# Problem Overview We are analyzing the compression process of carbon dioxide (\( \mathrm{CO_2} \)) in a piston-cylinder assembly. The initial and final states are given, and the process follows a specific relation. Our goal is to find: - **Part A:** The work done during the process (in kJ). **Part B:** The heat transfer the process ( kJ). # Given Data | Parameter | Value | |------------------------------|-------------------------------------| | Initial pressure, \( p_1 \) | .7 bar | | Initial temperature, \( T_1 \) | 280 K | | Final pressure, \( p_2 \) | 10 bar | | Initial volume, \( V_1 \) | .2 m³ | | Process relation | \( p V^{\frac{5}{2}} = \text{constant} \) | | Gas behavior | Ideal gas | | Specific heats (constant) | \( C_p \), \( C_v \) at 300 K (from Table A-23) | --- # Step-by-Step Solution ## Part A: Find the work done, \( W \) ### Step 1: Find initial and final states - Convert initial pressure from bar to kPa: \[ p_1 = .7\, \text{bar} = 70\, \text{kPa} \] - Final pressure: \[ p_2 = 10\, \text{bar} = 100\, \text{kPa} \] - Use ideal gas law: \[ p V = R T \] where \( R \) for \( \mathrm{CO_2} \) (molar gas constant): \[ R_{CO_2} = .1889\, \text{kJ/(kg·K)} \] ### Step 2: Find initial specific volume, \( v_1 \) \[ v_1 = \frac{V_1}{m} \] But since \( V_1 \) is given, and \( p_1, T_1 \) are known: \[ p_1 v_1 = R T_1 \] \[ v_1 = \frac{R T_1}{p_1} = \frac{.1889 \times 280}{70} \approx .756\, \text{m}^3/\text{kg} \] ### Step 3: Find the final specific volume, \( v_2 \) Given the process relation: \[ p v^{\frac{5}{2}} = \text{constant} \] \[ p_1 v_1^{\frac{5}{2}} = p_2 v_2^{\frac{5}{2}} \] Solve for \( v_2 \): \[ v_2 = v_1 \left( \frac{p_1}{p_2} \right)^{\frac{2}{5}} \] Calculate: \[ v_2 = .756 \times \left( \frac{70}{100} \right)^{.4} \] \[ \left( \frac{70}{100} \right)^{.4} \approx (.07)^{.4} \approx .265 \] \[ v_2 \approx .756 \times .265 \approx .2\, \text{m}^3/\text{kg} \] ### Step 4: Find the mass of the gas \[ m = \frac{V_1}{v_1} = \frac{.2}{.756} \approx .264\, \text{kg} \] --- ### Step 5: Calculate the work done For a polytropic process: \[ W = \frac{p_2 V_2 - p_1 V_1}{1 - n} \] where \( n = \frac{5}{2} = 2.5 \). But it's easier to express in specific volume: \[ W = m \times \frac{p_2 v_2 - p_1 v_1}{1 - n} \] Convert pressures to kPa: \[ W = .264 \times \frac{100 \times .2 - 70 \times .756}{1 - 2.5} \] Calculate numerator: \[ 100 \times .2 = 200\, \text{kPa·m}^3 \] \[ 70 \times .756 \approx 52.92\, \text{kPa·m}^3 \] Difference: \[ 200 - 52.92 \approx 147.08\, \text{kPa·m}^3 \] Since the denominator: \[ 1 - 2.5 = -1.5 \] Finally: \[ W = .264 \times \frac{147.08}{-1.5} \approx .264 \times (-98.05) \approx -25.87\, \text{kJ} \] **Note:** Negative work indicates work done **on** the gas (compression). --- ## **Final Answer for Part A:** \[ \boxed{ W \approx \mathbf{-25.87\, \text{kJ}} } \] --- ## Part B: Find the heat transfer, \( Q \) ### Step 1: Use the first law of thermodynamics: \[ Q = \Delta U + W \] where: \[ \Delta U = m C_v (T_2 - T_1) \] ### Step 2: Find \( T_2 \) Using ideal gas law at state 2: \[ T_2 = \frac{p_2 v_2}{R} = \frac{100 \times .2}{.1889} \approx 1058\, \text{K} \] ### Step 3: Find \( C_v \) From Table A-23 (assuming typical values for \( \mathrm{CO_2} \) at 300 K): \[ C_v \approx .658\, \text{kJ/(kg·K)} \] \[ C_p = C_v + R \approx .658 + .1889 \approx .847\, \text{kJ/(kg·K)} \] ### Step 4: Calculate \( \Delta U \) \[ \Delta U = m C_v (T_2 - T_1) = .264 \times .658 \times (1058 - 280) \] \[ = .264 \times .658 \times 778 \approx .264 \times 510.724 \approx 134.9\, \text{kJ} \] ### Step 5: Calculate \( Q \) \[ Q = \Delta U + W = 134.9 + (-25.87) \approx 109.03\, \text{kJ} \] **Positive \( Q \)** indicates heat **absorbed** by the system. --- # **Final answers:** ### **Part A: Work done** \[ \boxed{ W \approx -25.87\, \text{kJ} } \] ### **Part B: Heat transfer** \[ \boxed{ Q \approx 109.03\, \text{kJ} } \] --- # Summary - The compression process involves approximately **26 kJ** of work **done on** the gas. - About **109 kJ** of heat is **absorbed** by the system during this process. If you need further clarification or details, feel free to ask!