Let's solve each part step by step, as requested. **Given Data:** - \( n = 2.25 \) moles - \( C_{V,m} = \frac{3R}{2} \) - Initial: \( T_1 = 310\,K \), \( P_1 = 1.00\,\text{bar} \) - \( R = 0.08314\ \text{L·bar·K}^{-1}\text{·mol}^{-1} \) (for consistency in units) --- ## (a) Heated to 675 K at constant external pressure of 1.00 bar #### **Step 1: Find Initial and Final Volumes** \[ P_1 V_1 = nRT_1 \implies V_1 = \frac{nRT_1}{P_1} \] \[ V_1 = \frac{2.25 \times 0.08314 \times 310}{1.00} = 58.05\ \text{L} \] At constant pressure, final volume: \[ V_2 = \frac{nRT_2}{P_1} = \frac{2.25 \times 0.08314 \times 675}{1.00} = 126.22\ \text{L} \] --- #### **Step 2: Calculate \(q\) (heat)** At constant pressure, \[ q = nC_{P,m}\Delta T \] For a monatomic ideal gas, \(C_{P,m} = C_{V,m} + R = \frac{3R}{2} + R = \frac{5R}{2}\): \[ q = 2.25 \times \frac{5 \times 0.08314}{2} \times (675 - 310) \] \[ = 2.25 \times 0.20785 \times 365 = 2.25 \times 75.865 = 170.7\ \text{L·bar} \] --- #### **Step 3: Calculate \(w\) (work)** At constant pressure: \[ w = -P_{ext} \Delta V = -1.00 \times (126.22 - 58.05) = -68.17\ \text{L·bar} \] --- #### **Step 4: Calculate \(\Delta U\) (internal energy change)** \[ \Delta U = nC_{V,m} \Delta T = 2.25 \times \frac{3 \times 0.08314}{2} \times 365 \] \[ = 2.25 \times 0.12471 \times 365 = 2.25 \times 45.92 = 103.3\ \text{L·bar} \] --- #### **Step 5: Calculate \(\Delta H\) (enthalpy change)** \[ \Delta H = nC_{P,m} \Delta T = q = 170.7\ \text{L·bar} \] --- #### **Step 6: Calculate \(\Delta S\) (entropy change)** \[ \Delta S = nC_{P,m} \ln\left(\frac{T_2}{T_1}\right) \] \[ = 2.25 \times 0.20785 \times \ln\left(\frac{675}{310}\right) \] \[ = 2.25 \times 0.20785 \times 0.775 = 2.25 \times 0.1612 = 0.3627\ \text{L·bar·K}^{-1} \] --- ### **Summary of Answers for (a):** - **q = 170.7 L·bar** - **w = -68.17 L·bar** - **ΔU = 103.3 L·bar** - **ΔH = 170.7 L·bar** - **ΔS = 0.3627 L·bar·K⁻¹** --- **Would you like to proceed with part (b) in the same detailed way?**

Let's proceed with parts (b) and (c) step by step, providing detailed calculations and explanations for each. --- ## **Given Data Recap:** - Number of moles: \( n = 2.25 \) mol - Molar heat capacities: \( C_{V,m} = \frac{3R}{2} \), \( C_{P,m} = C_{V,m} + R = \frac{5R}{2} \) - Gas constant: \( R = 0.08314\, \text{L·bar·K}^{-1} \) - Initial conditions: - Temperature: \( T_1 = 310\,K \) - Pressure: \( P_1 = 1.00\, \text{bar} \) - Volume: \( V_1 = \frac{nRT_1}{P_1} = \frac{2.25 \times 0.08314 \times 310}{1.00} \approx 58.05\, \text{L} \) --- # **Part (b): Heating at constant volume to 675 K** ### **Step 1: Final temperature and volume** - Final temperature: \( T_2 = 675\,K \) - Since volume is constant: \( V_2 = V_1 = 58.05\,L \) ### **Step 2: Find the change in internal energy \(\Delta U\)** \[ \Delta U = n C_{V,m} \Delta T \] \[ = 2.25 \times \frac{3R}{2} \times (T_2 - T_1) \] \[ = 2.25 \times 0.12471 \times (675 - 310) = 2.25 \times 0.12471 \times 365 \] \[ = 2.25 \times 45.52 \approx 102.4\, \text{L·bar} \] *Explanation:* Internal energy depends only on temperature for an ideal gas, so the change is proportional to \(\Delta T\). --- ### **Step 3: Calculate heat transferred \(q\)** At constant volume, \(q = \Delta U\): \[ q = 102.4\, \text{L·bar} \] *Explanation:* Since volume is constant, the heat added equals the change in internal energy. --- ### **Step 4: Calculate work \(w\)** Work done at constant volume: \[ w = 0 \] *Explanation:* No volume change, so no work is done. --- ### **Step 5: Enthalpy change \(\Delta H\)** \[ \Delta H = n C_{P,m} \Delta