# How to get equations of state by differentiating a thermodynamic potential?

For an ideal gas in a closed system, the thermodynamic potential U, the internal energy, is given by

$$U(V,S) = U_0 \left( \frac{V_0}{V} \right)^{2/3} \text{e}^{\frac{S-S_0}{C_V}},\quad C_V = \frac{3}{2}Nk_B$$

cf. for instance Eq. (5.6) of this document. Due to it being a thermodynamic potential, I should be able to retrieve state functions, i.e. equations of state from $$U(V,S)$$ by differentiating it.

I should find the ideal gas law and the caloric equation of state by using

\begin{align}\left(\frac{\partial U}{\partial V}\right)_{S} = -p \quad &\stackrel{?}\Rightarrow\quad pV=Nk_BT \\ \left(\frac{\partial U}{\partial S}\right)_{V} = T \quad &\stackrel{?}\Rightarrow\quad U=\frac{3}{2}Nk_BT \end{align}

Taking the derivative of $$U$$ yields

\begin{align} \left(\frac{\partial U}{\partial V}\right)_{S} &= -U_0 \frac{2}{3}V_0^{2/3}\frac{1}{V^{5/3}} \text{e}^{\frac{S-S_0}{C_V}} = -p\\ \left(\frac{\partial U}{\partial S}\right)_{V} &= U_0 \left( \frac{V_0}{V} \right)^{2/3} \text{e}^{\frac{S-S_0}{C_V}} \frac{1}{C_V} = T \end{align}

I am now faced with two options:

1. Eliminate $$U_0$$, $$V_0$$ and $$S_0$$, which yields the ideal gas law, $$pV=Nk_BT$$. Where is the caloric e.o.s?
2. Let $$V\to V_0$$ and $$S\to S_0$$, which yields both equations of state as expected. However, this $$X\to X_0$$ seems like a hand-waving argument to me.

Why do I only get one equation of state in option 1? What is the reason for performing the substitution as described in option 2?

The answer is to resubstitute $$U(S,N)$$. First, for the caloric equation:
$$\left(\frac{\partial U}{\partial S}\right)_V = U_0 \left(\frac{V_0}{V}\right)^{2/3} \text{e}^{(S-S_0)/C_V} \frac{1}{C_V} = \frac{U}{C_V} \stackrel{!}=T \quad\to\quad U=C_VT$$
$$\left(\frac{\partial U}{\partial V}\right)_S = U_0 V_0^{2/3}(-2/3) \frac{1}{V^{5/3}} \text{e}^{(S-S_0)/C_V} = -\frac{2}{3}\frac{U}{V} \stackrel{!}=-p \quad\to\quad pV=\frac{2}{3}U\stackrel{\text{caloric e.o.s.}}=Nk_BT$$