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Solution_8_recitation

# Solution_8_recitation - Solution to the problems in...

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Solution to the problems in recitation 8 Problem 1 Known: Ideal gas in two compartments of a closed, rigid container. Find: p 2B , T 2B and S 2B -S 1B ; p 2A , T 2A and S 2A -S 1A ; Q 12 (kJ); Change of S for the universe. Schematic and Given Data: A B 2 A B 1 C v = 0.75 kJ/kg.K, C p /C v = 1.4, R = 0.300 kJ/kg.K p 1A = p 1B = 180 kPa, T 1A = T 1B = 300 K, V 1A = V 1B = 0.5 m 3 , V2A = 0.75 m 3 , V2B = 0.25 m 3 Assumptions: 1. gas in each compartment is a closed system 2. part B undergoes a quasi-static process 3. Heat transfer occurs at reservoir temperature 4. Constant specific heat 5. No change in KE and PE Analysis: a) For gas in compartment B, it undergoes an isentropic (adiabatic and reversible) process. C p /C v = 1.4 p 1B (V 1B ) 1.4 = p 2B (V 2B ) 1.4 p 2B = p 1B (V 1B /V 2B ) 1.4 = 180 kPa * (0.5 m 3 /0.25 m 3 ) 1.4 p 2B = 475 kPa m B = (p 1B V 1B )/(RT 1B ) = 1 kg From ideal gas law, T 2B /T 1B = (p 2B V 2B )/(p 1B V 1B ) T 2B = T 1B (p 2B V 2B )/(p 1B V 1B ) = 300 K (475 kPa * 0.25 m 3 )/(180 kPa * 0.5 m 3 ) = 395.83 K Entropy change should be zero for part B, since there is no heat transfer, and there is no internal irreversibility for part B. according to entropy change for ideal gas S 2B -S 1B = m B [C v ln(T 2B /T 1B ) + R ln(V 2B /V 1B )] = 1 (kg) [0.75 (kJ/kg.K) ln(395.83/300) + 0.3 (kJ/kg.K) ln(0.25/0.5)]

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