### Step-by-Step Solution To solve for \( \frac{Q}{m} \) and \( \frac{W}{m} \) (heat and work per unit mass) for each process, we need details about the cycle, such as **pressure, volume, and temperature at each state**, and the **type of processes** (isothermal, isobaric, isochoric, adiabatic). Since the image does **not provide those values or process types**, I'll give you a **general method** for each process in a typical thermodynamic cycle (like the air-standard Otto, Diesel, or Brayton cycles). Let’s proceed with generic equations for **ideal gas** and **closed system** (as stated). --- ## **Process 1-2** #### **General Equations** - **Heat transferred:** \( \frac{Q_{12}}{m} = u_2 - u_1 + W_{12}/m \) - **Work done:** \( \frac{W_{12}}{m} = \int_{1}^{2} P\,dv \) If process type is known (e.g., isochoric, isobaric, isothermal, or adiabatic), formulas simplify: - **Isochoric (constant volume):** - \( \frac{Q_{12}}{m} = u_2 - u_1 \) - \( \frac{W_{12}}{m} = \) - **Isobaric (constant pressure):** - \( \frac{Q_{12}}{m} = h_2 - h_1 \) - \( \frac{W_{12}}{m} = P (v_2 - v_1) \) - **Isothermal (ideal gas):** - \( \frac{Q_{12}}{m} = \frac{W_{12}}{m} = R T \ln\left(\frac{v_2}{v_1}\right) \) - **Adiabatic (isentropic, ideal gas):** - \( \frac{Q_{12}}{m} = \) - \( \frac{W_{12}}{m} = u_1 - u_2 \) (if reversible, \( k = c_p/c_v \)) --- ## **Process 2-3** _Similar logic as above for appropriate process:_ - **Isochoric:** \( \frac{Q_{23}}{m} = u_3 - u_2 \), \( \frac{W_{23}}{m} = \) - **Isobaric:** \( \frac{Q_{23}}{m} = h_3 - h_2 \), \( \frac{W_{23}}{m} = P (v_3 - v_2) \) - **Isothermal:** \( \frac{Q_{23}}{m} = \frac{W_{23}}{m} = R T \ln\left(\frac{v_3}{v_2}\right) \) - **Adiabatic:** \( \frac{Q_{23}}{m} = \), \( \frac{W_{23}}{m} = u_2 - u_3 \) --- ## **Process 3-4** _Repeat as above for the process type:_ - **Isochoric:** \( \frac{Q_{34}}{m} = u_4 - u_3 \), \( \frac{W_{34}}{m} = \) - **Isobaric:** \( \frac{Q_{34}}{m} = h_4 - h_3 \), \( \frac{W_{34}}{m} = P (v_4 - v_3) \) - **Isothermal:** \( \frac{Q_{34}}{m} = \frac{W_{34}}{m} = R T \ln\left(\frac{v_4}{v_3}\right) \) - **Adiabatic:** \( \frac{Q_{34}}{m} = \), \( \frac{W_{34}}{m} = u_3 - u_4 \) --- ## **Process 4-1** - **Isochoric:** \( \frac{Q_{41}}{m} = u_1 - u_4 \), \( \frac{W_{41}}{m} = \) - **Isobaric:** \( \frac{Q_{41}}{m} = h_1 - h_4 \), \( \frac{W_{41}}{m} = P (v_1 - v_4) \) - **Isothermal:** \( \frac{Q_{41}}{m} = \frac{W_{41}}{m} = R T \ln\left(\frac{v_1}{v_4}\right) \) - **Adiabatic:** \( \frac{Q_{41}}{m} = \), \( \frac{W_{41}}{m} = u_4 - u_1 \) --- ## **Summary Table** | Process | \( \frac{Q}{m} \) (kJ/kg) | \( \frac{W}{m} \) (kJ/kg) | |---------|---------------------------|---------------------------| | 1-2 | [depends on process] | [depends on process] | | 2-3 | [depends on process] | [depends on process] | | 3-4 | [depends on process] | [depends on process] | | 4-1 | [depends on process] | [depends on process] | --- ### **Final Answer** You need to **identify the process type** for each stage and **plug in the values** for temperature, pressure, and specific volumes to use the correct equation from above. If you provide **state properties or process types**, I can give you the exact numerical answers. **Let me know the process types or properties, and I’ll solve it step-by-step with numbers!**