# 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 $$H$$:

$\oint dS =0$

$\oint dH =0$

$\oint dU =0$

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

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 $$T_2$$:

$\Delta S_{B+C} = nR\ln \dfrac{V_2}{V_1} \label{19.21}$

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.