Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1

Question

Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1 and 2 in the figure below. In process 1 the gas is heated at constant volume from an initial pressure of Pi = 113 kPa to a final pressure of Pf = 226 kPa. In process 2 the gas expands at constant pressure from an initial volume of Vi = 1.00 m3 to a final volume of Vf = 3.00 m3.

process 1 1
Your response differs from the correct answer by more than 10%. Double check your calculations. kJ process 2 2 kJ Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1 and 2 in the figure below. In process 1 the gas is heated at constant volume from an initial pressure of Pi = 113 kPa to a final pressure of Pf = 226 kPa. In process 2 the gas expands at constant pressure from an initial volume of Vi = 1.00 m^3 to a final volume of Vf = 3.00 m^3. (a) How much heat is added to the gas during these two processes? process 1 process 2 kJ (b) How much work does the gas do during this expansion? kJ (c) What is the change in the internal energy of the gas? KJ

Explanation / Answer

No.of moles n = 60 mol Process 1 : ----------- Initial pressure P = 113 kPa = 113 * 10 ^ 3 Pa final pressure P ' = 226* 10^3 pa Specific heat at constant volume Cv = 1.5 R = 1.5 * 8.314 = 12.471 J / mol K Specific heat at constant pressure Cp = 2.5 R = 2.5 * 8.314 = 20.785 J / mol K Initial volume V= 1 m^ 3 from the relation PV = nRT initial temperature T = PV / nR where R = gas constant = 8.314 J / mol K So, T = 226.52 K final volume V ' = 1 m^ 3 final temperature  T ' = P'V' / nR                                 = 453 K heat added in process 1 is Q = n*Cv * ( T ' - T )                                              = 169.5 *10^ 3 J work done W = 0    Since change in volume = 0 Change in internal energy U = Q-W = 169.5 * 10^3 J Process 2: ---------- final volume V " = 3 m^ 3 final Pressure P " = P ' = 226* 10 ^ 3 Pa final temperature T" = P"V" / nR                                = 1359.15 K heat added Q ' = n*Cp * ( T " -T ' )                          = 1.13*10^6 J work done W ' = P ' ( V " - V ' )                          = 452000 J change in internal energy U ' = Q ' - W '                                             = 678000 J (b). total work done = W + W ' = 452000 J (c).total internal energy = U + U ' = 847500 J

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