1. **Calculate denominator:** \[ \exp(\theta_{vib}/T) - 1 \approx \exp(9.063) - 1 \approx 8640 - 1 = 8639 \] 2. **Calculate numerator:** \[ R \theta_{vib} = 8.314 \times 2719 \approx 22,623\, \text{J/mol} \] 3. **Calculate the internal energy:** \[ U_{vib} = \frac{22,623}{8639} \approx 2.62\, \text{J/mol} \] **Explanation**: The vibrational energy is very small at room temperature because almost all molecules are in the ground state. --- ### **Final Answer, Part B:** \[ \boxed{U_{vib} = 2.62 \ \text{J/mol}} \] --- ## **Part C: Molar Vibrational Heat Capacity (\(C_{V,vib}\))** The heat capacity is: \[ C_{V,vib} = R \left( \frac{\theta_{vib}}{T} \right)^2 \frac{\exp(\theta_{vib}/T)}{[\exp(\theta_{vib}/T) - 1]^2} \] ### Step-by-step Solution 1. **Calculate \(\left(\frac{\theta_{vib}}{T}\right)^2\):** \[ (9.063)^2 \approx 82.18 \] 2. **Calculate \(\exp(\theta_{vib}/T)\):** \[ \exp(9.063) \approx 8640 \] 3. **Calculate denominator:** \[ [\exp(\theta_{vib}/T) - 1]^2 = (8639)^2 \approx 74,690,321 \] 4. **Calculate the whole fraction:** \[ \frac{8640}{74,690,321} \approx 0.0001157 \] 5. **Multiply all together:** \[ C_{V,vib} = 8.314 \times 82.18 \times 0.0001157 \approx 0.079 \ \text{J/mol·K} \] **Explanation**: The vibrational heat capacity is very low at room temperature because very few molecules are thermally excited to higher vibrational levels. --- ### **Final Answer, Part C:** \[ \boxed{C_{V,vib} = 0.079 \ \text{J/mol·K}} \] --- Would you like to continue with Parts D, E, and F? If yes, please specify if you want detailed explanations for those as well!