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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.