Let's solve the problem step by step: **Given:** - Initial pressure, \(P_1 = 100\) kPa - Initial temperature, \(T_1 = 27^\circ C = 300.15\) K - Final pressure, \(P_2 = 600\) kPa - Final temperature, \(T_2 = 77^\circ C = 350.15\) K - \(c_p = 1.005\) kJ/kg·K - \(c_v = 0.718\) kJ/kg·K - \(R = c_p - c_v = 1.005 - 0.718 = 0.287\) kJ/kg·K **Formula for entropy change of air (ideal gas):** \[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \] Let's plug in the values: ### Step 1: Calculate \(\ln\left(\frac{T_2}{T_1}\right)\) \[ \ln\left(\frac{T_2}{T_1}\right) = \ln\left(\frac{350.15}{300.15}\right) = \ln(1.1667) = 0.1543 \] ### Step 2: Calculate \(\ln\left(\frac{P_2}{P_1}\right)\) \[ \ln\left(\frac{P_2}{P_1}\right) = \ln\left(\frac{600}{100}\right) = \ln(6) = 1.7918 \] ### Step 3: Substitute in the formula \[ \Delta s = 1.005 \times 0.1543 - 0.287 \times 1.7918 \] \[ \Delta s = 0.1551 - 0.5143 \] \[ \Delta s = -0.3592 \text{ kJ/kg·K} \] ### Step 4: Match with the options The closest option is: - **-0.3591 kJ/kg·K** --- **Final Answer:** \[ \boxed{-0.3591 \text{ kJ/kg·K}} \]