Let's analyze the problem step-by-step to evaluate the efficiency (\(\eta\)) of the cycle. --- ## Step 1: Understand the cycle The cycle is a **p-v** (pressure-volume) diagram with the following processes: - **Process 1→3**: Isothermal expansion from point 1 (p=3 bar, v=1) to point 2 (p=?, v=3) - **Process 3→2**: Isothermal compression from point 3 (p=1 bar, v=1) to point 2 (p=?, v=3) The given points: - \(1: p=3\, \text{bar}, v=1\, \text{m}^3/\text{kg}\) - \(2: p=?, v=3\, \text{m}^3/\text{kg}\) - \(3: p=1\, \text{bar}, v=1\, \text{m}^3/\text{kg}\) The process 1→2 is an *isothermal expansion*, and 3→2 is an *isothermal compression*. --- ## Step 2: Find the pressure at point 2 Since processes 1→2 and 3→2 are isothermal: - For 1→2: \(p_1 v_1 = p_2 v_2\) \[ p_2 = p_1 \times \frac{v_1}{v_2} = 3\, \text{bar} \times \frac{1}{3} = 1\, \text{bar} \] - For 3→2: \(p_3 v_3 = p_2 v_2\) \[ p_2 = p_3 \times \frac{v_3}{v_2} = 1\, \text{bar} \times \frac{1}{3} = \frac{1}{3}\, \text{bar} \] But this conflicts with the earlier calculation. Since the pressure at point 2 must be the same for both processes, the pressure at 2 is: \[ p_2 = 1\, \text{bar} \] **Therefore:** - \(p_2 = 1\, \text{bar}\) --- ## Step 3: Calculate work done during each process ### Process 1→2 (Isothermal expansion): Work done: \[ W_{1\rightarrow 2} = nRT \ln \frac{v_2}{v_1} \] But since we are assuming ideal gas behavior, and the process is isothermal, the work can also be expressed as: \[ W_{1 \rightarrow 2} = p_1 v_1 \ln \frac{v_2}{v_1} \] Given \(p_1 v_1 = RT\) (for ideal gas): \[ W_{1 \rightarrow 2} = p_1 v_1 \ln \frac{v_2}{v_1} = (3\, \text{bar} \times 1\, \text{m}^3/\text{kg}) \times \ln 3 \] We need to convert pressure to absolute units for consistency: \[ 1\, \text{bar} = 100\, \text{kPa} \] So, \[ p_1 v_1 = 3 \times 100\, \text{kPa} \times 1\, \text{m}^3/\text{kg} = 300\, \text{kPa} \times 1\, \text{m}^3/\text{kg} = 300\, \text{kJ} \] (Using \(pV = nRT\), but in specific units, the work is in kJ). Similarly, \[ W_{1 \rightarrow 2} = 300\, \text{kJ} \times \ln 3 \approx 300 \times 1.0986 \approx 329.58\, \text{kJ} \] ### Process 3→2 (Isothermal compression): Work done: \[ W_{3 \rightarrow 2} = p_3 v_3 \ln \frac{v_2}{v_3} \] Given \(p_3=1\, \text{bar}=100\, \text{kPa}\), \(v_3=1\, \text{m}^3/\text{kg}\): \[ W_{3 \rightarrow 2} = 100\, \text{kPa} \times 1\, \text{m}^3/\text{kg} \times \ln 3 \approx 100 \times 1.0986 \approx 109.86\, \text{kJ} \] --- ## Step 4: Determine heat transfer during each process For an **ideal gas**: - Heat transfer during isothermal process: \[ Q = W \] Thus, during process 1→2: \[ Q_{1 \rightarrow 2} = W_{1 \rightarrow 2} \approx 329.58\, \text{kJ} \] And during process 3→2: \[ Q_{3 \rightarrow 2} = W_{3 \rightarrow 2} \approx 109.86\, \text{kJ} \] Note: Since these are isothermal processes, the heat added or removed equals the work done. --- ## Step 5: Calculate net work and heat input **Net work done per cycle:** \[ W_{net} = W_{1 \rightarrow 2} - W_{3 \rightarrow 2} = 329.58 - 109.86 = 219.72\, \text{kJ} \] **Heat input:** - Heat added during expansion (1→2): \(Q_{in} = 329.58\, \text{kJ}\) - Heat rejected during compression (3→2): \(Q_{out} = 109.86\, \text{kJ}\) --- ## Step 6: Calculate thermal efficiency \[ \eta = \frac{\text{Net work output}}{\text{Heat input}} = \frac{W_{net}}{Q_{in}} \] \[ \eta = \frac{219.72}{329.58} \approx .6667 \] Expressed as a percentage: \[ \eta \approx 66.67\% \] --- ## **Final Answer:** \[ \boxed{ \eta \approx 66.7\% } \] *Note:* The provided value in the question was 15.5%, but based on the assumptions and calculations above, the efficiency is approximately **66.7%**. If the question expects a different value, please verify the assumptions or provide additional details.