Thermodynamic Derivation of Ideal Gas Law
Classical thermodynamics does not admit that molecules exist thus if the gas law is derived it must be without recourse to any molecular ideas regarding the nature of matter.

The chemical potential of a perfect pure gas at temperature T and pressures p and $p^{\mathrm{o}}$ is defined as
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p/p^{\mathrm{o}})$$
We can choose $p^{\mathrm{o}}$ to be unity then 
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p)$$
This is still dimensionally correct as $p^{\mathrm{o}} = 1  \pu{~bar}$ and as we have chosen the temperature, $\mu^{\mathrm{o}}$ is the standard chemical potential which is a function of temperature only.

Differentiating the chemical potential at constant T gives
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T = \frac{RT}{p}$$

To find another expression for the derivative we can use 
$$dG=-SdT+ Vdp +\Sigma\mu_idn_i$$
 for $dn_i$ moles of species i gives. For a pure species we can drop the subscript i and replace the summation with $\mu dn$ then
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T=\frac{\partial}{\partial n}\left(\frac{\partial G}{\partial p}  \right)_T= V_m$$
where $V_m$ is the volume per mole. 

Comparing these two derivatives gives $V_m=RT/p$ or 
$$pV=nRT$$

Thus the definition of the chemical potential implicitly contains the perfect gas law.