--- title: "Cylinder with movable piston: specific volumes" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:cylinder-piston-specific-volumes-8" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/cylinder-piston-specific-volumes-8 site: BCT Engineering --- # Cylinder with movable piston: specific volumes Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2074 Bhadra, 6 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/cylinder-piston-specific-volumes-8 - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3 ## Question A cylinder with a total volume of 2 m³ has a movable piston as shown in figure below, when the piston is at one fourth of the length, both sides have the same specific volume of 8 m³/kg. Determine the specific volumes of both sides when the piston is at middle of the cylinder. [Figure: horizontal cylinder of length L divided by a movable piston into side A (left) and side B (right); piston at x = L/4 from the left end.] ## Short answer Initial volumes from length fractions: 0.5 m³ and 1.5 m³ with v=8 m³/kg, giving masses 0.0625 kg and 0.1875 kg. When piston is at middle, both volumes are 1 m³, so specific volumes are 16 m³/kg and 5.33 m³/kg. ## Answer **Given:** total volume V = 2 m³; the piston at one fourth of the length (x = L/4) from side A; v_A = v_B = 8 m³/kg. The piston is free, so the masses on the two sides remain constant. **Step 1: Volumes at the initial position** (constant cross-section, so volume ∝ length) V_A1 = (1/4)(2) = 0.5 m³ V_B1 = (3/4)(2) = 1.5 m³ **Step 2: Masses** m_A = V_A1 / v_A = 0.5/8 = 0.0625 kg m_B = V_B1 / v_B = 1.5/8 = 0.1875 kg **Step 3: Piston at the middle** (x = L/2) V_A2 = V_B2 = 1 m³ **Step 4: Specific volumes** v_A2 = V_A2/m_A = 1/0.0625 = **16 m³/kg** v_B2 = V_B2/m_B = 1/0.1875 = **5.33 m³/kg** **Answer: v_A = 16 m³/kg and v_B = 5.33 m³/kg.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the masses on each side be taken as constant? A. The piston is free and no mass crosses it, so each side's mass is fixed B. Because the pressure is constant C. Because specific volumes are equal D. Because the temperature is constant Answer: A. Each side is a closed system; only its volume and specific volume change. (Key point: Mass constant) 2. What are the initial volumes of A and B with the piston at L/4? A. 1 m3 and 1 m3 B. 0.5 m3 and 1.5 m3 C. 1.5 m3 and 0.5 m3 D. 0.25 m3 and 0.75 m3 Answer: B. For constant cross-section, volume is proportional to length, so VA = 2/4 = 0.5 and VB = 3/4 x 2 = 1.5 m3. (Key point: Initial volumes) 3. How is mass of each side found? A. m = V x v: mA = 4 kg, mB = 12 kg B. m = v/V C. m = V/v: mA = 0.0625 kg, mB = 0.1875 kg D. Both sides have equal mass of 0.125 kg Answer: C. Using m = V1/v with v = 8 m3/kg gives 0.0625 and 0.1875 kg. (Key point: Mass calculation) 4. With the piston at middle, what are vA and vB? A. vA = 8 and vB = 8 m3/kg B. vA = 5.33 and vB = 16 m3/kg C. vA = 0.0625 and vB = 0.1875 m3/kg D. vA = 16 and vB = 5.33 m3/kg Answer: D. Each side has V = 1 m3, so v = 1/0.0625 = 16 and 1/0.1875 = 5.33 m3/kg. 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