Let's break down the problem step-by-step! --- # **Given Data** - **Process:** Polytropic, \( n = 1.4 \) - **Initial pressure:** \( p_1 = 1) bar \( 100 \ kPa **Initial temperature \( T_1 = 295 \) K - **Final pressure:** \( p_2 = 9 \) bar \( = 900 \) kPa - **Gas:** Air (ideal gas, neglect kinetic and potential energy changes) - **Assume:** \( c_v \) is constant at 300 K --- ## **Step 1: Find Final Temperature (\( T_2 \))** The polytropic relation for an ideal gas: \[ \frac{T_2}{T_1} = \left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} \] Plug in the values: \[ \frac{T_2}{295} = \left(\frac{900}{100}\right)^{\frac{1.4-1}{1.4}} \] \[ = (9)^{.4/1.4} \] \[ = 9^{.2857} \] \[ = \exp(.2857 \cdot \ln(9)) \] \[ = \exp(.2857 \cdot 2.197) \] \[ = \exp(.6277) \] \[ = 1.873 \] \[ T_2 = 295 \times 1.873 = 552.3 \, \text{K} \] --- ## **Step 2: Calculate Work Done (per kg)** The work for a polytropic process for an ideal gas: \[ W = \frac{R (T_2 - T_1)}{1 - n} \] For air, \( R = .287 \) kJ/kg·K \[ W = \frac{.287 \times (552.3 - 295)}{1 - 1.4} \] \[ = \frac{.287 \times 257.3}{-.4} \] \[ = \frac{73.86}{-.4} \] \[ = -184.65 \, \text{kJ/kg} \] The negative sign means work is **done on the air** (compression). --- ## **Step 3: Calculate Heat Transfer (per kg)** First, find \( c_v \) for air at 300K: - \( c_v \) (air) ≈ .718 kJ/kg·K - \( c_p \) (air) ≈ 1.005 kJ/kg·K Change in internal energy: \[ \Delta u = c_v (T_2 - T_1) = .718 \times (552.3 - 295) = .718 \times 257.3 = 184.81 \, \text{kJ/kg} \] From the first law: \[ Q = \Delta u + W \] \[ Q = 184.81 + (-184.65) \] \[ Q = .16 \, \text{kJ/kg} \] --- # **Summary Table** | Quantity | Value (kJ/kg) | Significance | |------------|:-------------:|---------------| | **Work** (\(W\)) | -184.65 | Work done **on** air (compression) | | **Heat** (\(Q\)) | .16 | Small heat added to air | --- ## **Final Answers (Constant \( c_v \), Part 1):** - **Work done per kg:** \(\boxed{-184.7 \ \text{kJ/kg}}\) - **Heat transfer per kg:** \(\boxed{.16 \ \text{kJ/kg}}\) --- ## **Step 4: For Variable Specific Heats (Part 2)** For most practical purposes, using air tables (property tables) at the given \( T_1 \) and \( T_2 \) is best (not required for your current part). --- ### **Let me know if you need the next steps or help with the variable specific heats part!**