Question

3. Two kilograms of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are 2 m'and 1 bar, respectively. Heat transfer to the oxygen occurs at constant pressure until the volume is doubled. Assuming the ideal gas behavior determine the heat transfer for the process, in kJ. Kinetic and potential energy effects can be ignored. Given Cy = 0.742 kJ/kg-K. Oz V = 2V ! ! Patm = I am m = 2 kg Vi = 2 m P = 1 atm 112

Answer 1

Heat Transfer for the process = 781.428 KJ

Question 3 : Solution zatm= 101.325 kPa Y-2m Pi= Latm m=2Kg constant Pressure Exchansion Paccess Calculation of Gas constant of air : • Here Mos is molecular moss of os.com Substitute Mos = -32. Kete kan King R- 8:314 kep R= 0.2598 KI • Calculation of shocific heat for Oz at consland Pressure : Cp - Cv = R Here Cp is specific heat of De at constant pressure. Subotitulo cerpt Cu = 0.742 KJ and R = 0.2 Cp-D-742 = 0.2598 Cp = 1.0018 KI • Colculation of Temperature at stale 1 using ideal gas relation, Ti = Piv MR Subsolulute Pi= 101.325 kPa ( tatm = 10/-395*Pa], =2M3 and M=2Kg and R=0-2598 Ti = 101.325 x 2 2 x 0.2598

Question 3 : Solution zatm= 101.325 kPa Y-2m Pi= Latm m=2Kg constant Pressure Exchansion Paccess Calculation of Gas constant of air : • Here Mos is molecular moss of os.com Substitute Mos = -32. Kete kan King R- 8:314 kep R= 0.2598 KI • Calculation of shocific heat for Oz at consland Pressure : Cp - Cv = R Here Cp is specific heat of De at constant pressure. Subotitulo cerpt Cu = 0.742 KJ and R = 0.2 Cp-D-742 = 0.2598 Cp = 1.0018 KI • Colculation of Temperature at stale 1 using ideal gas relation, Ti = Piv MR Subsolulute Pi= 101.325 kPa ( tatm = 10/-395*Pa], =2M3 and M=2Kg and R=0-2598 Ti = 101.325 x 2 2 x 0.2598