20.3: Unlike heat, Entropy is a State Function

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 ).

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:

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

• 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.