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20.3: Unlike heat, Entropy is a State Function

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    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 \(H\):

    \[\oint dS =0\]

    \[\oint dH =0\]

    \[\oint dU =0\]

    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)

    Along adiabat B and isochore C:

    • There is no heat transfer along adiabat B:

    \[q_{rev,B} = 0\]

    • There is no work along isochore C:

    \[\delta w=0\]

    • But the temperature changes from \(T_2\) back to \(T_1\). This requires heat:

    \[q_{rev,C}=C_V\Delta T\]

    Along the isotherm A:

    • We have seen that

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

    The quantities qrev,A, qrev,B, and qrev,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:

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

    \[\int dS_{B+C} = \int \dfrac{dq_{rev,B+C}}{T} = \int _{T_2}^{T_1} \dfrac{C_v}{T} dT \]

    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:

    \[q_{rev,A} = nRT \ln\frac{V_2}{V_1}\]

    \(T\) is a constant so we can just divide \(q_{rec,A}\) by \(T\) to get \(\Delta S_A\):

    \[\Delta S_A = nR\ln \dfrac{V_2}{V_1}\]

    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.

    20.3: Unlike heat, Entropy is a State Function is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.