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Chemistry LibreTexts

20.3: Unlike heat, Entropy Is a State Function

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 shown in section 20-3 we can use this fact to revisit the isotherm + isochore + adiabat circular path

A: isothermal expansion, B:adiabat, C:isochore
 

On the adiabat B

  • there is no heat so

\[q_{rev} = 0\]

On the isochore C:

\[δw=0\]

  • 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 \(T_2\) (See Eq. 19.21):

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

(Notice the + sign instead of the - sign in 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.