Let’s solve the problem step by step: --- ## **Step 1: Initial Data and Constants** - **Mass of \( O_2 \), \( m \):** 5 kg - **Initial Temperature, \( T_1 \):** 200 K - **Initial Pressure, \( P_1 \):** 80 kPa = 80,000 Pa - **Gas Constant for Oxygen, \( R \):** 0.2598 kJ/kg·K = 259.8 J/kg·K - **Specific heat at constant pressure, \( C_p \):** For diatomic gases (like \( O_2 \)), \( C_p = \frac{7}{2} R \) \( C_p = 7/2 \times 259.8 = 909.3 \) J/kg·K --- ## **Step 2: Initial State** Use the ideal gas law: \[ P_1 V_1 = m R T_1 \] \[ V_1 = \frac{m R T_1}{P_1} = \frac{5 \times 259.8 \times 200}{80,000} = \frac{259,800}{80,000} = 3.2475~m^3 \] --- ## **Step 3: Process 1 (1→2): Isobaric Expansion (Volume Doubles)** - \( P_1 = P_2 \) - \( V_2 = 2 V_1 \) Use ideal gas law for state 2: \[ P_2 V_2 = m R T_2 \implies T_2 = \frac{P_2 V_2}{m R} = \frac{P_1 \times 2 V_1}{m R} \] But \( P_1 V_1 = m R T_1 \implies V_1 = \frac{m R T_1}{P_1} \), plug this in: \[ T_2 = \frac{P_1 \times 2 \times \frac{m R T_1}{P_1}}{m R} = 2 T_1 = 2 \times 200 = 400~K \] **Enthalpy change for isobaric process:** \[ \Delta H_{1\to2} = m C_p (T_2 - T_1) = 5 \times 909.3 \times (400-200) = 5 \times 909.3 \times 200 = 909,300~J = \boxed{909.3~kJ} \] --- ## **Step 4: Process 2→3: Isochoric (Volume Constant), Pressure Doubles** - \( V_3 = V_2 = 2 V_1 \) - \( P_3 = 2 P_2 = 2 P_1 = 160~kPa \) Use ideal gas law for state 3: \[ P_3 V_3 = m R T_3 \implies T_3 = \frac{P_3 V_3}{m R} \] But \( P_3 = 2 P_1 \), \( V_3 = 2 V_1 \): \[ T_3 = \frac{2 P_1 \times 2 V_1}{m R} = 4 \frac{P_1 V_1}{mR} = 4 T_1 = 800~K \] **Enthalpy change:** \[ \Delta H_{2\to3} = m C_p (T_3 - T_2) = 5 \times 909.3 \times (800-400) = 5 \times 909.3 \times 400 = 1,818,600~J = \boxed{1,818.6~kJ} \] --- ## **Step 5: Process 3→4: Isothermal Expansion (Temperature Constant at \( T_3 \)), Pressure drops to initial value** - \( T_4 = T_3 = 800~K \) - \( P_4 = P_1 = 80~kPa \) Use ideal gas law to find \( V_4 \): \[ P_4 V_4 = m R T_4 \implies V_4 = \frac{m R T_4}{P_4} \] Plug in values: \[ V_4 = \frac{5 \times 259.8 \times 800}{80,000} = \frac{1,039,200}{80,000} = 12.99~m^3 \] **Enthalpy change for isothermal process:** \[ \Delta H_{3\to4} = m C_p (T_4 - T_3) = 0 \] --- ## **Step 6: Process 4→1: Adiabatic Compression (Returns to Initial Volume, so \( V_1 \)), Find Final Temperature** - \( V_4 \to V_1 \) - \( T_4 = 800~K \) - \( V_1 = 3.2475~m^3 \) - \( V_4 = 12.99~m^3 \) For adiabatic process for an ideal gas: \[ T V^{\gamma-1} = \text{constant} \] where \( \gamma = \frac{C_p}{C_v} = \frac{7}{5} = 1.4 \) for diatomic gases. So, \[ T_4 V_4^{\gamma-1} = T_1 V_1^{\gamma-1} \implies T_1 = T_4 \left( \frac{V_4}{V_1} \right)^{\gamma-1} \] But since \( V_4 > V_1 \), and returning to \( V_1 \): \[ T_1 = T_4 \left( \frac{V_4}{V_1} \right)^{1-\gamma} \] But since we started at \( T_1 = 200~K \), let's check consistency: \[ T_1 = 800 \times \left( \frac{3.2475}{12.99} \right)^{0.4} \] Calculate \( \frac{3.2475}{12.99} = 0.25 \) Then \( 0.25^{0.4} \): \[ \ln(0.25) = -1.3863 \implies -1.3863 \times 0.4 = -0.5545 \implies e^{-0.5545} = 0.574 \] So, \[ T_1 = 800 \times 0.574 = 459~K \] But this does **not** return us to the initial temperature. However, we're told the process returns to the initial volume (not necessarily temperature), so the enthalpy change is: \[ \Delta H_{4\to1} = m C_p (T_1 - T_4) = 5 \times 909.3 \times (200 - 800) = 5 \times 909.3 \times (-600) = -2,727,900~J = \boxed{-2,727.9~kJ} \] But our calculation above for \( T_1 \) says it returns to 459 K, not 200 K. This suggests one of two things: - The process does not return to the initial temperature (only the initial volume), - Or, something is inconsistent in the problem statement. Let's proceed with the calculation as per the correct adiabatic relation: \[ T_1 = T_4 \left( \frac{V_4}{V_1} \right)^{1-\gamma} = 800 \times (4)^{-0.4} \] \( 4^{-0.4} = e^{-0.4 \ln 4} = e^{-0.4 \times 1.3863} = e^{-0.5545} = 0.574 \) So, \[ T_1 = 800 \times 0.574 = 459~K \] So, the final temperature is 459 K. Thus, \[ \Delta H_{4\to1} = m C_p (T_1 - T_4) = 5 \times 909.3 \times (459 - 800) = 5 \times 909.3 \times (-341) = -1,551,196~J = \boxed{-1,551.2~kJ} \] --- ## **Step 7: Total Enthalpy Change** Sum the enthalpy changes for each process: 1. \( 1 \to 2 \): \( +909.3~kJ \) 2. \( 2 \to 3 \): \( +1,818.6~kJ \) 3. \( 3 \to 4 \): \( 0~kJ \) 4. \( 4 \to 1 \): \( -1,551.2~kJ \) \[ \Delta H