Calculation of the free energy of an ideal gas

1.17 Calculation of the free energy of an ideal gas

We can calculate the Free Energy starting with the state function and the inner energy of an ideal gas

\begin{equation*} p V = N k T \qquad , \label{ideal_gas_equation} \end{equation*}

(1.35)

\begin{equation*} U= 3/2 N k T \qquad . \label{ideal_gas_inner_energy} \end{equation*}

(1.36)

(Note: In this notation \(U\) is not a potential, since \(T\) is not a coordinate of \(U\)!). Using Eq. (1.35) we get

\begin{equation*} p= -\frac{\partial F}{\partial V}=\frac{NkT}{V} \qquad , \end{equation*}

(1.37)

i.e.

\begin{equation*} F(V,N,T) = - NkT (ln(V)+K(N,T)) \qquad. \end{equation*}

(1.38)

The function \(K(N,T)\) must still be calculated. Combining Eq. (1.36),

\begin{equation*} U = F + TS \mbox{, and } S = - dF/dT \end{equation*}

(1.39)

we find

\begin{equation*} \begin{array}{rcccl} \frac{3}{2}NkT & = & U & = &-NkT \left(\ln(V) + K(N, T)\right) - T \left[-N k \left(\ln(V) + K(N, T)\right) - NkT \frac{\partial K(T, N)}{\partial T} \right] \\ &&& = & NkT T \frac{\partial K(N, T)}{\partial T} \qquad. \end{array} \end{equation*}

(1.40)

Consequently

\begin{equation*} K(N,T) = 3/2 \ln(T) + K'(N) \qquad, \end{equation*}

(1.41)

leading to

\begin{equation*} F(V,N,T) = - NkT \left[ \ln(V) + 3/2 \ln(T) + K'(N) \right] \qquad. \end{equation*}

(1.42)

Successively integrating the state functions of a system allows to calculate the thermodynamic potential.
This procedure is necessary because in contrast to an electrical potential there is no way of measuring a thermodynamic potential directly. We therefor have to measure all ”forces” in each state, determine thus the state functions which allow us to calculate the potential.