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Chemistry LibreTexts

Basic Thermodynamics

In the microcanonical ensemble, the entropy \(S\) is a natural function of \(N\),\(V\) and \(E\), i.e., \(S=S(N,V,E)\). This can be inverted to give the energy as a function of \(N\), \(V\), and \(S\), i.e., \(E=E(N,V,S)\). Consider using Legendre transformation to change from \(S\) to \(T\) using the fact that

\[T= \left(\frac {\partial E}{\partial S}\right)_{N,V}\]

The Legendre transform \(\tilde{E}\) of \(E(N,V,S)\) is

\[ \tilde {E} (N, V, T ) = E (N,V,S(T)) - S \frac {\partial E}{\partial S}\]

\[ = E(N,V,S(T)) - TS \]

The quantity \(\tilde{E}(N,V,T)\) is called the Hemlholtz free energy and is given the symbol \(A(N,V,T)\) and is the fundamental energy in the canonical ensemble.
The differential of \(A\) is

\[dA = \left( \partial A \over \partial T \right)_{N,V}dT + \left( \partial A \over \partial V \right)_{N,T} dV +\left( \partial A \over \partial N \right)_{T,V} dN\]

However, from \(A = E - TS \), we have

\[ dA = dE - TdS - SdT \]

From the first law, \(dE\) is given by

\[ dE = TdS - PdV + \mu dN \]

Thus,

\[ dA = - PdV - S dT + \mu dN \]

Comparing the two expressions, we see that the thermodynamic relations are

\[ S =  -\left(\frac {\partial A}{\partial T}\right)_{N,V}\]

\[ P =  -\left(\frac {\partial A}{\partial V}\right)_{N,T}\]

\[ \mu =  -\left(\frac {\partial A}{\partial N}\right)_{V,T}\]