Circular integrals Because entropy is a state function, it integrates to zero over any circular path going back to initial conditions, just like \(U\) and \(H\): \[\oint dS =0 \nonumber \] \[\oint dH =0 \nonumber \] \[\oint dU =0 \nonumber \] As discussed previously, we can use this fact to revisit the isotherm + isochore + adiabat circular path (Figure 20.3.1 ). Along adiabat B and isochore C: There is no heat transfer along adiabat B: \[q_{rev,B} = 0 \nonumber \] There is no work along isochore C: \[\delta w=0 \nonumber \] But the temperature changes from \(T_2\) back to \(T_1\). This requires heat: \[q_{rev,C}=C_V\Delta T \nonumber \] Along the isotherm A: We have seen that \[q_{rev,A} = nRT \ln \dfrac{V_2}{V_1} \nonumber \] The quantities q rev ,A , q rev ,B , and q rev ,C are not the same, which once again underlines that heat is a path function. How about entropy? First, consider the combined paths of B and C: \[q_{rev,B+C} = \int _{T_2}^{T_1} C_v dT \nonumber \] \[\int dS_{B+C} = \int \dfrac{dq_{rev,B+C}}{T} = \int _{T_2}^{T_1} \dfrac{C_v}{T} dT \nonumber \] We had seen this integral before from Section 19-6, albeit from \(T_1\) to \(T_2\): \[\Delta S_{B+C} = nR\ln \dfrac{V_2}{V_1} \label{19.21} \] (Notice sign in Equation \ref{19.21} is positive) Along the isotherm A: \[q_{rev,A} = nRT \ln\frac{V_2}{V_1} \nonumber \] \(T\) is a constant so we can just divide \(q_{rec,A}\) by \(T\) to get \(\Delta S_A\): \[\Delta S_A = nR\ln \dfrac{V_2}{V_1} \nonumber \] We took two different paths to get start and end at the same points. Both paths had the same change in entropy. Clearly entropy is a state function while \(q_{rev}\) is not.