First Law of Thermodynamics: Closed Systems
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1 utorial # First Law of hermodynamics: Closed Systems Problem -7 A 0.-m tank contains oxygen initially at 00kPa and 7 C. A paddle wheel within the tank is rotated until the pressure inside rise to 50kPa. During the process KJ of heat is lost to the surroundings. Determine the paddle-wheel work done. Neglect the energy stored in the paddle wheel. Solution: Step : Draw a schematic diagram to represent the system ank Q Paddle Wheel Step : What to determine? he work done by paddle-wheel work, W e Step : he information given in the problem statement. utorial # Page of 5
2 Volume of the tank, V = 0. m, and Volume remains constant during the whole process: V V = V =. Initial condition of the Oxygen in the tank: P = 00 kpa, = 7 C; Final condition of the Oxygen in the tank: P = 50 kpa; Heat loss to the surroundings, Q =! KJ. Step 4: able of all known values and properties Oxygen Pressure (kpa Volume (m emperature (K Initial condition Final condition Heat loss (KJ - Step 5: Assumptions It s a closed system and no mass loss in the whole process; Consider the Oxygen as ideal gas for given conditions; Neglect the energy stored in the paddle wheel; Change in kinetic and potential energy is negligible. Step 6: Solve. According the ideal gas equation, PV = mr From the initial condition, the mass of the Oxygen is determined by PV (00kPa " (0.m m = = = 0.849( kg R (0.598kPa! m / kg! K(00K. he final temperature of the Oxygen can be determined from PV PV = utorial # Page of 5
3 So, PV (50kPa! (0.m = =! (00K = 450( K PV (00kPa! (0.m. he energy conservation equation of this closed system can be described as: where, Q " W =! U +! KE +! PE! KE and! PE are both zero according to the assumptions. As for the work, W = W + W + W e Due to the volume remains constant, the boundary work is zero and there is no other works indicated in the problem. hus the above energy conservation equation became, b Q " We =! U Because it s a constant-volume process, the above equation can be concluded, Q! W other = m( u! u = Cvm( e! Here, the specific heat of Oxygen at the average temperature of avg = ( / = 75( K is, = KJ /( kg (from the able A-b. C v, avg! K So, rearrange the above equation and substitute the values, W = Q! C m( e v! = (! KJ! (0.6745KJ / kg # K " (0.849kg " (450K! 00K =!40.94KJ Step 7: Conclusion statement he work done by the paddle-wheel is KJ. utorial # Page of 5
4 Problem -8 A piston-cylinder device contains 5kg of argon at 400kPa and 0 C. During a quasi-equilibrium, isothermal expansion process, 5KJ of boundary work is done by the system, and KJ of paddle-wheel work is done on the system. Determine the heat transfer for this process. Solution: Step : Draw a schematic diagram to represent the problem W b Q W e Initial Condition Final Condition Step : What to determine? he heat transfer between the system and the surroundings, Q Step : he information given in the problem statement.. Argon in the cylinder: m=5kg, P=400kPa and =0V;. A boundary work done by the system, W b =5KJ;. he paddle-wheel work done on the system, W e =-KJ utorial # Page 4 of 5
5 Step 4: Assumptions. It s a quasi-equilibrium, isothermal expansion process, which means that the temperature remains constant in the whole process;. For the argon in the piston-cylinder system, it s a closed system, no mass enters or leaves.. Change in kinetic and potential energy is negligible. Step 6: Solve We take the argon in the piston-cylinder system as our system, and the energy conservation equation of this closed system can be described as: where, Q " W =! U +! KE +! PE! KE and! PE are both zero according to the assumptions. It s a quasiequilibrium, isothermal expansion process. he temperature remains constant, so is the total internal energy. he change of the total internal energy is zero during the whole process, which give us So, Substituting the works,!u = 0 Q!W = 0 Q = W = We + Wb = (! KJ + (5KJ = KJ he sign is positive, which means that heat transfers from the surroundings to the system. Step 7: Conclusion statement In this process, the argon in the piston-cylinder device will absorb heat with an amount of KJ from the surroundings. utorial # Page 5 of 5
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