# 20.3: Unlike heat, Entropy is a State Function

- Page ID
- 13717

## 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 \(\PageIndex{1}\)).

**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 q_{rev,A} and q_{rev,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 \(T_2\):

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

(Notice the + sign instead of the - sign in Equation \ref{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.