A heat engine operates on a cycle with the following state points and given data. Calculate the work done during adiabatic compression (process 1-2) and the heat rejected during isochoric cooling (pro

Here are the solutions to the questions based on the provided pressure-volume diagram:

Given Data:

• Adiabatic index, $\gamma = 1.4$
• Mass of gas, $m = 1$ kg
• Specific entropy at point 1, $s_1 = 0.515$ kJ/kg.K

State Points from Diagram:

• Point 1: $P_1 = 100$ kPa, $V_1 = 0.082$ m$^3$
• Point 2: $V_2 = 0.0054$ m$^3$
• Point 3: $P_3 = 6200$ kPa, $V_3 = 0.0054$ m$^3$
• Point 4: $P_4 = 6200$ kPa, $V_4 = 0.0125$ m$^3$
• Point 5: $V_5 = 0.082$ m$^3$

---

1.1 The work done during adiabatic compression
The adiabatic compression process is from point 1 to point 2.

Step 1: Calculate the pressure at point 2 ($P_2$) using the adiabatic relation $P_1V_1^\gamma = P_2V_2^\gamma$.

$$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma$$ $$P_2 = 100 \text{ kPa} \times \left(\frac{0.082 \text{ m}^3}{0.0054 \text{ m}^3}\right)^{1.4}$$ $$P_2 = 100 \text{ kPa} \times (15.185)^{1.4}$$ $$P_2 = 100 \text{ kPa} \times 36.790$$ $$P_2 = 3679.0 \text{ kPa}$$

Step 2: Calculate the work done during the adiabatic process 1-2. For an adiabatic process, the work done by the system is given by:

$$W_{1-2} = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$$ $$$$