Here we will show how the entropy can be related to the partition function without resorting to Helmholtz free energy  F . From the first law

 dE = T dS

 dS = \frac{dE}{T} = \frac{1}{T} \frac{\partial E}{\partial T} dT = \frac{1}{T} \frac{\partial}{\partial T} \left( kT^2 \frac{\partial}{\partial T} \ln Z \right) dT

In the last transition we used the relation between the partition function and the energy. Integrating the last equation yields

 S = \int \frac{1}{T} \frac{\partial}{\partial T} \left( kT^2 \frac{\partial}{\partial T} \ln Z \right) dT  =

We use integration by parts to simplify the last integral

 = \int \frac{\partial}{\partial T} \left( kT \frac{\partial}{\partial T} \ln Z \right) dT - \int \frac{\partial}{\partial T} \left(\frac{1}{T} \right) \left( kT^2 \frac{\partial}{\partial T} \ln Z \right) dT  =

 = kT \frac{\partial}{\partial T} \ln Z  + k \int \left( \frac{\partial}{\partial T} \ln Z \right) dT = kT \frac{\partial}{\partial T} \ln Z + k \ln Z =

 = \frac{\partial}{\partial T} \left( kT \ln Z \right) = -\frac{\partial F}{\partial T}

which reproduces the known result.

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