Helmholtz Free Energy Representation of Ideal Gas

Question

It is asked to find the fundamental equation of a monoatomic ideal gas in Helmholtz Representation, where

$$F = U - TS$$

My Attempt ::

$S$ of a monoatomic ideal gas is given by:

$$S = Ns_0 + NR\ln(\frac{T}{T_0})^{3/2}+NR\ln(\frac{V}{V_0})-NR\ln(\frac{N_0}{N})$$

Thus,

$$S = Ns_0 + NR\ln[(\frac{T}{T_0})^{3/2}(\frac{V}{V_0})(\frac{N_0}{N})^{-1}]$$

and

$$U = \frac{3}{2}NRT$$

Therefore,

$$F = \frac{3}{2}NRT - T(Ns_0 + NR\ln[(\frac{T}{T_0})^{3/2}(\frac{V}{V_0})(\frac{N_0}{N})^{-1}])$$

So,

$$F = \frac{3}{2}NRT - TNs_0 - NRT\ln[(\frac{T}{T_0})^{3/2}(\frac{V}{V_0})(\frac{N_0}{N})^{-1}])$$

Here, I am stuck. The wanted solution is in the form

$$F = NRT \{\frac{F_0}{N_0RT_0}-\ln[(\frac{T}{T_0})^{3/2}(\frac{V}{V_0})(\frac{N_0}{N})^{-1}]\}$$

And I cannot resolve what $F_0$ is.

So, I've also attempted to solve for $U$ in the equation of $S$ in another form where,

$$S = Ns_0 + NR\ln[(\frac{U}{U_0})^{3/2}(\frac{V}{V_0})(\frac{N}{N_0})^{-5/2}]$$

But the form becomes ridiculously complicated if I were to try

$$T = \frac{\partial U}{\partial S}$$

and plug it in to

$$F = U - TS$$

In short, how do I resolve $F_0$?

Answer 1

$F_0$ is the Helmholtz free energy at $T=T_0$, $V=V_0$, $N=N_0$. So $F_0=\frac {3}{2}N_0RT_0-T_0N_0s_0$ and $\frac {F_0}{N_0RT_0}= \frac {3}{2}- \frac {s_0}{R} $. You can show
$$F = NRT \{\frac{F_0}{N_0RT_0}-\ln[(\frac{T}{T_0})^{3/2}(\frac{V}{V_0})(\frac{N_0}{N})^{-1}]\} =NRT( \frac{3}{2} - \frac{s_0}{R})+ NR\ln[(\frac{T}{T_0})^{3/2}(\frac{V}{V_0})(\frac{N_0}{N})^{-1}]= \frac{3}{2}NRT - T(Ns_0 + NR\ln[(\frac{T}{T_0})^{3/2}(\frac{V}{V_0})(\frac{N_0}{N})^{-1}] )$$ and you derived the formula correctly.