(12) = + and raising both sides to the power, 1/Cv’ Cp’ Cv’ R*, (13) g = Cp’/Cv’. For the isothermal leg from B to C with = = constant, from the gas law we T T1 have (14) For the leg from C to D we have from (13) adiabatic (15) For the isothermal leg from D to A with = = constant, from the ideal gas law we T T2 have (16) Combining (15) and (16) yields (17) Combining (13) and (14) yields (18) such that from (17) and (18) we have and finally (19) So this us a relationship between the volume of the gas at the four corners of the diagram. The heat, absorbed from the warm reservoir is absorbed at constant temperature. DQ1, Therefore the internal energy of the gas does not change and all of the energy absorbed goes solely into work. Therefore (20) Rewrite pressure in terms of the ideal gas law,
(12) Cp’ = Cv’ + R*, and raising both sides to the power, 1/Cv’ (13) g = Cp’/Cv’. For the isothermal leg from B to C with T = T1 = constant, from the gas law we have (14) For the adiabatic leg from C to D we have from (13) (15) For the isothermal leg from D to A with T = T2 = constant, from the ideal gas law we have (16) Combining (15) and (16) yields (17) Combining (13) and (14) yields (18) such that from (17) and (18) we have and finally (19) So this us a relationship between the volume of the gas at the four corners of the diagram. The heat, DQ1, absorbed from the warm reservoir is absorbed at constant temperature. Therefore the internal energy of the gas does not change and all of the energy absorbed goes solely into work. Therefore (20) Rewrite pressure in terms of the ideal gas law,
CarnotCycle.docx