--- title: "Free-moving piston between two chambers" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:free-piston-two-chambers-100-50" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/free-piston-two-chambers-100-50 site: BCT Engineering --- # Free-moving piston between two chambers Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2069 Poush, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/free-piston-two-chambers-100-50 - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3 ## Question The device shown in figure below has a free moving piston between the two chambers. The initial total volumes of A and B are equal with VA = 100 m³/kg and VB = 50m³/kg. If the piston is moved so that x is one-fourth of the entire length, determine the final specific volumes of chambers A and B. [Figure: a horizontal cylinder with chambers A (left) and B (right) separated by a piston; x is the distance of the piston from the left end; total volume marked 1 m³.] ## Short answer Masses: mA = 0.5/100 = 0.005 kg, mB = 0.5/50 = 0.010 kg. With piston at one‑fourth length, VA = 0.25 m³, VB = 0.75 m³. Final specific volumes: vA = 0.25/0.005 = 50 m³/kg, vB = 0.75/0.010 = 75 m³/kg. ## Answer **Given:** chambers A and B, separated by a free piston; total volume 1 m³; initially V_A = V_B (each 0.5 m³); v_A = 100 m³/kg, v_B = 50 m³/kg. Finally the piston is at x = one-fourth of the length. We assume x is measured from the A end, so V_A = ¼ of the total volume. No mass crosses the piston, so m_A and m_B remain constant. **Step 1: Masses** m_A = 0.5/100 = 0.005 kg m_B = 0.5/50 = 0.010 kg **Step 2: Final volumes** (constant cross-section, so volume ∝ length) V_A2 = ¼ × 1 = 0.25 m³ V_B2 = ¾ × 1 = 0.75 m³ **Step 3: Final specific volumes** v_A2 = V_A2/m_A = 0.25/0.005 = **50 m³/kg** v_B2 = V_B2/m_B = 0.75/0.010 = **75 m³/kg** **Answer: v_A = 50 m³/kg and v_B = 75 m³/kg.** (If x were measured from the B end, V_A = 0.75 m³ and V_B = 0.25 m³, giving v_A = 150 m³/kg and v_B = 25 m³/kg.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the masses in chambers A and B be treated as constant while the piston moves? A. No mass crosses the piston, so each chamber is a closed system B. The piston is heated so mass is generated C. Pressure is equal on both sides D. The total volume of 1 m³ is constant, so each mass is constant automatically Answer: A. Nothing leaks across the piston, so m_A and m_B stay fixed; only volumes (and so specific volumes) change. (Key point: Mass constant) 2. What are the masses in A and B (each initially 0.5 m³)? A. m_A = 0.010 kg, m_B = 0.005 kg B. m_A = 0.005 kg, m_B = 0.010 kg C. m_A = 200 kg, m_B = 100 kg D. m_A = m_B = 0.0075 kg Answer: B. m = V/v = 0.5/100 = 0.005 kg and 0.5/50 = 0.010 kg. (Key point: Initial masses) 3. With x = 1/4 of the length measured from the A end (constant cross-section), the final volumes are: A. V_A = 0.75 m³ and V_B = 0.25 m³ B. V_A = 0.25 m³ and V_B = 0.25 m³ C. V_A = 0.25 m³ and V_B = 0.75 m³ D. V_A = 0.5 m³ and V_B = 0.5 m³ Answer: C. Volume is proportional to length, so A gets 1/4 and B gets 3/4 of the 1 m³. (Key point: Final volumes) 4. What are the final specific volumes of A and B? A. v_A = 75 m³/kg, v_B = 50 m³/kg B. v_A = 25 m³/kg, v_B = 75 m³/kg C. v_A = 100 m³/kg, v_B = 50 m³/kg D. v_A = 50 m³/kg, v_B = 75 m³/kg Answer: D. v = V/m: 0.25/0.005 = 50 and 0.75/0.010 = 75 m³/kg. 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