Let's break down the question: **Given:** - Net work done, \( W_{net} = 418.17 \) kJ/kg - Compression ratio, \( r = 8 \) - Initial temperature, \( T_1 = 27^\circ C = 300 \) K - Initial pressure, \( P_1 = 100 \) kPa **Asked:** Calculate the mean effective pressure (MEP) for the Otto cycle. --- ### Step 1: Formula for Mean Effective Pressure (MEP) The mean effective pressure (MEP) is given by: \[ \text{MEP} = \frac{W_{net}}{v_{max} - v_{min}} \] Where: - \( W_{net} \) = Net work done per cycle per kg - \( v_{max} \) = Maximum specific volume - \( v_{min} \) = Minimum specific volume \[ v_{max} - v_{min} = v_1 - v_2 \] For Otto cycle: \[ \frac{v_1}{v_2} = r \implies v_2 = \frac{v_1}{r} \] So, \[ v_1 - v_2 = v_1 - \frac{v_1}{r} = v_1 \left(1 - \frac{1}{r}\right) \] --- ### Step 2: Find \( v_1 \) (Initial Specific Volume) Use the ideal gas law: \[ v_1 = \frac{RT_1}{P_1} \] For air, \( R = 0.287 \) kJ/kg·K. \[ v_1 = \frac{0.287 \times 300}{100} = \frac{86.1}{100} = 0.861\, \text{m}^3/\text{kg} \] --- ### Step 3: Calculate \( v_1 - v_2 \) \[ v_1 - v_2 = v_1 \left(1 - \frac{1}{r}\right) \] \[ = 0.861 \times \left(1 - \frac{1}{8}\right) = 0.861 \times \left(\frac{7}{8}\right) = 0.861 \times 0.875 = 0.754\, \text{m}^3/\text{kg} \] --- ### Step 4: Calculate MEP \[ \text{MEP} = \frac{418.17}{0.754} = 554.62 \text{ kPa} \] --- ### Step 5: Closest Option The closest answer to **554.62 kPa** is **547 kPa**. --- **Final Answer:** \[ \boxed{547 \text{ kPa}} \]