Let's solve the problem step by step: ### **Given:** - \( n = 2.50 \) mol (monatomic ideal gas) - Initial temperature: \( T_1 = 300 \) K - Initial pressure: \( P_1 = 1.00 \) atm - **Step 1:** Pressure increases proportionally with volume until \( P_2 = 2.00 \) atm. - **Step 2:** Pressure reduced to \( P_3 = 1.00 \) atm at constant \( V \). - **Step 3:** Volume decreases at constant pressure (to initial state). --- ## **Step 1: Find states and draw PV diagram** ### **Initial State (1):** - \( P_1 = 1.00 \) atm, \( T_1 = 300 \) K Use ideal gas law: \[ P_1 V_1 = nRT_1 \] \[ V_1 = \frac{nRT_1}{P_1} = \frac{2.50 \times 0.0821 \times 300}{1.00} = 61.575 \text{ L} \] --- ### **State 2 (after step 1):** - \( P_2 = 2.00 \) atm Since \( P \propto V \) (i.e., \( P = kV \)), we can write: \[ \frac{P_2}{P_1} = \frac{V_2}{V_1} \] \[ \frac{2.00}{1.00} = \frac{V_2}{61.575} \] \[ V_2 = 2 \times 61.575 = 123.15 \text{ L} \] Now, calculate \( T_2 \): \[ P_2 V_2 = nRT_2 \implies T_2 = \frac{P_2 V_2}{nR} = \frac{2.00 \times 123.15}{2.50 \times 0.0821} = \frac{246.3}{0.20525} \approx 1200 \text{ K} \] --- ### **State 3 (after step 2):** - At constant \( V_2 \), \( P_3 = 1.00 \) atm Calculate \( T_3 \): \[ P_3 V_3 = nRT_3 \] But \( V_3 = V_2 \), so: \[ T_3 = \frac{P_3 V_2}{nR} = \frac{1.00 \times 123.15}{2.50 \times 0.0821} = \frac{123.15}{0.20525} \approx 600 \text{ K} \] --- ### **State 1 (Return):** - At constant \( P_1 = 1.00 \) atm, return to \( V_1 \), \( T_1 = 300 \) K --- ## **(a) PV Diagram** Draw the three states: - 1: (61.6 L, 1 atm) - 2: (123.2 L, 2 atm) - 3: (123.2 L, 1 atm) - Back to 1: (61.6 L, 1 atm) Connect: 1 → 2: Straight line (since \( P \propto V \)), 2 → 3: Vertical line (isochoric), 3 → 1: Horizontal line (isobaric). --- ## **(b) Calculate the efficiency of the cycle** ### **Step 1 (1 → 2): \( P \propto V \) (polytropic)** - Work done (\(W_{12}\)): \[ P = kV \implies P_1 = kV_1,\, P_2 = kV_2 \implies k = \frac{P_1}{V_1} = \frac{P_2}{V_2} \] \[ W_{12} = \int_{V_1}^{V_2} P\,dV = \int_{V_1}^{V_2} kV\,dV = k\int_{V_1}^{V_2} V\,dV = k\left[\frac{V^2}{2}\right]_{V_1}^{V_2} \] \[ k = \frac{P_1}{V_1} \] \[ W_{12} = \frac{P_1}{V_1} \cdot \frac{1}{2} (V_2^2 - V_1^2) \] Plug in the numbers: \[ W_{12} = \frac{1}{61.575} \times \frac{1}{2} (123.15^2 - 61.575^2) \] \[ = 0.01624 \times 0.5 \times (15170.6 - 3792.7) \] \[ = 0.00812 \times 11377.9 = 92.37 \text{ L·atm} \] Convert to Joules (\(1 \text{ L·atm} = 101.325 \text{ J}\)): \[ W_{12} = 92.37 \times 101.325 = 9360 \text{ J} \] --- ### **Step 2 (2 → 3): Isochoric (V constant)** - Work done \( W_{23} = 0 \) - Heat transferred: \( Q_{23} = nC_V \Delta T = \frac{3}{2} nR (T_3 - T_2) \) \[ Q_{23} = \frac{3}{2} \times 2.50 \times 8.314 \times (600 - 1200) = 3.75 \times 8.314 \times (-600) = 31.18 \times (-600) = -18,708 \text{ J} \] (Negative, so heat is **released**.) --- ### **Step 3 (3 → 1): Isobaric (P constant)** - Work done: \( W_{31} = P_1 (V_1 - V_3) \) \[ V_1 = 61.575 \text{ L},\, V_3 = V_2 = 123.15 \text{ L} \] \[ W_{31} = 1.00 \times (61.575 - 123.15) = -61.575 \text{ L·atm} = -6237 \text{ J} \] - Heat transferred: \( Q_{31} = nC_P \Delta T = \frac{5}{2} nR (T_1 - T_3) \) \[ Q_{31} = \frac{5}{2} \times 2.50 \times 8.314 \times (300 - 600) = 6.25 \times 8.314 \times (-300) = 51.96 \times (-300) = -15,588 \text{ J} \] (Negative, so heat is **released**.) --- ### **Net Work Done by the Cycle:** \[ W_{cycle} = W_{12} + W_{23} + W_{31} = 9360 + 0 + (-6237) = 3123 \text{ J} \] --- ### **Net Heat Input:** #### **Find the heat absorbed (positive Q):** - Step 1 (\( Q_{12} \)): Use first law: \[ \Delta U_{12} = nC_V (T_2 - T_1) = \frac{3}{2} \times 2.50 \times 8.314 \times (1200 - 300) = 3.75 \times 8.314 \times 900 = 31.18 \times 900 = 28,062 \text{ J} \] \[ Q_{12} = \Delta U_{12} + W_{12} = 28,062 + 9360 = 37,422 \text{ J} \] - Step 2 (\( Q_{23} \)) is negative (heat released). - Step 3 (\( Q_{31} \)) is negative (heat released). So, only heat **absorbed** is \( Q_{12} \). ### **Efficiency:** \[ e = \frac{W_{cycle}}{Q_{in}} = \frac{3123}{37422} = 0.0835 \text{ or } 8.35\% \] --- ## **(c) Carnot Efficiency** The highest temperature in the cycle is \( T_H = 1200 \) K, the lowest is \( T_L = 300 \) K. \[ e_{Carnot} = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{1200} = 1 - 0.25 = 0.75 \text{ or } 75\% \] --- ## **Final Answers** ### **(a) PV Diagram** - (See description above: a triangle with points at (61.6, 1), (123.2, 2), and (123.2, 1) in L and atm.) ### **(b) Efficiency of the cycle** \[ \boxed{8.35\%} \] ### **(c) Carnot Efficiency** \[ \boxed{75\%} \] If you’d like the diagram or more detail on any step, let me know!

Certainly! Let's go through each part in detailed steps, including all calculations, with clear explanations. --- ## **Given Data:** - Number of moles, \( n = 2.50 \) mol - Initial temperature, \( T_1 = 300\,K \) - Initial pressure, \( P_1 = 1.00\,atm \) --- ## **Part (a): Plotting the processes on PV diagram** --- ### **Step 1: Find initial volume \( V_1 \)** Using the ideal gas law: \[ PV = nRT \] where \( R = 0.0821\, L\, atm\, mol^{-1}\, K^{-1} \). \[ V_1 = \frac{nRT_1}{P_1} = \frac{2.50 \times 0.0821 \times 300}{1.00} \] \[ V_1 = 2.50 \times 0.0821 \times 300 = 2.50 \times 24.63 = 61.58\, L \] --- ### **Step 2: State 2 (after step 1)** **Process:** Pressure increases proportionally with volume until \( P_2 = 2.00\, atm \). Since \( P \propto V \): \[ \frac{P_2}{P_1} = \frac{V_2}{V_1} \] \[ V_2 = V_1 \times \frac{P_2}{P_1} = 61.58 \times \frac{2.00}{1.00} = 123.16\, L \] **Calculate \( T_2 \):** \[ P_2 V_2 = nRT_2 \] \[ T_2 = \frac{P_2 V_2}{n R} = \frac{2.00 \times 123.16}{2.50 \times 0.0821} \] \[ T_2 = \frac{246.32}{0.20525} \approx 1200\, K \] --- ### **Step 3: State 3 (after step 2)** **Process:** Pressure reduces back to \( P_3 = 1.00\, atm \) at constant volume \( V_2 \). Calculate \( T_3 \): \[ P_3 V_3 = n R T_3 \] \[ V_3 = V_2 = 123.16\, L \] \[ T_3 = \frac{P_3 V_3}{n R} = \frac{1.00 \times 123.16}{0.20525} \approx 600\, K \] --- ### **Step 4: Return to initial state** At initial state: \[ V_1 = 61.58\, L, \quad T_1 = 300\,K \] --- ### **Summary of states:** | State | \( P \) (atm) | \( V \) (L) | \( T \) (K) | |---------|--------------|--------------|--------------| | 1 | 1.00 | 61.58 | 300 | | 2 | 2.00 | 123.16 | 1200 | | 3 | 1.00 | 123.16 | 600 | --- ### **Plotting PV diagram:** - **Point 1:** (61.58 L, 1 atm) - **Point 2:** (123.16 L, 2 atm) - **Point 3:** (123.16 L, 1 atm) - **Back to:** (61.58 L, 1 atm) **Process paths:** - 1 → 2: Along the line \( P \propto V \) (straight line with \( P V = \text{constant} \)) - 2 → 3: Vertical