To solve this, let's break the problem into steps.

Step 1: Definitions and Assumptions

• Given:

  • Initial pressure, $P_1 = 29 \, \text{psi} = \frac{29}{14.5} \, \text{bar} = 2 \, \text{bar}$
  • Initial temperature, $T_1 = 27 \, \text{°C} = 300.15 \, \text{K}$
  • Initial volume, $V_1 = 5000 \, \text{liters} = 5 \, \text{m}^3$
  • Final pressure after compression, $P_2 = 87 \, \text{psi} = \frac{87}{14.5} \, \text{bar} = 6 \, \text{bar}$
  • Adiabatic index, $\gamma = 1.3$

• The first process is polytropic compression with $PV^{1.3} = C$.

• The second process is adiabatic expansion to $P_3 = 2 \, \text{bar}$.

Step 2: Volume After Compression

Using the polytropic equation: $P_1 V_1^{1.3} = P_2 V_2^{1.3}$ Rearrange to find $V_2$: $V_2 = V_1 \left( \frac{P_1}{P_2} \right)^{1 / 1.3}$ Substitute values: $V_2 = 5 \left( \frac{2}{6} \right)^{1 / 1.3} = 5 \cdot (0.333)^{0.769}$ $V_2 \approx 2.43 \, \text{m}^3$

Step 3: Work Done During Compression

The work done during polytropic compression is given by: $W_{12} = \frac{P_2 V_2 - P_1 V_1}{1 - n}$ Substitute $n = 1.3$, $P_1 = 2 \, \text{bar}$, $P_2 = 6 \, \text{bar}$, $V_1 = 5 \, \text{m}^3$, $V_2 = 2.43 \, \text{m}^3$: $W_{12} = \frac{(6 \cdot 2.43) - (2 \cdot 5)}{1 - 1.3}$ $W_{12} = \frac{14.58 - 10}{-0.3} = \frac{4.58}{-0.3} \approx -15.27 \, \text{kJ}$ (The negative sign indicates work input.)

Step 4: Adiabatic Expansion

For adiabatic processes, $PV^\gamma = C$. The volume after expansion $V_3$ can be found using: $P_2 V_2^\gamma = P_3 V_3^\gamma$ Rearrange to find $V_3$: $V_3 = V_2 \left( \frac{P_2}{P_3} \right)^{1 / \gamma}$ Substitute values: $V_3 = 2.43 \left( \frac{6}{2} \right)^{1 / 1.3} = 2.43 \cdot (3)^{0.769}$ $V_3 \approx 5.01 \, \text{m}^3$

Step 5: Work Done During Expansion

The work done during adiabatic expansion is: $W_{23} = \frac{P_3 V_3 - P_2 V_2}{1 - \gamma}$ Substitute $\gamma = 1.3$, $P_2 = 6 \, \text{bar}$, $P_3 = 2 \, \text{bar}$, $V_2 = 2.43 \, \text{m}^3$, $V_3 = 5.01 \, \text{m}^3$: $W_{23} = \frac{(2 \cdot 5.01) - (6 \cdot 2.43)}{1 - 1.3}$ $W_{23} = \frac{10.02 - 14.58}{-0.3} = \frac{-4.56}{-0.3} \approx 15.20 \, \text{kJ}$

Step 6: Net Work Done

The net work done over the two processes is: $W_{\text{net}} = W_{12} + W_{23}$ $W_{\text{net}} = -15.27 + 15.20 \approx -0.07 \, \text{kJ}$

Step 7: Heat Transfer and Entropy Change

• Compression: In polytropic compression, heat transfer is: $Q_{12} = W_{12} + \Delta U$ $\Delta U$ depends on the change in temperature, which can be calculated using the ideal gas law and specific heats.
• Expansion: In adiabatic expansion, $Q_{23} = 0$.

Entropy change can be calculated using: $\Delta S = \int \frac{\delta Q}{T}$

Do you want me to compute these values step by step, including detailed entropy and heat transfer calculations? Let me know!

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Here are 5 questions to expand understanding:

1. What is the significance of the polytropic index in compression/expansion processes?
2. How is entropy affected in reversible vs. irreversible processes?
3. How does the adiabatic process differ from an isothermal process thermodynamically?
4. Why is the net work done nearly zero in this cycle?
5. What role does the gas constant play in determining internal energy changes?

Tip: Always track the units in thermodynamic problems; converting consistently between systems (e.g., bar and psi) ensures accuracy.