The four reversible transformations all arise from the same condition ΔStot=0\Delta S_\text{tot} = 0, distinguished by the constraint imposed. The table collects, for each one, the characteristic constraint, the work done by the gas, the heat exchanged and the change in entropy of the gas: it is the reference scheme to keep in mind when solving a cycle made up of several segments.

ProcessConstraintLgasL_\text{gas}QQΔSgas\Delta S_\text{gas}
IsochoricVB=VAV_B = V_A00nCvΔTnC_v\,\Delta TnCvlnTBTAnC_v\ln\dfrac{T_B}{T_A}
IsobaricPB=PAP_B = P_APΔVP\,\Delta VnCpΔTnC_p\,\Delta TnCplnTBTAnC_p\ln\dfrac{T_B}{T_A}
IsothermalTB=TAT_B = T_AnRTlnVBVAnRT\ln\dfrac{V_B}{V_A}=Lgas= L_\text{gas}nRlnVBVAnR\ln\dfrac{V_B}{V_A}
AdiabaticQ=0Q = 0ΔEint-\Delta E_\text{int}0000

Topics: Thermodynamics Concepts: Thermodynamic transformations · Entropy Skills: Solving a thermodynamic cycle · Reading graphs Objects: Ideal gas

Related exercises: Worked exercise — reversible vs irreversible isothermal expansion · Problem — Carnot cycle in the T-S plane · Ranking ΔS in three expansions