Let's break down the question: **Given Data:** - Net work done per cycle, \( W_{net} = 418.17 \) kJ/kg - Compression ratio, \( r = 8 \) - Beginning of process air: \( T_1 = 17^\circ C = 290 \) K, \( P_1 = 100 \) kPa - Find: **Mean effective pressure (MEP)** --- ### Step 1: Mean Effective Pressure Formula \[ MEP = \frac{W_{net}}{v_{d}} \] where \( v_{d} \) is the **specific displacement volume**. ### Step 2: Finding Specific Displacement Volume, \( v_d \) For an Otto cycle: \[ v_d = v_{max} - v_{min} \] For 1 kg of air, using the ideal gas law: \[ v_1 = \frac{RT_1}{P_1} \] where \( R = .287 \) kJ/kg·K \[ v_1 = \frac{.287 \times 290}{100} = \frac{83.23}{100} = .8323 \text{ m}^3/\text{kg} \] At the start, this is the maximum volume, \( v_{max} \). \[ v_{max} = v_1 = .8323 \text{ m}^3/\text{kg} \] Compression ratio: \[ r = \frac{v_{max}}{v_{min}} \implies v_{min} = \frac{v_{max}}{r} = \frac{.8323}{8} = .104 \text{ m}^3/\text{kg} \] So, \[ v_d = v_{max} - v_{min} = .8323 - .104 = .7283 \text{ m}^3/\text{kg} \] --- ### Step 3: Calculate MEP \[ MEP = \frac{418.17}{.7283} = 574.1 \text{ kPa} \] --- ### Step 4: Choose the Correct Option \[ \boxed{574 \text{ kPa}} \] This matches one of the given options.