Answer to Question #96946 in Mechanical Engineering for Michael

Question #96946
the following equation gives the internal energy of a certain substance U=3.64PV + 90 where U is in kj/kg,P is in kPa and V is in m^3/kg. A system composed of 3.5kg of this substance expand from an initial pressure of 500kPa and a volume of 0.25m^3 to fixed pressure 100kPa in a process in which pressure and volume are related by PV^1.25 = constant.if the expansion is quasi-static,find Q,DU and N for the another process the same system expand according to the same pressure volume relationship and from the same initial state to the same final state but the heat transfer in this case is 32kj.find the work transfer for this process. explain the difference in work transfer
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