VIPSolutions logo ✨ VIPSolutions

1.02kg of R-134a fills a 0.14m3 weighted piston cylinder device at a tempreture of -26.40C The container is now heated untill tempreture is 100.0C Determine the final volume of R134. Use data from the refrigrant table.

Question:

1.02kg of R-134a fills a 0.14m3 weighted piston cylinder device at a tempreture of -26.40C The container is now heated untill tempreture is 100.0C Determine the final volume of R134. Use data from the refrigrant table.

This is AI generated Response.
Asked by: Rajveer Prajapat
Created at: 2025-10-10 11:40:20

Final Answer:

Rendered Raw Text
Did you like this answer?
# Final Volume of R-134a in a Piston-Cylinder Device ## Given Data - **Mass of R-134a,** \( m = 1.02\, \text{kg} \) - **Initial volume,** \( V_1 = .14\, \text{m}^3 \) - **Initial temperature,** \( T_1 = -26.4^\circ\text{C} \) - **Final temperature,** \( T_2 = 100.^\circ\text{C} \) - **Device:** Weighted piston-cylinder (constant pressure) ## Step 1: Find Initial State ### 1. Calculate Initial Specific Volume \[ v_1 = \frac{V_1}{m} = \frac{.14}{1.02} = .1373\, \text{m}^3/\text{kg} \] ### 2. Use R-134a Property Tables At \( T_1 = -26.4^\circ\text{C} \): - From R-134a tables, the saturation pressure \( P_{\text{sat}} \approx 100\, \text{kPa} \). - Saturated liquid specific volume \( v_f \approx .000757\, \text{m}^3/\text{kg} \). - Saturated vapor specific volume \( v_g \approx .3666\, \text{m}^3/\text{kg} \). Since \( v_1 \) is much greater than \( v_f \) and less than \( v_g \), the refrigerant is a **mixture** at -26.4°C. ### 3. Find Initial Pressure Since the volume is fixed with a weighted piston, the pressure is constant. At \( T_1 \), the pressure is the saturation pressure at \( -26.4^\circ\text{C} \): \[ P_1 = P_{\text{sat, at } -26.4^\circ\text{C}} = 100\, \text{kPa} \] ## Step 2: Final State (After Heating to 100°C, at Constant Pressure) - **Final temperature:** \( T_2 = 100.^\circ\text{C} \) - **Pressure:** \( P_2 = P_1 = 100\, \text{kPa} \) Check the phase at final state: From R-134a tables, at \( P = 100\, \text{kPa} \): - \( T_{\text{sat}} = -26.4^\circ\text{C} \) - At \( T_2 = 100^\circ\text{C} \), the refrigerant is **superheated** at 100 kPa. ## Step 3: Find Final Specific Volume From R-134a superheated tables: - At \( P = 100\, \text{kPa} \), \( T = 100^\circ\text{C} \): - \( v \approx .8845\, \text{m}^3/\text{kg} \) ## Step 4: Calculate Final Volume \[ V_2 = m \cdot v_2 = 1.02 \times .8845 = .902\, \text{m}^3 \] ## **Summary Table** | State | T (\( ^\circ\)C) | P (kPa) | v (\( \text{m}^3/\text{kg} \)) | V (\( \text{m}^3 \)) | Phase | |-------|------------------|---------|-------------------------------|----------------------|-----------| | 1 | -26.4 | 100 | .1373 | .14 | Mixture | | 2 | 100. | 100 | .8845 | .902 | Superheated | ## **Final Answer** \[ \boxed{V_2 = .90\, \text{m}^3} \] (Rounded to two decimal places) --- **References:** R-134a refrigerant property tables (superheated and saturated). **Note:** Actual values might vary slightly depending on the refrigerant property source used.

📢 Instructions:

  • Ask anything relatd to above question only.
  • For new Question goto :

Study Help, Fast Answers

VIPSolutions focuses on clear, step‑by‑step explanations so you can learn quickly. Whether you need an AI‑generated walkthrough or a short hint to get unstuck, each solution is organized for fast reading and easy review later.

Search similar questions, compare approaches, and bookmark the best answers for revision. Our goal is simple: quick, reliable study help that feels natural—not noisy.