T = 2.25 \times 0.20785 \times 365 \approx 2.25 \times 75.92 = 171\, \text{L·bar} \] *Explanation:* Enthalpy change depends on temperature change and molar heat capacity at constant pressure. --- ### **Step 6: Entropy change \(\Delta S\)** \[ \Delta S = n C_{P,m} \ln \left( \frac{T_2}{T_1} \right) = 2.25 \times 0.20785 \times \ln \left( \frac{675}{310} \right) \] \[ = 0.468 \times \ln(2.177) \approx 0.468 \times 0.776 = 0.363\, \text{L·bar·K}^{-1} \] *Explanation:* Entropy increases with temperature; since volume is constant, the entropy change is due solely to temperature. --- ## **Summary for Part (b):** | Quantity | Value | Explanation | |------------|---------|--------------| | \( q \) | **102.4 L·bar** | Heat added at constant volume equals \(\Delta U\) | | \( w \) | **0** | No work done at constant volume | | \( \Delta U \) | **102.4 L·bar** | Internal energy change due to temperature increase | | \( \Delta H \) | **171 L·bar** | Enthalpy change for temperature increase | | \( \Delta S \) | **0.363 L·bar·K\(^{-1}\)** | Entropy increase due to temperature rise | --- # **Part (c): Reversible isothermal expansion at 310 K until pressure is one-third of initial** ### **Step 1: Initial conditions** - Initial pressure: \( P_1 = 1.00\, \text{bar} \) - Final pressure: \( P_2 = \frac{1}{3} P_1 = \frac{1}{3} \times 1.00 = 0.333\, \text{bar} \) - Temperature: \( T = 310\,K \) ### **Step 2: Find final volume \( V_2 \)** Since the process is isothermal: \[ P_1 V_1 = P_2 V_2 \Rightarrow V_2 = V_1 \times \frac{P_1}{P_2} = 58.05 \times \frac{1}{\frac{1}{3}} = 58.05 \times 3 = 174.15\, \text{L} \] *Explanation:* Volume increases as pressure decreases at constant temperature for an ideal gas. --- ### **Step 3: Calculate work \(w\)** For isothermal expansion: \[ w = - nRT \ln \left( \frac{V_2}{V_1} \right) \] Alternatively, in terms of pressure: \[ w = - nRT \ln \left( \frac{P_2}{P_1} \right) \] Using pressure: \[ w = - 2.25 \times 0.08314 \times 310 \times \ln \left( \frac{0.333}{1} \right) \] \[ = - 2.25 \times 25.8 \times (-1.0986) \approx - 58.05 \times (-1.0986) = 63.8\, \text{L·bar} \] *Explanation:* Work done by the gas during expansion is positive (since the gas expands). --- ### **Step 4: Calculate heat \(q\)** For isothermal process: \[ \Delta U = 0 \] \[ q = -w = 63.8\, \text{L·bar} \] *Explanation:* Since internal energy doesn't change, the heat supplied equals the work done. --- ### **Step 5: Calculate entropy change \(\Delta S\)** \[ \Delta S = n R \ln \left( \frac{V_2}{V_1} \right) = n R \ln \left( \frac{P_1}{P_2} \right) \] \[ = 2.25 \times 0.08314 \times \ln(3) \approx 2.25 \times 0.08314 \times 1.0986 \approx 2.25 \times 0.0913 = 0.205\, \text{L·bar·K}^{-1} \] *Explanation:* Entropy increases as the gas expands at constant temperature. --- ## **Summary for Part (c):** | Quantity | Value | Explanation | |------------|---------|--------------| | \( w \) | **63.8 L·bar** | Work done by the gas during expansion | | \( q \) | **63.8 L·bar** | Heat absorbed during the process | | \( \Delta U \) | **0** | No change in internal energy (isothermal) | | \( \Delta H \) | **0** | Enthalpy remains unchanged at constant T | | \( \Delta S \) | **0.205 L·bar·K\(^{-1}\)** | Entropy increases due to expansion | --- **Final notes:** - All calculations assume ideal gas behavior. - Units are consistently in L·bar for energy-related quantities. - For real applications, consider unit conversions if necessary. --- If you'd like, I can also convert these values into other units (e.g., Joules) or provide additional explanations.

