Calculation of the free energy of an ideal gas

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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

$p V = N k T,$

(1.35)

$U = 3 ∕ 2 N k T .$

(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

$p = - \frac{\partial F}{\partial V} = \frac{N k T}{V},$

(1.37)

i.e.

$F (V, N, T) = - N k T (l n (V) + K (N, T)) .$

(1.38)

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

$U = F + T S, and S = - d F ∕ d T$

(1.39)

we find

$\begin{matrix} \frac{3}{2} N k T & = & U & = & - N k T (ln (V) + K (N, T)) - T [- N k (ln (V) + K (N, T)) - N k T \frac{\partial K (T, N)}{\partial T}] \\ = & N k T T \frac{\partial K (N, T)}{\partial T} . \end{matrix}$

(1.40)

Consequently

$K (N, T) = 3 ∕ 2 ln (T) + K^{'} (N),$

(1.41)

leading to

$F (V, N, T) = - N k T [ln (V) + 3 ∕ 2 ln (T) + K^{'} (N)] .$

(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.

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