Let's **solve step-by-step**: --- ## **Given Data** - **Process:** Isothermal reversible compression of ammonia - ** Pressure (\(P_1\**: \(1 \times 10^5\) Pa - **Final Pressure (\(P_2\))**: \(5 \times 10^6\) Pa - **Temperature (T):** 100°C = 373.15 K (isothermal) - **Equation of State (EOS):** \[ Z = 1 + \frac{\alpha P}{RT^2} + \frac{bP}{RT} \] where: - \(\alpha = .4 \, \mathrm{m}^3/\mathrm{mol} \cdot \mathrm{K}\) - \(b = 3.37 \times 10^{-5} \, \mathrm{m}^3/\mathrm{mol}\) --- # **a) Derive the Energy Equation** ### **Assumptions** - Isothermal (\(dT = \)) - Steady-state, reversible process - Neglect changes in kinetic and potential energies ### **General Energy Balance (Steady Flow)** \[ \Delta h = q + w \] For a **reversible, isothermal compression** in a flow system, the change in enthalpy is: \[ \Delta h = \int_{P_1}^{P_2} v \, dP \] Because the process is isothermal, \(T\) is constant. --- ### **Step 1: Find the Molar Volume (\(v\))** From the EOS: \[ Z = \frac{P v}{RT} \implies v = \frac{Z RT}{P} \] Plug the expression for \(Z\): \[ v = \frac{RT}{P} \left[1 + \frac{\alpha P}{RT^2} + \frac{bP}{RT}\right] \] \[ v = \frac{RT}{P} + \frac{\alpha}{T} + b \] --- ### **Step 2: Write the Enthalpy Change Equation** Isothermal process (\(dT=\)): \[ \Delta h = \int_{P_1}^{P_2} v \, dP \] Plug in \(v\): \[ \Delta h = \int_{P_1}^{P_2} \left[ \frac{RT}{P} + \frac{\alpha}{T} + b \right] dP \] --- # **b) Compute the Molar Enthalpy Change (\(\Delta H\))** Let's **integrate each term**: \[ \Delta h = RT \int_{P_1}^{P_2} \frac{1}{P} dP + \frac{\alpha}{T} \int_{P_1}^{P_2} dP + b \int_{P_1}^{P_2} dP \] \[ \Delta h = RT \ln\left(\frac{P_2}{P_1}\right) + \frac{\alpha}{T}(P_2 - P_1) + b(P_2 - P_1) \] --- ## **Substitute Values** - \(R = 8.314 \, \mathrm{J/(mol \cdot K)}\) - \(T = 373.15 \, \mathrm{K}\) - \(P_1 = 1 \times 10^5 \, \mathrm{Pa}\) - \(P_2 = 5 \times 10^6 \, \mathrm{Pa}\) - \(\alpha = .4 \, \mathrm{m}^3/\mathrm{mol} \cdot \mathrm{K}\) - \(b = 3.37 \times 10^{-5} \, \mathrm{m}^3/\mathrm{mol}\) --- ### **Calculate Each Term** #### 1. **First Term** \[ RT \ln\left(\frac{P_2}{P_1}\right) = 8.314 \times 373.15 \times \ln\left(\frac{5 \times 10^6}{1 \times 10^5}\right) \] \[ = 3102.9 \times \ln(50) \approx 3102.9 \times 3.912 \] \[ \approx 12146.5 \ \text{J/mol} \] #### 2. **Second Term** \[ \frac{\alpha}{T} (P_2 - P_1) = \frac{.4}{373.15} \times (5 \times 10^6 - 1 \times 10^5) \] \[ = .001072 \times 4.9 \times 10^6 \approx 5251 \ \text{J/mol} \] #### 3. **Third Term** \[ b (P_2 - P_1) = 3.37 \times 10^{-5} \times 4.9 \times 10^6 \] \[ = 3.37 \times 4.9 \times 10^{1} \approx 16.5 \ \text{J/mol} \] --- ## **Sum All Terms** \[ \Delta h = 12146.5 + 5251 + 16.5 = \boxed{17,414 \ \text{J/mol}} \] --- # **Summary of Answers** ### **a) Energy Equation** \[ \Delta h = RT \ln\left(\frac{P_2}{P_1}\right) + \frac{\alpha}{T}(P_2 - P_1) + b(P_2 - P_1) \] --- ### **b) Molar Enthalpy Change** \[ \Delta h = \boxed{17,414 \ \text{J/mol}} \] Let me know if you need further explanation or details!