The problem involves analyzing thermodynamic processes of an ideal gas using fundamental principles and formulas. The key concepts applied include the ideal gas law, which relates pressure, volume, and temperature (PV = nRT), allowing calculation of initial and final states. For each process, the first law of thermodynamics (ΔU = q + w) is employed, where the change in internal energy depends solely on temperature for an ideal gas. In isobaric heating (part a), the heat added (q) is calculated from the product of the molar heat capacity at constant pressure (C_P,m) and the temperature change, while work done is derived from pressure and volume change. For constant-volume heating (part b), the internal energy change equals heat added, with no work performed. In the isothermal expansion (part c), since temperature is constant, internal energy remains unchanged, and the work done by the gas equals the heat absorbed, calculated from the natural logarithm of the volume or pressure ratio. Entropy changes are evaluated using the relation ΔS = nC_P,m ln(T₂/T₁) for temperature variations and ΔS = nR ln(V₂/V₁) for volume changes, reflecting the disorder or dispersal of energy during the processes. These concepts collectively facilitate the quantitative analysis of heat, work, internal energy, enthalpy, and entropy changes during each transformation.

Certainly! Here's a concise summary of the final answers for each part without a table: **Part (a):** When 2.25 moles of the ideal gas is heated from 310 K to 675 K at constant pressure of 1.00 bar, the heat added (q) is approximately **170.7 L·bar**, the work done (w) is about **-68.17 L·bar** (work done by the gas), the change in internal energy (ΔU) is roughly **103.3 L·bar**, the enthalpy change (ΔH) is approximately **170.7 L·bar**, and the entropy change (ΔS) is about **0.3627 L·bar·K⁻¹**. **Part (b):** Heating the same amount of gas from 310 K to 675 K at constant volume results in a heat input (q) of around **102.4 L·bar**, no work is performed (w = 0), the internal energy increases by approximately **102.4 L·bar**, the enthalpy change (ΔH) is about **171 L·bar**, and the entropy increases by approximately **0.363 L·bar·K⁻¹**. **Part (c):** During a reversible isothermal expansion from 310 K where pressure drops to one-third (0.333 bar), the work done by the gas is approximately **63.8 L·bar**, the heat absorbed (q) is also about **63.8 L·bar**, the internal energy remains unchanged (ΔU = 0), and the entropy increases by roughly **0.205 L·bar·K⁻¹**. These results are based on ideal gas assumptions and standard thermodynamic relations.

Part (a): Heat added (q) ≈ 170.7 L·bar Work done (w) ≈ -68.17 L·bar Change in internal energy (ΔU) ≈ 103.3 L·bar Change in enthalpy (ΔH) ≈ 170.7 L·bar Change in entropy (ΔS) ≈ 0.3627 L·bar·K⁻¹ Part (b): Heat added (q) ≈ 102.4 L·bar Work done (w) = 0 L·bar Change in internal energy (ΔU) ≈ 102.4 L·bar Change in enthalpy (ΔH) ≈ 171 L·bar Change in entropy (ΔS) ≈ 0.363 L·bar·K⁻¹ Part (c): Work done by gas (w) ≈ 63.8 L·bar Heat absorbed (q) ≈ 63.8 L·bar Change in internal energy (ΔU) = 0 Change in enthalpy (ΔH) = 0 Change in entropy (ΔS) ≈ 0.205 L·bar·K⁻¹