Active mass,Pressure and concentration

The concentration is connected to the activity via $$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$ where the standard concentration is $c^\circ = 1\:\mathrm{mol/L}$. At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as $$\lim_{c(\ce{A})\to0\:\mathrm{mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$

The partial pressure is connected to the activity via $$a(\ce{A}) = \frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$ with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=1\:\mathrm{bar}$ or traditional use of $p^\circ=1\:\mathrm{atm}$. For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since $$\lim_{p\to0\:\mathrm{bar}}\left(\phi_{\ce{A}}\right)=1,\hspace{10ex} p(\ce{A})=y_{\ce{A}}\cdot{}p.$$ Now my question is since we can approximate activity to partial pressure as well as concentration therefore pressure must approximately equal to concentration. However we know that $p=cRT$, where $c$ is the concentration, hence we definitely cannot approximate pressure and concentration. So how is this possible?