Let's solve ** A** step by step. --- ## **Given Data** - Initial Pressure, \( p_1 = 1 \) bar = \( 100 \) kPa - Final Pressure, \( p_2 = 9 \) bar = \( 900 \) kPa - Initial Temperature, \( T_1 = 295 \) K - Polytropic exponent, \( n = 1.4 \) - Air is an ideal gas. --- ## **Step 1: Find the Final Temperature (\(T_2\))** For a **polytropic process**: \[ \frac{T_2}{T_1} = \left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} \] Plug in the values: \[ \frac{T_2}{295} = \left(\frac{900}{100}\right)^{\frac{1.4-1}{1.4}} \] \[ \frac{T_2}{295} = (9)^{.4/1.4} \] \[ \frac{T_2}{295} = (9)^{.2857} \] Calculate the exponent: \[ (9)^{.2857} \approx e^{.2857 \cdot \ln 9} \approx e^{.2857 \cdot 2.1972} \approx e^{.6276} \approx 1.872 \] So, \[ T_2 = 295 \times 1.872 \approx 552 \text{ K} \] --- ## **Step 2: Find the Work Done per kg (W/m)** The work per unit mass for a polytropic process: \[ w = \frac{R (T_2 - T_1)}{1 - n} \] Where \( R \) is the specific gas constant for air: \[ R = .287 \, \text{kJ/kg·K} \] Now plug in the numbers: \[ w = \frac{.287 \times (552 - 295)}{1 - 1.4} \] \[ w = \frac{.287 \times 257}{-.4} \] \[ w = \frac{73.759}{-.4} \] \[ w = -184.4 \ \text{kJ/kg} \] **The negative sign indicates work is done *on* the gas (compression).** --- ## **Final Answer (Part A):** \[ \boxed{-184 \ \text{kJ/kg}} \] --- Let me know when you're ready for Part B (heat transfer)!