Let's break down the problem and solve it step by step. --- ## **Given:** - **Mixture Mass:** \( m = 1.5 \) lbm - **Mixture Analysis (by mass):** - N₂: 30% - CD: 40% - O₂: 30% - **Initial Pressure:** \( P_1 = 1 \) atm - **Initial Temperature:** \( T_1 = 540^\circ \)R (\( = 540 \) Rankine) - **Final Pressure:** \( P_2 = 5 \) atm - **Final Temperature:** \( T_2 = 1000 \)R (\( = 1000 \) Rankine) - **Compression is Adiabatic** (no heat transfer) - **Find:** - (a) The **work of compression** (magnitude) - (b) The **entropy produced** --- ## **Step 1: Identify the Gases and Molar Masses** Assume "CD" is a typo or a placeholder. Commonly, CD could refer to CO₂ or CO; for this calculation, let's assume it's **CO₂** unless clarified otherwise. - For calculations, we'll use: - N₂: \( M_{N_2} = 28.01 \) lbm/lbmol - CO₂: \( M_{CO_2} = 44.01 \) lbm/lbmol - O₂: \( M_{O_2} = 32.00 \) lbm/lbmol --- ## **Step 2: Calculate moles of each component** \[ \text{Total mass} = 1.5 \text{ lbm} \] \[ \text{Mass fraction: } x_{N_2} = 0.3, \; x_{CO_2} = 0.4, \; x_{O_2} = 0.3 \] \[ n_{N_2} = \frac{0.3 \times 1.5}{28.01} = 0.01607 \text{ lbmol} \] \[ n_{CO_2} = \frac{0.4 \times 1.5}{44.01} = 0.01363 \text{ lbmol} \] \[ n_{O_2} = \frac{0.3 \times 1.5}{32.00} = 0.01406 \text{ lbmol} \] \[ n_{total} = 0.01607 + 0.01363 + 0.01406 = 0.04376 \text{ lbmol} \] --- ## **Step 3: Gas Constant and Specific Heat for the Mixture** ### **a. Gas Constant \( R \)** \[ R_{univ} = 1.986 \ \text{Btu/(lbmol·R)} \] \[ R_{mix} = \frac{R_{univ}}{M_{mix}} \] First, find the average molar mass \( M_{mix} \): \[ M_{mix} = \frac{1.5}{n_{total}} = \frac{1.5}{0.04376} = 34.28 \ \text{lbm/lbmol} \] \[ R_{mix} = \frac{1.986}{34.28} = 0.05797 \ \text{Btu/(lbm·R)} \] --- ### **b. Specific Heats (\( C_p \) and \( C_v \))** Typical values (approximate, in Btu/(lbm·R)): - N₂: \( C_{p,N_2} = 0.248 \), \( C_{v,N_2} = 0.177 \) - O₂: \( C_{p,O_2} = 0.219 \), \( C_{v,O_2} = 0.156 \) - CO₂: \( C_{p,CO_2} = 0.204 \), \( C_{v,CO_2} = 0.156 \) Weighted average for mixture: \[ C_{p,mix} = x_{N_2} C_{p,N_2} + x_{CO_2} C_{p,CO_2} + x_{O_2} C_{p,O_2} \] \[ = 0.3 \times 0.248 + 0.4 \times 0.204 + 0.3 \times 0.219 \] \[ = 0.0744 + 0.0816 + 0.0657 = 0.2217 \ \text{Btu/(lbm·R)} \] \[ C_{v,mix} = C_{p,mix} - R_{mix} \] \[ = 0.2217 - 0.05797 = 0.1637 \ \text{Btu/(lbm·R)} \] \[ k = \frac{C_p}{C_v} = \frac{0.2217}{0.1637} = 1.354 \] --- ## **Step 4: Calculate Work Done (Adiabatic Compression of Ideal Gas)** The **work per unit mass** for adiabatic (irreversible) with change in entropy: \[ w = \Delta u + \Delta (Pv) \] But for a **reversible adiabatic process** (isentropic): \[ w_{rev} = \frac{C_{v,mix}(T_2 - T_1)}{} \] But since \( \Delta Q = 0 \), the first law gives: \[ w = \Delta h - \Delta (Pv) \] But for an **adiabatic process** \( Q=0 \): \[ \Delta u = w + \Delta (Pv) \] But more conveniently: \[ w = m \cdot \frac{C_{v,mix}}{R_{mix}} \left[ P_2 v_2 - P_1 v_1 \right] \] Alternatively, for an **ideal gas**: \[ w = m \cdot C_{v,mix} (T_2 - T_1) \] But let's use: \[ w = m \cdot (h_2 - h_1) \] Because for adiabatic (irreversible) compression, unless specified as reversible, we use: \[ w = \Delta h \] \[ h = C_{p,mix} T \] So: \[ w = m \cdot C_{p,mix} (T_2 - T_1) \] Insert values: - \( m = 1.5 \) lbm - \( C_{p,mix} = 0.2217 \) Btu/(lbm·R) - \( T_2 = 1000 \)R, \( T_1 = 540 \)R \[ w = 1.5 \cdot 0.2217 \cdot (1000 - 540) \] \[ = 1.5 \cdot 0.2217 \cdot 460 \] \[ = 1.5 \cdot 102.982 = 154.47 \text{ Btu} \] **But this is the change in enthalpy. For a constant pressure process, this would be the work. For a constant volume process, it's \( \Delta u \). For a general adiabatic process:** For an **adiabatic process (with possible irreversibility):** \[ \Delta u = w \] \[ u = C_{v,mix} T \] \[ w = m \cdot C_{v,mix} (T_2 - T_1) \] \[ w = 1.5 \cdot 0.1637 \cdot (1000 - 540) \] \[ = 1.5 \cdot 0.1637 \cdot 460 = 1.5 \cdot 75.302 = 112.95 \text{ Btu} \] So, **the magnitude of the work of compression is:** \[ \boxed{112.95 \text{ Btu}} \] --- ## **Step 5: Entropy Produced** The entropy change of the system is: \[ \Delta s = m \left[ C_{v,mix} \ln \left( \frac{T_2}{T_1} \right) + R_{mix} \ln \left( \frac{v_2}{v_1} \right) \right] \] But for an **irreversible adiabatic process**, \( \Delta s_{gen} = \Delta s_{system} \), since \( Q = 0 \). First, find \( v_1 \) and \( v_2 \): \[ PV = nRT \implies v = \frac{RT}{P} \] Where: - \( R = R_{univ}/M_{mix} = 0.05797 \) Btu/(lbm·R) - \( T_1

Let's carefully revisit the calculations to ensure accuracy and reconcile with the provided answer of approximately 122 Btu for the work. --- ## **Step-by-step correction and detailed explanation** ### **Given Data Recap:** - Mass of mixture: \( m = 1.5\, \text{lbm} \) - Composition (by mass): - N₂: 30% - CO₂ (assumed "CD"): 40% - O₂: 30% - Initial pressure: \( P_1 = 1\, \text{atm} \) - Initial temperature: \( T_1 = 540\,^\circ R \) - Final pressure: \( P_2 = 5\, \text{atm} \) - Final temperature: \( T_2 = 1000\,^\circ R \) - Process: **Adiabatic compression** (possibly irreversible) --- ### **Step 1: Moles of each component** Mass fractions: \[ x_{N_2} = 0.3,\quad x_{CO_2} = 0.4,\quad x_{O_2} = 0.3 \] Moles: \[ n_{N_2} = \frac{0.3 \times 1.5}{28.01} = 0.01607\, \text{lbmol} \] \[ n_{CO_2} = \frac{0.4 \times 1.5}{44.01} = 0.01363\, \text{lbmol} \] \[ n_{O_2} = \frac{0.3 \times 1.5}{32.00} = 0.01406\, \text{lbmol} \] Total moles: \[ n_{total} = 0.01607 + 0.01363 + 0.01406 = 0.04376\, \text{lbmol} \] Average molar mass of the mixture: \[ M_{mix} = \frac{\text{Mass}}{\text{moles}} = \frac{1.5\, \text{lbm}}{0.04376\, \text{lbmol}} \approx 34.3\, \text{lbm/lbmol} \] --- ### **Step 2: Gas constant for the mixture** Universal gas constant: \[ R_{univ} = 1.986\, \text{Btu/(lbmol·R)} \] Specific gas constant: \[ R_{mix} = \frac{R_{univ}}{M_{mix}} = \frac{1.986}{34.3} \approx 0.0579\, \text{Btu/(lbm·R)} \] --- ### **Step 3: Specific heats** Using typical values: | Gas | \( C_{p} \) (Btu/(lbm·R)) | \( C_{v} \) (Btu/(lbm·R)) | |-------|------------------------|------------------------| | N₂ | 0.248 | 0.177 | | O₂ | 0.219 | 0.156 | | CO₂ | 0.204 | 0.156 | Weighted averages: \[ C_{p,\,mix} = 0.3 \times 0.248 + 0.4 \times 0.204 + 0.3 \times 0.219 \] \[ = 0.0744 + 0.0816 + 0.0657 = 0.2217\, \text{Btu/(lbm·R)} \] \[ C_{v,\,mix} = C_{p,\,mix} - R_{mix} = 0.2217 - 0.0579 = 0.1638\, \text{Btu/(lbm·R)} \] --- ### **Step 4: Work of the process** For an **adiabatic process**, the work per unit mass can be derived from the enthalpy change: \[ w = m \left[ h_2 - h_1 \right] \] where: \[ h = C_{p,\,mix} \times T \] Thus, \[ w = m \times C_{p,\,mix} \times (T_2 - T_1) \] Insert known values: \[ w = 1.5\, \text{lbm} \times 0.2217\, \text{Btu/(lbm·R)} \times (1000\, R - 540\, R) \] \[ = 1.5