line (constant volume, pressure decreases) - 3 → 1: Horizontal line (constant pressure, volume decreases) --- ## **Part (b): Calculate the efficiency of the cycle** ### **Step 1: Work done during process 1→2** Process 1→2: \( P \propto V \). Since \( P = kV \), the work is: \[ W_{12} = \int_{V_1}^{V_2} P\, dV = \int_{V_1}^{V_2} kV\, dV = k \left[ \frac{V^2}{2} \right]_{V_1}^{V_2} \] Calculate \( k \): \[ k = \frac{P_1}{V_1} = \frac{1.00}{61.58} \approx 0.01624\, atm/L \] Compute: \[ W_{12} = 0.01624 \times \frac{1}{2} (V_2^2 - V_1^2) \] \[ V_2^2 = (123.16)^2 \approx 15170.6,\quad V_1^2 = (61.58)^2 \approx 3792.7 \] \[ W_{12} = 0.01624 \times 0.5 \times (15170.6 - 3792.7) = 0.00812 \times 11377.9 \approx 92.37\, \text{L·atm} \] Convert to Joules: \[ 1\, L\, atm = 101.325\, J \] \[ W_{12} = 92.37 \times 101.325 \approx 9360\, J \] --- ### **Step 2: Heat absorbed during 1→2** Change in internal energy: \[ \Delta U_{12} = n C_V (T_2 - T_1) \] For a monatomic ideal gas: \[ C_V = \frac{3}{2} R = 1.5 \times 8.314 = 12.471\, J/(mol\,K) \] \[ \Delta U_{12} = 2.50 \times 12.471 \times (1200 - 300) = 2.50 \times 12.471 \times 900 \] \[ = 2.50 \times 12.471 \times 900 \approx 2.50 \times 11223.9 \approx 28,059.8\, J \] Heat input during 1→2: \[ Q_{in} = \Delta U_{12} + W_{12} = 28,059.8 + 9360 \approx 37,420\, J \] --- ### **Step 3: Work during 2→3** Process 2→3: constant volume, pressure decreases from 2.00 atm to 1.00 atm. Work: \[ W_{23} = 0 \quad (\text{no volume change}) \] Heat exchanged: \[ Q_{23} = n C_V (T_3 - T_2) \] \[ T_3 = 600\, K, \quad T_2 = 1200\, K \] \[ Q_{23} = 2.50 \times 12.471 \times (600 - 1200) = 31.18 \times (-600) = -18,708\, J \] (Heat released) --- ### **Step 4: Work during 3→1** Process 3→1: constant pressure \( P = 1\, atm \), volume decreases from 123.16 L to 61.58 L. Work: \[ W_{31} = P \times (V_{final} - V_{initial}) = 1 \times (61.58 - 123.16) = -61.58\, L\, atm \] Convert: \[ W_{31} = -61.58 \times 101.325 \approx -6237\, J \] Heat exchanged: \[ Q_{31} = n C_P (T_1 - T_3) \] For monatomic ideal gas: \[ C_P = C_V + R = 12.471 + 8.314 = 20.785\, J/(mol\,K) \] \[ Q_{31} = 2.50 \times 20.785 \times (300 - 600) = 51.96 \times (-300) = -15,588\, J \] --- ### **Step 5: Total work and heat input** **Net work:** \[ W_{total} = W_{12} + W_{23} + W_{31} = 9360 + 0 - 6237 = 3123\, J \] **Total heat input (only during 1→2):** \[ Q_{in} \approx 37,420\, J \] ### **Efficiency:** \[ \boxed{ \eta = \frac{W_{total}}{Q_{in}} = \frac{3123}{37420} \approx 0.0835 \text{ or } 8.35\% } \] --- ## **Part (c): Carnot efficiency** The maximum thermal efficiency between two reservoirs at temperatures \( T_H \) and \( T_L \): \[ \eta_{Carnot} = 1 - \frac{T_L}{T_H} \] - Highest temperature \( T_H = T_2 = 1200\,K \) - Lowest temperature \( T_L = T_1 = 300\,K \) \[ \eta_{Carnot} = 1 - \frac{300}{1200} = 1 - 0.25 = 0.75 \] **Final answer:** \[ \boxed{ \text{Carnot efficiency} = 75\% } \] --- ## **Summary:** | Part | Description | Final Result | |---------|----------------|--------------| | **(a)** | PV diagram points: (61.58 L, 1 atm), (123.16 L, 2 atm), (123.16 L, 1 atm) | Diagram as described | | **(b)** | Cycle efficiency | **8.35%** | | **(c)** | Carnot efficiency | **75%** | If you'd like a visual sketch of the PV diagram or further clarification, feel free to ask!