The work done during this process is minus nR, times T2 minus T1, times natural log of V2 over V1, or minus the total of heat absorbed and heat discharged, calculated by applying Boyle's law for two of the phases and a modification of Boyle's law with ratio of heat capacities for the other two phases.
The work done during this process is minus nR, times T2 minus T1, times natural log of V2 over V1, or minus the total of heat absorbed and heat discharged, calculated by applying Boyle's law for two of the phases and a modification of Boyle's law with ratio of heat capacities for the other two phases. The energy change is zero, as is the entropy change for this reversible process. For 10 points, one minus the ratio of reservoir and source temperatures gives the efficiency of what process, an upper bound for real cycles, which reverses a heat pump and contains pairs of isotherms and adiabats?
The work done during this process is minus nR, times T2 minus T1, times natural log of V2 over V1, or minus the total of heat absorbed and heat discharged, calculated by applying Boyle's law for two of the phases and a modification of Boyle's law with ratio of heat capacities for the other two phases.
{Carnot} cycle or engine