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20.3: Unlike heat, Entropy is a State Function

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  • Circular integrals

    Because it is a state function it integrates to zero over any circular path going back to initial conditions, just like \(U\) and \(H\)

    \[\oint dS =0\]

    \[\oint dH =0\]

    \[\oint dU =0\]

    As discussed previously, we can use this fact to revisit the isotherm + isochore + adiabat circular path (Figure 20.3.1 ).

    Figure 20.3.1 : A: isothermal expansion, B:adiabat, C:isochore (CC BY-NC; Ümit Kaya via LibreTexts)

    On the adiabat B

    • there is no heat so

    \[q_{rev} = 0\]

    On the isochore C:


    • and the temperature changes from \(T_2\) back to \(T_1\) this requires heat \(c_VΔT\).

    Along the isotherm A

    • we have seen that

    \[q_{rev,A} = +nRT \ln \left[\dfrac{V_2}{V_1}\right]\]

    The two quantities qrev,A and qrev,B+C are not the same, which once again underlines that heat is a path function. How about entropy?

    First consider the path B+C:

    \[q_{rev}^{B+C} = \int _{T_2}^{T_1} c_v dT \]

    \[\int _{B+C} dS = \int _{B+C} \dfrac{dq_{rev}}{T} = \int _{C} \dfrac{dq_{rev}}{T} = \int _{T_2}^{T_1} \dfrac{c_v}{T} dT \]

    We had seen this integral before, albeit from \(T_1\) to \(T_2\):

    \[ΔS_{B+C} = + nR\ln \dfrac{V_2}{V_1} \label{19.21}\]

    (Notice the + sign instead of the - sign in Equation \ref{19.21})

    Along the isotherm A:

    \[q_{rev,A} = +nRT \ln[V_2/V_1].\]

    \(T\) is a constant so that we can just divide and \(ΔS_A\) is also \(+nR\ln \left[\dfrac{V_2}{V_1}\right]\). Clearly entropy is a state function where \(q_{rev}\) is not.