# Detailed Solution for Isothermal Compression of Ammonia ## Given Data - **Initial Pressure (\(P_1\))**: \(1 \times 10^5\) Pa - **Final Pressure (\(P_2\))**: \(5 \times 10^6\) Pa - **Temperature (\(T\))**: 100°C = 373.15 K - **Equation of State (EOS)**: \[ Z = 1 + \frac{\alpha P}{RT^2} + \frac{bP}{RT} \] where: - \(\alpha = 0.4 \, \mathrm{m}^3/\mathrm{mol} \cdot \mathrm{K}\) - \(b = 3.37 \times 10^{-5} \, \mathrm{m}^3/\mathrm{mol}\) ## a) Derivation of Energy Equation ### Assumptions - The process is **isothermal** (constant temperature). - The process is **reversible** and at **steady-state**. - Changes in kinetic and potential energies are negligible. ### General Energy Balance The energy balance for a steady flow process can be expressed as: \[ \Delta h = q + w \] For isothermal conditions, \(q = 0\) (no heat exchange), so: \[ \Delta h = w \] ### Step 1: Find the Molar Volume (\(v\)) Using the EOS: \[ Z = \frac{Pv}{RT} \implies v = \frac{ZRT}{P} \] Substituting for \(Z\): \[ v = \frac{RT}{P} \left[1 + \frac{\alpha P}{RT^2} + \frac{bP}{RT}\right] \] This simplifies to: \[ v = \frac{RT}{P} + \frac{\alpha}{T} + b \] ### Step 2: Write the Enthalpy Change Equation The change in enthalpy (\(\Delta h\)) is given by: \[ \Delta h = \int_{P_1}^{P_2} v \, dP \] Substituting for \(v\): \[ \Delta h = \int_{P_1}^{P_2} \left[ \frac{RT}{P} + \frac{\alpha}{T} + b \right] dP \] ## b) Compute the Molar Enthalpy Change (\(\Delta H\)) ### Integration of Each Term 1. **First Term:** \[ \Delta h_1 = RT \int_{P_1}^{P_2} \frac{1}{P} dP = RT \ln\left(\frac{P_2}{P_1}\right) \] 2. **Second Term:** \[ \Delta h_2 = \frac{\alpha}{T} \int_{P_1}^{P_2} dP = \frac{\alpha}{T} (P_2 - P_1) \] 3. **Third Term:** \[ \Delta h_3 = b \int_{P_1}^{P_2} dP = b (P_2 - P_1) \] ### Final Enthalpy Change Equation Combining all three terms: \[ \Delta h = \Delta h_1 + \Delta h_2 + \Delta h_3 \] ### Substitute Values - \(R = 8.314 \, \mathrm{J/(mol \cdot K)}\) - \(T = 373.15 \, \mathrm{K}\) - \(P_1 = 1 \times 10^5 \, \mathrm{Pa}\) - \(P_2 = 5 \times 10^6 \, \mathrm{Pa}\) ### Calculate Each Term 1. **First Term:** \[ \Delta h_1 = 8.314 \times 373.15 \times \ln\left(\frac{5 \times 10^6}{1 \times 10^5}\right) \] \[ \approx 3102.9 \times 3.912 \approx 12146.5 \, \text{J/mol} \] 2. **Second Term:** \[ \Delta h_2 = \frac{0.4}{373.15} \times (5 \times 10^6 - 1 \times 10^5) \] \[ \approx 0.001072 \times 4.9 \times 10^6 \approx 5251 \, \text{J/mol} \] 3. **Third Term:** \[ \Delta h_3 = 3.37 \times 10^{-5} \times (5 \times 10^6 - 1 \times 10^5) \] \[ \approx 3.37 \times 4.9 \times 10^1 \approx 16.5 \, \text{J/mol} \] ### Total Molar Enthalpy Change \[ \Delta h = 12146.5 + 5251 + 16.5 \approx 17414 \, \text{J/mol} \] ### Final Answer \[ \Delta h = \boxed{17,414 \, \text{J/mol}} \] This provides a comprehensive solution to the problem, detailing each step and calculation for clarity. Let me know if you need further assistance!