# 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 discussed previously, we can use this fact to revisit the isotherm + isochore + adiabat circular path (Figure $$\PageIndex{1}$$).

• 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 $$T_2$$:

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

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